Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap19 Constrained particles and rigid objects 21 Elementary introduction to 3D rigid-object dynamics

Chapter 20 Fixed-axis rotation in 3D

Download this chapter by section

20.0Introductionstudy PDF
20.13-D description of circular motionstudy PDF
20.2Dynamics of fixed-axis rotationstudy PDF
20.3Moment of inertia matrices $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ study PDF
20.4Using the $[\boldsymbol{I}]$ matrixstudy PDF
20.5Dynamic balancestudy PDF

All chapters and individual parts

The ideas of chapter 10 on 2D circular motion are extended to fixed axis rotation in 3D. The key difference here is the non-trivial use of the cross product for calculating velocities and accelerations. Fixed axis rotation is the simplest motion with which one can introduce the full moment of inertia matrix, where the diagonal terms are analogous to the scalar 2D moment of inertia and the off-diagonal terms have a “centripetal” interpretation. The main new application is dynamic balance.

Filename:tfigure8-rel-ang-vel

Figure 20.1: A car crankshaft is a complex three-dimensional object which is well approximated for many purposes as rotating about a fixed axis. The relative timing of the reciprocating pistons is controlled by this complex shape. “Connecting rods” are pinned to the cylinders at one end and to the short offset cylinders on the crankshaft at the other.

In chapter 10 we discussed the motions of particles and rigid bodies that lie in a plane and rotate about a fixed axis perpendicular to that plane. In this chapter we again are going to think about fixed axis rotation, but now in three dimensions. The axis of rotation might be in any skew direction and the rotating bodies might be arbitrarily complicated three-dimensional shapes. As a cartoon, imagine a rigid body skewered with a rigid rod and then turned by a motor that speeds up and slows down. More practically think of the crankshaft in a car engine (fig. 20.1). Other applications include accelerating or decelerating shafts of all kinds, gears, turbines, flywheels, pendula, and swinging doors.

To understand this motion we need to take a little more care with the kinematics because it now involves three dimensions, although in some sense the basic ideas are unchanged from the previous two-dimensional chapter. The three-dimensional mechanics naturally gets more involved.

This one special motion, rotation about a fixed axis, serves as our introduction to three-dimensional rigid body mechanics.

As for all motions of all systems, the momentum balance equations apply to any system or any part of a system that has fixed axis rotation. So our mechanics results will be based on these familiar equations:

\mbox{Linear momentum balance:}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{i}=\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}},

\mbox{Angular momentum balance:}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{i/O}=\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}.

and

\mbox{Power balance:}\qquad P={\dot{E}_{\mathrm{K}}}.

As always, we will evaluate the left hand sides of the momentum equations using the forces and moments in the free-body diagram. We evaluate the right hand sides of these equations using our knowledge of the velocities and accelerations of the various mass points.

The chapter starts with a discussion of kinematics. Then we consider the mechanics of systems with fixed axis rotation. The moment of inertia matrix is then introduced followed by a section where the moment of inertia is used as a shortcut in the evaluation of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ . Finally we discuss dynamic balance, an important genuinely three-dimensional topic in machine design.

20.1 3-D description of circular motion

Let’s first assume each particle is going in circles around the $z$ axis, as in the previous chapter. The figure below shows this two-dimensional situation first two-dimensionally (left) and then as a two-dimensional motion in a three-dimensional world (right).

Filename:tfigure4-2-const-rate3D

Filename:tfigure8-ang-vel-ex

Figure 20.2: When a point P is going in circles about the

z

axis, we define the unit vector

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}

to be pointed from the axis to the point P. We define the unit vector

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}

to be tangent to the circle at P. Both of these vectors change in time as the point moves along its circular path.

Filename:tfigure8-ang-vel

Figure 20.3: The acceleration is the sum of two components. One directed towards the center of the circle in the

-\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}

direction, and one tangent to the circle in the

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}

direction.

Either way, the velocity and acceleration are the same:

•

the velocity is tangent to the circle it is going around and is proportional in magnitude to the radius of the circle and also to its angular speed. That is, the direction of the velocity is in the direction $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ and has magnitude $\omega R$ , where $\omega$ is the angular rate of rotation and $R$ is the radius of the circle that the particle is going around.
•

the acceleration can be constructed as the sum of two vectors. One is pointed to the center of the circle and proportional in magnitude to both the square of the angular speed and to the radius. The other vector is tangent to the circle and equal in magnitude to the rate of increase of speed.

These two ideas are summarized by the following formulas:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\omega R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\qquad\mbox{and}\qquad{\accentset{\rightharpoonup}{% \boldsymbol{a}}}=-\underbrace{\omega^{2}R}_{v^{2}/R}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\underbrace{R\ddot{\theta}}_{\dot{v}}\hat{\boldsymbol{e% }}_{\scriptscriptstyle\!\theta}

(20.1)

with

v=\omega R.

(20.2)

The axis of rotation might not be the $z$ -axis of a convenient $x y z$ coordinate system. So the $x y$ plane of circles might not be the $x y$ plane of the coordinate system you might want to use for some other reasons. Fortunately, we can write the formulas 20.1 in a way that rids us of these problems.

Here are some formulas which are equivalent to the formulas 20.1 but which do not make use of the polar coordinate base vectors.

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}},$		(20.3)
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}})+\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}.$		(20.4)

To check that equations 20.3 and 20.4 are really equivalent to 20.1 we need to verify that the vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ is equal to $\omega R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , that the vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})$ is equal to the vector $-\omega^{2}R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ , and that $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ is equivalent to $R\ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ .

First, define ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ as the position of the point of interest relative to any point on the axis of rotation. If this point happens to be the center of the circle then ${\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{R}}}$ . But, in general, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\neq{\accentset{\rightharpoonup}{% \boldsymbol{R}}}$ .

Now look at ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ with respect to fig. 20.2. Using the right hand rule, it is clear that the direction of the cross product ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ is in fact the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ direction. What about the magnitude? The magnitude $|{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}|=|{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}||{\accentset{\rightharpoonup}{\boldsymbol{r}}}|\sin\phi$ . But $|{\accentset{\rightharpoonup}{\boldsymbol{r}}}|\sin\phi=R$ . So the magnitude of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ is $\omega R$ . That is,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}$	$\displaystyle=$	$\displaystyle\mbox{($\|{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times% {\accentset{\rightharpoonup}{\boldsymbol{r}}}\|$)}\cdot\mbox{(unit vector in % the direction of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}})$}$	(20.5)
	$\displaystyle=$	$\displaystyle(\omega R)\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	(20.6)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	(20.7)

So ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is correct. The check of the second term in the acceleration formula follows the same reasoning. But the check of the first term involves the triple cross product.

Triple cross product

The formula for acceleration of a point on a rigid body includes the centripetal term ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})$ . This expression is a special case of the general vector expression

{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{C}}})

which is sometimes called the ‘vector triple product’ because its value is a vector (as opposed to the scalar value of the ‘scalar triple product’). The primary useful identity with vector triple products (see (17.45) ) is:

{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{C}}})=({\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{C}}}){\accentset{\rightharpoonup}{% \boldsymbol{B}}}-({\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{B}}}){\accentset{\rightharpoonup}{% \boldsymbol{C}}}.\par

(20.8)

This formula may be remembered by the semi-mnemonic device ‘cab minus bac’ since ${\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{C}}}={\accentset{\rightharpoonup}{\boldsymbol{C}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{A}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{B}}}={\accentset{\rightharpoonup}{\boldsymbol{B}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{A}}}$ . This formula is discussed in box 17.2 on page 17.2.

So now we can write

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times(% \omega R)\mathord{\cdot}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-% \omega^{2}\underbrace{R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}}_{{% \accentset{\rightharpoonup}{\boldsymbol{R}}}}=-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{R}}}={\accentset{\rightharpoonup}{\boldsymbol{a}}}.

(20.9)

Note that equations 20.3 and 20.4 are vector equations. They do not make use of any coordinate system. So, for example, we can use them even if ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is not in the $z$ direction.

Angular velocity of a rigid body in 3D

Filename:tfigure8-ang-accel

Figure 20.4: A rigid body spinning about the

z

axis. Every point on the body, like point

P

{\accentset{\rightharpoonup}{\boldsymbol{r}}}

, is going in circles. All of these circles have centers on the axis of rotation. All the points are going around at the same angular rate,

\dot{\theta}=\omega

If a rigid body is constrained to rotate about an axis then all points on the body have the same angular rate about that axis. Hence one says that the body has an angular velocity. So the measure of rotation rate of a three-dimensional rigid body is the body’s angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . At any instant in time a given body has one and only one angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . Although we only discuss fixed axis rotation in this chapter, a given body has a unique angular velocity for general motion.

For rotations about the $z$ -axis, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . $\omega$ is the $\dot{\theta}$ shown in fig. 20.4. Since all points of a rigid body have the same $\dot{\theta}$ , even if they have different $\theta$ ’s, the definition is not ambiguous.

We would like to make this idea precise enough to be useful for calculations. Why, one may ask, do we talk about rotation rate $\omega$ or ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ instead of just using the derivative of an angle $\theta$ , namely $\dot{\theta}$ ? The answer is that for a rigid body one would have trouble deciding what angle $\theta$ to measure.

First recall the situation for a two-dimensional rigid body. Consider all possible $\theta_{1},\ \theta_{2},\ \theta_{3},\dots$ , the angles that all possible lines marked on a body could make with the positive $x$ -axis, the positive $y$ axis, or any other fixed line that does not rotate. As the body rotates all of these angles increment by the same amount. Therefore, each of these angles increases at the same rate. Because all these angular rates are the same, one need not define $\dot{\theta}_{1}=\omega_{1}$ , $\dot{\theta}_{2}=\omega_{2}$ , $\dot{\theta}_{3}=\omega_{3}$ , etc. for each of the lines. Every line attached to the body rotates at the same rate and we call this rate $\omega$ . So $\dot{\theta}_{1}=\omega$ , $\dot{\theta}=\omega$ , $\dot{\theta}_{3}=\omega$ , etc. Rather than say the lengthy phrase ‘the rate of rotation of every line attached to the rigid body is $\omega$ ’, we instead say ‘the rigid body has angular velocity $\omega$ ’. For use in vector equations, we define the angular velocity vector of a two-dimensional rigid body as ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ and for a 3-D body rotating about an axis in the $\hat{\boldsymbol{\lambda}}$ direction as ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \lambda}}$ .

Filename:tfigure8-ang-accel-ex

Figure 20.5: The shadows of lines marked in a 3-D rigid body are shown on a plane perpendicular to the axis of rotation. The shadows rotate on the plane at the rate

\dot{\theta}_{1}=\dot{\theta}_{2}=\omega

. The angular velocity vector is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \lambda}}

What do we mean by these angles $\theta_{i}$ for crooked lines in a three-dimensional body? We simply look at shadows of lines drawn in or on the body of interest onto a plane perpendicular to the axis of rotation; i.e., perpendicular to $\hat{\boldsymbol{\lambda}}$ . See fig. 20.5. The rate of change of their orientation ( $\dot{\theta}_{1}=\dot{\theta}_{2}=\dot{\theta}_{3}$ ) is $\omega$ , and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is therefore $\omega\hat{\boldsymbol{\lambda}}$ . This intuitive geometric definition of $\omega$ in terms of the rotation of shadows has run its course. It gives you a picture but is not very convenient for developing formulas.

Example: What are the velocity and acceleration of one corner of a cube that is spinning about a diagonal?

Filename:tfigure8-ang-accel1

Figure 20.6: A spinning cube.

A one foot cube is spinning at $60\,{\rm rpm}$ about the diagonal $O C$ . What are the velocity and acceleration of point B?

First let’s find the velocity using ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ :

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}$
	$\displaystyle=$	$\displaystyle(60\,{\rm rpm}\ \boldsymbol{\lambda}_{OC})\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{OB}$
	$\displaystyle=$	$\displaystyle\left(2\pi\,\mathrm{s}^{-1}{(\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}}+\hat{\boldsymbol{k}})\over\sqrt{3}}\right)\times(1\,% \mathrm{ft}(\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}))$
	$\displaystyle=$	$\displaystyle(2\pi/\sqrt{3})(-\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}})% \,{\mathrm{ft/s}}.$

Now of course this equation could have been worked out with the first of equations 20.1 but it would have been quite tricky to find the vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ , and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ ! More simply, to find the acceleration we just plug in the formula ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}})$ as follows:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times[{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}]$
	$\displaystyle=$	$\displaystyle(60\,{\rm rpm}\ \boldsymbol{\lambda}_{OC})\times[(60\,{\rm rpm}\ % \boldsymbol{\lambda}_{OC})\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {OB}]$
	$\displaystyle=$	$\displaystyle(2\pi\,\mathrm{s}^{-1}{(\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}}+\hat{\boldsymbol{k}})\over\sqrt{3}})$
		$\displaystyle\qquad\qquad\times\left[({2\pi\,\mathrm{s}^{-1}(\hat{\boldsymbol{% \imath}}+\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}})\over\sqrt{3}})\times(% 1\,\mathrm{ft}(\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}})\right]$
	$\displaystyle=$	$\displaystyle(2\pi/\sqrt{3})^{2}(2\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{% \jmath}}-\hat{\boldsymbol{k}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}.$

The last line of calculation is eased by the calculation of velocity above where the term in square brackets, the velocity, was already calculated.

Relative motion of points on a rigid body

The relative velocity of two points $A$ and $B$ is defined to be

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}\equiv% {\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}-{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}

So, the relative velocity of two points glued to one rigid body, as observed from a Newtonian frame, is given by

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B/A}}$	$\displaystyle\equiv$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}-{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm% {A}}$	(20.10)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/O}}-{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{A/O}}$	(20.11)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/O}}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/O}})$	(20.12)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}},$	(20.13)

where point $O$ is a point in the Newtonian frame on the fixed axis of rotation. Clearly, since points $A$ and $B$ are fixed in the body $\cal B$ their velocities and hence their relative velocity as observed in a reference frame fixed to $\cal B$ is ${\accentset{\rightharpoonup}{\boldsymbol{0}}}$ . But, point $A$ has some absolute velocity that is different from the absolute velocity of point $B$ , as viewed from point $O$ in the fixed frame. The relative velocity of points $A$ and $B$ , the difference in absolute velocity of the two points, is due to the difference in their positions relative to point $O$ .

Similarly, the relative acceleration of two points glued to one rigid body spinning about a fixed axis is

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/A}}\equiv{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B}}-{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{A}}={\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}})+\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B/A}}.

(20.14)

Again, the relative acceleration is due to the difference in the points’ positions relative to the point $O$ fixed on the axis. Like their junior 2D cousins, these kinematics results, 20.13 and 20.14, are useful for calculating angular momentum relative to the center-of-mass as well as for the understanding of the motions of machines with moving connected parts.

Filename:sfig8-2-3

Figure 20.7: Two points,

A

and

B

on one body that has a fixed axis of rotation.

To repeat, for two points on one rigid body we have that

\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}{{}_{B/A}}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B/A}}.

(20.15)

Equation (20.15) is perhaps the most fundamental equation for those desiring an understanding of the motions of rigid bodies. Unless one desires to pursue matrix representations of rotation, equation (20.15) is the defining equation for ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . There is always exactly one vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ so that equation (20.15) is true for every pair of points on a rigid body.

Equation (20.15) is not so simple a defining equation as one would hope for such an intuitive concept as spinning. But, besides the pictorial definition with shadows, it’s the simplest definition we have.

Relative velocity and acceleration using rotating frames

If we glued a coordinate system $x^{\prime}y^{\prime}$ to a rotating rigid body $\cal C$ , we would have what is called a rotating frame as shown in fig. 20.8.

Filename:sfig8-2-3a

Figure 20.8: A rotating rigid body

\cal C

with rotating frame

x^{\prime}y^{\prime}

attached.

The base vectors in this frame change in time the same way as did $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ in section 15.1. That is

\frac{d}{dt}\hat{\boldsymbol{\imath}}^{\prime}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal C}\times\hat{\boldsymbol{\imath}}^{\prime}\quad% \mbox{and}\quad\frac{d}{dt}\hat{\boldsymbol{\jmath}}^{\prime}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times\hat{\boldsymbol{\jmath}}% ^{\prime}.

If we now write the relative position of B to A in terms of $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ , we have

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}=x^{\prime}\hat{% \boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}.

Since the coordinates $x^{\prime}$ and $y^{\prime}$ rotate with the body to which A and B are attached, they are constant with respect to that body,

\dot{x}^{\prime}=0\mbox{ and }\dot{y}^{\prime}=0.

$\displaystyle\frac{d}{dt}({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{B/A}})$	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^% {\prime}\hat{\boldsymbol{\jmath}}^{\prime}\right)$
	$\displaystyle=$	$\displaystyle\underbrace{\dot{x}^{\prime}}_{0}\hat{\boldsymbol{\imath}}^{% \prime}+x^{\prime}\frac{d}{dt}\hat{\boldsymbol{\imath}}^{\prime}+\underbrace{% \dot{y}^{\prime}}_{0}\hat{\boldsymbol{\jmath}}^{\prime}+y^{\prime}\frac{d}{dt}% \hat{\boldsymbol{\jmath}}^{\prime}$
	$\displaystyle=$	$\displaystyle x^{\prime}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \cal C}\times\hat{\boldsymbol{\imath}}^{\prime})+y^{\prime}({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times\hat{\boldsymbol{\jmath}}% ^{\prime})$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times% \underbrace{(x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime})}_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{B/A}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}.$

If we now try to calculate the rate of change of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}$ ,

$\displaystyle\frac{d}{dt}({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{B/A}})$	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left({\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}_{\cal C}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A% }}\right)$
	$\displaystyle=$	$\displaystyle\frac{d{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C% }}{dt}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times\frac{d{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}}{dt}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/A}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}).$

Mechanics

Now that we know the velocity and acceleration of every point in the system we are ready, in principle, to find $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ in terms of the angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , its rate of change $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , and the position of all the mass in the system. This we do in the next section.

SAMPLE 20.1 Kinematics in 3-D—some basic questions: The following questions are about the velocity and acceleration formulae for the non-constant rate circular motion about a fixed axis:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}})$

1.

In the formulae above, what is ⇀ r ? How is it different from ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}$ used in the formulae ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=R\dot{\theta}\hat{\boldsymbol{e}% }_{\scriptscriptstyle\!\theta}$ ?
2.

What is the difference between $\dot{\theta}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , and $\ddot{\theta}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ ?
3.

Are the parentheses around the ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ term necessary in the acceleration formula?
4.

Under what condition(s) can a particle have only tangential acceleration?

Solution

1.

In the formulae for velocity and acceleration, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ refers to a vector from any point on the axis of rotation to the point of interest. Usually the origin of a coordinate system located on the axis of rotation is a convenient point to take as the base point for ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ . You can, however, choose any other point on the axis of rotation as the base point.

The vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is different from ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ in that ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ is the position vector of the point of interest with respect to the center of the circular path that the point traces during its motion. See Fig. 5.2 of the text.
2.

$\dot{\theta}$ and $\ddot{\theta}$ are the magnitudes of angular velocity and angular acceleration, respectively, in planar motion, i.e., ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\theta}\hat{% \boldsymbol{k}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\ddot{\theta}\hat{% \boldsymbol{k}}$ . We have introduced these notations to highlight the simple nature of planar circular motion. Of course, you are free to use ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\alpha\hat{\boldsymbol{% k}}$ if you wish.
3.

Yes, the parentheses around ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ in the acceleration formula are mandatory. The parentheses imply that this term has to be calculated before carrying out the cross product with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ in the formula. Since the term in the parentheses is the velocity, you may also write the acceleration formula as

${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\dot{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}.$

Even if the formula is clear in your mind and you know which cross product to carry out first, it is a good idea to put the parentheses.
4.

First of all let us identify the tangential and the normal (or radial) components of the acceleration:

${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\overbrace{\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}}^{\rm tangential}+\overbrace{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}).}^{\rm radial/normal}$

Clearly, for a particle to have only tangential acceleration, the second term must be zero. For the second term to be zero we must have either ${\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ or ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}={\accentset{\rightharpoonup% }{\boldsymbol{0}}}$ . But if ${\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ , then the tangential acceleration also becomes zero; the particle is on the axis of rotation and hence has no acceleration. Thus the condition that allows only tangential acceleration to survive is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}={\accentset{\rightharpoonup% }{\boldsymbol{0}}}$ . Now remember that $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is not zero. Therefore, the condition we have found can be true only momentarily. This disappearance of the radial acceleration happens at start-up motions and in direction-reversing motions. ^†^†margin: In all start-up motions, the velocity is zero but the acceleration is not zero at the start up ( $t=0$ ). In direction-reversing motions, such as that of the washing machine drum during the wash-cycle, just at the moment when the direction of motion reverses, velocity becomes zero but the acceleration is non-zero.

SAMPLE 20.2 For a particle in circular motion, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=2\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}},\;{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=4\,% \mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}=3\,\mathrm{m}(\cos 30^{{}^{% \circ}}\hat{\boldsymbol{\imath}}+\sin 30^{{}^{\circ}}\hat{\boldsymbol{\jmath}})$ , where ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$ is the position vector of the particle. Find (i) ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ , (ii) the tangential acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}$ , and (iii) the radial acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{r}$ , and show the resulting vectors.

Solution

Filename:sfig8-2-3b

Figure 20.9:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}=2\,\mathrm{rad}/\mathrm{s}\hat{% \boldsymbol{k}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$	$\displaystyle=$	$\displaystyle\alpha\hat{\boldsymbol{k}}=4\,\mathrm{rad}/\mathrm{s}^{2}\hat{% \boldsymbol{k}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$	$\displaystyle=$	$\displaystyle r_{G_{x}}\hat{\boldsymbol{\imath}}+r_{G_{y}}\hat{\boldsymbol{% \jmath}}=3\,\mathrm{m}(\cos 30^{{}^{\circ}}\hat{\boldsymbol{\imath}}+\sin 30^{% {}^{\circ}}\hat{\boldsymbol{\jmath}}).$

(i) From the given formulae, the linear velocity

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times(r_{G_{x}}\hat{\boldsymbol{\imath% }}+r_{G_{y}}\hat{\boldsymbol{\jmath}})=\omega r_{G_{x}}\hat{\boldsymbol{\jmath% }}+\omega r_{G_{y}}(-\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 6\,\mathrm{m}/\mathrm{s}(\cos 30^{{}^{\circ}}\hat{\boldsymbol{% \jmath}}-\sin 30^{{}^{\circ}}\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 3\,\mathrm{m}/\mathrm{s}(-\hat{\boldsymbol{\imath}}+\sqrt{3}\hat% {\boldsymbol{\jmath}}).$

The velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ is perpendicular to both ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$ . These vectors are shown in Fig. 20.9. You should use the right hand rule to confirm the direction of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ .

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=3\,\mathrm{m}/\mathrm{s}(-\hat{% \boldsymbol{\imath}}+\sqrt{3}\hat{\boldsymbol{\jmath}})$

(ii) The tangential acceleration

Filename:sfig8-3-1

Figure 20.10:

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}={\accentset{\rightharpoonup}% {\boldsymbol{\alpha}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$
	$\displaystyle=$	$\displaystyle\alpha\hat{\boldsymbol{k}}\times(r_{G_{x}}\hat{\boldsymbol{\imath% }}+r_{G_{y}}\hat{\boldsymbol{\jmath}})=\alpha r_{G_{x}}\hat{\boldsymbol{\jmath% }}-\alpha r_{G_{y}}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle 8\,\mathrm{m}/\mathrm{s}^{2}(\cos 30^{{}^{\circ}}\hat{% \boldsymbol{\jmath}}-\sin 30^{{}^{\circ}}\hat{\boldsymbol{\imath}}).$

Since ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ are in the same direction, calculation of ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}$ is similar to that of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}$ has to be in the same direction as ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ . This vector is shown in Fig. 20.10. Once again, just as in the case of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ we could easily check that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}$ is perpendicular to both ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$ .

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{t}=4\,\mathrm{m}/\mathrm{s}^{2}% (3\,\mathrm{m}/\mathrm{s}(-\hat{\boldsymbol{\imath}}+\sqrt{3}\hat{\boldsymbol{% \jmath}})$

(iii) Finally, the radial acceleration

Filename:sfig8-3-1a

Figure 20.11:

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{r}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}% }\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G})

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{r}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{G})$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times(\omega r_{G_{x}}\hat{\boldsymbol% {\jmath}}-\omega r_{G_{y}}\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle\omega^{2}r_{G_{x}}(-\hat{\boldsymbol{\imath}})-\omega^{2}r_{G_{y% }}\hat{\boldsymbol{\jmath}}=-\omega^{2}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{G}$
	$\displaystyle=$	$\displaystyle 12\,\mathrm{m}/\mathrm{s}^{2}(-\cos 30^{{}^{\circ}}\hat{% \boldsymbol{\imath}}-\sin 30^{{}^{\circ}}\hat{\boldsymbol{\jmath}}).$

This cross product is illustrated in Fig. 20.11. Both from the illustration as well as the calculation you should be able to see that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{r}$ is in the direction of $-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$ . In fact, you could show that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{r}=-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{G}$ .

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{r}=6\,\mathrm{m}/\mathrm{s}^{2}% (-\sqrt{3}\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}})$

SAMPLE 20.3

Filename:sfig8-5-wiper

Figure 20.12:

Filename:sfig8-5-wiper-a

Figure 20.13:

Simple 3-D circular motion. A system with two point masses A and B is mounted on a rod OC which makes an angle $\theta=45^{{}^{\circ}}$ with the horizontal. The entire assembly rotates about the y-axis with constant angular speed $\omega=3\,\mathrm{rad}/\mathrm{s}$ , maintaining the angle $\theta$ . Find the velocity of point A. What is the radius of the circular path that A describes? Assume that at the instant shown, AB is in the xy plane.

Solution The angular velocity of the system is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \jmath}}=3\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{\jmath}}.

Let ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}$ be the position vector of point A. Then the velocity of point A is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}\qquad({\mbox{ and % }{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{A/C}}})$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{\jmath}}\times(\underbrace{l\cos\theta\hat% {\boldsymbol{\imath}}+l\sin\theta\hat{\boldsymbol{\jmath}}}_{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}}+\underbrace{d\cos\theta\hat{% \boldsymbol{\jmath}}-d\sin\theta\hat{\boldsymbol{\imath}}}_{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/C}}})$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{\jmath}}\times[(l\cos\theta-d\sin\theta)% \hat{\boldsymbol{\imath}}+(l\sin\theta+d\cos\theta)\hat{\boldsymbol{\jmath}}]$
	$\displaystyle=$	$\displaystyle-(\omega l\cos\theta-\omega d\sin\theta)\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle-[3\,\mathrm{rad}/\mathrm{s}(1\,\mathrm{m}\mathord{\cdot}\cos{45^% {{}^{\circ}}}-0.5\,\mathrm{m}\mathord{\cdot}\sin{45^{{}^{\circ}}})]$
	$\displaystyle=$	$\displaystyle-1.06\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{k}}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=-1.06\,\mathrm{m}/\mathrm{s}\hat% {\boldsymbol{k}}.$

Filename:sfig8-5-2disks

Figure 20.14: Circular trajectory of point A as seen by looking down along the

y

-axis. At the instant shown,

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\hat{\boldsymbol{k}}

We can find the radius of the circular path of A by geometry. However, we know that the velocity of A is also given by

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\omega R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}

where $R$ is the radius of the circular path. At the instant of interest, $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\hat{\boldsymbol{k}}$ (see figure 20.14).

\mbox{Thus}\quad\quad{\accentset{\rightharpoonup}{\boldsymbol{v}}}=-\omega R% \hat{\boldsymbol{k}}.

Comparing with the answer obtained above, we get

$\displaystyle-1.06\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle-\omega R\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\quad R$	$\displaystyle=$	$\displaystyle\frac{1.06\,\mathrm{m}/\mathrm{s}}{3\,\mathrm{rad}/\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 0.35\,\mathrm{m}.$

Answer: $R=0.35\,\mathrm{m}$

SAMPLE 20.4

Filename:sfig8-4-1

Figure 20.15:

Velocity and acceleration in 3-D: The rod shown in the figure rotates about the $y$ -axis at angular speed $10\,\mathrm{rad}/\mathrm{s}$ and accelerates at the rate of $2\,\mathrm{rad}/\mathrm{s}^{2}$ . The dimensions of the rod are $L=h=2\,\mathrm{m}$ and $r=1\,\mathrm{m}$ . There is a small mass $P$ glued to the rod at its free end. At the instant shown, the three segments of the rod are parallel to the three axes.

1.

Find the velocity of point P at the instant shown.
2.

Find the acceleration of point P at the instant shown.

Solution We are given:

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \jmath}}=10\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{\jmath}}\hbox to36.135pt{% \hfil and\hfil}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{% \omega}\hat{\boldsymbol{\jmath}}=2\,\mathrm{rad}/\mathrm{s}^{2}\hat{% \boldsymbol{\jmath}}.

1.

The velocity of point P is

${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}.$

At the instant shown, the position vector of point P (the vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm P/O}$ ) seems to be a good choice for ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ . ^†^†margin: An even better choice, perhaps, is the vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm P/A}$ . Remember, the only requirement on ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is that it must start at some point on the axis of rotation and must end at the point of interest. Thus,

${\accentset{\rightharpoonup}{\boldsymbol{r}}}\equiv{\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{\rm P/O}=L\hat{\boldsymbol{\imath}}+h\hat{\boldsymbol{% \jmath}}+r\hat{\boldsymbol{k}}.$

Therefore,

Filename:sfig8-4-1a

Figure 20.16: By drawing the velocity vector at point P (a vector tangent to the path) we see that ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ must have a positive $x$ -component and a negative $z$ -component.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ $\displaystyle=$ $\displaystyle\omega\hat{\boldsymbol{\jmath}}\times(L\hat{\boldsymbol{\imath}}+% h\hat{\boldsymbol{\jmath}}+r\hat{\boldsymbol{k}})$

$\displaystyle=$ $\displaystyle\omega(-L\hat{\boldsymbol{k}}+r\hat{\boldsymbol{\imath}})$

$\displaystyle=$ $\displaystyle 10\,\mathrm{rad}/\mathrm{s}\cdot(-2\,\mathrm{m}\hat{\boldsymbol{% k}}+1\,\mathrm{m}\hat{\boldsymbol{\imath}})$

$\displaystyle=$ $\displaystyle(20\hat{\boldsymbol{\imath}}-10\hat{\boldsymbol{k}})\,\mathrm{m}/% \mathrm{s}.$

As a check, we look down the $y$ -axis and draw a velocity vector at point P (tangent to the circular path at point P) without paying attention to the answer we got. From the top view in Fig. 20.16 we see that at least the signs of the components of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ seem to be correct.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=(20\hat{\boldsymbol{\imath}}-10% \hat{\boldsymbol{k}})\,\mathrm{m}/\mathrm{s}$

The acceleration of point P is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}})$
	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{\jmath}}\times(L\hat{\boldsymbol{% \imath}}+h\hat{\boldsymbol{\jmath}}+r\hat{\boldsymbol{k}})+\omega\hat{% \boldsymbol{\jmath}}\times\underbrace{\omega(-L\hat{\boldsymbol{k}}+r\hat{% \boldsymbol{\imath}})}_{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}}$
	$\displaystyle=$	$\displaystyle\dot{\omega}(-L\hat{\boldsymbol{k}}+r\hat{\boldsymbol{\imath}})+% \omega^{2}(-L\hat{\boldsymbol{\imath}}-r\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{rad}/\mathrm{s}^{2}(-2\,\mathrm{m}\hat{\boldsymbol{k}% }+1\,\mathrm{m}\hat{\boldsymbol{\imath}})-100(\,\mathrm{rad}/\mathrm{s})^{2}(2% \,\mathrm{m}\hat{\boldsymbol{\imath}}+1\,\mathrm{m}\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle-(98\hat{\boldsymbol{\imath}}+104\hat{\boldsymbol{k}})\,\mathrm{m% }/\mathrm{s}^{2}.$

We can check the sign of the components of ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ also. Note that the tangential acceleration, $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ , is much smaller than the centripetal acceleration, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})$ . Therefore, the total acceleration is almost in the same direction as the centripetal acceleration, that is, directed from point P to A. If you draw a vector from P to A, you should be able to see that it has negative components along both the $x$ - and $z$ -axes. Thus the answer we have got seems to be correct, at least in direction.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-(98\hat{\boldsymbol{\imath}}+10% 4\hat{\boldsymbol{k}})\,\mathrm{m}/\mathrm{s}^{2}$

Problems for 20.1 3-D description of circular motion

20.1.1 If ${\accentset{\rightharpoonup}{\boldsymbol{A}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{B}}}$ are perpendicular, what is $|{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{A}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{B}}})|$ in terms of $|{\accentset{\rightharpoonup}{\boldsymbol{A}}}|$ and $|{\accentset{\rightharpoonup}{\boldsymbol{B}}}|$ ? (Hint: make a neat 3-D sketch.)

20.1.2 A circular disk of radius $r=100\,{\rm mm}$ rotates at constant speed $\omega$ about a fixed but unknown axis. At the instant when the disk is in the $x y$ -plane, the angular velocity of each point on the diameter $A B$ (which is parallel to the $x$ -axis) is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=-2\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{k}}$ . At the same instant, the magnitude of the acceleration of the center of mass $C$ is $50\,\mathrm{m}/\mathrm{s}^{2}$ .

(a)

Find the angular velocity of the disk.
(b)

Find the location of the axis of rotation.
(c)

Find the acceleration of points $A$ and $B$ .

Filename:sfig8-6-2

Figure 20.17

20.1.3 In the expression for normal acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{n}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}% }\times{\accentset{\rightharpoonup}{\boldsymbol{r}}})$ , the parenthesis around ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ is important because vector product is nonassociative. Showing ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})\neq({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}})\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ is almost trivial. (why?) Take ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}=\omega_{1}\hat{% \boldsymbol{k}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{2}=\omega_{2}\hat{% \boldsymbol{\imath}}+\omega_{1}\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=a\hat{\boldsymbol{\imath}}-b\hat% {\boldsymbol{\jmath}}$ and show that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{2}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})\neq({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}% \times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{2})\times{\accentset% {\rightharpoonup}{\boldsymbol{r}}}$ .

20.1.4

For the given angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and the position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ below, Compute the velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ in each of the three cases. Draw the circular orbit, and show the vectors ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}},\;{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ .

(a)

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=2\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}},\;\;{\accentset{\rightharpoonup}{\boldsymbol{r}}}=3\,% \mathrm{m}\hat{\boldsymbol{\imath}}+4\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ .
(b)

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=4.33\,\mathrm{rad}/\mathrm{% s}\hat{\boldsymbol{\imath}}+2.5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{% \jmath}},\;\;{\accentset{\rightharpoonup}{\boldsymbol{r}}}=0.5\,\mathrm{m}\hat% {\boldsymbol{\imath}}-0.87\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ .
(c)

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=2\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{\imath}}+2\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{\jmath}}+% 2.83\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}},\;\;{\accentset{% \rightharpoonup}{\boldsymbol{r}}}=3.5\,\mathrm{m}\hat{\boldsymbol{\imath}}+3.5% \,\mathrm{m}\hat{\boldsymbol{\jmath}}-4.95\,\mathrm{m}\hat{\boldsymbol{k}}.$

20.1.5 A shell moves down a barrel. At the instant of interest the barrel is being raised at a constant angular speed $\omega_{1}$ . Simultaneously the gun turret (the structure which holds the barrel) rotates about the vertical axis at a constant rate $\omega_{2}$ . At the instant of interest, the barrel is tipped up an angle $\theta$ and the mass $m$ is moving at speed $v$ along the frictionless barrel. Neglecting gravity, find the force of the barrel on the shell.

Filename:sfig8-6-2a

Figure 20.18

A shell moves along a cannon barrel.

20.1.6 The following questions are about the velocity and acceleration formulae for the non-constant rate circular motion about a fixed axis:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}})$

(a)

For planar circular motion about the $z$ -axis, take

$\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\alpha\hat{\boldsymbol{% k}},\quad{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{% \boldsymbol{k}},\mbox{ and }{\accentset{\rightharpoonup}{\boldsymbol{r}}}=R% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$

and show that the velocity and acceleration formulae reduce to the formulae from elementary physics: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\omega R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\omega^{2}R\hat{\boldsymbol{e}}% _{\scriptscriptstyle\!R}+R\alpha\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ .
(b)

Draw a schematic picture of a particle (or a point on a rigid body) going in circles about a fixed axis in 3-D. Take two distinct points on the axis and draw vectors ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{2}$ from these two points to the particle of interest. Show, using the formulae for ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ above, that the acceleration and velocity of the particle are the same irrespective of which ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is used and thus conclude that the vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ can be taken from any point on the axis of rotation.

20.1.7 A rectangular plate with width $w=1\,\mathrm{m}$ and length $\ell=2\,\mathrm{m}$ is rigidly attached to a shaft along one of its edges. The shaft rotates about the $x$ -axis. The center of mass of the plate goes around at a constant linear speed $v=2.5\,\mathrm{m}/\mathrm{s}$ .

(a)

Find the angular velocity of the plate.
(b)

At the instant shown, find the velocity of point $P$ .

Filename:efig1-2-28

Figure 20.19

20.1.8 The solid cylinder shown in the figure has an angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ of magnitude 40 $\,\mathrm{rad}/\mathrm{s}$ . The vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ lies in the $y z$ -plane. The origin O of the coordinate frame is at the center of the cylinder and is at rest. Find the following quantities:

(a)

The vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ .
(b)

The velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A/B}$ of point $A$ relative to point $B$ .
(c)

The acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$ of point $A$ .

Filename:pfigure-blue-118-2

Figure 20.20

20.1.9 A rigid body rotates about a fixed axis with an angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=3\pi^{2}(\hat{\boldsymbol{% \imath}}+1.5\ \hat{\boldsymbol{\jmath}}+1.5\ \hat{\boldsymbol{k}})\,{\rm rad}/% \,\mathrm{s}^{2}$ . At a certain instant, the $x$ and $y$ components of the velocity of a point A of the body are $v_{x}=3\,\mathrm{m}/\mathrm{s}$ and $v_{y}=-4\,\mathrm{m}/\mathrm{s}$ , respectively. Find the magnitude $v$ of the velocity of point A.

20.1.10 Particle attached to a shaft, no gravity. A shaft of negligible mass spins at a constant rate. The motor at A imposes this rate if needed. The bearing at A prevents translation of its end of the shaft in any direction but causes no torques other than that of the motor (which causes a torque in the $\hat{\boldsymbol{k}}$ direction). The frictionless bearing at C holds that end of the shaft from moving in the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ directions but allows slip in the $\hat{\boldsymbol{k}}$ direction. A bar of negligible mass is welded at right angles to the shaft. A small sphere, considered as a particle, of mass $1\,\mathrm{kg}$ is attached to the free end of the bar. There is no gravity. The shaft is spinning counterclockwise (when viewed from the positive $\hat{\boldsymbol{k}}$ direction) at $3\,\mathrm{rad}/\mathrm{s}$ . In the configuration of interest the rod BD is parallel to the $\hat{\boldsymbol{\imath}}$ direction.

(a)

What is the position vector of the mass in the given coordinate system at the instant of interest?
(b)

What is angular velocity vector of the shaft?
(c)

What are the velocity and acceleration of the particle at the instant of interest?

Filename:pfigure-s95f3a

Figure 20.21

20.1.11 Consider the particle on a spinning shaft of problem 20.1 again.

(a)

What is the velocity of a point halfway down the rod $B D$ relative to the velocity of the mass at $D$ ?
(b)

What is the acceleration of a point halfway down the rod $B D$ relative to the acceleration of the mass at $D$ ?

20.1.12 Particle attached to a spinning and accelerating shaft, no gravity. Consider the particle on a spinning shaft of problem 20.1 again. At the instant of interest, the shaft is spinning counterclockwise (when viewed from the positive $\hat{\boldsymbol{k}}$ direction) at $\omega=3\,\mathrm{rad}/\mathrm{s}$ and $\dot{\omega}=5\,\mathrm{rad}/\mathrm{s}^{2}$ . In the configuration of interest the rod BD is parallel to the $\hat{\boldsymbol{\imath}}$ direction. In the coordinate system shown,

(a)

What is the position vector of the particle at the instant of interest?
(b)

What are angular velocity and angular acceleration vectors of the shaft?
(c)

What are the velocity and acceleration of the particle at the instant of interest?

Filename:Danef94s1q2

Figure 20.22

20.1.13 Particle attached to a spinning and accelerating shaft, still no gravity, relative motion. Reconsider the particle on the spinning and accelerating shaft of problem 20.1. What is the velocity and acceleration of a point halfway down the rod $B D$ relative the velocity and acceleration of the particle?

20.1.14 A crooked rod spinning at a constant rate. A one meter long uniform rod is welded at its center to a shaft at an angle $\phi$ as shown. The shaft is supported by bearings at its ends and spins at a constant rate $\omega$ . At the instant of interest, the crooked rod lies in the $x y$ -plane. In the coordinate system shown,

(a)

What is the position vector of the point $P$ on the rod at the instant of interest? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O}=0.5\,\mathrm{m}\left[(\cos% \phi)\hat{\boldsymbol{\imath}}+\sin\phi\hat{\boldsymbol{\jmath}}\right]$ .
(b)

What are the velocity and acceleration of point $P$ at the instant of interest?

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=(0.5\omega\sin\phi\hat{% \boldsymbol{k}})\,\mathrm{m}/\mathrm{s}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=(-0.5\omega^{2}\sin\phi\hat{% \boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}$ .

Filename:pfigure-blue-123-1

Figure 20.23

A crooked rod spins on a shaft.

20.1.15 The spinning crooked rod, again. Reconsider the crooked rod on a shaft in problem 20.23. What are the velocity and acceleration of point $P$ relative to point $Q$ ?

Filename:pfigure-blue-119-2

Figure 20.24

A crooked rod spins on a shaft.

20.1.16 A crooked rod spinning at a variable rate. Consider the crooked rod shown in problem 20.23 again. The shaft is spinning counterclockwise (when viewed from the positive $x$ direction) at a constant angular acceleration $\dot{\omega}=5\,\mathrm{rad}/\mathrm{s}^{2}$ . At the instant of interest, the crooked rod lies in the $x y$ -plane and its angular speed is $\omega=10\,\mathrm{rad}/\mathrm{s}$ . In the coordinate system shown,

(a)

Find the position vector of the point P on the rod?
(b)

What are angular velocity and angular acceleration vectors of the rod and shaft?
(c)

What are the velocity and acceleration of point P?
(d)

What is the velocity and acceleration of point P relative to point Q?

20.1.17 A rigid rod OP of length $\ell=500\,{\rm mm}$ is welded to a shaft AB at an angle $\theta=30^{\circ}$ . The shaft rotates about its longitudinal axis at a constant rate $\omega=120\,{\rm rpm}$ . At the instant shown, the rod OP is in the the $x z$ -plane.

(a)

Neatly draw the path of point $P$ . At the instant shown, draw the basis vectors $\hat{\boldsymbol{e}}_{r}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ and express them in terms of the basis vectors $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , $\hat{\boldsymbol{k}}$ .
(b)

Find the radius of the circular path of point $P$ and calculate the velocity and acceleration of the point using planar circular motion formulae ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{\theta}\hat{\boldsymbol{e}}% _{\scriptscriptstyle\!\theta}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\dot{\theta}^{2}R\hat{% \boldsymbol{e}}_{r}$ .
(c)

Find the position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\equiv{\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{P/O}$ of point $P$ . Use this position vector to compute the velocity and acceleration of point $P$ using the general formulae ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ . Show that the answers obtained here are the same as those in (b).

Filename:pfigure-s95q14

Figure 20.25

20.1.18 A uniform rectangular plate $A B C D$ of width $\ell$ and length $2\ell$ rotates about its diagonal $A C$ at a constant angular speed $\omega=5\,\mathrm{rad}/\mathrm{s}$ . At the instant shown, the plate is in the $x z$ -plane.

(a)

Find the velocity and acceleration of point $B$ at the instant shown.
(b)

Find the velocity and acceleration of point $B$ when the plate is in the $y z$ -plane.

Filename:Danef94s3q2

Figure 20.26

20.1.19 A rectangular plate of width $w=100\,{\rm mm}$ and length $\ell=200\,{\rm mm}$ rotates about one of its diagonals at a constant angular rate $\omega=3\,\mathrm{rad}/\mathrm{s}$ . There is a fixed coordinate system $x y z$ with the origin at the center of the plate and the $x$ -axis aligned with the axis of rotation. Another set of axes $x^{\prime}y^{\prime}z^{\prime}$ , aligned with the principle axes of the plate, has the same origin but is attached to the plate and therefore, rotates with the plate. At the instant of interest, the plate is in the $x y$ -plane and the two axes $z$ and $z^{\prime}$ coincide. Find the velocity of the corner point $P$ as follows:

(a)

Write the angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ of the plate in terms of its components in the two coordinate systems.
(b)

Find the velocity of point P in both coordinate systems using the expressions for ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ found in part (a).
(c)

Find the perpendicular distance $d$ of point P from the axis of rotation, and show that the linear speed of point P calculated from velocities found in part (b) is $\omega d$ in each case.

Filename:bikefork1-ang-accel

Figure 20.27

20.1.20 A cone of angle $\beta$ sits on its side on a horizontal disk of radius $r=100\,{\rm mm}$ as shown in the figure. The cone is attached to shaft AB that turns in its bearing. The disk rotates at a constant angular speed $\omega=120\,{\rm rpm}$ . Assume that the cone rolls without slip on the disk. Thus, the disk drives the cone-and-shaft assembly causing it to turn in the bearing. Find the angular velocity of the cone as a function of the cone angle $\beta$ . In particular, find the angular velocity of the cone for

(a)

$\beta=30^{\circ}$ , and
(b)

$\beta=90^{\circ}$ (or $90^{\circ}$ minus a hair if that makes you more comfortable).

Filename:bikefork-ang-accel

Figure 20.28

20.2 Dynamics of fixed-axis rotation

We now address mechanics questions concerning objects which are known a priori to spin about a fixed axis. We would like to calculate forces and moments if the motion is known. And we would like to determine the details of the motion, the angular acceleration in particular, if the applied forces and moments are known. Once the angular acceleration is known (as a function of some combination of time, angle and angular rate) the angular rate and angular position can be found by integration or solution of an ordinary differential equation.

The full content of the subject follows from the basic mechanics equations

\mbox{linear momentum balance,}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{i}=\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}

\mbox{angular momentum balance,}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{i/C}=\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}},

\mbox{and power balance:}\qquad P={\dot{E}_{\mathrm{K}}}+{\dot{E}_{\mathrm{P}}% }+\dot{E}_{\mathrm{int}}.

The quantities $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}$ are defined in terms of the position and acceleration of the system’s mass (see the second page of the inside cover). To evaluate $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}$ for fixed-axis rotation we can use the kinematics relations from the previous chapter which determine velocity and acceleration of points on a body spinning about a fixed axis in terms of the position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ of the point of interest relative to any point on the axis.

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}},$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}})+\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}.$

For fixed-axis rotation ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \lambda}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\omega}\hat{% \boldsymbol{\lambda}}$ with $\hat{\boldsymbol{\lambda}}$ a constant unit vector along the axis of rotation.

To solve problems we draw a free-body diagram, write the equations of linear and angular momentum balance, and evaluate the terms using the kinematics relations. In general this will lead to the evaluation of a sum or an integral. A short cut, the moment of inertia matrix, will be introduced in later sections.

Before proceeding to more difficult three-dimensional problems, let’s review a simple 2D problem.

Example: Spinning disk

The round flat uniform disk in fig. 20.29 is in the $x y$ plane spinning at the constant rate ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ about its center. It has mass $m_{\mathrm{tot}}$ and radius $R_{0}$ . What force is required to cause this motion? What torque? What power?

From linear momentum balance we have:

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}\ =\ \dot{\accentset{% \rightharpoonup}{\boldsymbol{L}}}\ =\ m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{cm}\ =\ {\accentset{\rightharpoonup}{% \boldsymbol{0}}},\\

which we could also have calculated by evaluating the integral $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}\equiv\int{\accentset{% \rightharpoonup}{\boldsymbol{a}}}\,dm$ instead of using the general result that $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{cm}$ . From angular momentum balance we have:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{i/O}$	$\displaystyle=$	$\displaystyle{\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}}_{/O}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{M}}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}\,dm$
	$\displaystyle=$	$\displaystyle\int_{0}^{R_{0}}\int_{0}^{2\pi}(R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R})\times(-R\omega^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R})\,\underbrace{{m_{\mathrm{tot}}\over\pi R_{O}^{2}}\,% \overbrace{R\,d\theta\,dR}^{dA}}_{dm}$
	$\displaystyle=$	$\displaystyle\int\int{\accentset{\rightharpoonup}{\boldsymbol{0}}}\ d\theta\,dR$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Filename:bikefork1-alt

Figure 20.29: A uniform disk turned by a motor at a constant rate.

Power balance is of limited use for constant rate circular motion. If all parts of a system move at constant angular rate at a constant radius then they all have constant speed. Thus the kinetic energy of the system is constant. So the power balance equation just says that the net power into the system is the amount dissipated inside (assuming no energy storage).

Example: Spinning disk with power balance

Consider the spinning disk from fig. 20.29 and the previous example. The power balance equation III gives

P\ =\ {\dot{E}_{\mathrm{K}}}+\underbrace{{\dot{E}_{\mathrm{P}}}}_{0}+% \underbrace{\dot{E}_{\mathrm{int}}}_{0}\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}P=\ \int{\accentset{\rightharpoonup}{\boldsymbol{v}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{a}}}\,dm\ =\ \int 0\,dm\ =\ 0.

(20.16)

In this example there is no force or torque acting on the disk so the power $P$ must turn out to be zero. In other constant rate problems the force and moment will not turn out to be zero, but the kinetic energy of the system will still be constant and so, assuming no energy storage or dissipation, we will still have $P=0$ .

A stick sweeps out a cone

Now we consider a genuinely three-dimensional problem involving fixed-axis, rigid body rotation. Consider a long narrow stick swinging in circles so that it sweeps out a cone (fig. 20.30). Each point on the stick is moving in circles around the $z$ -axis at a constant rate $\omega$ . What is the relation between $\omega$ and the angle of the stick $\phi$ ? The approach to this problem is, as usual, to draw a free-body diagram, write momentum balance equations, evaluate the left and right hand sides, and then solve for quantities of interest. The hard part of this problem is evaluating the right hand side of the angular momentum balance equations.

Filename:bikefork-alt

Figure 20.30: A spherical rigid body pendulum (uniform thin rod) going in circles at constant rate

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}

Filename:pfigure-blue-128-1

Figure 20.31: Free-body diagram of the rod.

To simplify calculation, we look at the pendulum at the instant it passes through the $y z$ -plane, assuming the $x y z$ axes are fixed in space.

The free-body diagram shown in fig. 20.31 shows the gravity force at the center-of-mass, the reaction force at point $O$ , and, consistent with the shown construction of the hinge, the moments at $O$ perpendicular to the hinge.

Because we are interested in the relation between $\phi$ and $\omega$ and not the reaction force, at least for now, we look at angular momentum balance about point $O$ .

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

First, we show and discuss the results of evaluating the equation of angular momentum balance. Then, we will show the details of calculating $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$ and the details of several methods for calculating $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ .

Evaluation of $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$

To find $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$ , we had to find the moment of the gravity force. The most direct method is to use the definition

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}$
	$\displaystyle=$	$\displaystyle\frac{\ell}{2}[\cos\phi(-\hat{\boldsymbol{k}})+\sin\phi\hat{% \boldsymbol{\jmath}}]\times(-mg\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle-\frac{mg\ell\sin\phi}{2}\hat{\boldsymbol{\imath}}$

Filename:tfigure8-syst-bods

Figure 20.32: The rod is shown on the

y z

plane. We use this figure to locate the bit of mass

d m

corresponding to the bit of length of rod

d s

One could also ‘slide’ the gravity force to the level of O (a force displaced in its direction of action is mechanically equivalent). Then you can see from the figure that the force is perpendicular to its position relative to $O$ . Moment is then force ( $m g$ ) times distance ( $\frac{\ell}{2}\sin\phi$ ) in the direction given by the right hand rule ( $-\hat{\boldsymbol{\imath}}$ ).

Evaluation of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$

We now evaluate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ by adding up the contribution to the sum from each bit of mass.

\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

\displaystyle=

\displaystyle\int\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}% }^{\text{${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}=-s\cos\phi\hat{% \boldsymbol{k}}+s\sin\phi\hat{\boldsymbol{\jmath}}$}}\times\underbrace{{% \accentset{\rightharpoonup}{\boldsymbol{a}}}}_{\text{For constant rate % circular motion, \@@eqnarray{\accentset{\rightharpoonup}{\boldsymbol{a}}}&=&{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O})% \\ &=&(\omega\hat{\boldsymbol{k}})\times\left[(\omega\hat{\boldsymbol{k}})\times% \right.\\ &&\left.s(\sin\phi\hat{\boldsymbol{\jmath}}-\cos\phi\hat{\boldsymbol{k}})% \right]\\ &=&-\omega^{2}s\sin\phi\hat{\boldsymbol{\jmath}} }}\par\overbrace{dm}^{\text{$dm=\rho\ ds$, where \hfil\break\mbox{ }$\rho$ = % mass per unit length}}

\displaystyle=

\displaystyle\int_{0}^{\ell}\underbrace{(-s\cos\phi\hat{\boldsymbol{k}}+s\sin% \phi\hat{\boldsymbol{\jmath}})}_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{/O}}\times\underbrace{(-\omega^{2}s\sin\phi\hat{\boldsymbol{\jmath}})}_{{% \accentset{\rightharpoonup}{\boldsymbol{a}}}}\,(\rho\,ds)

\displaystyle=

\displaystyle\int_{0}^{\ell}-s^{2}\cos\phi\sin\phi\ \omega^{2}\hat{\boldsymbol% {\imath}}\rho ds\mbox{ \qquad(evaluating the cross product)}

\displaystyle=

\displaystyle-\cos\phi\sin\phi\ \omega^{2}\rho\hat{\boldsymbol{\imath}}\int_{0% }^{\ell}s^{2}\,ds\mbox{\qquad($\phi,\rho$, and $\omega$ do not vary with $s$)}

\displaystyle=

\displaystyle-\cos\phi\sin\phi\ \omega^{2}(\rho\frac{\ell^{3}}{3})\hat{% \boldsymbol{\imath}}\mbox{\qquad\qquad(evaluating the integral)}

\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

\displaystyle=

\displaystyle-\cos\phi\sin\phi\ \omega^{2}(\frac{m\ell^{2}}{3})\hat{% \boldsymbol{\imath}}\mbox{ \qquad\qquad(because $m=\rho\ell$).}

We could have taken a short-cut in the calculation of acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ . Instead of using ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}})$ , we could have used ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{R}}}$ where ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ is the radius of the circle each particle is traveling on. It is evident from the picture that the appropriate radius is ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=s\sin\phi\hat{\boldsymbol{\jmath}}$ , so ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\omega^{2}s\sin\phi\hat{% \boldsymbol{\jmath}}$ .

We will show two more methods for calculating $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ in section 20.4 on page 20.4 using the moment of inertia matrix.

The results for the conically swinging stick

We can now evaluate the terms in the angular momentum balance equation as

\overbrace{-mg\sin\phi\frac{\ell}{2}\hat{\boldsymbol{\imath}}+M_{O_{y}}\hat{% \boldsymbol{\jmath}}+M_{O_{z}}\hat{\boldsymbol{k}}}^{\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{O}$}}}}=\overbrace{-\sin\phi\cos\phi\frac{m\ell^{2}}{3}\omega^{2}\hat{% \boldsymbol{\imath}}}^{\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/O}}.

(20.18)

We can get three scalar equations from eqn. 20.18 by dotting it with $\hat{\boldsymbol{\jmath}}$ , $\hat{\boldsymbol{k}}$ , and $\hat{\boldsymbol{\imath}}$ to get

\framebox{$M_{O_{y}}=0$}\hbox to36.135pt{\hfil and\hfil}\framebox{$M_{O_{z}}=0$}

and

\framebox{$\omega^{2}=\frac{3g}{2\ell\cos\phi}$}.

Note that $M_{O_{y}}=M_{O_{z}}=0$ . That is, for this special motion, the hinge joint at $O$ could be replaced with a ball-and-socket joint.

Note that the solution for a point mass spherical pendulum is $\omega^{2}=\frac{g}{\ell\cos\phi}$ . That is, this stick would rotate at the same rate and angle as a point mass at the end of a rod of length $\frac{2\ell}{3}$ . One could not easily anticipate this result. We point it out here to emphasize that the analysis of this rigid-body problem cannot be reduced a priori to any simple particle mechanics problem.

Filename:sfig8-7-2

Figure 20.33: Plot of non-dimensional rotational speed

\frac{\omega^{2}\ell}{g}

versus hang angle

\phi

. For

\omega^{2}\ell/g<3/2

the only solution is

\phi=0

(hanging straight down). At or very close to

\omega^{2}\ell/g=3/2

a range of

\phi

’s is possible. As

\phi\rightarrow\pi/2

and the rod becomes close to horizontal, the spin rate

\omega

goes to infinity.

In fig. 20.33, non-dimensional rotational speed $\frac{\omega^{2}\ell}{g}$ is plotted versus hang angle $\phi$ . As one might expect intuitively unless $\omega$ is high enough, ( $\omega^{2}>\frac{3g}{2\ell}$ ), the only solution is hanging straight down ( $\phi=0$ ).

At the critical speed ( $\omega^{2}=\frac{3g}{2\ell}$ ), the curve is nearly flat, implying that a range of hang angles $\phi$ is possible all with nearly the same angular velocity. As is also intuitively plausible, as the bar gets close to the horizontal (close to $\frac{\pi}{2}$ ), the spin rate goes to infinity.

The scalar equations governing rotation about an axis

For two-dimensional motion of flat hinged objects we had the simple relation “ $M=I\alpha$ ”. This formula captures our simple intuitions about angular momentum balance. When you apply torque to a body its rate of rotation increases. It turns out that, for three-dimensional motion of a rigid body about a fixed axis, the same result applies if we interpret the terms correctly.^†^†margin: Caution: For more general three-dimensional motion than rotation about a fixed axis the equation $M=I\alpha$ does not apply. Trying to vectorize by underlining various terms gives the wrong answer.

If the axis of rotation goes through C and is in the direction $\hat{\boldsymbol{\lambda}}$ we can define $M=\hat{\boldsymbol{\lambda}}\cdot{\accentset{\rightharpoonup}{\boldsymbol{M}}}% _{\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{% $\textstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{C}$}}}$ as the moment about the axis of rotation. We can similarly look at the $\hat{\boldsymbol{\lambda}}$ component of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}$ (assume, for definiteness, that the system is continuous).

$\displaystyle\hat{\boldsymbol{\lambda}}\cdot\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\lambda}}\cdot\int{\accentset{\rightharpoonup}{% \boldsymbol{r}}}\times{\accentset{\rightharpoonup}{\boldsymbol{a}}}\,dm$	(20.19)
	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\lambda}}\cdot\int{\accentset{\rightharpoonup}{% \boldsymbol{r}}}\times\left[(\omega\hat{\boldsymbol{\lambda}})\times\left((% \omega\hat{\boldsymbol{\lambda}})\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}\right)+(\dot{\omega}\hat{\boldsymbol{\lambda}})\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}\right]\,dm.$
	$\displaystyle=$	$\displaystyle\dot{\omega}\int R^{2}\,dm,$

where $R$ is the distance of the mass points from the axis. The last line follows from the previous most simply by paying attention to directions and magnitudes when using the right-hand rule and the geometric definition of the cross product. We thus have derived the result that

M=I\alpha,

if by $M$ we mean moment about the fixed axis and by $I$ we mean $\int R^{2}\,dm$ . Actually, the scalar we call $I$ in the above equation is a manifestation of a more general matrix $[\boldsymbol{I}]$ that we will explore in the next section.

SAMPLE 20.5

Filename:sfig8-7-2a

Figure 20.34: A carnival ride rotating at a constant speed

Going on a carnival ride at a constant rate. A carnival ride with roof AB and carriage BC is rotating about the vertical axis with constant angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \jmath}}$ . If the carriage with its occupants has mass $m=100\,\mathrm{kg}$ , find the tension in the inextensible and massless rod BC when $\theta=30^{o}$ . What is the required angular speed $\omega$ (in revolutions/minute) to maintain this angle?

^†^†margin:

Solution

The free-body diagram of the carriage is shown in Fig. 20.35(a). The geometry of motion of the carriage is shown in Fig. 20.35(b). The carriage goes around a circle of radius $r=O^{\prime}C$ with constant speed $v=\omega r$ . The only acceleration that the carriage has is the centripetal acceleration and at the instant of interest ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\omega^{2}r\hat{\boldsymbol{% \imath}}$ .

Filename:sfig8-7-2again

Figure 20.35:

The linear momentum balance ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{L}}}$ ) for the carriage gives:

	$\displaystyle T\hat{\boldsymbol{\lambda}}_{CB}-mg\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
	$\displaystyle\hbox to36.135pt{or\hfil}T(-\sin\theta\hat{\boldsymbol{\imath}}+% \cos\theta\hat{\boldsymbol{\jmath}})-mg\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle-m\omega^{2}r\hat{\boldsymbol{\imath}}$		(20.20)

Scalar equations from eqn. (20.20) are:

$\displaystyle\mbox{[eqn.~{}(\ref{eq3.1.1})]}\cdot\hat{\boldsymbol{\jmath}}% \hbox to36.135pt{\hfil$\Rightarrow$\hfil}T\cos\theta-mg$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T$	$\displaystyle=$	$\displaystyle\frac{mg}{\cos\theta}$
	$\displaystyle=$	$\displaystyle\frac{100\,\mathrm{kg}\cdot 9.8\,\mathrm{m}/\mathrm{s}^{2}}{\sqrt% {3}/{2}}$
	$\displaystyle=$	$\displaystyle 1133\,{\mbox{N}}.$
$\displaystyle\mbox{[eqn.~{}(\ref{eq3.1.1})]}\cdot\hat{\boldsymbol{\imath}}% \hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad-T\sin\theta$	$\displaystyle=$	$\displaystyle-m\omega^{2}r$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega^{2}$	$\displaystyle=$	$\displaystyle\frac{T\sin\theta}{mr}$
	$\displaystyle=$	$\displaystyle\frac{T\sin\theta}{m(R+l\sin\theta)}$
	$\displaystyle=$	$\displaystyle\frac{1133\,{\mbox{N}}\cdot\frac{1}{2}}{100\,\mathrm{kg}(4\,% \mathrm{m}+5\,\mathrm{m}\cdot\frac{1}{2})}$
	$\displaystyle=$	$\displaystyle 0.87\frac{1}{\,\mathrm{s}^{2}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega$	$\displaystyle=$	$\displaystyle 0.93\frac{\,{\rm rad}}{\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 0.93\frac{1}{\cancel{\,\mathrm{s}}}\cdot\frac{1\mbox{rev}}{2\pi}% \cdot\frac{60\cancel{\,\mathrm{s}}}{1\,{\rm min}}$
	$\displaystyle=$	$\displaystyle 8.9\,{\rm rpm}.$

Answer: $T=1133\,{\mbox{N}},\;\omega=8.9\,{\rm rpm}$

Alternatively,

we could also find the angular speed using angular momentum balance. The angular momentum balance about point B gives

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/B}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/B}

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/B}\times(-mg\hat% {\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-mgl\sin\theta\hat{\boldsymbol{k}}$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/B}\times(-m% \omega^{2}r\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle-m\omega^{2}rl\cos\theta\hat{\boldsymbol{k}}$

Equating the two quantities, we get

$\displaystyle\cancel{m}\omega^{2}r\cancel{l}\cos\theta$	$\displaystyle=$	$\displaystyle\cancel{m}g\cancel{l}\sin\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega^{2}$	$\displaystyle=$	$\displaystyle\frac{g}{r}\tan\theta$
	$\displaystyle=$	$\displaystyle\frac{g\tan\theta}{R+l\sin\theta}$
	$\displaystyle=$	$\displaystyle\frac{9.8\,\mathrm{m}/\mathrm{s}^{2}\cdot 0.577}{4\,\mathrm{m}+5% \,\mathrm{m}\frac{1}{2}}$
	$\displaystyle=$	$\displaystyle 0.87\frac{1}{\,\mathrm{s}^{2}}$
$\displaystyle\omega$	$\displaystyle=$	$\displaystyle 0.93\,\mathrm{s}^{-1}=8.9\,{\rm rpm}$

which is the same value as we found using the linear momentum balance .

SAMPLE 20.6

Filename:sfig8-7-2disks

Figure 20.36: A bar held by two strings rotates in 3-D.

A crooked bar rotating with a shaft in space. A uniform rod CD of mass $m=2\,\mathrm{kg}$ and length $\ell=1\,\mathrm{m}$ is fastened to a shaft AB by means of two strings: AC of length $R_{1}=30\,\mathrm{cm}$ , and BD of length $R_{2}=50\,\mathrm{cm}$ . The shaft is rotating at a constant angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=5\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ . There is no gravity. At the instant shown, find the tensions in the two strings.

^†^†margin:

Solution The free-body diagram of the rod is shown in Fig. 20.37. The linear momentum balance ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ ) for the rod gives:

T_{1}+T_{2}=m\omega^{2}r_{G}.

(20.21)

Filename:sfig8-4-4

Figure 20.37:

Usually, linear momentum balance gives us two scalar equations in 2-D and three scalar equations in 3-D. Unfortunately, in this case, it gives only one equation for two unknowns $T_{1}$ and $T_{2}$ . Therefore, we need one more equation.

The angular momentum balance about point D gives:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/D}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/D},$
$\displaystyle\mbox{where}\qquad\sum{\accentset{\rightharpoonup}{\boldsymbol{M}% }}_{/D}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/D}\times(-T_{1}% \hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\ell(-\sin\theta\hat{\boldsymbol{\jmath}}+\cos\theta\hat{% \boldsymbol{k}})\times(-T_{1}\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\ell\,T_{1}\cos\theta\hat{\boldsymbol{\imath}},$
and
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/D}$	$\displaystyle=$	$\displaystyle\int_{m}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/D}\times% (-\omega^{2}R_{P}\hat{\boldsymbol{\jmath}})dm$
	$\displaystyle=$	$\displaystyle\int_{0}^{\ell}\overbrace{l(-\sin\theta\hat{\boldsymbol{\jmath}}+% \cos\theta\hat{\boldsymbol{k}})}^{{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{P/D}}\times(-\omega^{2}\overbrace{(R_{2}-l\sin\theta)}^{R_{P}}\hat{% \boldsymbol{\jmath}})\,\overbrace{\frac{m}{\ell}\,dl}^{dm}$
	$\displaystyle=$	$\displaystyle\frac{m\omega^{2}}{\ell}\left(R_{2}\cos\theta\int_{0}^{\ell}l\,dl% -\cos\theta\sin\theta\int_{0}^{\ell}l^{2}\,dl\right)\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\frac{m\omega^{2}}{\ell}\left(R_{2}\cos\theta\frac{\ell^{2}}{2}-% \cos\theta\sin\theta\frac{\ell^{3}}{3}\right)\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle m\omega^{2}\ell\cos\theta(\frac{1}{2}R_{2}-\frac{1}{3}\ell\sin% \theta)\hat{\boldsymbol{\imath}}.$

Thus,

	$\displaystyle\cancel{\ell}\,T_{1}\cancel{\cos\theta}$	$\displaystyle=$	$\displaystyle m\omega^{2}\cancel{\ell}\cancel{\cos\theta}(\frac{1}{2}R_{2}-% \frac{1}{3}\ell\sin\theta)$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad T_{1}$	$\displaystyle=$	$\displaystyle m\omega^{2}(\frac{1}{2}R_{2}-\frac{1}{3}\ell\sin\theta).$

Substituting in (20.21) we get

T_{2}=m\omega^{2}(r_{G}-\frac{1}{2}R_{2}+\frac{1}{3}\ell\sin\theta).

Plugging in the given numerical values and noting that $r_{G}=(R_{1}+R_{2})/2=40\,\mathrm{cm}$ and $\ell\sin\theta=R_{2}-R_{1}=20\,\mathrm{cm}$ , we get

$\displaystyle T_{1}$	$\displaystyle=$	$\displaystyle 2\,\mathrm{kg}\cdot(5\frac{1}{s})^{2}\cdot(0.4\,\mathrm{m}-\frac% {1}{2}0.5\,\mathrm{m}+\frac{1}{3}0.2\,\mathrm{m})$
	$\displaystyle=$	$\displaystyle 9.17\frac{\,\mathrm{kg}\cdot\,\mathrm{m}}{{\,\mathrm{s}}^{2}}=9.% 17\,{\mbox{N}}$
$\displaystyle\hbox to36.135pt{and\hfil}T_{2}$	$\displaystyle=$	$\displaystyle 10.83\,{\mbox{N}}.$

Answer: $T_{1}=9.1\,{\mbox{N}},\quad T_{2}=10.9\,{\mbox{N}}$

SAMPLE 20.7

Filename:sfig8-4-4a

Figure 20.38: A rectangular plate, mounted rigidly at an angle

\phi

on a shaft, wobbles as the shaft rotates at a constant speed.

A crooked plate rotating with a shaft in space. A rectangular plate of mass $m$ , length $\ell$ , and width b is welded to a shaft AB in the center. The long edge of the plate is parallel to the shaft axis but is tipped by an angle $\phi$ with respect to the shaft axis. The shaft rotates with a constant angular speed $\omega$ . The end B of the shaft is free to move in the $z$ -direction. Assume there is no gravity. Find the reactions at the supports.

Solution A simple line sketch and the Free-Body Diagram of the system are shown in Fig. 20.39ab.

Filename:sfig8-4-4b

Figure 20.39:

The linear momentum balance equation for the shaft and the plate system is:

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m_{total}\,{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{cm}.

Since the center-of-mass is on the axis of rotation, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}={\accentset{\rightharpoonup% }{\boldsymbol{0}}}$ . Therefore,

(A_{x}+B_{x})\hat{\boldsymbol{\imath}}+(A_{y}+B_{y})\hat{\boldsymbol{\jmath}}+% A_{z}\hat{\boldsymbol{k}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad A_{x}+B_{x}=0,\quad A_{y}+B_{y}% =0,\quad A_{z}=0.

(20.22)

The angular momentum balance about the center-of-mass O is:

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

•

Calculation of $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$ :

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{A}+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/O}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{B}+% M_{z}\hat{\boldsymbol{k}}$	(20.23)
	$\displaystyle=$	$\displaystyle-\frac{\ell}{2}\hat{\boldsymbol{k}}\times(A_{x}\hat{\boldsymbol{% \imath}}+A_{y}\hat{\boldsymbol{\jmath}}+A_{z}\hat{\boldsymbol{k}})+\frac{\ell}% {2}\hat{\boldsymbol{k}}\times(B_{x}\hat{\boldsymbol{\imath}}+B_{y}\hat{% \boldsymbol{\jmath}})+M_{z}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle\frac{\ell}{2}(A_{y}-B_{y})\hat{\boldsymbol{\imath}}+\frac{\ell}{% 2}(B_{x}-A_{x})\hat{\boldsymbol{\jmath}}+M_{z}\hat{\boldsymbol{k}}$

•

Calculation of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ : $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ can be computed in various ways. ^†^†margin: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ could also be computed using the moment of inertia matrix of the body. See the next two text sections. Here, to compute $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ , we use

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=\int_{M}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{dm/O}\times{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{dm% }\;dm,

the formula which we have used so far. To carry out this integration for the plate, we take, as usual, an infinitesimal mass $d m$ of the body, calculate its angular momentum about O, and then integrate over the entire mass of the body:

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=\int_{M}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{dm/O}\times{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{dm% }\;dm

We need to write carefully each term in the integrand. Let us define an axis $w$ (don’t confuse this dummy variable $w$ with $\omega$ ) along the length of the plate (see Fig. 20.40(a)). We take an area element $dA=dw\,dy$ on the plate as our infinitesimal mass. Fig. 20.40(b) shows this element and its coordinates.

Filename:sfig8-4-4c

Figure 20.40: Calculation of

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}

: (a) mass element

d m

is shown on the plate, (b) the mass element as an area element and its geometry.

$\displaystyle dm$	$\displaystyle=$	$\displaystyle\rho\,dA=\frac{m}{\ell b}dw\,dy\quad(\rho=\mbox{mass per unit % area})$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{dm}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{dm/O})$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{dm/O}$	$\displaystyle=$	$\displaystyle x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}+z\hat{% \boldsymbol{k}}$

where

x=w\sin\phi,\quad y=y,\quad z=w\cos\phi.

(20.24)

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}$	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times(\omega\hat{\boldsymbol{k}}\times% (w\sin\phi\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}+w\cos\phi\hat{% \boldsymbol{k}}))$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times(\omega\,w\sin\phi\hat{% \boldsymbol{\jmath}}-\omega\,y\hat{\boldsymbol{\imath}})=-\omega^{2}(w\sin\phi% \hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}),$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{dm/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{dm}$	$\displaystyle=$	$\displaystyle(w\sin\phi\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}+w% \cos\phi\hat{\boldsymbol{k}})\times[-\omega^{2}(w\sin\phi\hat{\boldsymbol{% \imath}}+y\hat{\boldsymbol{\jmath}})]$
	$\displaystyle=$	$\displaystyle\omega^{2}(-w^{2}\sin\phi\cos\phi\hat{\boldsymbol{\jmath}}+wy\cos% \phi\hat{\boldsymbol{\imath}}).$

Thus,

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\int_{-b/2}^{b/2}\int_{-\ell/2}^{\ell/2}\overbrace{\omega^{2}(-w^% {2}\sin\phi\cos\phi\hat{\boldsymbol{\jmath}}+wy\cos\phi\hat{\boldsymbol{\imath% }})}^{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{dm/O}\times{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{dm}}\overbrace{\frac{m}{\ell b}dw\,dy}^{dm}$	(20.25)
	$\displaystyle=$	$\displaystyle\frac{m}{\ell b}\omega^{2}\int_{-b/2}^{b/2}\left(\int_{-\ell/2}^{% \ell/2}(-w^{2}\sin\phi\cos\phi\hat{\boldsymbol{\jmath}}+wy\cos\phi\hat{% \boldsymbol{\imath}})dw\right)dy$
	$\displaystyle=$	$\displaystyle\frac{m}{\ell b}\omega^{2}\int_{-b/2}^{b/2}\left(-\sin\phi\cos% \phi\left.\frac{w^{3}}{3}\right\|_{-\ell/2}^{\ell/2}+y\cos\phi\underbrace{\left% .\frac{w^{2}}{2}\right\|_{-\ell/2}^{\ell/2}}_{0}\right)dy$
	$\displaystyle=$	$\displaystyle-\frac{m\omega^{2}\ell^{2}}{12}\sin\phi\cos\phi\hat{\boldsymbol{% \jmath}}.$

•

Now, back to angular momentum balance: Now equating (20.23) and (20.25) and dotting both sides with $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ we get

$A_{y}-B_{y}=0,\quad B_{x}-A_{x}=-\frac{m\omega^{2}\ell}{6}\sin\phi\cos\phi,% \quad M_{z}=0,$ (20.26)

respectively. Solving (20.22) and (20.26) simultaneously we get

$A_{y}=B_{y}=0,\quad A_{x}=\frac{m\omega^{2}\ell}{12},\quad\sin\phi\cos\phi B_{% x}=-\frac{m\omega^{2}\ell}{12}\sin\phi\cos\phi.$

Answer: $A_{x}=\frac{m\omega^{2}\ell}{12}\sin\phi\cos\phi,\;B_{x}=-\frac{m\omega^{2}% \ell}{12}\sin\phi\cos\phi,\;A_{y}=B_{y}=A_{z}=M_{z}=0$

SAMPLE 20.8

Filename:pfigure-s94h13p2

Figure 20.41: A short rod swings in 3-D.

A short rod as a 3-D pendulum. A uniform rod AB of mass $m$ and length $2\ell$ is welded to a massless, inextensible thin rod OA at point A. Rod OA is attached to a ball and socket joint at point O. The rods are going around in a circle with constant speed maintaining a constant angle $\theta$ with the vertical axis. Assume that $\theta$ is small.

1.

How many revolutions does the system make in one second?

Solution

The free-body diagram of the system (rod OA + rod AB) is shown in Fig. 20.42. Let $\omega$ be the angular speed of the system. Then, the number of revolutions in one second is $n=\omega/(2\pi)$ . Therefore, to find the answer we need to calculate $\omega$ .

Filename:pfigure-f93f5

Figure 20.42:

The angular momentum balance about point O gives $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$ . Now,

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times(-mg\hat% {\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle 4\ell(\sin\theta\hat{\boldsymbol{\jmath}}-\cos\theta\hat{% \boldsymbol{k}})\times(-mg\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle-4\ell\,mg\sin\theta\hat{\boldsymbol{\imath}}.$
$\displaystyle\hbox to36.135pt{and\hfil}\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\int_{m}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{dm/O}% \times{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{dm}\,dm$
	$\displaystyle=$	$\displaystyle\int_{3\ell}^{5\ell}\overbrace{\ell(\sin\theta\hat{\boldsymbol{% \jmath}}-\cos\theta\hat{\boldsymbol{k}})}^{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{dm/O}}\times\overbrace{(-\omega^{2}l\sin\theta\hat{% \boldsymbol{\jmath}})}^{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{dm}}\,% \overbrace{\frac{m}{2\ell}\,d\ell}^{dm}$
	$\displaystyle=$	$\displaystyle-\frac{m}{2\ell}\,\omega^{2}\sin\theta\cos\theta\hat{\boldsymbol{% \imath}}\;\int_{3\ell}^{5\ell}\ell^{2}\,d\ell$
	$\displaystyle=$	$\displaystyle-\frac{49}{3}\ell^{2}m\omega^{2}\sin\theta\cos\theta\hat{% \boldsymbol{\imath}}.$

By equating the two quantities ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$ ), we get

	$\displaystyle-4\ell\,mg\sin\theta\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-\frac{49}{3}\ell^{2}m\omega^{2}\sin\theta\cos\theta\hat{% \boldsymbol{\imath}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega^{2}$	$\displaystyle=$	$\displaystyle\frac{12g}{49\ell\cos\theta}.$

But for small $\theta$ , $\cos\theta\approx 1$ . Therefore,

\omega=\sqrt{\frac{12g}{49\ell}}=\frac{2\sqrt{3}}{7}\sqrt{\frac{g}{\ell}}

and the number of revolutions per unit time is $n=\frac{2\sqrt{3}}{14\pi}\sqrt{\frac{g}{\ell}}.$

Answer: $n=\frac{2\sqrt{3}}{14\pi}\sqrt{\frac{g}{\ell}}$

2.

Note, the natural frequency of this rod swinging back and forth as a simple pendulum turns out to be the same as the angular speed $\omega$ of the rotating system above for small $\theta$ .

Problems for 20.2 Dynamics of fixed axis rotation

20.2.1 Let $P(x,y,z)$ be a point mass on an abstract massless body that rotates at a constant rate $\omega$ about the $z$ -axis. Let $C(0,0,z)$ be the center of the circular path that $P$ describes during its motion.

(a)

Show that the rate of change of angular momentum of mass P about the origin $O$ is given by

$\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=-m\omega^{2}Rz\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$

where $R=\sqrt{x^{2}+y^{2}}$ is the radius of the circular path of point P and $m$ is its mass.
(b)

Is there any point on the axis of rotation ( $z$ -axis) about which $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ is zero. Is this point unique?

Filename:pfigure-s94h13p3

Figure 20.43

20.2.2 System of Particles going in a circle. Assume for the problems below that you have two mass points going in circles about a common axis at a fixed common angular rate.

(a)

2-D (particles in the plane, axis of rotation normal to the plane). Show, by calculating the sum in $\dot{\bf H}_{C}$ that the result is the same when using the two particles or when using a single particle at the center of mass (with mass equal to the sum of the two particles). You may do the general problem or you may pick particular appropriate locations for the masses, particular masses for the masses, a specific point C, a specific axis and a particular rotation rate.
(b)

3-D (particles not in a common plane normal to the axis of rotation). Using particular mass locations that you choose, show that 2 particles give a different value for $\dot{\bf H}_{C}$ than a single particle at the center of mass. [Moral: In general, $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}\neq{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{cm/C}\times m_{tot}{\accentset{\rightharpoonup}{\boldsymbol{% a}}}_{cm}$ .]

20.2.3 The rotating two-mass system shown in the figure has a constant angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=12\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{\imath}}$ . Assume that at the instant shown, the masses are in the $x y$ -plane. Take $m=1\,\mathrm{kg}$ and $\ell=1\,\mathrm{m}$ .

(a)

Find the linear momentum ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and its rate of change $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ for this system.
(b)

Find the angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ about point $O$ and its rate of change $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ for this system.

Filename:p-s96-p3-3

Figure 20.44

20.2.4 Particle attached to a shaft, no gravity. A shaft of negligible mass spins at constant rate. The motor at A imposes this rate if needed. The bearing at A prevents translation of its end of the shaft in any direction but causes no torques other than that of the motor (which causes a torque in the $\hat{\boldsymbol{k}}$ direction). The frictionless bearing at C holds that end of the shaft from moving in the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ directions but allows slip in the $\hat{\boldsymbol{k}}$ direction. The shaft is welded orthogonally to a bar, also of negligible mass, which is attached to a small ball of (non-negligible) mass of $1\,\mathrm{kg}$ . There is no gravity. The shaft is spinning counterclockwise (when viewed from the positive $\hat{\boldsymbol{k}}$ direction) at $3\,\mathrm{rad}/\mathrm{s}$ . In the configuration of interest the rod BD is parallel with the $\hat{\boldsymbol{\imath}}$ direction.

(a)

What is the tension in the rod BD? Answer: $T_{DB}=0.9\,{\mbox{N}}$ .
(b)

What are all the reaction forces at A and C? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{A}=-0.3\,{\mbox{N}}\hat{% \boldsymbol{\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}=-0.6\,{\mbox{N}}\hat{% \boldsymbol{\imath}}$
(c)

What is the torque that the motor at A causes on the shaft? Answer: $\mbox{Torque}_{motor}=0$ .
(d)

There are two ways to do problems (b) and (c): one is by treating the shaft, bar and mass as one system and writing the momentum balance equations. The other is to use the result of (a) with action-reaction to do statics on the shaft BC. Repeat problems (b) and (c) doing it the way you did not do it the first time.

Filename:bikefork1-ang-mom

Figure 20.45

20.2.5 Particle attached to a shaft, with gravity. Consider the same situation as in problem 20.2 with the following changes: There is gravity (pointing in the $\bf-i$ direction). At the time of interest the rod BD is parallel to the $\bf+j$ direction.

(a)

What is the force on D from the rod BD? (Note that the rod BD is not a ‘two-force’ member. Why not?)
(b)

What are all the reaction forces at A and C?
(c)

What is the torque that the motor at A causes on the shaft?
(d)

Do problems (b) and (c) again a different way.

20.2.6 A turntable rotates at the constant rate of $\omega=2\,\mathrm{rad}/\mathrm{s}$ about the $z$ axis. At the edge of the $2\,\mathrm{m}$ radius turntable swings a pendulum which makes an angle $\phi$ with the negative $z$ direction. The pendulum length is $\ell=10\,\mathrm{m}$ . The pendulum consists of a massless rod with a point mass $m=2\,\mathrm{kg}$ at the end. The gravitational constant is $g=10\,\mathrm{m}/\mathrm{s}^{2}$ pointing in the $-z$ direction. A perspective view and a side view are shown.

(a)

What possible angle(s) could the pendulum make in a steady motion. In other words, find $\phi$ for steady circular motion of the system. It turns out that this quest leads to a transcendental equation. You are not asked to solve this equation but you should have the equation clearly defined. Answer: Angles $\phi$ for steady circular motion are solutions to $10\tan\phi=8+40\sin\phi$ .
(b)

How many solutions are there? [hint, sketch the functions involved in your final equation.] Answer: $\phi=[1.3627\,{\rm rad},-0.2735\,{\rm rad},-1.2484\,{\rm rad},-2,9813\,{\rm rad}]$ , four solutions, found numerically. Graphically, the solutions are the intersections of the curves $f_{1}(\phi)=10\tan\phi$ and $f_{2}(\phi)=8+40\sin\phi$ between $-\pi$ and $\pi$ .
(c)

Sketch very approximately any equilibrium solution(s) not already sketched on the side view in the picture provided.

Filename:bikefork-ang-mom

Figure 20.46

20.2.7 Mr. X swings his daughter Y about the vertical axis at a constant rate $\omega=1\,\mathrm{rad}/\mathrm{s}$ . Y is merely one year old and weighs $12\,\mathrm{kg}$ . Assuming Y to be a point mass located at the center of mass, estimate the force on each of her shoulder joints.

Filename:summer95f-5-a

Figure 20.47

20.2.8 A block of mass $m=2\,\mathrm{kg}$ supported by a light inextensible string at point $O$ sits on the surface of a rotating cone as shown in the figure. The cone rotates about its vertical axis of symmetry at constant rate $\omega$ . Find the value of $\omega$ in $\,{\rm rpm}$ at which the block loses contact with the cone.

Filename:pfigure4-2-rp10

Figure 20.48

20.2.9 A small rock swings in circles at constant angular rate $\omega$ . It is hung by a string with length $\ell$ the other end of which is at the origin of a fixed coordinate system. The rock has mass $m$ and the gravitational constant is $g$ . The string makes an angle $\phi$ with the negative $z$ -axis. The mass rotates so that if you follow it with the fingers of your right hand, your right thumb points up the $z$ -axis. At the instant of interest the mass is passing through the $y z$ -plane.

(a)

How long does the rock take to make one revolution? How does this compare with the period of oscillation of a simple pendulum. Answer: Period of revolution $=\frac{2\pi}{\omega}$ , Period of simple pendulum (for small oscillations) $=2\pi\sqrt{\frac{L}{g}}$ , where $\ell$ is the length of the pendulum.
(b)
Find the following quantities in any order that pleases you. All solutions should be in terms of $m$ , $g$ , $\ell$ , $\phi$ , $\omega$ and the unit vectors $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ .
- •
  
  The tension in the string ( $T$ ).
- •
  
  The velocity of the rock ( ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ ).
- •
  
  The acceleration of the rock ( ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ ).
- •
  
  The rate of change of the angular momentum about the + $x$ axis $(dH_{x}/dt)$ .
Answer: $T=m\omega^{2}L$ (strange that $\phi$ drops out!), ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{rock}=-\omega L\sin\phi\hat{% \boldsymbol{\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{rock}=-\omega^{2}L\sin\phi\hat{% \boldsymbol{\jmath}}$ , $\frac{dH_{x}}{dt}=-m\omega L^{2}\sin\phi\cos\phi$ .

Filename:pfigure-blue-125-2

Figure 20.49

20.2.10 Consider the configuration in problem 20.2. What is the relation between the angle (measured from the vertical) at which the rock hangs and the speed at which it moves? Do an experiment with a string, weight, and stopwatch to check your calculation. Answer: $\phi=\cos^{-1}({\frac{g}{\omega^{2}L}})$ .

20.2.11 A small block of mass $m=5\,\mathrm{kg}$ sits on an inclined plane which is supported by a bearing at point $O$ as shown in the figure. At the instant of interest, the block is at a distance $\ell=0.5\,\mathrm{m}$ from where the surface of the incline intersects the axis of rotation. By experiments, the limiting rate of rotation of the turntable before the block starts sliding up the inclined plane is found to be $\omega=125\,{\rm rpm}$ . Find the coefficient of friction between the block and the incline.

Filename:pfigure-blue-68-1

Figure 20.50

20.2.12 Particle sliding in circles in a parabolic bowl. As if in a James Bond adventure in a big slippery radar bowl, a particle-like human with mass $m$ is sliding around in circles at speed $v$ . The equation describing the bowl is $z=CR^{2}=C(x^{2}+y^{2}$ ).

(a)

Find $v$ in terms of any or all of $R$ , $g$ , and $C$ .
(b)

Now say you are given $\omega,C$ and $g$ . Find $v$ and $R$ if you can. Explain any oddities.

Filename:pfigure-blue-58-1

Figure 20.51

20.2.13 Mass on a rotating inclined platform. A block of mass $m=5\,\mathrm{kg}$ is held on an inclined platform by a rod $B C$ as shown in the figure. The platform rotates at a constant angular rate $\omega$ . The coefficient of friction between the block and the platform is $\mu$ . $\theta=30^{o}$ , $\ell=1\,\mathrm{m}$ , and $g=9.81\,\mathrm{m}/\mathrm{s}^{2}$ . (a) Assuming $\omega=5\,\mathrm{rad}/\mathrm{s}$ and $\mu=0$ find the tension in the rod. (b) If $\mu=.3$ what is the range of $\omega$ for which the block would stay on the ramp without slipping even if there were no rod? Answer: (a) $T=-69.2\,{\mbox{N}}$ , (b) $1.64\,\mathrm{rad}/\mathrm{s}<\omega<3.47\,\mathrm{rad}/\mathrm{s}$

Filename:pfigure-blue-157-1

Figure 20.52

20.2.14 A circular plate of radius $r=10\,\mathrm{cm}$ rotates in the horizontal plane about the vertical axis passing through its center $O$ . Two identical point masses hang from two points on the plate diametrically opposite to each other as shown in the figure. At a constant angular speed $\omega$ , the two masses maintain a steady state angle $\theta$ from their vertical static equilibrium.

(a)

Find the linear speed and the magnitude of the centripetal acceleration of either of the two masses.
(b)

Find the steady state $\omega$ as a function of $\theta$ . Do the limiting values of $\theta$ , $\theta\rightarrow 0$ and $\theta\rightarrow\pi/2$ give sensible values of $\omega$ ?

Filename:summer95f-5

Figure 20.53

20.2.15 For the rod and mass system shown in the figure, assume that $\theta=60^{\circ}$ , $\omega\equiv\dot{\theta}=5\,\mathrm{rad}/\mathrm{s}$ , and $\ell=1\,\mathrm{m}$ . Find the acceleration of mass $B$ at the instant shown by

(a)

calculating ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ where ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is the position vector of point $B$ with respect to point $O$ , i.e., ${\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/O}$ , and by
(b)

calculating ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-R\dot{\theta}^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}$ where $R$ is the radius of the circular path of mass $B$ .

Filename:pfigure-blue-127-2

Figure 20.54

20.2.16 A spring of relaxed length $\ell_{0}$ (same as the diameter of the plate) is attached to the mid point of the rods supporting the two hanging masses.

(a)

Find the steady angle $\theta$ as a function of the plate’s rotation rate $\omega$ and the spring constant $k$ . (Your answer may include other given constants.)
(b)

Verify that, for $k=0$ , your answer for $\theta$ reduces to that in problem 20.53.
(c)

Assume that the entire system sits inside a cylinder of radius $20\,\mathrm{cm}$ and the axis of rotation is aligned with the longitudinal axis of the cylinder. The maximum operating speed of the rotating shaft is $200\,{\rm rpm}$ (i.e., $\omega_{max}=200\,{\rm rpm}$ ). If the two masses are not to touch the walls of the cylinder, find the required spring stiffness $k$ .

Filename:pfigure-s94h13p4

Figure 20.55

20.2.17 Flyball governors are used to control the flow of working fluid in steam engines, diesel engines, steam turbines, etc. A spring-loaded flyball governor is shown schematically in the figure. As the governor rotates about the vertical axis, the two masses tend to move outwards. The rigid and massless links that connect the outer arms to the collar, in turn, try to lift the central collar (mass $m/4$ ) against the spring. The central collar can move along the vertical axis but the top of the governor remains fixed. The mass of each ball is $m=5$ kg, and the length of each link is $\ell=0.25$ m. There are frictionless hinges at points $A$ , $B$ , $C$ , $D$ , $E$ , $F$ where the links are connected. Find the amount the string is elongated, $\Delta x$ , when the governor rotates at 100 rpm, given that the spring constant $k=500$ N/m, and that in this steady state, $\theta=90^{\rm o}$ . Answer: $\Delta x=-0.554\,\mathrm{m}$

Filename:pfigure-blue-90-2

Figure 20.56

20.2.18 Crooked rod. A $2\ell$ long uniform rod of mass $m$ is welded at its center to a shaft at an angle $\phi$ as shown. The shaft is supported by bearings at its ends and spins at a constant rate $\omega$ .

(a)

What are $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ ?
(b)

Calculate $\dot{\bf H}_{/O}$ using the integral definition of angular momentum. Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=-\frac{1}{12}m\omega^{2}\ell^{2}\sin\phi% \cos\phi\hat{\boldsymbol{k}}$ .
(c)

What are the six reaction components if there is no gravity? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{A}=(\frac{1}{12}m\omega^{2}\ell% ^{2}/d\sin\phi\cos\phi)\hat{\boldsymbol{\jmath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{B}=-{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{A}$ .
(d)

Repeat the calculations, including gravity $g$ . Which reactions change and by how much? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{A}=(\frac{1}{12}m\omega^{2}\ell% ^{2}/d\sin\phi\cos\phi+\frac{1}{2}mg)\hat{\boldsymbol{\jmath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{B}\not=-{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{A}$ .

Filename:s92f1p7

Figure 20.57

20.2.19 For the configuration in problem 20.2, find two positions along the rod where concentrating one-half the mass at each of these positions gives the same $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ .

20.2.20 Rod on a shaft. A uniform rod is welded to a shaft which is connected to frictionless bearings. The rod has mass $m$ and length $\ell$ and is attached at an angle $\gamma$ at its midpoint to the midpoint of the shaft. The length of the shaft is $2d$ . A motor applies a torque $M_{motor}$ along the axis of the shaft. At the instant shown in the figure, assume $m$ , $\ell$ , $d$ , $\omega$ and $M_{motor}$ are known and the rod lies in the $y z$ -plane.

(a)

Find the angular acceleration of the shaft.
(b)

The moment that the shaft applies to the rod.
(c)

The reaction forces at the bearings.

Filename:twodisks-ang-mom

Figure 20.58

Rod on a shaft.

20.2.21 Rod spins on a shaft. A uniform rod $\mathcal{R}$ with length $R$ and mass $m$ spins at constant rate $\omega$ about the $z$ axis. It is held by a hinge at A and a string at B. Neglect gravity. Find the tension in the string in terms of $\omega$ , $R$ and $m$ . Answer: $T=\frac{\sqrt{2}m\omega^{2}R}{6}$ .

Filename:pfigure-blue-90-1

Figure 20.59

20.2.22 A uniform narrow shaft of length $2\ell$ and mass $2M$ is bent at an angle $\theta$ half way down its length. It is spun at constant rate $\omega$ .

(a)

What is the magnitude of the force that must be applied at $O$ to keep the shaft spinning?
(b)

(harder) What is the magnitude of the moment that must be applied at $O$ to keep the shaft spinning?

Filename:pfigure-blue-110-1

Figure 20.60

Spinning rod. No gravity.

20.2.23 For the configuration in problem 20.2, find two positions along the bent portion of the shaft where concentrating one-half its mass at each of these positions gives the same $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ .

20.2.24 Rod attached to a shaft. A uniform $1\,\mathrm{kg}$ , $\sqrt{2}\,\mathrm{m}$ rod is attached to a shaft which spins at a constant rate of $1\,\mathrm{rad}/\mathrm{s}$ . One end of the rod is connected to the shaft by a ball-and-socket joint . The other end is held by two strings (or pin jointed rods) that are connected to a rigid cross bar. The cross bar is welded to the main shaft. Ignore gravity.

(a)

Draw a free-body diagram of the rod AB, including a bit of the strings.
(b)
Find the tension in the string DB. Two approaches:
- (i)
  
  Write the six equations of linear and angular momentum balance (about any point of your choosing) and solve them (perhaps using a computer) for the tension $T_{DB}$ .
- (ii)
  
  Write the equation of angular momentum balance about the axis AE. (That is, take the equation of angular momentum balance about point A and dot both sides with a vector in the direction of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{E/A}$ .) Use this calculation to find $T_{DB}$ and compare to method ii.
(c)

Find the reaction force (a vector) at A.

Filename:pfigure-blue-107-1

Figure 20.61

Uniform rod attached to a shaft.

20.2.25 3D, crooked bar on shaft falls down. A massless, rigid shaft is connected to frictionless hinges at A and B. A bar with mass $M_{b}$ and length $\ell$ is welded to the shaft at $45^{o}$ . At time $t=0$ the assembly is tipped a small angle $\theta_{0}$ from the $y z$ -plane but has no angular velocity. It then falls under the action of gravity. What is $\theta$ at $t=t_{1}$ (assuming $t_{1}$ is small enough so that $\theta_{1}$ is much less than $1$ )?

Filename:pfigure-s94h14p4

Figure 20.62

A crooked rod welded to a shaft falls down.

20.3 Moment of inertia matrices $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$

We now know how to find the velocity and acceleration of every bit of mass on a rigid body as it spins about a fixed axis. It is just a matter of doing integrals or sums to calculate the various motion quantities (momenta, energy) of interest. As the body moves and rotates the region of integration and the values of the integrands change. So, in principle, in order to analyze a rigid body one has to evaluate a different integral or sum at every different configuration. But there is a shortcut. A big sum (over all atoms, say), or a difficult integral is reduced to a simple multiplication using the moment of inertia. In three-dimensions this multiplication is a matrix multiplication.

$\bf[I]$ , the moment of inertia matrix^†^†margin: In fact the moment of inertia matrix for a given object depends on what reference point is used. Most commonly when people say ‘the’ moment of inertia they mean to use the center-of-mass as the reference point. For clarity this moment of inertia matrix is often written as $[\boldsymbol{I}^{\mathrm{cm}}]$ in this book. If a different reference point, say point $O$ is used, the matrix is notated as $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ . , is defined for the purpose of simplifying the expressions for the angular momentum, the rate of change of angular momentum, and the energy of a system which moves like a rigid body.

First review the situation for flat objects in planar motion. A flat object spinning with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ in the $x y$ plane has a mass distribution which gives a polar moment of inertia $I^{cm}_{zz}$ or just ‘ $I$ ’ so that:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle I\omega\hat{\boldsymbol{k}}$	(20.27)
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	(20.28)
$\displaystyle{E_{\mathrm{K}}}_{/cm}$	$\displaystyle=$	$\displaystyle{1\over 2}\omega^{2}I.$	(20.29)

Now, for a rigid body spinning in 3-D about a fixed axis with the angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ we need matrix multiplication, where the determination of the needed matrix is the central topic of this section.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle[I^{cm}]\cdot{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}$	(20.30)
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times% \underbrace{[I^{cm}]\cdot{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}}% _{{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}}+[I^{cm}]\cdot{% \dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}$	(20.31)
$\displaystyle{E_{\mathrm{K}}}_{/cm}$	$\displaystyle=$	$\displaystyle{1\over 2}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot% ([I^{cm}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}})={1\over 2}{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}.\par$	(20.32)

In detail, for example,

\left[{\begin{matrix}H_{x/cm}\cr H_{y/cm}\cr H_{z/cm}\cr\end{matrix}}\right]\ % =\ \left[{\begin{matrix}I^{cm}_{xx}&I^{cm}_{xy}&I^{cm}_{xz}\cr I^{cm}_{xy}&I^{% cm}_{yy}&I^{cm}_{yz}\cr I^{cm}_{xz}&I^{cm}_{yz}&I^{cm}_{zz}\cr\end{matrix}}% \right]_{xyz}\cdot\left[{\begin{matrix}{\omega}_{x}\cr{\omega}_{y}\cr{\omega}_% {z}\cr\end{matrix}}\right]

(20.33)

where ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}=H_{x/cm}\hat{% \boldsymbol{\imath}}+H_{y/cm}\hat{\boldsymbol{\jmath}}+H_{z/cm}\hat{% \boldsymbol{k}}$ . Note that the 2-D results are a special case of the 3-D results because, as you will soon see, for 2-D objects $I_{xz}=I_{yz}=0$ . We postpone the use of these equations until section 20.4.

The moment of inertias in 3-D: $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$

For the study of three-dimensional mechanics, including the simple case of constant rate rotation about a fixed axis, one often makes use of the moment of inertia matrix, defined below and motivated by the box 20.3 on page 20.3.

The distances $x, y, z$ in the formulas below are the $x, y, z$ components of the position of mass relative to a coordinate system which has either the center-of-mass ( $c m$ ) or the point O as its origin. ^†^†margin: Caution: While we have $I_{xy}=-\int xy\,dm$ , some old books define $I_{xy}=\int xy\,dm$ . They then have minus signs in front of the off-diagonal terms in the moment of inertia matrix. They would say $I_{12}=-I_{xy}$ . The numerical values in the matrix they write is the same as in the one we write. They just have a different sign convention in the definition of the components.

	$\displaystyle[\boldsymbol{I}]\$	$\displaystyle=$	$\displaystyle\ \left[{\begin{matrix}I_{xx}&I_{xy}&I_{xz}\cr I_{xy}&I_{yy}&I_{% yz}\cr I_{xz}&I_{yz}&I_{zz}\cr\end{matrix}}\right]$		(20.34)
		$\displaystyle=$	$\displaystyle\left[{\begin{matrix}\int(y^{2}+z^{2})\,dm&-\int xy\,dm&-\int xz% \,dm\cr-\int xy\,dm&\int(x^{2}+z^{2})\,dm&-\int yz\,dm\cr-\int xz\,dm&-\int yz% \,dm&\int(x^{2}+y^{2})\,dm\cr\end{matrix}}\right]$		(20.35)

If all mass is on the $x y$ plane then it is clear that $I_{xz}=I_{yz}=0$ since $z=0$ for the whole $x y$ plane. If rotation is also about the $z$ -axis then ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ .

Applying the formulas above we find that, if all of the mass is in the $x y$ -plane and rotation is about the $z$ -axis, the only relevant non-zero term in $[\boldsymbol{I}]$ is $I_{zz}=\int(x^{2}+y^{2})\ dm$ . And, $I_{xx}$ , $I_{yy}$ , and $I_{xy}$ don’t contribute to ${\accentset{\rightharpoonup}{\boldsymbol{H}}}$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ , or ${E_{\mathrm{K}}}$ . In this manner you can check that the three dimensional equations, when applied to two-dimensional bodies, give the same results that we found directly for two-dimensional bodies.

Example: Moment of inertia matrix for a uniform sphere

A sphere is a special shape which is, naturally enough, spherically symmetric. Therefore,

I_{xx}^{cm}=I_{yy}^{cm}=I_{zz}^{cm}

and

I_{xy}^{cm}=I_{xz}^{cm}=I_{yz}^{cm}=0.

So, all we need is $I_{xx}^{cm}$ or $I_{yy}^{cm}$ or $I_{zz}^{cm}$ .

Filename:pfigure-blue-112-1

Figure 20.63:

Here is the trick:

$\displaystyle I_{xx}^{cm}$	$\displaystyle=$	$\displaystyle\frac{1}{3}(I_{xx}^{cm}+I_{yy}^{cm}+I_{zz}^{cm})$
	$\displaystyle=$	$\displaystyle\frac{1}{3}\left[\int(y^{2}+z^{2})dm\right.+\int(x^{2}+z^{2})dm+% \left.\int(x^{2}+y^{2})dm\right]$
	$\displaystyle=$	$\displaystyle\frac{2}{3}\int(x^{2}+y^{2}+z^{2})dm$
	$\displaystyle=$	$\displaystyle\frac{2}{3}\int r^{2}\ dm$
	$\displaystyle=$	$\displaystyle\frac{2}{3}\int_{0}^{R}r^{2}(4\rho\pi r^{2}\ dr)$
	$\displaystyle=$	$\displaystyle\frac{8}{3}\rho\pi\int_{0}^{R}r^{4}\ dr$
	$\displaystyle=$	$\displaystyle\frac{8}{15}\rho\pi R^{5}$
	$\displaystyle=$	$\displaystyle\frac{2}{5}mR^{2}\,\,\,\,\,\,\,\,\,\,\ (m=\frac{4}{3}\rho\pi R^{3% }).$

So,

[\boldsymbol{I}^{\mathrm{cm}}]=\frac{2}{5}mR^{2}\left[\begin{array}[]{ccc}1&0&% 0\\ 0&1&0\\ 0&0&1\end{array}\right].

The parallel axis theorem for rigid bodies in three dimensions

The 3-D parallel axis theorem is stated below and in the table on the inside back cover. It is derived in box 20.3 on page 20.3. The parallel axis theorem for rigid bodies in three dimensions is the equation

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=[\boldsymbol{I}^{\mathrm{cm}}]% +m\left[\begin{array}[]{ccc}y_{cm/o}^{2}+z_{cm/o}^{2}&-x_{cm/o}y_{cm/o}&-x_{cm% /o}z_{cm/o}\\ -x_{cm/o}y_{cm/o}&x_{cm/o}^{2}+z_{cm/o}^{2}&-y_{cm/o}z_{cm/o}\\ -x_{cm/o}z_{cm/o}&-y_{cm/o}z_{cm/o}&x_{cm/o}^{2}+y_{cm/o}^{2}\\ \end{array}\right]

(20.36)

In this equation, $x_{cm/o}$ , $y_{cm/o}$ , and $z_{cm/o}$ are the $x$ , $y$ , and $z$ coordinates, respectively, of the center-of-mass defined with respect to a coordinate system whose origin is located at some point O not at the center-of-mass $c m$ . That is, if you know $[\boldsymbol{I}^{\mathrm{cm}}]$ , you can find $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ without doing any more integrals or sums. Like the 2-D parallel axis theorem, the primary utility of the 3-D parallel axis theorem is for the determination of $\bf[I]$ for an object that is a composite of simpler objects. Such are not beyond the scope of this book in principle. But in fact, given the finite time available for calculation, we do not leave much time for practice of this tedious but routine calculation.

Matrices and tensors

We have just introduced the 3 by 3 moment of inertia matrix $[\boldsymbol{I}]$ . We will find it in expressions having to do with angular momentum sitting next to either a vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ or a vector ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ : $\quad[\boldsymbol{I}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \quad\mbox{or}\quad[\boldsymbol{I}]\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\alpha}}}$ . What we mean by this expression is the three element column vector that comes from matrix multiplication of the matrix $[\boldsymbol{I}]$ and the column vector for ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , $\left[\begin{array}[]{c}\omega_{x}\\ \omega_{y}\\ \omega_{z}\end{array}\right]$ , an expression that only makes sense if everyone knows what bases are being used.

More formally, and usually only in more advanced treatments, people like to define a coordinate-free quantity called the tensor $\underline{\underline{\boldsymbol{I}}}$ . Then we would have

\underline{\underline{\boldsymbol{I}}}\mathord{\cdot}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}

by which we would mean the vector whose components would be found by $[\boldsymbol{I}]\cdot\left[\begin{array}[]{c}\omega_{x}\\ \omega_{y}\\ \omega_{z}\end{array}\right]$ .

Eigenvectors and Eigenvalues

A square matrix $\left[\boldsymbol{A}\right]$ when multiplied by a column vector $\left[\boldsymbol{v}\right]$ yields a new vector $\left[\boldsymbol{w}\right]$ . A given matrix has a few special vectors, somehow characteristic of that matrix, called eigenvectors. The vector ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ is an eigenvector of $\left[\boldsymbol{A}\right]$ if

\left[\boldsymbol{A}\right]\cdot\left[\boldsymbol{v}\right]\mbox{ is parallel % to }\left[\boldsymbol{v}\right].

In other words, if

\left[\boldsymbol{A}\right]\cdot\left[\boldsymbol{v}\right]=\lambda\left[% \boldsymbol{v}\right]

for some $\lambda$ .

The scalar $\lambda$ is called the eigenvalue associated with the eigenvector $\left[\boldsymbol{v}\right]$ of the matrix $\left[\boldsymbol{A}\right]$ . The eigen-values and eigen-vectors of a matrix are found with a single command in many computer math programs. In statics you had little or no use for eigen-values and eigen-vectors. In dynamics, eigenvectors and eigenvalues are useful for understanding dynamic balance, 3-D rigid body rotations, and normal mode vibrations.

Eigenvectors of $[\boldsymbol{I}]$

The moment of inertia matrix is always a symmetric matrix. This symmetry means that $[\boldsymbol{I}]$ always has a set of three mutually orthogonal eigenvectors. The importance of the eigenvectors of $[\boldsymbol{I}]$ will be discussed in section 20.5 on dynamic balance. Sometimes a pair of the eigenvalues are equal to each other implying that any vector in the plane of the corresponding eigenvectors is also an eigenvector.

If the physical object has any natural symmetry directions these directions will usually manifest themselves in the dynamics of the body as being in the directions of the eigenvectors of the object’s moment of inertia matrix. For example, the dotted lines on fig. 2.34 are all in directions of eigenvectors for the objects shown. But even if an object is wildly asymmetric in shape, its moment of inertia matrix is always symmetric and thus all objects have moment of inertia matrices with at least three different eigenvectors at least three of which are mutually orthogonal.

Properties of $[\boldsymbol{I}]$

For those with experience with linear algebra various properties of the moment of inertia matrix $[\boldsymbol{I}]$ are worth noting (although not worth proving here). Unless all mass is distributed on one straight line, the moment of inertia matrix is invertible (it is non-singular and has rank 3). Further, when invertible it is positive definite. In the special case that all the mass is on some straight line, the moment of inertia matrix is non-invertible and only positive semi-definite. The positive (semi) definiteness of the moment of inertia matrix is equivalent to the statement that the rotational kinetic energy of a body is always equal to or greater than zero. Finally, the eigenvalues of the moment of inertia matrix are all positive and have the property that no one can be greater than the sum of the other two (the same inequalities are satisfied by the lengths of the sides of a triangle, the “triangle inequality”).

Box 20.1 Discovering the moment of inertia matrix

Derivation 1

Here we present a direct derivation of the moment of inertia matrix; that is, a derivation in which the moment of inertia matrix arises as a convenient short hand. Assume a rigid body is moving in such a way that point $O$ is fixed (i.e., it is either on the line of a hinge or a ball-and-socket joint).

Filename:tfigure4-inertia-fig0

The most basic kinematic relation for a rigid body is that

{\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}

where ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}=x\hat{\boldsymbol{\imath}}+% y\hat{\boldsymbol{\jmath}}+z\hat{\boldsymbol{k}}$ is the position of a point on the body relative to $O$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , the angular velocity of the body.

Now, we tediously calculate and arrange the terms in the angular momentum about point $O$ ,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}dm$
	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{/O})dm$
	$\displaystyle=$	$\displaystyle\int(x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}+z\hat{% \boldsymbol{k}})\times$
		$\displaystyle\left[(\omega_{x}\hat{\boldsymbol{\imath}}+\omega_{y}\hat{% \boldsymbol{\jmath}}+\omega_{z}\hat{\boldsymbol{k}})\times(x\hat{\boldsymbol{% \imath}}+y\hat{\boldsymbol{\jmath}}+z\hat{\boldsymbol{k}})\right]dm$
	$\displaystyle=$	$\displaystyle\int(x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}+z\hat{% \boldsymbol{k}})\times$
		$\displaystyle\left[(\omega_{y}z-\omega_{z}y)\hat{\boldsymbol{\imath}}+(\omega_% {z}x-\omega_{x}z)\hat{\boldsymbol{\jmath}}+(\omega_{x}y-\omega_{y}x)\hat{% \boldsymbol{k}}\right]dm$
	$\displaystyle=$	$\displaystyle\int\left[\left(y(\omega_{x}y-\omega_{y}x)-z(\omega_{z}x-\omega_{% x}z)\right)\hat{\boldsymbol{\imath}}\right.$
		$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ +\left(z(\omega_{y}z-\omega% _{z}y)-x(\omega_{x}y-\omega_{y}x)\right)\hat{\boldsymbol{\jmath}}$
		$\displaystyle\left.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ +\left(x(\omega_{z}x-% \omega_{x}z)-y(\omega_{y}z-\omega_{z}y)\right)\hat{\boldsymbol{k}}\right]dm$
	$\displaystyle=$	$\displaystyle\int\left[\left((y^{2}+z^{2})\omega_{x}-xy\omega_{y}-xz\omega_{z}% \right)\hat{\boldsymbol{\imath}}\right.$
		$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ +\left(-yx\omega_{x}+(z^{2}+x% ^{2})\omega_{y}-yz\omega_{z}\right)\hat{\boldsymbol{\jmath}}$
		$\displaystyle\left.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ +\left(-zx\omega_{x}-xy% \omega_{y}+(x^{2}+y^{2})\omega_{z}\right)\hat{\boldsymbol{k}}\right]dm$

Since the integral is over the mass and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is constant over the body, we can pull ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ out of the integral so that we may write the equation in matrix form. Writing ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O}$ as a column vector, we can rewrite the last equation as

\begin{array}[]{ll}\left[{\begin{matrix}H_{O_{x}}\cr H_{O_{y}}\cr H_{O_{z}}\cr% \end{matrix}}\right]=&\\ &\\ \underbrace{\left[{\begin{matrix}\int(y^{2}+z^{2})\,dm&-\int xy\,dm&-\int xz\,% dm\cr-\int xy\,dm&\int(x^{2}+z^{2})\,dm&-\int yz\,dm\cr-\int xz\,dm&-\int yz\,% dm&\int(x^{2}+y^{2})\,dm\cr\end{matrix}}\right]}_{[\boldsymbol{I}^{% \scriptscriptstyle\mathrm{O}}]}\cdot\left[{\begin{matrix}{\omega}_{x}\cr{% \omega}_{y}\cr{\omega}_{z}\cr\end{matrix}}\right].&\end{array}

Finally, defining $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ by the matrix above, we can compactly write

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O% }}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}

assuming $O$ is a fixed point on the body where we represent ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ in terms of $x$ , $y$ , and $z$ components.

Center-of-mass inertia matrix

For any system moving, distorting, and rotating any crazy way, we have the general result that

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=\overbrace{{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{cm/O}\times m_{tot}{\accentset{\rightharpoonup}{\boldsymbol{% v}}}_{cm}}}^{\text{Contribution of the system to ${\accentset{\rightharpoonup}% {\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}% }$ if treated as a particle at the system center-of-mass}}+\overbrace{\int({\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{/cm}\times\text{${\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{/cm}=({\accentset{\rightharpoonup}{% \boldsymbol{v}}}-{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm})$}{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{/cm})dm}^{{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}}

as you can verify by substituting ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{cm}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{/cm}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{cm}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/cm}$ into the general definition of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=\int{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{/O}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}dm$ . For a rigid body, we have

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{/cm}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{/cm}.

So, by a derivation essentially identical to that for ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ , we get

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}=[\boldsymbol{I}^{% \mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}

with $[\boldsymbol{I}^{\mathrm{cm}}]$ being defined using $x$ , $y$ , and $z$ , as the distances from the center of mass rather than from point $O$ . So, for a rigid body in general motion, we can find the angular momentum by

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{cm/O}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}m_{tot}+[% \boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}

(20.37)

Comment (aside)

In the special case that the body is rotating about point $O$ , we also have

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O% }}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}.

(20.38)

You will see, if you look at the parallel axis theorem, that these two expressions 20.37 and 20.38 do in fact agree.

Derivation 2

Here, we present a less direct but perhaps more intuitive derivation of the moment of inertia matrix. We start with the special case of a 3-D rigid body spinning in circles at constant rate about a fixed axis.

To see from where the moment of inertia matrix comes, we will first calculate the angular momentum about point $O$ of a general 3-D rigid body spinning about the $z$ -axis with constant rate $\dot{\theta}\equiv const.=\omega_{z}$ or ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega_{z}\hat{\boldsymbol{% k}}$ . We will refer to this case as (1).

Filename:tfigure4-inertia-fig2

Starting with the definition of angular momentum, we get

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}dm$
		$\displaystyle=$	$\displaystyle\int(z\hat{\boldsymbol{k}}+{\accentset{\rightharpoonup}{% \boldsymbol{R}}})\times(\dot{\theta}R\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta})dm.$

Filename:tfigure4-inertia-fig1

Looking down the $z$ -axis in the figure, we see that

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{R}}}$	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\overbrace{R\cos% \theta}^{x}\hat{\boldsymbol{\imath}}+\overbrace{R\sin\theta}^{y}\hat{% \boldsymbol{\jmath}},\quad{\rm or}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{R}}}$	$\displaystyle=$	$\displaystyle x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}.$

where $R=\sqrt{x^{2}+y^{2}}$ .

To compute the cross product in the integrand, we need

	$\displaystyle\hat{\boldsymbol{k}}\times\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{k}}\times(-\sin\theta\hat{\boldsymbol{\imath}}+% \cos\theta\hat{\boldsymbol{\jmath}})=-\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!R}\,\,\,\,{\rm and}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{R}}}\times\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}=R\hat{\boldsymbol{k}}.$

Therefore, we now have

\begin{array}[]{ll}{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O}=\int-z% \dot{\theta}R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}dm+\int\dot{\theta}R^% {2}\hat{\boldsymbol{k}}dm&\\ &\\ \hskip 14.45377pt=\dot{\theta}\left[-\int z(\underbrace{R\cos\theta}_{x}\hat{% \boldsymbol{\imath}}+\underbrace{R\sin\theta}_{y}\hat{\boldsymbol{\jmath}})dm+% \int(x^{2}+y^{2})\hat{\boldsymbol{k}}dm\right]&\\ &\\ \hskip 14.45377pt=\dot{\theta}\left[\left(-\int zx\ dm\right)\hat{\boldsymbol{% \imath}}+\left(-\int zy\ dm\right)\hat{\boldsymbol{\jmath}}+\left(\int(x^{2}+y% ^{2})dm\right)\hat{\boldsymbol{k}}\right].&\end{array}

To ‘un-clutter’ this expression, let’s define the following:

$\displaystyle I^{O}_{xz}$	$\displaystyle=$	$\displaystyle-\int xz\ dm$
$\displaystyle I^{O}_{yz}$	$\displaystyle=$	$\displaystyle-\int yz\ dm$
$\displaystyle I^{O}_{zz}$	$\displaystyle=$	$\displaystyle\int(x^{2}+y^{2})\ dm$

So, now, we have for case (1)

({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O})_{1}=I^{O}_{xz}\omega_{z}% \hat{\boldsymbol{\imath}}+I^{O}_{yz}\omega_{z}\hat{\boldsymbol{\jmath}}+I^{O}_% {zz}\omega_{z}\hat{\boldsymbol{k}}.

The substitutions we have defined form the elements of the third column of the inertia matrix, as we will see below in the general case. Let’s now move on to general 3-D rigid body motion and infer the first and second columns of the inertia matrix.

In general, the angular velocity of a rigid body is given by

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega_{x}\hat{\boldsymbol{% \imath}}+\omega_{y}\hat{\boldsymbol{\jmath}}+\omega_{z}\hat{\boldsymbol{k}}

So far, we have considered the special case above, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega_{z}\hat{\boldsymbol{% k}}$ . But, we could have looked at ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega_{x}\hat{\boldsymbol{% \imath}}$ , case (2), and, similarly, would obtain instead the following angular momentum about point $O$

({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O})_{2}=I^{O}_{xx}\omega_{x}% \hat{\boldsymbol{\imath}}+I^{O}_{yx}\omega_{x}\hat{\boldsymbol{\jmath}}+I^{O}_% {zx}\omega_{x}\hat{\boldsymbol{k}}.

Likewise, for ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega_{y}\hat{\boldsymbol{% \jmath}}$ , case (3), we would obtain

({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O})_{3}=I^{O}_{xy}\omega_{y}% \hat{\boldsymbol{\imath}}+I^{O}_{yy}\omega_{y}\hat{\boldsymbol{\jmath}}+I^{O}_% {zy}\omega_{y}\hat{\boldsymbol{k}}.

Finally, for ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega_{x}\hat{\boldsymbol{% \imath}}+\omega_{y}\hat{\boldsymbol{\jmath}}+\omega_{z}\hat{\boldsymbol{k}}$ , we obtain

\begin{array}[]{lll}{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O}&=&({% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{O})_{1}+({\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{O})_{2}+({\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{O})_{3}\\ &=&\phantom{+\ }I^{O}_{xz}\omega_{z}\hat{\boldsymbol{\imath}}+I^{O}_{yz}\omega% _{z}\hat{\boldsymbol{\jmath}}+I^{O}_{zz}\omega_{z}\hat{\boldsymbol{k}}\\ &&+\ I^{O}_{xx}\omega_{x}\hat{\boldsymbol{\imath}}+I^{O}_{yx}\omega_{x}\hat{% \boldsymbol{\jmath}}+I^{O}_{zx}\omega_{x}\hat{\boldsymbol{k}}\\ &&+\ I^{O}_{xy}\omega_{y}\hat{\boldsymbol{\imath}}+I^{O}_{yy}\omega_{y}\hat{% \boldsymbol{\jmath}}+I^{O}_{zy}\omega_{y}\hat{\boldsymbol{k}}.\end{array}

Collecting components, we get

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O}=H_{O_{x}}\hat{\boldsymbol{% \imath}}+H_{O_{y}}\hat{\boldsymbol{\jmath}}+H_{O_{z}}\hat{\boldsymbol{k}}

where

	$\displaystyle H_{O_{x}}=I^{O}_{xx}\omega_{x}+I^{O}_{xy}\omega_{y}+I^{O}_{xz}% \omega_{z}$
	$\displaystyle H_{O_{y}}=I^{O}_{yx}\omega_{x}+I^{O}_{yy}\omega_{y}+I^{O}_{yz}% \omega_{z}$
	$\displaystyle H_{O_{z}}=I^{O}_{zx}\omega_{x}+I^{O}_{zy}\omega_{y}+I^{O}_{zz}% \omega_{z}.$

We can combine the above results into a matrix representation. Representing ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{O}$ as a column vector, we can re-write the above set of three equations as a product of a matrix and the angular velocity written as a column vector.

\left[{\begin{matrix}H_{O_{x}}\cr H_{O_{y}}\cr H_{O_{z}}\cr\end{matrix}}\right% ]\ =\ \left[{\begin{matrix}I^{O}_{xx}&I^{O}_{xy}&I^{O}_{xz}\cr I^{O}_{yx}&I^{O% }_{yy}&I^{O}_{yz}\cr I^{O}_{zx}&I^{O}_{zy}&I^{O}_{zz}\cr\end{matrix}}\right]% \cdot\left[{\begin{matrix}{\omega}_{x}\cr{\omega}_{y}\cr{\omega}_{z}\cr\end{% matrix}}\right]

We define the coefficient matrix above to be the moment of inertia matrix about point $O$

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=\left[{\begin{matrix}I^{O}_{xx% }&I^{O}_{xy}&I^{O}_{xz}\cr I^{O}_{yx}&I^{O}_{yy}&I^{O}_{yz}\cr I^{O}_{zx}&I^{O% }_{zy}&I^{O}_{zz}\cr\end{matrix}}\right].

whose components are

\begin{array}[]{ll}I^{O}_{xx}=\int(y^{2}+z^{2})dm\,\ I^{O}_{xy}=-\int xy\ dm\,% \,\,\,\,\,\,\,\,\ I^{O}_{xz}=-\int xz\ dm\\ I^{O}_{yx}=-\int yx\ dm\,\,\,\,\,\,\,\,\,\ I^{O}_{yy}=\int(x^{2}+z^{2})dm\,\ I% ^{O}_{yz}=-\int yz\ dm\\ I^{O}_{zx}=-\int zx\ dm\,\,\,\,\,\,\,\,\,\ I^{O}_{zy}=-\int zy\ dm\,\,\,\,\,\,% \,\,\,\ I^{O}_{zz}=\int(x^{2}+y^{2})dm.\end{array}

By inspection, one can see that $I^{O}_{xy}=I^{O}_{yx}$ , $I^{O}_{xz}=I^{O}_{zx}$ , and $I^{O}_{yz}=I^{O}_{zy}$ . Thus, the inertia matrix is symmetric; i.e., $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]^{T}=[\boldsymbol{I}^{% \scriptscriptstyle\mathrm{O}}]$ . So, there are always at most only six, not nine, independent components in the inertia matrix to compute.

Box 20.2 3-D parallel axis theorem

In three dimensions, the two matrices $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ and $[\boldsymbol{I}^{\mathrm{cm}}]$ are related to each other in a way similar to the two-dimensional case. Since the inertia matrix has six independent entries in it, the derivation involves six integrals. Let’s look at a typical term on the diagonal, say, $I^{O}_{zz}$ and a typical off-diagonal term, say, $I^{O}_{xy}$ . The calculation for the other terms is similar with a simple change of letters in the subscript notation.

First, $I^{O}_{zz}=\int(x_{/O}^{2}+y_{/O}^{2})dm=I^{cm}_{zz}+m(x_{cm/O}^{2}+y_{cm/O}^{% 2})$ by exactly the same reasoning used to derive the 2-D parallel axis theorem. We cannot do the last line in that derivation, however, since $r_{cm/O}^{2}=x_{cm/O}^{2}+y_{cm/O}^{2}+z_{cm/O}^{2}\neq x_{cm/O}^{2}+y_{cm/O}^% {2}$ , because now, for three-dimensional objects, $z_{cm/O}\neq 0$ .

Now, let’s look at an off-diagonal term.

$\displaystyle I^{O}_{xy}$	$\displaystyle=$	$\displaystyle-\int x_{/O}y_{/O}dm$
	$\displaystyle=$	$\displaystyle-\int\overbrace{(x_{cm/O}+x_{/cm}}^{x_{/O}})(\overbrace{y_{cm/O}+% y_{/cm}}^{y_{/O}})dm$
	$\displaystyle=$	$\displaystyle-x_{cm/O}y_{cm/O}\underbrace{\int dm}_{m}-y_{cm/O}\underbrace{% \int x_{/cm}dm}_{0}$
		$\displaystyle-x_{cm/O}\underbrace{\int y_{/cm}dm}_{0}-\underbrace{\int x_{/cm}% y_{/cm}dm}_{-I^{cm}_{xy}}$
	$\displaystyle=$	$\displaystyle-mx_{cm/O}y_{cm/O}+I^{cm}_{xy}$

Similarly, we can calculate the other terms to get the whole 3-D parallel axis theorem.

\begin{array}[]{ll}[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=[% \boldsymbol{I}^{\mathrm{cm}}]+&\\ &\\ m\left[\begin{array}[]{lll}y_{cm/o}^{2}+z_{cm/o}^{2}&-x_{cm/o}y_{cm/o}&-x_{cm/% o}z_{cm/o}\\ -x_{cm/o}y_{cm/o}&x_{cm/o}^{2}+z_{cm/o}^{2}&-y_{cm/o}z_{cm/o}\\ -x_{cm/o}z_{cm/o}&-y_{cm/o}z_{cm/o}&x_{cm/o}^{2}+y_{cm/o}^{2}\\ \end{array}\right].&\end{array}

Again, one can think of this result as follows. The moment of inertia matrix about point $O$ is the same as that for parallel axes through the center-of-mass plus the moment of inertia matrix for a point mass at the center-of-mass.

Relation between 2-D and 3-D parallel axis theorems

The (3,3) (lower right corner) element in the matrix of the 3-D parallel axis theorem is the 2-D parallel axis theorem.

SAMPLE 20.9

Filename:tfigure8-alt-app2c

Figure 20.64:

For the dumbbell shown in Figure 20.64, take $m=0.5\,\mathrm{kg}$ and $\ell=0.4\,\mathrm{m}$ . Given that at the instant shown $\theta=30^{{}^{\circ}}$ and the dumbbell is in the $y z$ -plane, find the moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ , where O is the midpoint of the dumbbell.

Solution The dumbbell is made up of two point masses. Therefore we can calculate $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ for each mass using the formula from the table on the inside back cover of the text and then adding the two matrices to get $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ for the dumbbell.
Now, from Table 4.9 of the text,

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=m\left[\begin{array}[]{ccc}y^{% 2}+z^{2}&-xy&-xz\\ -xy&x^{2}+z^{2}&-yz\\ -xz&-yz&x^{2}+y^{2}\end{array}\right]

For mass 1 (shown in Figure 20.65)

x=0,\quad y=-\frac{\ell}{2}\cos\theta,\quad z=\frac{\ell}{2}\sin\theta.

Therefore,

Filename:sfig4-6-4a

Figure 20.65:

$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]_{\rm{mass1}}$	$\displaystyle=$	$\displaystyle m\left[\begin{array}[]{ccc}\frac{\ell^{2}}{4}&0&0\\ 0&\frac{\ell^{2}}{4}\sin^{2}\theta&\frac{\ell^{2}}{4}\cos\theta\sin\theta\\ 0&\frac{\ell^{2}}{4}\cos\theta\sin\theta&\frac{\ell^{2}}{4}\cos^{2}\theta\end{% array}\right]$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{kg}\left[\begin{array}[]{ccc}0.04\,\mathrm{m}^{2}&0% &0\\ 0&0.04\,\mathrm{m}^{2}\mathord{\cdot}\frac{1}{4}&0.04\,\mathrm{m}^{2}\mathord{% \cdot}\frac{\sqrt{3}}{4}\\ 0&0.04\,\mathrm{m}^{2}\mathord{\cdot}\frac{\sqrt{3}}{4}&0.04\,\mathrm{m}^{2}% \mathord{\cdot}\frac{3}{4}\end{array}\right]$
	$\displaystyle=$	$\displaystyle 0.02\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}\left[\begin{% array}[]{ccc}1&0&0\\ 0&\frac{1}{4}&\frac{\sqrt{3}}{4}\\ 0&\frac{\sqrt{3}}{4}&\frac{1}{4}\end{array}\right]$

Similarly for mass 2,

x=0,\quad y=\frac{\ell}{2}\cos\theta,\quad z=-\frac{\ell}{2}\sin\theta

\hbox to36.135pt{\hfil$\Rightarrow$\hfil}[\boldsymbol{I}^{\scriptscriptstyle% \mathrm{O}}]_{\rm{mass2}}=0.02\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}% \left[\begin{array}[]{ccc}1&0&0\\ 0&\frac{1}{4}&\frac{\sqrt{3}}{4}\\ 0&\frac{\sqrt{3}}{4}&\frac{1}{4}\end{array}\right].

Therefore,

	$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$	$\displaystyle=$	$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]_{\rm{mass1}}+[% \boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]_{\rm{mass2}}$
		$\displaystyle=$	$\displaystyle 0.04\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}\left[\begin{% array}[]{ccc}1&0&0\\ 0&\frac{1}{4}&\frac{\sqrt{3}}{4}\\ 0&\frac{\sqrt{3}}{4}&\frac{1}{4}\end{array}\right].$

SAMPLE 20.10

Filename:sfig4-6-3

Figure 20.66:

A uniform rod of mass $m=2\,\mathrm{kg}$ and length $\ell=\frac{1}{2}\,\mathrm{m}$ is pivoted at one of its ends. At the instant shown, the rod is in the $x y$ -plane and makes an angle $\theta=45^{{}^{\circ}}$ with the $x$ -axis. Find the moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ for the rod.

Solution The moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ for a continuous system is given by

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=\int_{\rm{over\;all\;mass}}% \left[\begin{array}[]{ccc}y^{2}+z^{2}&-xy&-xz\\ -xy&x^{2}+z^{2}&-yz\\ -xz&-yz&x^{2}+y^{2}\end{array}\right]dm.

Filename:sfig4-6-3a

Figure 20.67:

Thus we need to carry out the integrals for the rod to find each component of the inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ . Let us consider an infinitesimal length element $d l$ of the rod at distance $l$ from O (see Fig 20.67). The mass of this element is $dm=\underbrace{\frac{m}{\ell}}_{\rm{mass/length}}\mathord{\cdot}\;dl$ , where $m$ is the total mass.
The coordinates of this element are

x=l\cos\theta,\quad y=l\sin\theta,\quad z=0\quad\mbox{(since the rod is in the $xy$-plane)}.

Therefore,

	$\displaystyle I_{xx}$	$\displaystyle=$	$\displaystyle\int_{m}(y^{2}+\underbrace{z^{2}}_{0})dm=\int_{0}^{\ell}% \underbrace{l^{2}\sin^{2}\theta}_{y^{2}}\mathord{\cdot}\underbrace{\frac{m}{% \ell}dl}_{dm}$
		$\displaystyle=$	$\displaystyle\frac{m}{\ell}\sin^{2}\theta\int_{0}^{\ell}l^{2}dl=\frac{m}{\ell}% \sin^{2}\theta\mathord{\cdot}\frac{\ell^{3}}{3}=\frac{m\ell^{2}}{3}\sin^{2}\theta.$

Similarly,

$\displaystyle I_{yy}$	$\displaystyle=$	$\displaystyle\int_{m}(x^{2}+\underbrace{z^{2}}_{0})dm=\int_{0}^{\ell}l^{2}\cos% ^{2}\theta\frac{m}{\ell}dl=\frac{m\ell^{2}}{3}\cos^{2}\theta.$
$\displaystyle I_{zz}$	$\displaystyle=$	$\displaystyle\int_{m}(x^{2}+y^{2})dm=\int_{0}^{\ell}l^{2}\frac{m}{\ell}dl=% \frac{m\ell^{2}}{3}.$
$\displaystyle I_{xy}$	$\displaystyle=$	$\displaystyle-\int_{m}xy\;dm=-\int_{0}^{\ell}l^{2}\cos\theta\sin\theta\mathord% {\cdot}\frac{m}{\ell}dl=-\frac{m\ell^{2}}{3}\cos\theta\sin\theta.$
$\displaystyle I_{xz}$	$\displaystyle=$	$\displaystyle-\int_{m}x\underbrace{z}_{0}\;dm=0.\quad I_{yz}=-\int_{m}y% \underbrace{z}_{0}\;dm=0.$

Thus,

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=\frac{m\ell^{2}}{3}\left[% \begin{array}[]{ccc}\sin^{2}\theta&-\sin\theta\cos\theta&0\\ -\sin\theta\cos\theta&\cos^{2}\theta&0\\ 0&0&1\end{array}\right].

Substituting the values of $m$ , $\ell$ and $\theta$ , we get

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=0.083\,\mathrm{kg}\mathord{% \cdot}\,\mathrm{m}^{2}\left[\begin{array}[]{rrr}1&-1&0\\ -1&1&0\\ 0&0&1\end{array}\right].

Problems for 20.3 Moment of inertia matrices: $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$

20.3.1 Find the moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ of the system shown in the figure, given that $m=0.25\,\mathrm{kg}$ and $\ell=0.5\,\mathrm{m}$ . Which components in the matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ will change if one mass is removed and the other one is doubled?

Filename:pfigure4-4-rp12

Figure 20.68

20.3.2 A dumbbell (two masses connected by a massless rod of length $\ell$ ) lies in the $x y$ plane. Compute the moments and products of inertia of this body about the $x$ , $y$ , and $z$ axes. What are the principle axes of this body?

Filename:pfigure-blue-38-2

Figure 20.69

20.3.3 Equivalent dumbbell. It is a great fact that, for the purposes of dynamics, every rigid body can be replaced by a massless rigid structure holding six masses. The structure looks like a child’s jack. It has three orthogonal bars with three pairs of point masses along the bars. The center of the jack is at the center of mass. The bars are in the directions of the eigenvectors of the moment of inertia matrix of the body (the principle directions). For flat objects one only needs four point masses. Where should one put four masses on a massless disk to make it equivalent to a uniform and ‘massful’ disk in problem 20.3(a). That is, what set of four masses has the same total mass, the same center of mass location, and the same moment of inertia matrix as a uniform disk?

20.3.4 Three identical solid spheres of mass $m$ and radius $r$ are situated at $(b,0,0),(0,b,b$ ), and ( $-b,0,2b$ ) relative to the origin $O$ . Find all nine components of $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ of the system of three spheres.

Hint: $[\boldsymbol{I}^{\mathrm{cm}}]$ for each sphere is

\left[\begin{array}[]{ccc}\frac{2}{5}mr^{2}&0&0\\ 0&\frac{2}{5}mr^{2}&0\\ 0&0&\frac{2}{5}mr^{2}\end{array}\right].

Use the parallel axis theorem.

Filename:pfigure-blue-49-2

Figure 20.70

20.3.5 Find the moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ for the ‘L’-shaped rod shown in the figure. The mass of the rod is $m=1.5\,\mathrm{kg}$ and each leg is $\ell=0.4\,\mathrm{m}$ long. You may ignore the width of the rod.

Filename:pfigure4-4-rp13

Figure 20.71

20.3.6 Moment of Inertia Matrix. This problem concerns a uniform flat disk of radius $R_{0}$ with mass $M_{d}=M_{disc}$ . In the two parts of this problem, you are to calculate the moment of inertia matrix for the disk. In the first part, do it by direct integration and, in the second part, by using a change of basis.

(a)

Using a coordinate system that is lined up with the disk so that the $z$ axis is normal to the disk and the origin of the coordinate system is at the center of mass (the center) of the disk, calculate the moment of inertia matrix for the disk. [Hint: use polar coordinates, and use $dm=(M/\pi R_{O}^{2})\,dA,\hbox to36.135pt{\hfil and\hfil}dA=R\,dR\,d\theta.$ ]
(b)

Use a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ aligned as follows: the origin of the coordinate system is at the center of the disk, the $z^{\prime}$ axis makes an angle $\phi$ with the $z$ axis which is normal to the plane of the disk (measured from the $z$ axis towards the $z^{\prime}$ axis), the disk intersects the $x^{\prime}y^{\prime}$ plane along the $y^{\prime}$ axis. The $y$ axis and the $y^{\prime}$ axis are coincident. Calculate the moment of inertia matrix for the disk. Alternate method: use the solution from (a) above with a change of basis.

Filename:f92h7p1

Figure 20.72

20.3.7 For each of the shapes below, mark the approximate location of the center of mass and the principle axes about the center of mass

Filename:pfigure-blue-52-2

Figure 20.73

20.3.8 Find the moment of inertia matrix $[\boldsymbol{I}^{\mathrm{cm}}]$ for the uniform solid cylinder of length $\ell$ and radius $r$ shown in the figure.

Filename:pfigure4-4-rp16

Figure 20.74

20.3.9 An ‘L’-shaped structure consists of two thin uniform rectangular plates each of width $w=200\,{\rm mm}$ . The vertical plate has height $h=400\,{\rm mm}$ and the horizontal plate has depth $d=300\,{\rm mm}$ . The plates are welded at right angles along the side measuring $200\,{\rm mm}$ . The total mass of the structure is $m=0.7\,\mathrm{kg}$ . Find the moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ .

Filename:pfigure4-4-rp14

Figure 20.75

20.3.10 A solar panel consists of three rectangular plates each of mass $m=1.2\,\mathrm{kg}$ , length $\ell=1$ and width $w=0.5\,\mathrm{m}$ The two side panels are inclined at $30^{\circ}$ with the plane of the middle panel. Evaluate all components of the moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ of the solar panel.

Filename:pfigure4-4-rp15

Figure 20.76

20.4 Mechanics using the moment of inertia matrix

Once one knows the velocity and acceleration of all points in a system one can find all of the motion quantities in the equations of motion by adding or integrating using the defining sums from the inside cover. This addition or integration is an impractical task for many motions of many objects where the required sums may involve billions and billions of atoms or a difficult integral. Linear momentum and the rate of change of linear momentum can be calculated by just keeping track of the center-of-mass of the system of interest. One would like something so simple for the calculation of angular momentum.

We are in luck if we are only interested in the two-dimensional motion of two-dimensional rigid bodies, the scalar moment of inertia from Chapter 7 is all we need. The luck is not so great for 3-D rigid bodies but still there is some simplification^†^†margin: For general motion of non-rigid bodies, a topic not covered in this book, the simplification for linear momentum still holds; linear momentum and its rate of change are given by the system mass times the velocity and acceleration of the center-of-mass. But there is no general simplification for the sums needed to evaluate angular momentum and energy or their rates of change. .The simplification is to use the moment of inertia matrix $[\boldsymbol{I}^{\mathrm{cm}}]$ from the previous section. One may have to do a sum or integral to find $I\equiv I^{cm}_{zz}$ or $[\boldsymbol{I}^{\mathrm{cm}}]$ if an adequate table is not handy. But once you know $[\boldsymbol{I}^{\mathrm{cm}}]$ , say, you need not work with the integrals to evaluate angular momentum and its rate of change. Assuming that you are comfortable calculating and looking-up moments of inertia, we proceed to use it for the purposes of studying mechanics.

Let’s now consider again the conically swinging rod of fig. 20.30 on page 20.30. Method 1 for evaluating $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ was evaluating the sums directly. If we accept the formulae presented for rigid bodies in Table I at the back of the book, we can find all of the motion quantities by setting ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}={\accentset{\rightharpoonup% }{\boldsymbol{0}}}$ .

Method 2 of evaluating $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ : using the $x y z$ moment of inertia matrix about point O

In section 20.1 we examined the conical swinging of a straight uniform rod. Now let’s look at that example again using its moment of inertia matrix. For a rigid body in constant rate circular motion about an axis through $O$ ,

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

because the ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ vector rotates with the body. For a rigid body rotating about point $O$ ,

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O% }}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}.

We assumed at the outset that

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}=% \left[\begin{array}[]{c}0\\ 0\\ \omega\end{array}\right].

So, the only trick is to find $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ for a rod in the configuration shown. We recall, look up, or believe for now that

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=\left[{\begin{matrix}I^{O}_{xx% }&I^{O}_{xy}&I^{O}_{xz}\cr I^{O}_{yx}&I^{O}_{yy}&I^{O}_{yz}\cr I^{O}_{zx}&I^{O% }_{zy}&I^{O}_{zz}\cr\end{matrix}}\right].

$\displaystyle I^{O}_{xy}$	$\displaystyle=$	$\displaystyle-\int\underbrace{x}_{0}y\,dm=0\mbox{\qquad\qquad($x=0$, all mass % in the $yz$ plane)}$
$\displaystyle I^{O}_{xz}$	$\displaystyle=$	$\displaystyle-\int\underbrace{x}_{0}z\,dm=0\mbox{\qquad\qquad($x=0$, all mass % in the $yz$ plane)}$
$\displaystyle I^{O}_{yy}$	$\displaystyle=$	$\displaystyle\int(\underbrace{x^{2}}_{0}+z^{2})dm\mbox{\qquad\qquad($x=0$, all% mass in the $yz$ plane)}$
	$\displaystyle=$	$\displaystyle\int_{0}^{\ell}\underbrace{(-s\cdot\cos\phi)^{2}}_{z}\rho\,ds$
	$\displaystyle=$	$\displaystyle\cos^{2}\phi\,\rho\frac{\ell^{3}}{3}=\cos^{2}\phi\,m\frac{\ell^{2% }}{3}$
$\displaystyle I^{O}_{zz}$	$\displaystyle=$	$\displaystyle\int(\underbrace{x^{2}}_{0}+y^{2})dm\mbox{\qquad\qquad($x=0$, all mass in the $yz$ plane)}$
	$\displaystyle=$	$\displaystyle\int\underbrace{(s\cdot\sin\phi)^{2}}_{y}\rho\,ds$
	$\displaystyle=$	$\displaystyle\sin^{2}\phi\,\rho\frac{\ell^{3}}{3}=\sin^{2}\phi\,m\frac{\ell^{2% }}{3}$
$\displaystyle I^{O}_{xx}$	$\displaystyle=$	$\displaystyle\int(y^{2}+z^{2})dm\mbox{\qquad\qquad(both integrals have been % evaluated above)}$
	$\displaystyle=$	$\displaystyle I^{O}_{yy}+I^{O}_{zz}\mbox{\qquad\qquad\qquad(perpendicular axis theorem)}$
	$\displaystyle=$	$\displaystyle(\cos^{2}\phi+\sin^{2}\phi)m\frac{\ell^{2}}{3}$
	$\displaystyle=$	$\displaystyle m\frac{\ell^{2}}{3}$
$\displaystyle I^{O}_{yz}$	$\displaystyle=$	$\displaystyle-\int yz\,dm$
	$\displaystyle=$	$\displaystyle-\int(s\cdot\sin\phi)(-s\cdot\cos\phi)\,\rho\,ds$
	$\displaystyle=$	$\displaystyle\sin\phi\cos\phi\cdot\rho\frac{\ell^{3}}{3}=\sin\phi\cos\phi\cdot m% \frac{\ell^{2}}{3}.$

Putting these terms all together in the matrix, we get

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=m\frac{\ell^{2}}{3}\left[{% \begin{matrix}1&0&0\cr 0&\cos^{2}\phi&\cos\phi\sin\phi\cr 0&\cos\phi\sin\phi&% \sin^{2}\phi\cr\end{matrix}}\right].

Now we can calculate ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$
	$\displaystyle=$	$\displaystyle m\frac{\ell^{2}}{3}\underbrace{\left[{\begin{matrix}1&0&0\cr 0&% \cos^{2}\phi&\cos\phi\sin\phi\cr 0&\cos\phi\sin\phi&\sin^{2}\phi\cr\end{matrix% }}\right]}_{[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]}\cdot\underbrace{% \left[\begin{array}[]{c}0\\ 0\\ \omega\end{array}\right]}_{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}$
	$\displaystyle=$	$\displaystyle m\frac{\ell^{2}}{3}\left[\begin{array}[]{c}0\\ \omega\cos\phi\sin\phi\\ \omega\sin^{2}\phi\end{array}\right]$
	$\displaystyle=$	$\displaystyle m\omega\frac{\ell^{2}}{3}\sin\phi(\cos\phi\hat{\boldsymbol{% \jmath}}+\sin\phi\hat{\boldsymbol{k}}).$

You may notice, by the way, that for this problem, ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ (in the direction of $\cos\phi\hat{\boldsymbol{\jmath}}+\sin\phi\hat{\boldsymbol{k}}$ ) is perpendicular to the rod (in the direction of $\sin\phi\hat{\boldsymbol{\jmath}}-\cos\phi\hat{\boldsymbol{k}}$ ). So ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ is not in the direction of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . ^†^†margin: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ is only parallel to ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ if rotation is about one of the principal axes (eigen-vector directions) of the inertia matrix.

Now we calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ as

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$
	$\displaystyle=$	$\displaystyle(\omega\hat{\boldsymbol{k}})\times\left[\omega m\frac{\ell^{2}}{3% }(\cos\phi\sin\phi\hat{\boldsymbol{\jmath}}+\sin^{2}\phi\hat{\boldsymbol{k}})\right]$
	$\displaystyle=$	$\displaystyle-m\frac{\ell^{2}}{3}\cos\phi\sin\phi\omega^{2}\hat{\boldsymbol{% \imath}},$

the same result we got before.

Method 3 of calculating $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ : using $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ and a coordinate system lined up with the rod

Filename:tfigure4-spherical-rotaxis

Figure 20.77: The spherical pendulum using

x^{\prime}y^{\prime}z^{\prime}

axes aligned with the rod.

Here, as suggested above, we redo the problem using a rotated set of axes better aligned with the rod. This method makes calculation of the moment of inertia matrix quite a bit easier — we can even look it up in a table — but makes the determination of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ a little harder. We are stuck finding the components of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ in a rotated coordinate system.

Referring to the table of moment of inertias on the inside of the back cover and taking care because different coordinates are used, the moment of inertia matrix about point $O$ for a thin uniform rod, in terms of the rotated coordinates is

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]_{x^{\prime}y^{\prime}z^{\prime% }}=m\frac{\ell^{2}}{3}\left[{\begin{matrix}1&0&0\cr 0&0&0\cr 0&0&1\cr\end{% matrix}}\right].

and the angular velocity in rotated coordinates is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-\omega\cos\phi\hat{% \boldsymbol{\jmath}}^{\prime}+\omega\sin\phi\hat{\boldsymbol{k}}^{\prime}$ , which can be written in component form as

\ [{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}]_{x^{\prime}y^{\prime}z^% {\prime}}=\omega\left[\begin{array}[]{c}0\\ -\cos\phi\\ \sin\phi\end{array}\right].

We calculate the angular momentum about point $O$ as

	$\displaystyle\quad\qquad\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{H}% }}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}=[% \boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\cdot{\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}}^{\text{vector equation, independent of coordinates}}$
	$\displaystyle\underbrace{\left[\begin{array}[]{c}H_{/O_{x^{\prime}}}\\ H_{/O_{y^{\prime}}}\\ H_{/O_{z^{\prime}}}\end{array}\right]=m\frac{\ell^{2}}{3}\left[{\begin{matrix}% 1&0&0\cr 0&0&0\cr 0&0&1\cr\end{matrix}}\right]_{x^{\prime}y^{\prime}z^{\prime}% }\cdot\omega\left[\begin{array}[]{c}0\\ -\cos\phi\\ \sin\phi\end{array}\right]_{x^{\prime}y^{\prime}z^{\prime}}}_{\text{all % components in $x^{\prime}y^{\prime}z^{\prime}$ coordinate system}}$
	$\displaystyle\left[\begin{array}[]{c}H_{/O_{x^{\prime}}}\\ H_{/O_{y^{\prime}}}\\ H_{/O_{z^{\prime}}}\end{array}\right]=m\frac{\ell^{2}}{3}\omega\left[\begin{% array}[]{c}0\\ 0\\ \sin\phi\end{array}\right]_{x^{\prime}y^{\prime}z^{\prime}}.$

Finally, we calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ as

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$
	$\displaystyle=$	$\displaystyle[\omega(-\cos\phi\hat{\boldsymbol{\jmath}}^{\prime}+\sin\phi\hat{% \boldsymbol{k}}^{\prime}]\times(\omega m\frac{\ell^{2}}{3}\sin\phi\hat{% \boldsymbol{k}}^{\prime})$
	$\displaystyle=$	$\displaystyle-\sin\phi\cos\phi\ m\omega^{2}\frac{\ell^{2}}{3}\hat{\boldsymbol{% \imath}}^{\prime}.$

This answer is again the same as what we got before because the $x$ and $x^{\prime}$ axis are coincident so $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}}$ .

The multitude of ways to calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$

We just showed three ways to calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ for a conically swinging stick, but there are many more. You can get a sense of the possibilities by studying table I summarizing momenta and energy in the back of the book. Here are three basic choices.

	$\displaystyle\mbox{(1) }\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}\,dm,$
	$\displaystyle\mbox{(2) }\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}},\qquad\mbox{or}$

\displaystyle\qquad\qquad\mbox{(3) }\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}m_{tot}+\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}.

Choice (3) is the safest choice to make if you are in doubt, since it is the only one of the three choices that does not depend on point $O$ being a fixed point (which it was for this example). For option (2) above, we can calculate ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ various ways as

$\displaystyle\mbox{(a) }{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2% .0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt% \hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}\,dm\qquad\mbox{or,}$
$\displaystyle\mbox{(b) }{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2% .0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt% \hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\qquad\mbox{or,}$
$\displaystyle\mbox{(c) }{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2% .0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt% \hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}m_{tot}+{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}.$

For option (3) above, we can calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$ as

	$\displaystyle\mbox{(d) }\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \mathrm{cm}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/cm}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{/cm}dm\qquad\mbox{ or,}$
	$\displaystyle\mbox{(e) }\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \mathrm{cm}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}.$

For (c) and (e) above, we can calculate ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$ as

	$\displaystyle\mbox{(f) }{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm% {cm}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/cm}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{/cm}\,dm\qquad\mbox{ or,}$
	$\displaystyle\mbox{(g) }{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm% {cm}}$	$\displaystyle=$	$\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}.$

For either of options (b) or (g), we can calculate $[I]$ relative to the $x y z$ axes (as we did for the second method on the previous pages) or relative to some rotated axes better aligned with the rod (as we did for the third method on the previous pages.)

As you can surmise from all the choices above, the list of options for the calculation of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ is too long and boring to show here.

The pros and cons of these methods depend on the problem at hand. If you want to avoid integration you are pretty much stuck using either $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ or $[\boldsymbol{I}^{\mathrm{cm}}]$ with (e) and (g) above. Avoiding the integrals depends on your having a table of moments of inertia (like table IV in the back of this book).

Energy of things going in circles at variable rate

The energy only depends on the speeds of the parts of a system, not their accelerations. So, as with constant rate motion,

$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle{1\over 2}\int\underbrace{v^{2}}_{{\accentset{\rightharpoonup}{% \boldsymbol{v}}}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}}\,dm$
	$\displaystyle=$	$\displaystyle{1\over 2}\int({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}})\cdot({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})\,dm$
	$\displaystyle=$	$\displaystyle{1\over 2}\omega^{2}\int R\text{$R$ is the distance from the axis% to the mass}^{2}\,dm$
	$\displaystyle=$	$\displaystyle{1\over 2}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot% [\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$
	$\displaystyle=$	$\displaystyle\underbrace{{1\over 2}\omega^{2}I^{o}_{zz}}_{\text{For rotation % about the $z$ axis the 9 terms in the matrix formula reduce to this one simple term. }}$

The safest bet

The following are the most reliable (least prone to error) formulas for evaluating the motion quantities for a rigid body rotating about a fixed axes (they also apply to arbitrary motion of a rigid body).

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/o}}% \times m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0% .5mu\mathrm{cm}}+[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\cdot{{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}}$	(20.47)
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/o}}% \times m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{% cm}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\underbrace{[% \boldsymbol{I}^{\mathrm{cm}}]\cdot{{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}}}_{{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}}+[% \boldsymbol{I}^{\mathrm{cm}}]\cdot{\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}$	(20.48)
$\displaystyle\ \par{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle{1\over 2}m_{\mathrm{tot}}v_{cm}^{2}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\cdot([\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}})\par$	(20.49)

Filename:tfigure5-7

Figure 20.78: The two terms in the rate of change of angular momentum are shown.

Please survey table I at the back of the book which summarizes the ways of evaluating momenta and energy.

The geometry of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$

It is possible to understand the formula for the change of angular momentum geometrically. Here is one way of looking at it. For this example it is most direct to look at $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ calculated using $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ , but the same discussion would work with $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$ calculated using $[\boldsymbol{I}^{\mathrm{cm}}]$ .

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=\underbrace{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}}}_{\mbox{Contribution% from rotation of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$}}+\underbrace{[\boldsymbol{I}^{% \scriptscriptstyle\mathrm{O}}]\cdot{\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}}_{\mbox{Due to the changing length of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$}}\par

The first term in the equation describes the rotation of the angular momentum vector. The second term describes its rate of change of length.

Since the body is spinning about a fixed axis, the orientation of the axis of rotation is not only fixed relative to the Newtonian frame, the room environment, but also relative to the body. At one instant of time we use a coordinate system to calculate ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ . After the body has rotated a little, if we now used a coordinate system that rotated the same amount as the body, a coordinate system ‘glued’ to the body, we could calculate ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ again.

This new calculation of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ will be almost identical to the calculation before the small rotation, however, because the moment of inertia matrix does not change in time relative to a coordinate system that moves with the body. Also, the coordinates of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ will be unchanged except for possibly a multiplication by a constant because the direction of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ doesn’t change. Thus the only change of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ as represented in this rotated coordinate system, is a possible change in its length due to a change in the spinning rate.

But this new coordinate system is a rotated coordinate system. To find the actual change in ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ we need to take this rotation into account. To picture this rotation consider the special case when the rotation rate is constant. Then the vector ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ is constant in the rotating coordinate system. That is, the vector ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ rotates with the body.

So the net change in ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ is a change due to rotation of the body added to a change due to the change in the rotation rate. For small angular changes, the direction of the first term is the same as the tangent to the circle that is traced by the tip of the angular momentum vector ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ as drawn on the body. To approximate the first term, consider the following reasoning. First, let’s denote the first term by $(\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}})_{\rm rot}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}}$ , where

the subscript ‘rot’ indicates that this term is the contribution to $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ from the rotation of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ . For small angular changes, $(\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}})_{rot}=(d{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}% }/dt)_{rot}={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}}\approx(\Delta{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}% })_{rot}/\Delta t$ . Thus, the term due to rotation is approximately, for small $\Delta t$ , $(\Delta{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}})_{rot}=\Delta t({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{0}$ ).

The contribution to $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ due to change of length of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ is $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\cdot{\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}$ . Similarly, this term is approximately, for small $\Delta t$ , $\Delta t([\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\cdot{\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}})$ .

Rotation of a rigid body about a fixed axis: the general case

Consider a general rigid body of mass $m$ and moment of inertia matrix with respect to the center-of-mass $[\boldsymbol{I}^{\mathrm{cm}}]$ rotating about a fixed axis. Without loss of generality, let the axis of rotation be the $\hat{\boldsymbol{k}}$ axis.

Filename:tfigure5-gen-rigid-body

Figure 20.79: Free-body diagram of a general rigid body rotating about the

z

-axis.

Linear momentum balance

Referring to the free-body diagram of the rigid body, linear momentum balance gives

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{net}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}$	(20.50)
	$\displaystyle=$	$\displaystyle m[\omega\hat{\boldsymbol{k}}\times(\omega\hat{\boldsymbol{k}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O})+\dot{\omega}\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}].$	(20.50)

The first term on the right hand side of equation 20.50, the centripetal term, is directed from the center-of-mass (

Filename:center-of-mass

) through the axis of rotation; that is, it lies in the

x y

-plane (or has no

\hat{\boldsymbol{k}}

component). The second term on the right hand side of equation 20.50, the tangential term, is normal to the plane determined by the axis and the center-of-mass. It is tangent to the circle that the center-of-mass travels on. It is zero if the center of mass is on the axis.

Angular momentum balance

Angular momentum balance about point $O$ gives

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}$	$\displaystyle=$	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}% \times m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}}_{\mbox{term (i)}}$	(20.51)
		$\displaystyle+\underbrace{\omega\hat{\boldsymbol{k}}\times\left\{[\boldsymbol{% I}^{\mathrm{cm}}]\cdot\omega\hat{\boldsymbol{k}}\right\}}_{\mbox{term (ii)}}$
		$\displaystyle+\underbrace{[\boldsymbol{I}^{\mathrm{cm}}]\cdot(\dot{\omega}\hat% {\boldsymbol{k}})}_{\mbox{term (iii)}}.$

Let’s look at each of the three terms on the right hand side in turn.

term (i)

The first term (i) on the right hand side of equation 20.51 is

term (i)	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}\times m% \overbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}}^{\text{${% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}$=$\omega\hat{\boldsymbol{k}}% \times(\omega\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{cm/O})$ $\mbox{ }+\,\dot{\omega}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{cm/O}$}}$
	$\displaystyle=$	$\displaystyle m[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}\times(% \omega\hat{\boldsymbol{k}}\times(\omega\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{cm/O}))$
		$\displaystyle+\qquad{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}\times% (\dot{\omega}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{cm/O})]$

Now, let’s consider the two parts of term (i) in turn.

Filename:tfigure5-term1-a

Figure 20.80: The first part of term (i)

The first part of term (i) is in the direction of $-\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}$ . For example, if the center-of-mass is in the $x z$ plane, this contribution to ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}$ is in the $-\hat{\boldsymbol{\jmath}}$ direction and could be accommodated by reaction forces on the axis in the $\hat{\boldsymbol{\imath}}$ and $-\hat{\boldsymbol{\imath}}$ directions.

Filename:tfigure5-term1-b

Figure 20.81: The second part of term (i)

Now, the second part of term (i) can be decomposed into a part along the $\hat{\boldsymbol{k}}$ direction and a part perpendicular to $\hat{\boldsymbol{k}}$ (along $d$ ). The part along $\hat{\boldsymbol{k}}$ is

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O}\times(\dot{\omega}\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O})% \cdot\hat{\boldsymbol{k}}=md^{2}\dot{\omega}\hat{\boldsymbol{k}}.

The part of term (i) perpendicular to $\hat{\boldsymbol{k}}$ has magnitude $m\dot{\omega}dw$ .

term (ii)

Now, let’s look at the second term in equation 20.51, term (ii), and expand it.

term (ii)	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times[\boldsymbol{I}^{\mathrm{cm}}]% \cdot\omega\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times\left[{\begin{matrix}I^{cm}_{xx}&% I^{cm}_{xy}&I^{cm}_{xz}\cr I^{cm}_{xy}&I^{cm}_{yy}&I^{cm}_{yz}\cr I^{cm}_{xz}&% I^{cm}_{yz}&I^{cm}_{zz}\cr\end{matrix}}\right]\cdot\left[{\begin{matrix}0\cr 0% \cr\omega\cr\end{matrix}}\right]$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times\left[I^{cm}_{xz}\omega\hat{% \boldsymbol{\imath}}+I^{cm}_{yz}\omega\hat{\boldsymbol{\jmath}}+I^{cm}_{zz}% \omega\hat{\boldsymbol{k}}\right]$
	$\displaystyle=$	$\displaystyle I^{cm}_{xz}\omega^{2}\hat{\boldsymbol{\jmath}}-I^{cm}_{yz}\omega% ^{2}\hat{\boldsymbol{\imath}}\par$

So, the second term in equation 20.51, term (ii), has no $\hat{\boldsymbol{k}}$ component. The $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ components are due to the off-diagonal terms in $[\boldsymbol{I}^{\mathrm{cm}}]$ . The off-diagonal terms are responsible for dynamic imbalance.

term (iii)

Now, the third term in equation 20.51, term (iii), is

	$\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]\cdot\dot{\omega}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\left[{\begin{matrix}I^{cm}_{xx}&I^{cm}_{xy}&I^{cm}_{xz}\cr I^{cm% }_{xy}&I^{cm}_{yy}&I^{cm}_{yz}\cr I^{cm}_{xz}&I^{cm}_{yz}&I^{cm}_{zz}\cr\end{% matrix}}\right]\cdot\left[{\begin{matrix}0\cr 0\cr\dot{\omega}\cr\end{matrix}}\right]$
		$\displaystyle=$	$\displaystyle I^{cm}_{xz}\dot{\omega}\hat{\boldsymbol{\imath}}+I^{cm}_{yz}\dot% {\omega}\hat{\boldsymbol{\jmath}}+I^{cm}_{zz}\dot{\omega}\hat{\boldsymbol{k}}.$

Putting terms (i), (ii), and (iii) back together

Now, let’s put the three terms back together into the equation of angular momentum balance about point $O$ , equation 20.51. To help with the interpretation, assume the center-of-mass is in the $x z$ plane. So, we start with

{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{/O}$}}}.

Breaking this vector equation into components, we get

$\displaystyle M_{xx}={\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}\cdot% \hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}\cdot\hat{\boldsymbol{\imath}}$
$\displaystyle M_{yy}={\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}\cdot% \hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}\cdot\hat{\boldsymbol{\jmath}}$
$\displaystyle M_{zz}={\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}\cdot% \hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}\cdot\hat{\boldsymbol{k}}.$

Now, using all the results so far, we find the torques about the $x$ , $y$ , and $z$ axes. For the $z$ -axis,

M_{z}=\underbrace{[m\overbrace{d^{2}}^{\text{$d$ = distance of cm from axis of% rotation}}+I^{cm}_{zz}]}_{\text{sometimes called `$[\boldsymbol{I}^{% \scriptscriptstyle\mathrm{O}}]$' }}\dot{\omega}.

(20.52)

The only torque about the $z$ -axis, the axis of rotation, is due to the acceleration about that axis.

Next, for the $x$ -axis,

M_{x}=-I^{cm}_{yz}\omega^{2}+I^{cm}_{xz}\dot{\omega}+m\dot{\omega}dw.

(20.53)

There is a torque about the $x$ -axis due to off-diagonal terms in $[\boldsymbol{I}^{\mathrm{cm}}]$ . One term is the dynamic imbalance term $I^{cm}_{yz}$ and the other is associated with angular acceleration. (Recall, the center-of-mass is on the $x z$ plane.) There is also a term due to the center of mass tangential acceleration since the center-of-mass is on a plane that does not contain point $O$ .

Finally, the torque about the $y$ -axis is

M_{y}=I^{cm}_{xz}\omega^{2}+I^{cm}_{yz}\dot{\omega}+(-m\omega^{2}dw).

(20.54)

Here, we have nearly the same types of terms as in equation 20.53. In this case, though, the centripetal acceleration causes the center-of-mass motion to contribute to the torque.

So, if

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}% \mbox{ and }\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\omega% }\hat{\boldsymbol{k}}

then

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$% }}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\left(-I^{cm}_{yz}\omega^{2}+(I^{cm}_{xz}+mdw)\dot{\omega}\right)% \hat{\boldsymbol{\imath}}$
		$\displaystyle+((I^{cm}_{xz}-mdw)\omega^{2}+I^{cm}_{yz}\dot{\omega})\hat{% \boldsymbol{\jmath}}$
		$\displaystyle+(md^{2}+I^{cm}_{zz})\dot{\omega}\hat{\boldsymbol{k}}.$

SAMPLE 20.11 A scalar times a vector is a vector. A matrix times a vector is a vector. What is the difference? Find the vectors ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{1}=I_{G}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{2}=[\mbox{\bf I}_{G}]{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , if ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=2\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{\imath}}+3\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{\jmath}},% \;\;I_{G}=10\,\mathrm{kg}\,\mathrm{m}^{2}$ and $[\mbox{\bf I}_{G}]~{}=~{}\left[\begin{array}[]{rrr}5&0&0\\ 0&5&-2\\ 0&-2&10\end{array}\right]\,\mathrm{kg}\,\mathrm{m}^{2}$ . Draw ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}},\;{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{2}$ .

Solution

Filename:sfig1-2-12

Figure 20.82:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{1}$	$\displaystyle=$	$\displaystyle I_{G}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$
	$\displaystyle=$	$\displaystyle 10\,\mathrm{kg}\,\mathrm{m}^{2}(2\,\mathrm{rad}/\mathrm{s}\hat{% \boldsymbol{\imath}}+3\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(20\hat{\boldsymbol{\imath}}+30\hat{\boldsymbol{\jmath}})\,% \mathrm{kg}\,\mathrm{m}^{2}/\,\mathrm{s}.$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{2}$	$\displaystyle=$	$\displaystyle[\mbox{\bf I}_{G}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rrr}5&0&0\\ 0&5&-2\\ 0&-2&10\end{array}\right]\,\mathrm{kg}\,\mathrm{m}^{2}\cdot\left\{\begin{array% }[]{c}2\\ 3\\ 0\end{array}\right\}\,\mathrm{rad}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle(10\hat{\boldsymbol{\imath}}+15\hat{\boldsymbol{\jmath}}-6\hat{% \boldsymbol{k}})\,\mathrm{kg}\,\mathrm{m}^{2}/\,\mathrm{s}.$

These two vectors, ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{2}$ , are shown along with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ in Fig. 20.82. Note that ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{1}$ has the same direction as ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ but ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{2}$ does not. ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ are both in the $x y$ -plane but ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{2}$ is not; it is in 3-D.

Comments: Multiplying a vector by a scalar does not change the direction of the vector but multiplying by a matrix does change the direction, in general. Find the angles between ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{1}$ and between ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{2}$ to convince yourself.

SAMPLE 20.12

Filename:sfig4-6-5

Figure 20.83:

The composite rod OABCD shown in Figure 20.83 is made up of three identical rods OA, BC, CD of mass $0.5\,\mathrm{kg}$ and length $20\,\mathrm{cm}$ each, and the rod AB which is half of rod CD. The composite rod goes in circles about the $y$ -axis at a constant rate $\omega=5\,\mathrm{rad}/\mathrm{s}$ . Find the angular momentum and the rate of change of angular momentum of the rod at the instant shown (i.e., when the rod is in the $x y$ -plane).

1.

Are all the components of $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ necessary to compute ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ ? Find $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ or the necessary components of $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ .
2.

Find the angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ .
3.

Find the rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ .
4.

If the rod were rotating about the $z$ -axis instead, (i.e., if the motion were in the $x y$ -plane) which components of $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ would be required to find ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ ? What would be the value of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ in that case?

Solution Since the rod rotates about the $y$ -axis, and O is a fixed point on this axis,

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \jmath}}\quad\mbox{and}\quad{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{% \kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}^{% \scriptscriptstyle\mathrm{O}}]\mathord{\cdot}{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}.

Since

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{ccc}I_{xx}^{\rm{O}}&I_{xy}^{\rm{O}}&I_{xz}^% {\rm{O}}\\ I_{xy}^{\rm{O}}&I_{yy}^{\rm{O}}&I_{yz}^{\rm{O}}\\ I_{xz}^{\rm{O}}&I_{yz}^{\rm{O}}&I_{zz}^{\rm{O}}\end{array}\right]\left\{\begin% {array}[]{c}0\\ \omega\\ 0\end{array}\right\}$	(20.62)
	$\displaystyle=$	$\displaystyle I_{xy}^{\rm{O}}\omega\hat{\boldsymbol{\imath}}+I_{yy}^{\rm{O}}% \omega\hat{\boldsymbol{\jmath}}+I_{yz}^{\rm{O}}\omega\hat{\boldsymbol{k}}$	(20.63)
	$\displaystyle=$	$\displaystyle(I_{xy}^{\rm{O}}\hat{\boldsymbol{\imath}}+I_{yy}^{\rm{O}}\hat{% \boldsymbol{\jmath}}+I_{yz}^{\rm{O}}\hat{\boldsymbol{k}})\omega,$	(20.63)

we only need to find three components of $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ to compute ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ , namely $I_{xy}^{\rm{O}}$ , $I_{yy}^{\rm{O}}$ and $I_{yz}^{\rm{O}}$ .
We can compute these components by considering each rod individually. For any rod, the components can be calculated using the values of the components about the center-of-mass of the rod (see table IV in the back of the book) and then using the parallel axis theorem.

Filename:sfig4-6-5a

Figure 20.84:

$\displaystyle\mbox{\text@underline{Rod OA}:}\quad I_{yy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle I_{y^{\prime}y^{\prime}}^{\rm{cm}}+m\frac{l^{2}}{4}=\frac{1}{12}% ml^{2}+\frac{1}{4}ml^{2}$
$\displaystyle I_{xy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\underbrace{I_{x^{\prime}y^{\prime}}^{\rm{cm}}}_{0}+m(-x_{\rm{cm/% O}}\underbrace{y_{\rm{cm/O}}}_{0})=0$
$\displaystyle I_{yz}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\underbrace{I_{y^{\prime}z^{\prime}}^{\rm{cm}}}_{0}+m(-% \underbrace{y_{\rm{cm/O}}}_{0}\underbrace{z_{\rm{cm/O}}}_{0})=0.$

Filename:sfig4-6-5b

Figure 20.85:

$\displaystyle\mbox{\text@underline{Rod AB}:}\quad I_{yy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\underbrace{I_{y^{\prime}y^{\prime}}^{\rm{cm}}}_{0}+\frac{m}{2}(x% ^{2}_{\rm{cm/O}}+\underbrace{z^{2}_{\rm{cm/O}}}_{0})=\frac{ml^{2}}{2}$
$\displaystyle I_{xy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\underbrace{I_{x^{\prime}y^{\prime}}^{\rm{cm}}}_{0}+\frac{m}{2}(-% \underbrace{x_{\rm{cm/O}}}_{l}\underbrace{y_{\rm{cm/O}}}_{\frac{l}{4}})=-\frac% {ml^{2}}{8}$
$\displaystyle I_{yz}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\underbrace{I_{y^{\prime}z^{\prime}}^{\rm{O}}}_{0}+m(-y_{\rm{cm/O% }}\underbrace{z_{\rm{cm/O}}}_{0})=0.$

Filename:sfig4-6-5c

Figure 20.86:

$\displaystyle\mbox{\text@underline{Rod BC}:}\quad I_{yy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle I_{y^{\prime}y^{\prime}}^{\rm{cm}}+m(x^{2}_{\rm{cm/O}}+% \underbrace{z^{2}_{\rm{cm/O}}}_{0})=\frac{ml^{2}}{12}+m\frac{9l^{2}}{4}=\frac{% 7}{3}ml^{2}$
$\displaystyle I_{xy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\underbrace{I_{x^{\prime}y^{\prime}}^{\rm{cm}}}_{0}+m(-x_{\rm{cm/% O}}y_{\rm{cm/O}})=-m\frac{3l}{2}\mathord{\cdot}\frac{l}{2}=-\frac{3ml^{2}}{4}$
$\displaystyle I_{yz}^{\rm{O}}$	$\displaystyle=$	$\displaystyle 0.$

Filename:sfig4-6-5d

Figure 20.87:

$\displaystyle\mbox{\text@underline{Rod CD}:}\quad I_{yy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\underbrace{I_{y^{\prime}y^{\prime}}^{\rm{cm}}}_{0}+m(x^{2}_{\rm{% cm/O}}+\underbrace{z^{2}_{\rm{cm/O}}}_{0})=m(2l)^{2}=4ml^{2}$
$\displaystyle I_{xy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle I_{yz}^{\rm{O}}$	$\displaystyle=$	$\displaystyle 0.$

Thus for the entire rod,

$\displaystyle I_{yy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle\frac{1}{3}ml^{2}+\frac{1}{2}ml^{2}+\frac{7}{3}ml^{2}+4ml^{2}=% \frac{43}{6}ml^{2}$
$\displaystyle I_{xy}^{\rm{O}}$	$\displaystyle=$	$\displaystyle-\frac{1}{8}ml^{2}-\frac{3}{4}ml^{2}=-\frac{7}{8}ml^{2}$
$\displaystyle I_{yz}^{\rm{O}}$	$\displaystyle=$	$\displaystyle 0.$

Answer: $I_{yy}^{\rm{O}}=\frac{43}{6}ml^{2},\;I_{xy}^{\rm{O}}=-\frac{7}{8}ml^{2},\;I_{% yz}^{\rm{O}}=0.$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle(I_{yy}^{\rm{O}}\hat{\boldsymbol{\imath}}+I_{xy}^{\rm{O}}\hat{% \boldsymbol{\jmath}}+I_{yz}^{\rm{O}}\hat{\boldsymbol{k}})\omega\quad\mbox{(% from Eqn~{}\ref{eqn4.6.5.1})}$
	$\displaystyle=$	$\displaystyle ml^{2}\omega(\frac{43}{6}\hat{\boldsymbol{\jmath}}-\frac{7}{8}% \hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{kg}(0.2\,\mathrm{m})^{2}\mathord{\cdot}5\,\mathrm{% rad}/\mathrm{s}(7.17\hat{\boldsymbol{\jmath}}-0.87)\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle(0.717\hat{\boldsymbol{\jmath}}-0.087\hat{\boldsymbol{\imath}})\,% \mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}/\,\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=(0.717\hat{\boldsymbol{\jmath}}-0.087\hat{% \boldsymbol{\imath}})\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}/\,\mathrm{s}.$

3.

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=\omega\hat{\boldsymbol{\jmath}}\times\left([% \boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\mathord{\cdot}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\right)$

$\displaystyle=$ $\displaystyle 5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{\jmath}}\times(0.717% \hat{\boldsymbol{\jmath}}-0.087\hat{\boldsymbol{\imath}})\,\mathrm{kg}\mathord% {\cdot}\,\mathrm{m}^{2}/\,\mathrm{s}$

$\displaystyle=$ $\displaystyle 0.435\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}.$

Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=0.435\,{\rm N}\mathord{\cdot}{\rm m}\hat{% \boldsymbol{k}}.$
4.

If the rod were rotating in the $x y$ -plane, it would be planar circular motion. The only component of $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ required for the calculation of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=I_{zz}^{\rm{O}}{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}$ will be $I_{zz}^{\rm{O}}$ . Also,

$\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times I_{zz}^{\rm{O}}{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}=0$

since ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is parallel to $I_{zz}^{\rm{O}}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ .

SAMPLE 20.13

Filename:sfig4-5-6

Figure 20.88:

A 0.5 kg uniform rectangular solar panel rotates about an off-centered axis $z^{\prime}z^{\prime}$ with constant angular speed $\omega=1.5\,\mathrm{rad}/\mathrm{s}$ . Axis $z^{\prime}z^{\prime}$ is parallel to the transverse $z$ -axis of the plate and is $1.6\,\mathrm{m}$ away from the center-of-mass of the plate. The in-plane moments of inertia $I_{xx}^{\rm{cm}}$ and $I_{yy}^{\rm{cm}}$ of the plate are given: $I_{xx}^{\rm{cm}}=4.0\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$ and $I_{yy}^{\rm{cm}}=12.4\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$ . At the instant shown in Fig 20.88, calculate

1.

the linear momentum of the panel,
2.

the angular momentum of the panel, and
3.

the kinetic energy of the panel.

Solution Since the plate rotates about the $z^{\prime}z^{\prime}$ -axis which is parallel to the $z$ -axis,

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}=% 1.5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}.

1.

Linear momentum:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ $\displaystyle=$ $\displaystyle\,\mathrm{m}_{\rm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{v% }}}_{\rm{cm}}=m({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm{cm/O}})$

$\displaystyle=$ $\displaystyle 0.5\,\mathrm{kg}\mathord{\cdot}(1.5\,\mathrm{rad}/\mathrm{s}\hat% {\boldsymbol{k}}\times 1.6\,\mathrm{m}\hat{\boldsymbol{\imath}})$

$\displaystyle=$ $\displaystyle 1.2\,{\mathrm{kg}}\mathord{\cdot}{\mathrm{m}}/{\mathrm{s}}\hat{% \boldsymbol{\jmath}}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{L}}}=1.2\,{\mathrm{kg}}\mathord{\cdot% }{\mathrm{m}}/{\mathrm{s}}\hat{\boldsymbol{\jmath}}$

Filename:sfig4-5-6a

Figure 20.89:

We can easily check the direction of ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ since ${\accentset{\rightharpoonup}{\boldsymbol{L}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\rm{cm}}$ , it has to be in the same direction as ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm{cm}}$ . The center-of-mass goes in circles about O, therefore, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm{cm}}$ is tangential to the circular path, i.e. in the $y$ -direction (see Fig 20.89).

Angular momentum:

{\accentset{\rightharpoonup}{\boldsymbol{H}}}=I_{z^{\prime}z^{\prime}}^{\rm{O}% }\,\omega\hat{\boldsymbol{k}}

Thus to find ${\accentset{\rightharpoonup}{\boldsymbol{H}}}$ , we need to find $I_{z^{\prime}z^{\prime}}^{\rm{O}}$ . Since $z^{\prime}z^{\prime}\parallel zz$ , we can use the parallel axis theorem to find $I_{z^{\prime}z^{\prime}}^{\rm{O}}$ if we know $I_{zz}^{\rm{cm}}$ . We are given the in-plane moments of inertia $I_{xx}^{\rm{cm}}$ and $I_{yy}^{\rm{cm}}$ . Therefore, from the perpendicular axis theorem:

I_{zz}^{\rm{cm}}=I_{xx}^{\rm{cm}}+I_{yy}^{\rm{cm}}=(4.0+12.4)\,\mathrm{kg}% \mathord{\cdot}\,\mathrm{m}^{2}=16.4\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{% 2}.

Now using the parallel axis theorem,

$\displaystyle I_{z^{\prime}z^{\prime}}^{\rm{O}}$	$\displaystyle=$	$\displaystyle I_{zz}^{\rm{cm}}+Mr_{\rm{cm/O}}^{2}$
	$\displaystyle=$	$\displaystyle 16.4\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}+0.5\,\mathrm{kg% }(1.6\,\mathrm{m})^{2}=17.68\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}.$
$\displaystyle\mbox{Therefore,}\quad{\accentset{\rightharpoonup}{\boldsymbol{H}}}$	$\displaystyle=$	$\displaystyle 17.68\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}\mathord{\cdot}% (1.5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle 26.52\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}/\,\mathrm{s}% \hat{\boldsymbol{k}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}=26.52\,\mathrm{kg}\mathord{\cdot% }\,\mathrm{m}^{2}/\,\mathrm{s}\hat{\boldsymbol{k}}.$

3.

Kinetic energy:

$\displaystyle{E_{\mathrm{K}}}$ $\displaystyle=$ $\displaystyle\frac{1}{2}I_{z^{\prime}z^{\prime}}^{\rm{O}}\omega^{2}=\frac{1}{2% }(17.68\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2})(1.5\,\mathrm{rad}/\mathrm% {s})^{2}$

$\displaystyle=$ $\displaystyle 19.89\frac{\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}}{\,% \mathrm{s}^{2}}==19.89\,{\rm N}\mathord{\cdot}{\rm m}$

$\displaystyle=$ $\displaystyle 19.89\;\rm{J}$

Answer: ${E_{\mathrm{K}}}=19.89\;\rm{J}$

SAMPLE 20.14 The rotating crooked plate again. For the crooked plate considered in Sample 20.37 and shown again in Fig. 20.90,

Filename:sfig4-6-8

Figure 20.90: A rectangular plate of mass

m

rotates with shaft AB at a constant speed

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}

. A coordinate system

x^{\prime}y^{\prime}z^{\prime}

is aligned with the principle axes of the plate.

1.

compute the moment of inertia matrix $[\boldsymbol{I}]$ in the $x^{\prime}y^{\prime}z^{\prime}$ coordinate system,
2.

compute the rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ using the moment of inertia $[\boldsymbol{I}]_{x^{\prime}y^{\prime}z^{\prime}}$ computed above,
3.

express $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ as a vector in the $x y z$ coordinate system.

Solution A line sketch of the plate and the two coordinate systems attached to it are shown in Fig. 20.91. The set of basis vectors $(\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},\hat{\boldsymbol{k}})$ and $(\hat{\boldsymbol{\imath}}^{\prime},\hat{\boldsymbol{\jmath}}^{\prime},\hat{% \boldsymbol{k}}^{\prime})$ associated with coordinate systems $x y z$ and $x^{\prime}y^{\prime}z^{\prime}$ respectively are shown separately for the sake of clarity. From the diagram of the two basis vector sets, we may write

Filename:sfig4-6-8a

Figure 20.91: A line sketch of the rotating plate along with the two coordinate systems

x^{\prime}y^{\prime}z^{\prime}

and

x y z

. The basis vectors associated with the two systems are also shown.

\left.\begin{array}[]{l}\hat{\boldsymbol{\imath}}^{\prime}=\cos\phi\hat{% \boldsymbol{\imath}}-\sin\phi\hat{\boldsymbol{k}}\\ \hat{\boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}}\\ \hat{\boldsymbol{k}}^{\prime}=\sin\phi\hat{\boldsymbol{\imath}}+\cos\phi\hat{% \boldsymbol{k}}\end{array}\right\}.

(20.64)

Calculation of $[\boldsymbol{I}]_{x^{\prime}y^{\prime}z^{\prime}}$ :

[\boldsymbol{I}]_{x^{\prime}y^{\prime}z^{\prime}}=\left[\begin{array}[]{ccc}I_% {x^{\prime}x^{\prime}}&I_{x^{\prime}y^{\prime}}&I_{x^{\prime}z^{\prime}}\\ I_{x^{\prime}y^{\prime}}&I_{y^{\prime}y^{\prime}}&I_{y^{\prime}z^{\prime}}\\ I_{x^{\prime}z^{\prime}}&I_{y^{\prime}z^{\prime}}&I_{z^{\prime}z^{\prime}}\end% {array}\right].

Let us first consider the off diagonal terms of $[\boldsymbol{I}]_{x^{\prime}y^{\prime}z^{\prime}}$ .

Since $x^{\prime}=0$ on the entire plate (the plate is in the $y^{\prime}z^{\prime}$ plane and the origin is on the plate.),

I_{x^{\prime}y^{\prime}}=\int_{m}\underbrace{x^{\prime}}_{0}\;y^{\prime}\,dm=0% ,\quad I_{x^{\prime}z^{\prime}}=\int_{m}\underbrace{x^{\prime}}_{0}\;z^{\prime% }\,dm=0.

Filename:sfig4-6-8b

Figure 20.92: Symmetry about the

z^{\prime}

axis implies that

I_{y^{\prime}z^{\prime}}=0

. Same result holds if we consider the symmetry about the

y^{\prime}

axis.

How about $I_{y^{\prime}z^{\prime}}$ ? Well, you can calculate it two ways: (a) carry out the integration $I_{y^{\prime}z^{\prime}}=\int_{m}y^{\prime}z^{\prime}\,dm$ over the entire plate mass and find that $I_{y^{\prime}z^{\prime}}=0$ , or (b) realize that for every mass element $d m$ ( $=\frac{m}{\ell b}dA$ ) with coordinates $(+y^{\prime}z^{\prime})$ , there exists another element $d m$ at $(-y^{\prime}z^{\prime})$ such that the sum of their contributions to the integral is zero. Therefore, $I_{y^{\prime}z^{\prime}}=\int_{m}y^{\prime}z^{\prime}\,dm=0$ ^†^†margin: You can use a similar argument to find the off-diagonal terms in $[\boldsymbol{I}]$ whenever there is such symmetry with respect to the coordinate axes. Now the other terms:

$\displaystyle I_{z^{\prime}z^{\prime}}$	$\displaystyle=$	$\displaystyle\int_{m}(\underbrace{x^{\prime 2}}_{0}+y^{\prime 2})\,dm=\int_{-% \ell/2}^{\ell/2}\int_{-b/2}^{b/2}y^{\prime 2}\cdot\frac{m}{\ell b}dy^{\prime}% \;dz^{\prime}$
	$\displaystyle=$	$\displaystyle\frac{m}{\ell b}\int_{-\ell/2}^{\ell/2}\left(\left.\frac{y^{% \prime 3}}{3}\right\|_{-b/2}^{b/2}\right)\,dz^{\prime}=\frac{m}{\ell b}\cdot% \frac{b^{3}}{12}\int_{-\ell/2}^{\ell/2}dz^{\prime}$
	$\displaystyle=$	$\displaystyle\frac{m}{\ell b}\cdot\frac{b^{3}}{12}\cdot\ell=\frac{mb^{2}}{12},$

and similarly,

	$\displaystyle I_{y^{\prime}y^{\prime}}$	$\displaystyle=$	$\displaystyle\int_{m}(\underbrace{x^{\prime 2}}_{0}+z^{\prime 2})\,dm=\frac{m% \ell^{2}}{12},$
	$\displaystyle I_{x^{\prime}x^{\prime}}$	$\displaystyle=$	$\displaystyle\int_{m}(y^{\prime 2}+z^{\prime 2})\,dm=I_{y^{\prime}y^{\prime}}+% I_{z^{\prime}z^{\prime}}=\frac{m}{12}(\ell^{2}+b^{2}).$

Thus,

[\boldsymbol{I}]_{x^{\prime}y^{\prime}z^{\prime}}=\frac{m}{12}\left[\begin{% array}[]{ccc}\ell^{2}+b^{2}&0&0\\ 0&\ell^{2}&0\\ 0&0&b^{2}\end{array}\right].

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}\hbox to36.135pt{\hfil and\hfil}{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}]_{x^{\prime}y^{\prime}z^{% \prime}}\,\{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\}_{x^{\prime}y^% {\prime}z^{\prime}}.

The subscripts $x^{\prime}y^{\prime}z^{\prime}$ in $[\boldsymbol{I}]$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ have been used to denote that both $[\boldsymbol{I}]$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ are expressed in $x^{\prime}y^{\prime}z^{\prime}$ coordinate system. ^†^†margin: We can express $[\boldsymbol{I}]$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ in any coordinate system of our choice but both of them must be in the same system for their product to be valid. We have calculated $[I]_{x^{\prime}y^{\prime}z^{\prime}}$ above. Now we need to find $\{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\}_{x^{\prime}y^{\prime}z^% {\prime}}$ .

Let ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega_{x^{\prime}}\hat{% \boldsymbol{\imath}}^{\prime}+\omega_{y^{\prime}}\hat{\boldsymbol{\jmath}}^{% \prime}+\omega_{z^{\prime}}\hat{\boldsymbol{k}}^{\prime}$ . But, we can also write, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . So,

\omega\hat{\boldsymbol{k}}=\omega_{x^{\prime}}\hat{\boldsymbol{\imath}}^{% \prime}+\omega_{y^{\prime}}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{z^{% \prime}}\hat{\boldsymbol{k}}^{\prime}

(20.65)

Dotting both sides of Eqn. (20.65) with $\hat{\boldsymbol{\imath}}^{\prime}$ , $\hat{\boldsymbol{\jmath}}^{\prime}$ , and $\hat{\boldsymbol{k}}^{\prime}$ and using the relationships in (20.64) we get

\omega_{x^{\prime}}=\omega(\overbrace{\hat{\boldsymbol{k}}\cdot\hat{% \boldsymbol{\imath}}^{\prime}}^{-\sin\phi})=-\omega\sin\phi,\omega_{y^{\prime}% }=\omega(\overbrace{\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{\jmath}}^{\prime% }}^{0})=0,\omega_{k^{\prime}}=\omega(\overbrace{\hat{\boldsymbol{k}}\cdot\hat{% \boldsymbol{k}}^{\prime}}^{\cos\phi})=\omega\cos\phi.

Thus,

\{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\}_{x^{\prime}y^{\prime}z^% {\prime}}=-\omega\sin\phi\hat{\boldsymbol{\imath}}^{\prime}+\omega\cos\phi\hat% {\boldsymbol{k}}^{\prime}

So,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\frac{m}{12}\left[\begin{array}[]{ccc}\ell^{2}+b^{2}&0&0\\ 0&\ell^{2}&0\\ 0&0&b^{2}\end{array}\right]\left\{\begin{array}[]{c}-\omega\sin\phi\\ 0\\ \omega\cos\phi\end{array}\right\}$
	$\displaystyle=$	$\displaystyle\left\{\begin{array}[]{c}-\frac{m}{12}(\ell^{2}+b^{2})\omega\sin% \phi\\ 0\\ \frac{m}{12}b^{2}\omega\cos\phi\end{array}\right\}$
$\displaystyle\hbox to36.135pt{or\hfil}{\accentset{\rightharpoonup}{\boldsymbol% {H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\frac{m\omega}{12}[-(\ell^{2}+b^{2})\sin\phi\hat{\boldsymbol{% \imath}}^{\prime}+b^{2}\cos\phi\hat{\boldsymbol{k}}^{\prime}].$

Now we can easily compute $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ as follows:

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$
	$\displaystyle=$	$\displaystyle(-\omega\sin\phi\hat{\boldsymbol{\imath}}^{\prime}+\omega\cos\phi% \hat{\boldsymbol{k}}^{\prime})\times\frac{m\omega}{12}[-(\ell^{2}+b^{2})\sin% \phi\hat{\boldsymbol{\imath}}^{\prime}+b^{2}\cos\phi\hat{\boldsymbol{k}}^{% \prime}]$
	$\displaystyle=$	$\displaystyle\frac{m\omega^{2}}{12}b^{2}\sin\phi\cos\phi\hat{\boldsymbol{% \jmath}}^{\prime}-\frac{m\omega^{2}}{12}(\ell^{2}+b^{2})\sin\phi\cos\phi\hat{% \boldsymbol{\jmath}}^{\prime}$
	$\displaystyle=$	$\displaystyle-\frac{m\omega^{2}}{12}\ell^{2}\sin\phi\cos\phi\hat{\boldsymbol{% \jmath}}^{\prime}.$

3.

Now, back to the $x y z$ coordinate system: Since $\hat{\boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}}$ [Eqn (20.64)], we have

$\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=-\frac{m\omega^{2}}{12}\ell^{2}\sin\phi% \cos\phi\hat{\boldsymbol{\jmath}}$

which is the same result as obtained in Sample 20.37.

SAMPLE 20.15 The calculation of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ using the moment of inertia matrix. For the crooked plate considered in Sample 20.37, find the rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ , using the moment of inertia $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ , calculated in the $x y z$ coordinate system

Solution We calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ using the following formula:

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}^{\scriptscriptstyle\mathrm% {O}}]\cdot\underbrace{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_% {{\accentset{\rightharpoonup}{\boldsymbol{0}}}}\;+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times\overbrace{([\boldsymbol{I}^{\scriptscriptstyle% \mathrm{O}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}})^{{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}}.

Note that point O is the center-of-mass of the plate. Let us write the expression for the angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ in matrix form.

\left\{\begin{array}[]{c}H_{x/O}\\ H_{y/O}\\ H_{z/O}\end{array}\right\}=\left[\begin{array}[]{ccc}I_{xx}&I_{xy}&I_{xz}\\ I_{xy}&I_{yy}&I_{yz}\\ I_{xz}&I_{yz}&I_{zz}\end{array}\right]\left\{\begin{array}[]{c}0\\ 0\\ \omega\end{array}\right\}.

It is clear from the components of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ that we only need the last column of $[\boldsymbol{I}]$ matrix since other elements of $[\boldsymbol{I}]$ will multiply with zeros of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ vector. Carrying out the multiplication we get

\left\{\begin{array}[]{c}H_{0x}\\ H_{0y}\\ H_{0z}\end{array}\right\}=\left\{\begin{array}[]{c}I_{xz}\omega\\ I_{yz}\omega\\ I_{zz}\omega\end{array}\right\}

or, in vector form

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=I_{xz}\omega\hat{\boldsymbol{\imath}}+I_{yz}% \omega\hat{\boldsymbol{\jmath}}+I_{zz}\omega\hat{\boldsymbol{k}}.

Therefore,

	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times\omega(I_{xz}\hat{\boldsymbol{% \imath}}+I_{yz}\hat{\boldsymbol{\jmath}}+I_{zz}\hat{\boldsymbol{k}})$
		$\displaystyle=$	$\displaystyle-\omega^{2}I_{yz}\hat{\boldsymbol{\imath}}+\omega^{2}I_{xz}\hat{% \boldsymbol{\jmath}}.$

But,

I_{xz}=-\int_{m}xz\,dm\hbox to36.135pt{\hfil and\hfil}I_{yz}=-\int_{m}yz\,dm

and $x=w\sin\phi,\;y=y,\;z=w\cos\phi$ (see Fig. 20.40), hence

	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle-\omega^{2}\int_{m}(-wy\cos\phi\hat{\boldsymbol{\imath}}+w^{2}% \sin\phi\cos\phi\hat{\boldsymbol{\jmath}})\,dm$
		$\displaystyle=$	$\displaystyle\int_{-b/2}^{b/2}\int_{-\ell/2}^{\ell/2}\omega^{2}(-w^{2}\sin\phi% \cos\phi\hat{\boldsymbol{\jmath}}+wy\cos\phi\hat{\boldsymbol{\imath}})\frac{m}% {\ell b}dw\,dy$

which is the same integral as obtained in Sample 20.37 for $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ . Therefore, the result is also the same:

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=-\frac{m\omega^{2}\ell^{2}}{12}\sin\phi% \cos\phi\hat{\boldsymbol{\jmath}}.

Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=-\frac{m\omega^{2}\ell^{2}}{12}\sin\phi% \cos\phi\hat{\boldsymbol{\jmath}}$

SAMPLE 20.16 Direct application of formula: A rectangular plate is mounted on a massless shaft with the center-of-mass of the plate on the shaft axis. The shaft rotates about its axis with angular acceleration $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=0.5\,{\rm rpm}/\,% \mathrm{s}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$ . At the instant of interest, the angular velocity is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=100\,{\rm rpm}(\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$ and the components of the moment of inertia matrix of the plate are $I_{xx}=2\,{\rm kg}\mathord{\cdot}{\rm m}^{2},\;I_{yy}=4\,{\rm kg}\mathord{% \cdot}{\rm m}^{2},\;I_{zz}=6\,{\rm kg}\mathord{\cdot}{\rm m}^{2}$ and $I_{xy}=I_{yz}=I_{xz}=0$ .

1.

Find the angular momentum of the plate about its mass-center and show that it is not in the same direction as the angular velocity.
2.

Find the net moment acting on the plate.

Solution We are given the angular velocity, the angular acceleration, and the moment of inertia matrix of the plate:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle 100\,{\rm rpm}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath% }})=10.47\,\mathrm{rad}/\mathrm{s}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}})$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle 5\,{\rm rpm}/\,\mathrm{s}(\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}})=0.52\,\mathrm{rad}/\mathrm{s}^{2}(\hat{\boldsymbol{% \imath}}+\hat{\boldsymbol{\jmath}})$
$\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]$	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{ccc}2&0&0\\ 0&4&0\\ 0&0&6\end{array}\right]\,{\rm kg}\mathord{\cdot}{\rm m}^{2}.$

The angular momentum: The angular momentum of the plate is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm cm}$	$\displaystyle=$	$\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{ccc}2&0&0\\ 0&4&0\\ 0&0&6\end{array}\right]\,{\rm kg}\mathord{\cdot}{\rm m}^{2}\cdot\left\{\begin{% array}[]{c}10.47\\ 10.47\\ 0\end{array}\right\}\,\mathrm{rad}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle(20.94\hat{\boldsymbol{\imath}}+41.88\hat{\boldsymbol{\jmath}})% \underbrace{\,{\rm kg}\mathord{\cdot}{\rm m}^{2}\cdot{\,\mathrm{s}}^{-1}}_{\,{% \rm N}\mathord{\cdot}{\rm m}\cdot\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle(20.94\hat{\boldsymbol{\imath}}+41.88\hat{\boldsymbol{\jmath}})\,% {\rm N}\mathord{\cdot}{\rm m}\cdot\,\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm cm}=(20.94\hat{\boldsymbol{% \imath}}+41.88\hat{\boldsymbol{\jmath}})\,{\rm N}\mathord{\cdot}{\rm m}\cdot\,% \mathrm{s}$

There are many ways of showing that ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm cm}$ is not parallel to ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . We can simply draw the two vectors and show that they are not parallel

We can, alternatively, take the cross product of the two vectors:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm cm}\times{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle(20.94\hat{\boldsymbol{\imath}}+41.88\hat{\boldsymbol{\jmath}})\,% {\rm N}\mathord{\cdot}{\rm m}\cdot\,\mathrm{s}\times 10.47\,\mathrm{rad}/% \mathrm{s}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\left\|\begin{array}[]{ccc}\hat{\boldsymbol{\imath}}&\hat{% \boldsymbol{\jmath}}&\hat{\boldsymbol{k}}\\ 20.94&41.88&0\\ 10.47&10.47&0\end{array}\right\|\,{\rm N}\mathord{\cdot}{\rm m}$
	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{k}}(219.24-438.48)\,{\rm N}\mathord{\cdot}{\rm m% }=-20.94\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}$

which is not zero, implying that the two vectors are not parallel.

The net moment: The net moment on the plate can be found by applying angular momentum balance:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/\rm cm}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm cm}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\underbrace{[% \boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}}_{{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm cm}}+[% \boldsymbol{I}^{\mathrm{cm}}]\cdot\dot{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}$
		$\displaystyle=$	$\displaystyle\underbrace{20.94\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{% k}}}_{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\rm cm}}+\underbrace{1.04\,{\rm N}\mathord{% \cdot}{\rm m}\hat{\boldsymbol{\imath}}+2.08\,{\rm N}\mathord{\cdot}{\rm m}\hat% {\boldsymbol{\jmath}}}_{[\boldsymbol{I}^{\mathrm{cm}}]\cdot\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/\rm cm}=(1.04\hat{\boldsymbol{% \imath}}+2.08\hat{\boldsymbol{\jmath}}+20.94\hat{\boldsymbol{k}})\,{\rm N}% \mathord{\cdot}{\rm m}$

SAMPLE 20.17

Filename:sfig5-5-2

Figure 20.93: A rod, welded to a tipped shaft AB, swings around the shaft axis if it is tipped slightly from the vertical plane

y z

A rod swings around a tipped axis in 3-D. A rigid shaft of negligible mass is connected to frictionless hinges at A and B. The shaft is tipped from the horizontal plane ( $x y$ -plane) such that the shaft axis makes an angle $\gamma$ with the vertical axis. A uniform rod $O C$ of mass $m$ and length $\ell$ is welded to the shaft. At time $t=0$ , the rod is tipped by a small angle $\phi$ from its position $OC^{\prime}$ in the $y z$ (or $y^{\prime}z^{\prime}$ ) plane. Find the equation of motion of the rod.

^†^†margin:

Solution Let $\hat{\boldsymbol{\imath}}^{\prime},\;\hat{\boldsymbol{\jmath}}^{\prime},\;\hat% {\boldsymbol{k}}^{\prime}$ be the basis vectors associated with the $x^{\prime}y^{\prime}z^{\prime}$ coordinate system. Since $y^{\prime}z^{\prime}$ axes can be obtained by rotating $y z$ axes counterclockwise about the $x$ -axis by an angle $90^{o}-\gamma$ , we can relate the basis vectors of the two coordinate systems with the help of Fig. 20.94:

Filename:sfig5-5-2a

Figure 20.94:

y^{\prime}z^{\prime}

axes are obtained by rotating

y z

axes counterclockwise about the

x

-axis by an angle

90^{o}-\gamma

\begin{array}[]{lll}\hat{\boldsymbol{\imath}}^{\prime}&=&\hat{\boldsymbol{% \imath}},\\ \hat{\boldsymbol{\jmath}}^{\prime}&=&\sin\gamma\hat{\boldsymbol{\jmath}}+\cos% \gamma\hat{\boldsymbol{k}},\\ \hat{\boldsymbol{k}}^{\prime}&=&-\cos\gamma\hat{\boldsymbol{\jmath}}+\sin% \gamma\hat{\boldsymbol{k}}.\end{array}

(20.71)

The free-body diagram of the shaft with rod $O C$ is shown in Fig. 20.95. We can write angular momentum balance for this system about any point on the axis of rotation AB. However, rather than writing angular momentum balance about a point, let us write angular momentum balance about axis AB. This ‘trick’ will eliminate reactions ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}$ and ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{B}$ from our equations. Angular Momentum Balance about axis AB is:

	$\displaystyle\hat{\boldsymbol{\lambda}}_{AB}\cdot\big{[}\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}\big{]}$
	$\displaystyle\hbox to36.135pt{or\hfil}\hat{\boldsymbol{\jmath}}^{\prime}\cdot% \sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\jmath}}^{\prime}\cdot\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{/O}$}}}.$

Calculation of ( $\hat{\boldsymbol{\jmath}}^{\prime}\cdot\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{O}$}}}$ ) : Since ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}$ and ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{B}$ pass through axis AB, they do not produce any moment about this axis. Therefore,

Filename:sfig5-5-2b

Figure 20.95: Free-body diagram of the bar and shaft system. In the inset, only the weight of the bar,

m g

, is shown for clarity of geometry.

$\displaystyle\hat{\boldsymbol{\jmath}}^{\prime}\cdot\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\jmath}}^{\prime}\cdot[{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{G/O}\times mg(-\hat{\boldsymbol{k}})]$
	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\jmath}}^{\prime}\cdot\left[\frac{L}{2}(\sin\phi% \hat{\boldsymbol{\imath}}^{\prime}+\cos\phi\hat{\boldsymbol{k}}^{\prime})% \times mg(-\hat{\boldsymbol{k}})\right]$
	$\displaystyle=$	$\displaystyle(\overbrace{\sin\gamma\hat{\boldsymbol{\jmath}}+\cos\gamma\hat{% \boldsymbol{k}}}^{\hat{\boldsymbol{\jmath}}^{\prime}})\cdot[\frac{L}{2}mg(% \overbrace{\sin\phi\hat{\boldsymbol{\jmath}}+\cos\phi\,\cos\gamma\hat{% \boldsymbol{\imath}}}^{\mbox{using Eqn.(\ref{eq4.1.4.1})}})]$
	$\displaystyle=$	$\displaystyle\frac{\ell}{2}mg\sin\phi\,\sin\gamma$

Calculation of $\hat{\boldsymbol{\jmath}}^{\prime}\cdot\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$ : Since the rod rotates about axis AB, we may write

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\dot{\phi}\hat{\boldsymbol{\jmath}}^{\prime},$
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\ddot{\phi}\hat{\boldsymbol{\jmath}}^{\prime}.$

Now,

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}^{\scriptscriptstyle\mathrm% {O}}]\{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times[\boldsymbol{I}^{% \scriptscriptstyle\mathrm{O}}]\{{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\}

where

$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\{\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\}$	$\displaystyle=$	$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]_{x^{\prime}y^{% \prime}z^{\prime}}\,\{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\}% _{x^{\prime}y^{\prime}z^{\prime}}$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{ccc}I_{x^{\prime}x^{\prime}}&I_{x^{\prime}y% ^{\prime}}&I_{x^{\prime}z^{\prime}}\\ I_{x^{\prime}y^{\prime}}&I_{y^{\prime}y^{\prime}}&I_{y^{\prime}z^{\prime}}\\ I_{x^{\prime}z^{\prime}}&I_{y^{\prime}z^{\prime}}&I_{z^{\prime}z^{\prime}}\end% {array}\right]\left\{\begin{array}[]{c}0\\ \ddot{\phi}\\ 0\end{array}\right\}$
	$\displaystyle=$	$\displaystyle\left\{\begin{array}[]{c}I_{x^{\prime}y^{\prime}}\ddot{\phi}\\ I_{y^{\prime}y^{\prime}}\ddot{\phi}\\ I_{y^{\prime}z^{\prime}}\ddot{\phi}\end{array}\right\}.$
Similarly,
$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\}$	$\displaystyle=$	$\displaystyle\left\{\begin{array}[]{c}I_{x^{\prime}y^{\prime}}\dot{\phi}\\ I_{y^{\prime}y^{\prime}}\dot{\phi}\\ I_{y^{\prime}z^{\prime}}\dot{\phi}\end{array}\right\},$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times[\boldsymbol{I}^{% \scriptscriptstyle\mathrm{O}}]\{{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\}$	$\displaystyle=$	$\displaystyle\dot{\phi}\hat{\boldsymbol{\jmath}}^{\prime}\times[I_{x^{\prime}y% ^{\prime}}\dot{\phi}\hat{\boldsymbol{\imath}}^{\prime}+I_{y^{\prime}y^{\prime}% }\dot{\phi}\hat{\boldsymbol{\jmath}}^{\prime}+I_{y^{\prime}z^{\prime}}\dot{% \phi}\hat{\boldsymbol{k}}^{\prime}]$
	$\displaystyle=$	$\displaystyle-I_{x^{\prime}y^{\prime}}{\dot{\phi}}^{2}\hat{\boldsymbol{k}}^{% \prime}+I_{y^{\prime}z^{\prime}}{\dot{\phi}}^{2}\hat{\boldsymbol{\imath}}^{% \prime}.$

Therefore,

	$\displaystyle\hat{\boldsymbol{\jmath}}^{\prime}\cdot\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\jmath}}^{\prime}\cdot[(I_{x^{\prime}y^{\prime}}% \ddot{\phi}\hat{\boldsymbol{\imath}}^{\prime}+I_{y^{\prime}y^{\prime}}\ddot{% \phi}\hat{\boldsymbol{\jmath}}^{\prime}+I_{y^{\prime}z^{\prime}}\dot{\phi}\hat% {\boldsymbol{k}}^{\prime})+(-I_{x^{\prime}y^{\prime}}{\dot{\phi}}^{2}\hat{% \boldsymbol{k}}^{\prime}+I_{y^{\prime}z^{\prime}}{\dot{\phi}}^{2}\hat{% \boldsymbol{\imath}}^{\prime})]$
		$\displaystyle=$	$\displaystyle I_{y^{\prime}y^{\prime}}\ddot{\phi}.$

Now, setting $\hat{\boldsymbol{\jmath}}^{\prime}\cdot\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{O}$}}}=\hat{% \boldsymbol{\jmath}}^{\prime}\cdot\dot{\accentset{\rightharpoonup}{\boldsymbol% {H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$ , we get

$\displaystyle\frac{L}{2}mg\sin\phi\,\sin\gamma$	$\displaystyle=$	$\displaystyle I_{y^{\prime}y^{\prime}}\ddot{\phi}$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{\phi}$	$\displaystyle=$	$\displaystyle\frac{mg\,(L/2)\,\sin\gamma}{I_{y^{\prime}y^{\prime}}}\sin\phi$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{\phi}$	$\displaystyle=$	$\displaystyle C\sin\phi$

where

C=\frac{mg\,(L/2)\,\sin\gamma}{I_{y^{\prime}y^{\prime}}}.

For rod OC,

I_{y^{\prime}y^{\prime}}=\int_{m}(x^{\prime 2}+z^{\prime 2})\,dm=\int_{0}^{L}l% ^{2}\frac{m}{L}\,dl=\frac{1}{3}mL^{2}.

Therefore, the equation of motion of the rod is

	$\displaystyle\ddot{\phi}-\frac{mg\,(L/2)\,\sin\gamma}{(1/3)mL^{2}}\sin\phi$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\ddot{\phi}-\frac{3g\,\sin\gamma}{2L}\sin\phi$	$\displaystyle=$	$\displaystyle 0.$

Answer: $\ddot{\phi}-\frac{3g\,\sin\gamma}{2L}\sin\phi=0$

SAMPLE 20.18

Filename:sfig7-4-2

Figure 20.96:

Kinetic energy in 3-D rotation. A thin rod of mass $2\,\mathrm{kg}$ and length $L=\frac{1}{2}\,\mathrm{m}$ is welded to a massless shaft at an angle $\theta=45^{{}^{\circ}}$ . The shaft rotates about its longitudinal axis ( $y$ -axis) at $100\,{\rm rpm}$ . The moment of inertia matrix of the rod about the weld point O is

[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=0.08\,\mathrm{kg}\mathord{% \cdot}\,\mathrm{m}^{2}\left[\begin{array}[]{rrr}1&-1&0\\ -1&1&0\\ 0&0&2\end{array}\right].

Find the kinetic energy of the rod.

Solution The rod rotates about the fixed point O with angular velocity

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \jmath}}\quad\mbox{where}\quad\omega=100\,{\rm rpm}=10.47\,\mathrm{rad}/% \mathrm{s}.

The kinetic energy of a rigid body rotating about a fixed point O is given by

{E_{\mathrm{K}}}=\frac{1}{2}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \mathord{\cdot}\left[[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\mathord{% \cdot}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\right]

For the given problem, let us write the moment of inertia matrix $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ of the rod as ^†^†margin: This step is simply to facilitate computation. We carry out the multiplication and at the end substitute the value of $K_{0}$ .

$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$	$\displaystyle=$	$\displaystyle K_{0}\left[\begin{array}[]{rrr}1&-1&0\\ -1&1&0\\ 0&0&2\end{array}\right]$
where $K_{0}=0.08\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$ . Now
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{\jmath}}=\left\{0\quad\omega\quad 0\right% \}^{\rm T}.$
Therefore,
$\displaystyle[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]\mathord{\cdot}{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle K_{0}\left[\begin{array}[]{rrr}1&-1&0\\ -1&1&0\\ 0&0&2\end{array}\right]\left\{\begin{array}[]{c}0\\ \omega\\ 0\\ \end{array}\right\}$
	$\displaystyle=$	$\displaystyle K_{0}\left\{\begin{array}[]{r}-\omega\\ \omega\\ 0\\ \end{array}\right\}$
	$\displaystyle=$	$\displaystyle-K_{0}\omega\hat{\boldsymbol{\imath}}+K_{0}\omega\hat{\boldsymbol% {\jmath}}.$

Substituting the expression in the formula for ${E_{\mathrm{K}}}$ we get

$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\omega\hat{\boldsymbol{\jmath}}\mathord{\cdot}(-K_{0}% \omega\hat{\boldsymbol{\imath}}+K_{0}\omega\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\frac{1}{2}K_{0}\omega^{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\mathord{\cdot}0.08\,\mathrm{kg}\mathord{\cdot}\,% \mathrm{m}^{2}\mathord{\cdot}(10.47\,\mathrm{rad}/\mathrm{s})^{2}$
	$\displaystyle=$	$\displaystyle 4.38\,{\rm N}\mathord{\cdot}{\rm m}=4.38\rm{J}$

Answer: ${E_{\mathrm{K}}}=4.38\rm{J}$

Problems for 20.4 Mechanics using $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$

20.4.1

Find the angle between ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}=[\boldsymbol{I}^{\mathrm{cm}}]{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , if ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=3\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ and

[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}20&0&0\\ 0&20&0\\ 0&0&40\end{array}\right]\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}.

20.4.2 Find the projection of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}=[\boldsymbol{I}^{\mathrm{cm}}]{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ in the direction of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , if ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=(2\hat{\boldsymbol{\imath}}% -3\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/\mathrm{s}$ and $[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}5.4&0&0\\ 0&5.4&0\\ 0&0&10.8\end{array}\right]\,{\mathrm{lbm}}\,\mathrm{ft}^{2}$ .

20.4.3 Compute ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\bigl{(}[\boldsymbol{I% }^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\bigr{)}$ for the following two cases. In each case ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=2.5\,\mathrm{rad}/\mathrm{s% }\hat{\boldsymbol{\jmath}}$ .

(a)

$[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}8&0&0\\ 0&16&0\\ 0&0&8\end{array}\right]\,\mathrm{kg}\,\mathrm{m}^{2}$ , Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times([\boldsymbol{I}^{% \mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}})={% \accentset{\rightharpoonup}{\boldsymbol{0}}}\,{\rm kg}\mathord{\cdot}{\rm m}^{% 2}{\rm/s^{2}}$ .
(b)

$[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}8&0&0\\ 0&16&-4\\ 0&-4&8\end{array}\right]\,\mathrm{kg}\,\mathrm{m}^{2}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times([\boldsymbol{I}^{% \mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}})=-25\hat{% \boldsymbol{\imath}}\,{\rm kg}\mathord{\cdot}{\rm m}^{2}{\rm/s^{2}}$ .

20.4.4 Find the $z$ component of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}=[\boldsymbol{I}^{\mathrm{cm}}]{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ for ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=2\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ and

[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}10.20&-2.56&0\\ -2.56&10.20&0\\ 0&0&14.00\end{array}\right]\,{\mathrm{lbm}}\,\mathrm{ft}^{2}.

20.4.5 For ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=5\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{\lambda}}$ , find $\hat{\boldsymbol{\lambda}}$ such that the angle between ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}=[\boldsymbol{I}^{\mathrm{cm}}]% \cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is zero if $[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}3&0&2\\ 0&6&0\\ 2&0&3\end{array}\right]\,\mathrm{kg}\,\mathrm{m}^{2}$ .

20.4.6 Calculation of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ . You will consider a disc spinning about various axes through the center of the disc $O$ . For the most part, you are concerned with calculating $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ . But, you should keep in mind that the purpose of the calculation might be to calculate reaction forces and moments. Every point on the disk moves in circles about the same axis at the same number of radians per second. These problems can be done at least three different ways: using the $[I]$ ’s from problem 20.3, direct integration of the definition of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ , and using the equivalent dumbbell (with the definition of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ ). You should do them all three ways, first using the way you are most comfortable with.

(a)

Assume that the configuration is as in problem 20.3(a) and that the disk is spinning about the $z$ axis at the constant rate of $\omega\bf k$ . What are ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ ?
(b)

Assume that the configuration is as in problem 20.3(b) and that the disk is spinning about the $z$ axis at the constant rate of $\omega\bf k$ . What are ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ ? (In this problem there are two ways to use [I]: one is in the $x y z$ coordinate system, the other is in a coordinate system aligned nicely with the disk. )

20.4.7 What is the angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}$ for:

(a)

A uniform sphere rotating with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ (radius $R$ , mass $m$ )?
(b)

A rod shown in the figure rotating about its center of mass with these angular velocities: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \jmath}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega(\hat{\boldsymbol{% \imath}}+\hat{\boldsymbol{\jmath}})/\sqrt{2}$ ?

Filename:pfigure-blue-99-1

Figure 20.97

20.4.8 A hoop is welded to a long rigid massless shaft which lies on a diameter of the hoop. The shaft is parallel to the $x$ -axis. At the instant of interest the hoop is in the $x y$ -plane of a coordinate system which has its origin O at the center of the hoop. The hoop spins about the $x$ -axis at $1\,\mathrm{rad}/\mathrm{s}$ . There is no gravity.

(a)

What is $[\boldsymbol{I}^{\mathrm{cm}}]$ of the hoop?
(b)

What is ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ of the hoop?

Filename:pfigure-blue-47-2

Figure 20.98

20.4.9 Two thin rods, each of mass $m$ and length $d$ , are welded perpendicularly to an axle of mass $M$ and length $\ell$ , which is supported by a ball-and-socket joint at A and a journal bearing at B. A force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=-F\hat{\boldsymbol{\imath}}+G% \hat{\boldsymbol{k}}$ acts at the point P as shown in the figure. Determine the initial angular acceleration $\alpha$ and initial reactions at point B.

Filename:pfigure-blue-89-1

Figure 20.99

20.4.10 The thin square plate of mass $M$ and side $c$ is mounted vertically on a vertical shaft which rotates with angular speed $\omega_{0}$ . In the fixed $x, y, z$ coordinate system with origin at the center of mass (with which the plate is currently aligned) what are the vector components of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ ? What are the components of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{cm}$ ?

Filename:pfigure-blue-35-2

Figure 20.100

20.4.11 If you take any rigid body (with weight ), skewer it with an axle, hold the axle with hinges that cause no moment about the axle, and hold it so that it cannot slide along the axle, it will swing like a pendulum. That is, it will obey the equations:

	$\displaystyle\ddot{\theta}$	$\displaystyle=-C\sin\theta,\quad{\rm or}$	$\displaystyle(a)$
	$\displaystyle\ddot{\phi}$	$\displaystyle=+C\sin\phi,\quad$	$\displaystyle(b)$

where

: $\theta$ is the amount of rotation about the axis measuring zero when the center of mass is directly below the $z^{\prime}$ -axis; i. e., when the center of mass is in a plane containing the $z^{\prime}$ -axis and the gravity vector,
: $\phi$ is the amount of rotation about the axis measuring zero when the center of mass is directly above the $z^{\prime}$ -axis,
: $C$ $=M_{total}\ gd\sin\gamma/I_{z^{\prime}z^{\prime}}$ ,
: $M_{total}$ is the total mass of the body,
: $g$ is the gravitational constant,
: $d$ is the perpendicular distance from the axis to the center of mass of the object,
: $\gamma$ is the angle of tip of the axis from the vertical,
: $I_{z^{\prime}z^{\prime}}$ is the moment of inertia of the object about the skewer axis.

(a)

How many examples of such pendula can you think of and what would be the values of the constants? (e.g. a simple pendulum?, a broom upside down on your hand?, a door?, a box tipping on one edge?, a bicycler stopped at a red light and stuck in her new toe clips?, a person who got rigor mortis and whose ankles went totally limp at the same time?,…)
(b)

A door with frictionless hinges is misaligned by half an inch (the top hinge is half an inch away from the vertical line above the other hinge). Estimate the period with which it will swing back and forth. Use reasonable magnitudes of any quantities you need.
(c)

Can you verify the formula (a simple pendulum and a swinging stick are special cases)?

Filename:pfigure4-3Dpend

Figure 20.101

-3-D pendulum on a hinge.

20.4.12 A hanging rod going in circles. A uniform rod with length $\ell$ and mass $m$ hangs from a well greased ball and socket joint at $O$ . It swings in circles at constant rate ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\boldsymbol{k}$ while the rod makes an angle $\phi$ with the vertical (so it sweeps a cone).

(a)

Calculate ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ .
(b)

Using $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ and angular momentum balance, find the rotation rate $\omega$ in terms of $\phi$ , $g$ , $\ell$ , and $m$ . How long does it take to make one revolution? Answer: $\omega^{2}=\frac{3g}{2L\cos\phi}$ , $T_{rev}=\frac{2\pi}{\omega}$ .
(c)

If another stick rotates twice as fast while making the same angle $\phi$ , what must be its length compared to $\ell$ ?
(d)

Is there a simple way to describe what two uniform sticks of different length and different cone angle $\phi$ both have in common if their rate of rotation $\omega$ is the same? (If you accurately draw two such sticks, you should see the relation.)
(e)

Find the reaction force at the ball and socket joint. Is this force parallel to the stick?

Filename:pfigure-s94h6p3

Figure 20.102

20.4.13 From discrete to continuous. $n$ equal beads of mass $m/n$ each are glued to a rigid massless rod at equal distances $l=\ell/n$ from each other. For $n=4$ , the system is shown in the figure, but you are to consider the general problem with $n$ masses. The rod swings around the vertical axis maintaining angle $\phi$ .

(a)

Find the rate of rotation of the system as a function of $g,\,\ell,\,n,$ and $\phi$ for constant rate circular motion. [Hint: You may need these series sums: $1+2+3+\cdots+n={n(n+1)\over 2}$ and $1^{2}+2^{2}+3^{2}+\cdots+n^{2}={n(n+1)(2n+1)\over 6}$ .] Answer: $\omega^{2}={3g\,n\over L(2n+1)\cos\phi}$ .
(b)

Set $n=1$ in the expression derived in (a) and check the value with the rate of rotation of a conical pendulum of mass $m$ and length $\ell$ .
(c)

Now, put $n\rightarrow\infty$ in the expression obtained in part (a). Does this rate of rotation make sense to you? [Hint: How does it compare with the rate of rotation for the uniform stick of Problem 20.4?]

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Figure 20.103

20.4.14 Spinning dumbbell. Two point masses are connected by a rigid rod of negligible mass and length $2\sqrt{2}\ell$ . The rod is welded to a shaft that is along the $y$ axis. The shaft is spinning at a constant rate $\omega$ driven by a motor at A (Not shown). At the time of interest the dumbbell is in the $y z$ -plane. Neglect gravity.

(a)

What is the moment of inertia matrix about point O for the dumbbell at the instant shown? Answer:

$[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]=m\ell^{2}\left[\begin{array}[]% {rrr}4&0&0\\ 0&2&-2\\ 0&-2&2\par\par\end{array}\right],$

in terms of the coordinate system shown.
(b)

What is the angular momentum of the dumbbell at the instant shown about point O. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=2m\ell^{2}\omega(-\hat{\boldsymbol{\jmath}}+% \hat{\boldsymbol{k}}).$
(c)
What is the rate of change of angular momentum of the dumbbell at the instant shown? Calculate the rate of change of angular momentum two different ways:
- (i)
  
  by adding up the contribution of both of the masses to the rate of change of angular momentum (i.e. $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}=\sum{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{i}\times(m_{i}{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{i})$ ) and
- (ii)
  
  by calculating $I_{xy}$ and $I_{yz}$ , calculating ${\accentset{\rightharpoonup}{\boldsymbol{H}}}$ , and then calculating $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ from ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{H}}}$ . Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=-2m\ell^{2}\omega^{2}\hat{\boldsymbol{% \imath}}$ .
(d)
What is the total force and moment (at the CM of the dumbbell) required to keep this motion going? Do the calculation in the following ways:
- (i)
  
  Using angular momentum balance (about any point of your choice) for the shaft dumbbell system.
- (ii)
  
  By drawing free-body diagrams of the masses alone and calculating the forces on the masses. Then use action and reaction and draw a free-body diagram of the remaining (massless) part of the shaft-dumbbell system. Then use moment balance for this system. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{O}$}}}=-2mL^{2}\omega^{2}\hat{\boldsymbol{\imath}}$ .

Filename:pg84-3

Figure 20.104

Crooked spinning dumbbell.

20.4.15 A uniform rectangular plate of mass $m$ , height $h$ , and width $b$ spins about its diagonal at a constant angular speed $\omega$ , as shown in the figure. The plate is supported by frictionless bearings at $O$ and $C$ . The plate lies in the $Y Z$ -plane at the instant shown.

[Note: $\cos(\beta)=b/\sqrt{(h^{2}+b^{2})}$ , and $\sin(\beta)=h/\sqrt{(h^{2}+b^{2})}$ .]

(a)

Compute the angular momentum of the plate about G (with respect to the body axis $G x y z$ shown).
(b)

Ignoring gravity determine the net torque (moment) that the supports impose on the body at the instant shown. (Express your solution using base vectors and components associated with the inertial $X Y Z$ frame.)
(c)

Calculate the reactions at $O$ or $C$ for the configuration shown. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{O}=-\frac{m\omega^{2}bh(b^{2}-h% ^{2})}{12(b^{2}+h^{2})^{\frac{3}{2}}}\hat{\boldsymbol{k}}=-{\accentset{% \rightharpoonup}{\boldsymbol{R}}}_{C}$ .
(d)

In the case when $h=b=1\,\mathrm{m}$ what is the vertical reaction at $C$ (still neglecting gravity).

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Figure 20.105

20.4.16 A uniform rectangular plate spins about a fixed axis. The constant torque $M_{y}$ is applied about the $Y$ axis. The plate is held by bearings at the corners of the plate at A and B. At the instant shown the plate and the $y^{\prime}z^{\prime}$ axis attached to the plate are in the $Y Z$ plane. Neglect gravity. At $t=0$ the plate is at rest.

(a)

At $t=0^{+}$ (i.e., just after the start), what is the acceleration of point D? Give your answer in terms of any or all of $c,d,m,M_{y},\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},$ and $\hat{\boldsymbol{k}}$ (the base vectors aligned with the fixed $X Y Z$ axes), and $I^{cm}_{YY}$ [You may use $I^{cm}_{YY}$ in your answer. Or you could use this result from the change of coordinates formulae in linear algebra: $I^{cm}_{YY}=(\cos^{2}\beta){I}^{cm}_{y^{\prime}y^{\prime}}+(\sin^{2}\beta)I^{% cm}_{z^{\prime}z^{\prime}}$ to get an answer in terms of $c$ , $d$ , $m$ , and $M_{y}$ .] Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{D}=\frac{M_{y}}{I_{YY}^{cm}}% \frac{2dc}{\sqrt{c^{2}+d^{2}}}\hat{\boldsymbol{\imath}}$ .
(b)

At $t=0^{+}$ , what is the reaction at B? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{B}}}=-\frac{M_{y}I^{cm}_{YZ}}{2I_{YY}% ^{cm}\sqrt{c^{2}+d^{2}}}\hat{\boldsymbol{\imath}}$ .

Filename:pfigure-s95q8

Figure 20.106

20.4.17 A uniform semi-circular disk of radius $r=0.2\,\mathrm{m}$ and mass $m=1.5\,\mathrm{kg}$ is attached to a shaft along its straight edge. The shaft is tipped very slightly from a vertical upright position. Find the dynamic reactions on the bearings supporting the shaft when the disk is in the vertical plane $(xz)$ and below the shaft.

Filename:pfigure4-rpi

Figure 20.107

20.4.18 A thin triangular plate of base $b$ , height $h$ , and mass $m$ is attached to a massless shaft which passes through the bearings L and R. The plate is initially vertical and then, provoked by a teensy disturbance, falls, pivoting around the shaft. The bearings are frictionless. Given,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}$	$\displaystyle=$	$\displaystyle(0,\frac{2}{3}b,\frac{1}{3}h),$
$\displaystyle I^{O}_{xx}$	$\displaystyle=$	$\displaystyle\ \ \frac{m}{2}(b^{2}+h^{2}/3),$
$\displaystyle I^{O}_{yy}$	$\displaystyle=$	$\displaystyle\ \ mh^{2}/6,$
$\displaystyle I^{O}_{zz}$	$\displaystyle=$	$\displaystyle\ \ mb^{2}/2,$
$\displaystyle I^{cm}_{yy}$	$\displaystyle=$	$\displaystyle\ \ mh^{2}/18,$
$\displaystyle I^{O}_{zy}$	$\displaystyle=$	$\displaystyle-mbh/4,$

(a)

what is the plate’s angular velocity as it passes through the downward vertical position?
(b)

Compute the bearing reactions when the plate swings through the downward vertical position.

Filename:pfigure-blue-86-1

Figure 20.108

20.4.19 A welded structural tubing framework is proposed to support a heavy emergency searchlight. To select properly the tubing dimensions it is necessary to know the forces and moments at the critical section AA, when the framework is moving. The frame moves as shown.

(a)

Determine the magnitude and direction of the dynamic moments (bending and twisting) and forces (axial and shear) at section AA of the frame if, at the instant shown, the frame has angular velocity $\omega_{o}$ and angular acceleration $\alpha_{o}$ . Assume that the mass per unit length of the pipe is $\rho$ .

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Figure 20.109

20.4.20 A uniform cube with side $0.25\,\mathrm{m}$ is glued to a pipe along the diagonal of one of its faces $A B C D$ . The pipe rotates about the $y$ -axis at a constant rate $\omega=30\,{\rm rpm}$ . At the instant of interest, the face $A B C D$ is in the $x y$ -plane.

(a)

Find the velocity and acceleration of point $F$ on the cube.
(b)

Find the position of the center of mass of the cube and compute its acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}$ .
(c)

Suppose that the entire mass of the cube is concentrated at its center of mass. What is the angular momentum of the cube about point $A$ ?

Filename:pfigure4-2-rp3

Figure 20.110

20.4.21 A uniform square plate of mass $m$ and side length $b$ is welded to a narrow shaft at one corner. The shaft is perpendicular to the plate and has length $\ell$ . The upper end of the shaft is connected to a ball and socket joint at $O$ . The assembly is rotating about the $z$ axis through $O$ at the fixed rate ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ which happens to be just right to make the plate horizontal and the shaft vertical, as shown. The gravitational constant is $g$ .

(a)

What is $\omega$ in terms of $m$ , $\ell$ , $b$ , and $g$ ?
(b)

Is the reaction at $O$ on the rod-plate assembly in the direction of the line from $G$ to $O$ ? (Why?)

Filename:pfigure-s94q6p1

Figure 20.111

20.4.22 The uniform cube shown is rotating at constant angular velocity about the fixed axis OA with angular speed $10\,\mathrm{rad}/\mathrm{s}$ counterclockwise looking towards O from A. Find:

(a)

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$
(b)

${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}$
(c)

${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{C}}$
(d)

${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/O}}$
(e)

$\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/O}}$
(f)

The total torque applied to the block in the configuration shown during this motion.

Filename:pfigure-spinningbrick

Figure 20.112

20.4.23 A spinning box. A solid box $\cal B$ with dimensions $a$ , $a$ , and $2a$ (where $a=.1\,\mathrm{m}$ ) and mass = $1\,\mathrm{kg}$ . It is supported with ball and socket joints at opposite diagonal corners at $A$ and $B$ . It is spinning at constant rate of $(50/\pi)\,{{\rm rev}/{\rm sec}}$ and at the instant of interest is aligned with the fixed axis $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ . Neglect gravity.

(a)

What is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ ?
(b)

What is $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ ?
(c)

What is ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\cal B}/cm}$ ?
(d)

What is $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\cal B}/cm}$ ?
(e)

What is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{B}$ (where ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{B}$ is the reaction force on ${\cal B}$ at $B$ )? A dimensional vector is desired.
(f)

If the hinges were suddenly cut off when ${\cal B}$ was in the configuration shown, what would be $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ immediately afterwards?

Filename:pfigure-s94f1p2

Figure 20.113

20.4.24 Uniform rod attached to a shaft. A uniform $1\,\mathrm{kg}$ , $\sqrt{2}\,\mathrm{m}$ long rod is attached to a shaft which spins at a constant rate of $1\,\mathrm{rad}/\mathrm{s}$ . One end of the rod is connected to the shaft by a ball-and-socket joint. The other end is held by two strings (or pin jointed rods) that are connected to a rigid cross bar. The cross bar is welded to the main shaft. Ignore gravity. Find the reaction force at A.

Filename:s97f2

Figure 20.114

20.4.25 A round disk spins crookedly on a shaft. A thin homogeneous disk of mass $m$ and radius $r$ is welded to the horizontal axle AB. The normal to the plane of the disk forms an angle $\beta$ with the axle. The shaft rotates at the constant rate $\omega$ .

(a)

What is the $x$ component of the angular momentum of the disk about point G the center of mass of the disk?
(b)

What is the $x$ component of the rate of change of angular momentum of the disk about G?
(c)

At the given instant, what is the total force and moment (about G) required to keep the disk spinning at constant rate?

Filename:pg92-2

Figure 20.115

A round disk spins crookedly on a shaft.

20.4.26 A uniform disk of mass $m=2\,\mathrm{kg}$ and radius $r=0.2\,\mathrm{m}$ is mounted rigidly on a massless axle. The normal to the disk makes an angle $\beta=60^{\rm o}$ with the axle. The axle rotates at a constant rate $\omega=60\,{\rm rpm}$ . At the instant shown in the figure,

(a)

Find the angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ of the disk about point $O$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=(-0.108\hat{\boldsymbol{\imath}}^{\prime}+0.% 126\boldsymbol{k}^{\prime})\,{\rm kg}\mathord{\cdot}{\rm m}^{2}{\rm/s}$
(b)

Draw the angular momentum vector indicating its magnitude and direction. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ is in the $x^{\prime}z^{\prime}$ plane at the instant of interest and at an angle $\alpha\approx 41^{\circ}$ , clockwise from the $z^{\prime}$ -axis.
(c)

As the disk rotates (with the axle), the angular momentum vector changes. Does it change in magnitude or direction, or both? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ does not change in magnitude ( $m$ , $R$ , $\omega$ , $\beta$ are constants). ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ rotates with the disk; i.,e., it changes direction.
(d)

Find the direction of the net torque on the system at the instant shown. Note, only the direction is asked for. You can answer this question based on parts (b) and (c). Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=$ the rate of change of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ . The tip of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ goes in a circle. At the instant of interest, the change will be in the $\hat{\boldsymbol{\jmath}}^{\prime}$ or $\hat{\boldsymbol{\jmath}}$ direction. Thus, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$ will be in the $\hat{\boldsymbol{\jmath}}$ direction.

Filename:summer95p2-3

Figure 20.116

20.4.27 A uniform disk spins at constant angular velocity. The uniform disk of mass $m$ and radius $r$ spins about the z axis at the constant rate $\omega$ . Its normal $\hat{\boldsymbol{n}}$ is inclined from the $z$ axis by the angle $\beta$ and is in the $x z$ plane at the instant of interest. What is the total force and moment (relative to O) needed to keep this disk spinning at this rate? Answer in terms of $m$ , $r$ , $\omega$ , $\beta$ and any base vectors you need.

Filename:p-f96-p3-3

Figure 20.117

20.4.28 Spinning crooked disk. A rigid uniform disk of mass $m$ and radius $R$ is mounted to a shaft with a hinge. An apparatus not shown keeps the shaft rotating around the $z$ axis at a constant $\omega$ . At the instant shown the hinge axis is in the $x$ direction. A massless wire, perpendicular to the shaft, from the shaft at A to the edge of the disk at B keeps the normal to the plate (the $z^{\prime}$ direction) at an angle $\phi$ with the shaft.

(a)

What is the tension in the wire? Answer in terms of some or all of $R,m,\phi$ and $\omega$ .
(b)

What is the reaction force at the hinge at O?

Filename:s97p3-3

Figure 20.118

20.4.29 A disoriented disk rotating with a shaft. A uniform thin circular disk with mass $m=2\,\mathrm{kg}$ and radius $R=0.25\,\mathrm{m}$ is mounted on a massless shaft as shown in the figure. The shaft spins at a constant speed $\omega=3\,\mathrm{rad}/\mathrm{s}$ . At the instant shown, the disk is slanted at $30^{\circ}$ with the $x y$ -plane. Ignore gravity.

(a)

Find the rate of change of angular momentum of the disk about its center of mass. Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\,\mathrm{cm}}=\frac{9\sqrt% {3}}{128}\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{\jmath}}=0.122\,{\rm N% }\mathord{\cdot}{\rm m}\hat{\boldsymbol{\jmath}}$
(b)

Find the rate of change of angular momentum about point A. Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{A}=-26.89\,{\rm N}\mathord{% \cdot}{\rm m}\hat{\boldsymbol{\jmath}}$
(c)

Find the total force and moment required at the supports to keep the motion going. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}=-4.56\,{\mbox{N}}\hat{% \boldsymbol{\imath}},\,{\accentset{\rightharpoonup}{\boldsymbol{R}}}_{C}=-13.4% 39\,{\mbox{N}}\hat{\boldsymbol{\imath}},\,M_{z}=0$ .

Filename:pfigure-s94h8p1

Figure 20.119

20.5 Dynamic balance

Sometimes when something spins it can shake the structure that holds it. A familiar example is an ‘unbalanced’ car tire which makes a car shake. But rotor balance is important in all kinds of machines. Most often one seeks to eliminate or minimize the imbalance. For example the strange design of car crank-shafts is due in large part to an attempt to minimize its imbalance. But what does it mean, in the language of mechanics, to say a rotating body is ‘unbalanced’? It means that non-zero forces and/or torques are required to hold the axis still while the object spins at constant rate. Going back to the basic momentum balance equations:

\mbox{Linear momentum balance:}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{F}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}

and

\mbox{Angular momentum balance:}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{C}$}}}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}},

we see that forces and torques are required if $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}\neq{\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ or if $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}\neq{\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ .

So imbalance means $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and/or $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}$ is not zero. We break the concept of balance into the following two concepts:

(1)

An object is said to be statically balanced with respect to a given axis of rotation if $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ when it spins at constant rate about that axis.
(2)

An object is said to be dynamically balanced with respect to a given axis of rotation if both $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ for constant rate rotation about that axis.

The origin of the words ‘static’ balance and ‘dynamic’ balance is in how the imbalance can be measured. Static imbalance can be measured with a static test, dynamic imbalance requires a dynamic test.

Static balance

If the center-of-mass of a rigid body is on the fixed axis of rotation then it will not accelerate. Thus, $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{cm}m_{tot}={\accentset{\rightharpoonup}{\boldsymbol{0}}}$ and the object is statically balanced. An equivalent definition of static balance is that the net force on the spinning body is zero. Whether or not this net force is so can be tested with a statics experiment. Put the axis of rotation on good bearings and see if the mass hangs down in any preferred direction. In a tire shop, this kind of balancing is sometimes called bubble balancing because of the bubble in the level measuring device.

Dynamic balance

The first condition for dynamic balance of an object with respect to spinning about an axis is that the object be statically balanced. The center-of-mass must lie on the axis of rotation. The second condition for dynamic balance, that $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , is a little more subtle. To make things more specific, let’s calculate the rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ with respect to a point O that is on the axis of rotation. So, for constant rate rotation about a fixed axis we have:

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

(20.78)

because ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ spins with the body. $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ is evidently zero if the angular momentum, ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ , is parallel to the angular velocity, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . However, ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O% }}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ which means ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ is parallel to ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ only when ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is an eigenvector of $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ ^†^†margin: When you go to ACME garage and get your car tires balanced you can politely ask the mechanic: “Ma’am, could you make sure that one of the eigenvectors of my wheel’s moment of inertia matrix is parallel to the car axle? Thanks.” The mechanic, if she is a decent person, will make sure that it is the appropriate eigenvector that is parallel to the axle. Otherwise the wheel would point straight sideways with the axle piercing the rubber. .

So an object which is spinning about a fixed axis is dynamically balanced if its center-of-mass is on the axis (static balance) and the angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is an eigenvector of the moment of inertia matrix.

If we restrict ^†^†margin: Caution: A common misperception about angular momentum is that it is always parallel to angular velocity. In general, angular momentum is not parallel to angular velocity. When is angular momentum about the center-of-mass, say, parallel to angular velocity? When ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is an eigenvector of $[\boldsymbol{I}^{\mathrm{cm}}]$ . This is always the case for planar objects rotating about an axis perpendicular to the plane, but not generally in 3D. our attention to cases where the axis of rotation is the $z$ -axis then this condition is easy to recognize. It is when $I_{xz}=I_{yz}=0$ , that is, when the only non-zero element in the third column and in the third row of the $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ matrix is $I^{\scriptscriptstyle\mathrm{O}}_{zz}$ in the lower right corner.

Since these terms cause wobbling if one of the coordinate axis is an axis of rotation, the terms that are off the main diagonal in the moment of inertia matrix are sometimes called the ‘imbalance’ terms.

These terms lead to dynamic imbalance when the object is spun about the $x$ , $y$ , or $z$ -axes. The off diagonal terms are also sometimes called the ‘centrifugal’ terms. This naming has an intuitive basis. One way to understand dynamic imbalance is to think of it being due to the unbalanced centrifugal pull of bits of mass that are spinning in circles.

Often, but not always, you can tell by inspection if something is dynamically balanced for rotation about a certain axis. If, for every bit of mass that is not on the axis there is another equal bit of mass that is exactly opposite, with respect to the axis, then the object is balanced. Some objects that do not meet this symmetry condition are also dynamically balanced, however.

Example: Some common objects

All of the figures in fig. 2.34 on page 2.34 are dynamically balanced about all the axes shown and about any axis perpendicular to any pair of these axes.

Example: A cube = a sphere

Both a cube and a sphere have moment of inertia matrices proportional to the identity matrix

	$\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]_{\rm cube}$	$\displaystyle=$	$\displaystyle\frac{m\ell^{2}}{6}\left[\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right]$
	$\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]_{\rm sphere}$	$\displaystyle=$	$\displaystyle\frac{mD^{2}}{10}\left[\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right].$

So, for any ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , we get that ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$ is parallel to ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and, thus, both of these objects are dynamically balanced for rotation about any axis through their centers of mass.

Here is an example where the mathematics contradicts intuition; it does not seem like a cube should be balanced for rotation about a random skewed axis through its center-of-mass!

SAMPLE 20.19 Static and dynamic balance in 2-D motion.

Filename:sfig4-7-DH1

Figure 20.120:

A system with two equal masses at A and B rotates about point O at a constant rate $\omega=10\,{\rm rpm}$ . The system is shown in the figure. The rod connecting the two masses has negligible mass.

1.

Is the system statically balanced? If not, suggest a way to balance it.
2.

Is the system dynamically balanced? If not, suggest a way to balance it.

Solution

1.
Filename:sfig4-7-DH2

Figure 20.121:

Since the two masses are equal, the center-of-mass of the system is at the geometric center of rod AB, i.e., at a point O^′, halfway between A and B. Clearly, O^′ is not on the axis of rotation which passes through O and is perpendicular to the plane of motion. Therefore, the system is not statically balanced.

To balance the system, we must move the center-of-mass to O or pivot the system at O^′. Say we cannot move the pivot point. So, to move the center-of-mass O, we add mass $m^{\prime}$ to $m$ at A. From the definition of center-of-mass:

$\displaystyle(m+m^{\prime})r_{1}$ $\displaystyle=$ $\displaystyle mr_{2}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\quad\quad m^{\prime}$ $\displaystyle=$ $\displaystyle\frac{m^{\prime}(r_{2}-r_{1})}{r_{1}}$

$\displaystyle=$ $\displaystyle\frac{1\,\mathrm{kg}(0.5\,\mathrm{m}-0.3\,\mathrm{m})}{0.3\,% \mathrm{m}}$

$\displaystyle=$ $\displaystyle 0.67\,\mathrm{kg}.$

Answer: Add 0.67 kg to mass at A.

Filename:sfig4-7-DH3

Figure 20.122:

The system, as given, is not dynamically balanced since it is not statically balanced. Let us check if it is dynamically balanced after adding $m^{\prime}$ to A as suggested above.

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}.

Let us calculate $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ to see if it is zero:

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}\times m_{1}{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{1}\quad+\quad{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{2}\times m_{2}{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{2}$
	$\displaystyle=$	$\displaystyle(-r_{1}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})\times m_{1}(% \omega^{2}r_{1}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})\quad+\quad(r_{2}% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})\times(-\omega^{2}r_{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R})m_{2}$
	$\displaystyle=$	$\displaystyle-m_{1}\omega^{2}r_{1}^{2}(\underbrace{\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}}_{{% \accentset{\rightharpoonup}{\boldsymbol{0}}}})\quad-\quad m_{2}\omega^{2}r_{2}% ^{2}(\underbrace{\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}}_{{\accentset{\rightharpoonup}{% \boldsymbol{0}}}})$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

The net torque on the system is zero, therefore it is dynamically balanced. Note that $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{0% }}}$ irrespective of the values of $m_{1},m_{2},r_{1},$ and $r_{2}$ . Thus $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={\accentset{\rightharpoonup}{\boldsymbol{0% }}}$ even for the system as given. But the system, as given, is not dynamically balanced because static balance is a necessary condition for dynamic balance.

SAMPLE 20.20

Filename:sfig4-7-1

Figure 20.123: Imbalance of a rotor due to point masses spinning at a constant rate.

Imbalance of a rotor due to rotating masses. Two masses $m_{1}$ and $m_{2}$ are attached to a massless shaft AB by massless rigid rods of length $h$ each. The two masses are separated by distance $b$ along the shaft and are in the same plane but on the opposite sides of the shaft. The shaft is rotating with a constant angular speed $\omega$ . The shaft is free to move along the $z$ -axis at point B. Ignore gravity.

1.

Is the system statically balanced?
2.

What is the torque required to keep the motion going about the $z$ -axis (i.e., $M_{z}$ ) ?
3.

What are the reactions at the support points of the shaft?

Solution

A simple line sketch and the free-body diagram of the system are shown in Fig. 20.124. The linear momentum balance ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ ) for the system gives:

Filename:sfig4-7-1a

Figure 20.124: (a) A simple line diagram of the system. (b) Free-Body Diagram of the system.

Filename:sfig4-7-1b

Figure 20.125: Accelerations of

m_{1}

and

m_{2}

$\displaystyle(A_{x}+B_{x})\hat{\boldsymbol{\imath}}+(A_{y}+B_{y})\hat{% \boldsymbol{\jmath}}+A_{z}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle m_{1}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{1}+m_{2}{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{2}$
	$\displaystyle=$	$\displaystyle m_{1}\omega^{2}h(-\hat{\boldsymbol{\imath}})+m_{2}\omega^{2}h% \hat{\boldsymbol{\imath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad A_{x}+B_{x}$	$\displaystyle=$	$\displaystyle\omega^{2}h(m_{2}-m_{1})$	(20.81)
$\displaystyle A_{y}+B_{y}$	$\displaystyle=$	$\displaystyle 0$	(20.82)
$\displaystyle A_{z}$	$\displaystyle=$	$\displaystyle 0$

Clearly, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\neq{\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ which means the system is not statically balanced.

2.

$M_{z}$ can be easily calculated by writing angular momentum balance about axis AB. In fact, we do not even need to write the equation in this case; since all the forces pass through this axis and the accelerations of $m_{1}$ and $m_{2}$ also pass through it,

$\sum M_{/AB\mbox{(due to reaction forces)}}=0\hbox to36.135pt{\hfil and\hfil}% \dot{H}_{/AB}=0.$

But

$\displaystyle\sum M_{/AB}$ $\displaystyle=$ $\displaystyle\dot{H}_{/AB}$

$\displaystyle\hbox to36.135pt{or\hfil}M_{z}$ $\displaystyle+$ $\displaystyle\sum M_{/AB\mbox{(due to reaction forces)}}=\dot{H}_{/AB}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad M_{z}$ $\displaystyle=$ $\displaystyle 0.$

Answer: $M_{z}=0$

For the four unknown reactions at A and B (we have already found $A_{z}$ and $M_{z}$ ) we have two scalar equations so far (from Linear Momentum Balance ). Angular Momentum Balance about point A gives:

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/A}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/A}

Now,

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}\times\sum{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{B}$
	$\displaystyle=$	$\displaystyle(2\ell+b)\hat{\boldsymbol{k}}\times(B_{x}\hat{\boldsymbol{\imath}% }+B_{y}\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle B_{x}(2\ell+b)\hat{\boldsymbol{\jmath}}-B_{y}(2\ell+b)\hat{% \boldsymbol{\imath}}$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/A}\times m_{1}{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{1}+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{D/A}\times m_{2}{\accentset{\rightharpoonup}{\boldsymbol{a}}% }_{2}$
	$\displaystyle=$	$\displaystyle(\ell\hat{\boldsymbol{k}}+h\hat{\boldsymbol{\imath}})\times m_{1}% (-\omega^{2}h\hat{\boldsymbol{\imath}})+((\ell+b)\hat{\boldsymbol{k}}-h\hat{% \boldsymbol{\imath}})\times m_{2}(\omega^{2}h\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle\omega^{2}h(m_{2}(\ell+b)-m_{1}\ell)\hat{\boldsymbol{\jmath}}$

Equating $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/A}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}$ we get

B_{x}(2\ell+b)\hat{\boldsymbol{\jmath}}-B_{y}(2\ell+b)\hat{\boldsymbol{\imath}% }=\omega^{2}h(m_{2}(\ell+b)-m_{1}\ell)\hat{\boldsymbol{\jmath}}

Dotting both sides of the equation with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , we get

	$\displaystyle B_{y}$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{and\hfil}B_{x}$	$\displaystyle=$	$\displaystyle\frac{\omega^{2}h}{(2\ell+b)}[m_{2}(\ell+b)-m_{1}\ell].$

Substituting $B_{x}$ and $B_{y}$ in (20.81) and (20.82) we find

	$\displaystyle A_{x}$	$\displaystyle=$	$\displaystyle\frac{\omega^{2}h}{(2\ell+b)}[m_{2}\ell-m_{1}(\ell+b)]$
	$\displaystyle A_{y}$	$\displaystyle=$	$\displaystyle 0$

Answer: $\begin{array}[]{lll}A_{x}&=&\frac{\omega^{2}h}{(2\ell+b)}[m_{2}\ell-m_{1}(\ell% +b)]\\ A_{y}&=&A_{z}=0\\ B_{x}&=&\frac{\omega^{2}h}{(2\ell+b)}[m_{2}(\ell+b)-m_{1}\ell]\\ B_{y}&=&0\\ M_{z}&=&0\end{array}$

SAMPLE 20.21

Filename:sfig4-7-2

Figure 20.126: Imbalance of a rotor due to point masses spinning at a constant rate.

Balancing a rotor with spinning masses. Consider the same system as in the previous sample problem (Sample 20.120). Assume that $m_{1}=m_{2}=m$ .

1.

Set $m_{1}=m_{2}$ in the reactions calculated. Is the system statically balanced now? Explain.
2.

Is the system dynamically balanced? Explain.
3.

Balance the system dynamically by (a) adjusting the geometry, (b) adding two masses to the system.

Solution

Filename:sfig4-7-2a

Figure 20.127: Reaction forces at A and B are required to keep the motion going.
1.

Recall from the solution of Sample 20.120 that

$\displaystyle A_{x}$ $\displaystyle=$ $\displaystyle\frac{\omega^{2}h}{(2\ell+b)}[m_{2}\ell-m_{1}(\ell+b)]$

$\displaystyle B_{x}$ $\displaystyle=$ $\displaystyle\frac{\omega^{2}h}{(2\ell+b)}[m_{2}(\ell+b)-m_{1}\ell]$

$\displaystyle A_{y}$ $\displaystyle=$ $\displaystyle B_{y}=A_{z}=M_{z}=0$

Setting $m_{1}=m_{2}=m$ in the above expressions for the reactions, we get

$A_{x}=\frac{\omega^{2}h}{(2\ell+b)}(-mb)\hbox to36.135pt{\hfil and\hfil}B_{x}=% \frac{\omega^{2}h}{(2\ell+b)}(mb).$

Thus, $A_{x}+B_{x}=0$ . Therefore, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ which means the system is statically balanced.

Static Balance:

When the two masses are equal, the center-of-mass of the system is on the axis of rotation. Therefore, the static reactions are zero irrespective of the orientations of the two masses.

The requirement for static balance is $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ in any static orientation of the system. As long as the center-of-mass of the system lies on the axis of rotation, the system will be in static balance.

Dynamic Balance:

When the masses are rotating at constant speed, reaction forces at A and B are required to keep the motion going. Clearly, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}\neq{\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ now. Therefore, the system is not in dynamic balance.

At the instant shown the reaction forces are $A_{x}$ and $B_{x}$ as shown in Fig. 20.127, but they change directions as the position of the masses changes. For example, if the masses are in the y–z plane the reaction forces will be $A_{y}$ and $-B_{y}$ . The magnitudes of these forces will remain the same. Thus, we have rotating reaction forces at A and B. These rotary forces cause wear in the bearings and induce vibrations in the frame of the machine and the supporting structure. Therefore, these forces are undesirable.

If we can somehow make these dynamic reactions zero, the bearings, the machine frame, the supporting structure, the machine operator and the company will all be very happy. So, how do we do it? Read on.
2.
There are many ways in which the rotor can be dynamically balanced:
- •
  
  Trivial solution: Remove both the masses. Of course, there is nothing left to produce any non-zero $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ . Thus,
  
  $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mbox{/any point}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}.\quad\mbox{ Also, }\sum{% \accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$
  
  .
- •
  
  Adjust geometry: Set $b=0$ . The center-of-mass is still on the axis of rotation.
  
  $\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\sum{\accentset{% \rightharpoonup}{\boldsymbol{F}}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
  
  In addition,
  
  $\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
  
  $\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{/A}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$
- •
  Add two masses: If $b$ is required to be non-zero, we can balance the rotor by adding two masses in any two selected transverse (to the shaft-axis) planes to make $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ . ^†^†margin: Balancing a rotor by adding two masses in two selected transverse (to the shaft axis) planes is quite a general method. Even if the rotor has more than two spinning masses the net angular momentum due to all masses can be reduced to the angular momentum produced by an equivalent system with just two masses such as the one under consideration. For example:
  - –
    
    We can take two equal masses $m$ and $m$ , and place them in the opposite directions of the masses already on the shaft as shown in Fig. 20.128(a). It should be clear that the net angular momentum about any point on the shaft will be zero now.
    
    Filename:sfig4-7-2b
    
    Figure 20.128: Dynamic balancing of the rotor by adding two masses. (a) The two added masses (shown by dotted circles) are the same as the original masses on the shaft. (b) The two added masses are different from the original masses ( $m^{\prime}\neq m$ ).
  - –
    
    We can add two masses $m^{\prime}$ and $m^{\prime}$ different from the masses attached to the shaft and place them a distance $c$ apart on the shaft. Let the length of the connecting rods of the new masses be $h^{\prime}$ . For the net angular momentum to be zero we need
    
    $\displaystyle\overbrace{m^{\prime}\omega^{2}h^{\prime}c}^{\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mbox{\small due to $m^{\prime}$s}}}$ $\displaystyle=$ $\displaystyle\overbrace{m\omega^{2}hb}^{\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\mbox{\small due to $m$s}}}$
    
    $\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad m^{\prime}\,h^{% \prime}\,c$ $\displaystyle=$ $\displaystyle m\,h\,b$
    
    Thus, we have the freedom ^†^†margin: In practice, this freedom is usually restricted by geometric and space constraints. of selecting any combination of $m^{\prime}$ , $h^{\prime}$ and $c$ to give the required product. Here are two examples:
    
    (a)
    
    Let $m^{\prime}=m/2,\;\;c=b/2,\quad$ then $h^{\prime}=\frac{mhb}{m^{\prime}c}=4h.$
    
    (b)
    
    Let $m^{\prime}=m/2,\;\;c=2b,\quad$ then $h^{\prime}=\frac{mhb}{m^{\prime}c}=h.$
    
    Filename:sfig4-7-2c
    
    Figure 20.129: Dynamic balancing of the rotor by adding two masses. (a) $m^{\prime}=m/2$ , $c=b/2$ and $h^{\prime}=4h$ . (b) $m^{\prime}=m/2$ , $c=2b$ and $h^{\prime}=h$ .

Problems for 20.5 Dynamic balance

20.5.1 Is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=(2\hat{\boldsymbol{\imath}}% +5\hat{\boldsymbol{k}})\,\mathrm{rad}/\mathrm{s}$ an eigenvector of $[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}8&0&0\\ 0&16&-4\\ 0&-4&8\end{array}\right]\,\mathrm{kg}\,\mathrm{m}^{2}$ ?

20.5.2 Find the eigenvalues and eigenvectors of $[\boldsymbol{I}^{\mathrm{cm}}]=\left[\begin{array}[]{rrr}5&0&0\\ 0&5&-3\\ 0&-3&10\end{array}\right]\,\mathrm{kg}\,\mathrm{m}^{2}$ .

20.5.3 Three equal length ( $1\,\mathrm{m}$ ) massless bars are welded to the rigid shaft AB which spins at constant rate ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \imath}}$ . Three equal masses ( $1\,\mathrm{kg}$ ) are attached to the ends of the bars. At the instant of interest the masses are all in the $x y$ plane as shown. Where should what masses be added to make the system shown dynamically balanced? [The solution is not unique.]

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Figure 20.130

20.5.4 Dynamic balance of a system of particles. Three masses are mounted on a massless shaft AC with the help of massless rigid rods. Mass $m_{1}=1\,\mathrm{kg}$ and mass $m_{2}=1.5\,\mathrm{kg}$ . The shaft rotates with a constant angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=5\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ . At the instant shown, the three masses are in the $x z$ -plane. Assume frictionless bearings and ignore gravity.

(a)

Find the net reaction force (the total of all the forces) on the shaft. Is the system statically balanced?
(b)

Find the rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}$ , using $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}=\sum_{i}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{i/A}\times m_{i}{\accentset{\rightharpoonup% }{\boldsymbol{a}}}_{i}$ .
(c)

Find the angular momentum about point A, ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}$ . Use this intermediate result to find the rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}$ , using $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/A}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{/A}$ and verify the result obtained in (b).
(d)

What is the value of the resultant unbalanced torque on the shaft? Propose a design to balance the system by adding an appropriate mass in an appropriate location.

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Figure 20.131

20.5.5 For what orientation of the rod $\phi$ is the system shown dynamically balanced?

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Figure 20.132

20.5.6 A uniform square plate with $1\,\mathrm{m}$ sides and $1\,\mathrm{kg}$ mass is mounted (as shown) with a shaft that is in the plane of the square but not parallel to the sides. What is the net force and moment required to maintain rotation at a constant rate?

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Figure 20.133

20.5.7 An equilateral uniform triangular plate with sides of length $\ell$ plate spins at constant rate $\omega$ about an axis through its center of mass. The axis is coplanar with the triangle. The angle $\phi$ is arbitrary. When $\phi=0^{\circ}$ , the base of the triangle is parallel to the axis of rotation. Find the net force and moment required to maintain rotation of the plate at a constant rate for any $\phi$ . For what angle $\phi$ is the plate dynamically balanced?

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Figure 20.134

20.5.8 An abstract problem. This problem concerns a system rigidly attached to a massless rod spinning about a fixed axis at a fixed rate. That is, whether or not the system is a rigid body, it is moving as if it were a rigid body. Every point in the system is moving in circles about the same axis at the same number of radians per second. The position of a particle of the system relative to the axis is ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ . The position relative to some point C is ${{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{C}$ . In all cases, the body rotates about the axis shown. The total mass of all the objects is $M_{tot}$ . Point $O$ is at the center of mass of the system and is at a distance $d$ from the rotation axis and is on the $x$ -axis at the moment of interest. Assume there is no gravity. At the point $A$ , the rod is free to slide in the $z$ -direction. But, at point $B$ , the rod is constrained from moving in the $z$ -direction. The FBD of the system is shown in the figure. Referring to the free-body diagram, there are six scalar unknown reactions in this $3$ dimensional problem.

(a)

The general case: Imagine that you know $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=\dot{H}_{0x}{\hat{\boldsymbol{\imath}}}+% \dot{H}_{0y}{\hat{\boldsymbol{\jmath}}}+\dot{H}_{0z}{\hat{\boldsymbol{k}}}$ and also $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=\dot{L}_{x}{\hat{\boldsymbol% {\imath}}}+\dot{L}_{y}{\hat{\boldsymbol{\jmath}}}+\dot{L}_{z}{\hat{\boldsymbol% {k}}}$ . How would you find the six unknown reactions $A_{x}$ , $A_{y}$ , $M_{Az}$ , $B_{x}$ , $B_{y}$ , $B_{z}$ ? [Reduce the problem to one of solving six equations in six unknowns.]
(b)

The General Result: Is $M_{Az}$ necessarily equal to zero? Prove your result or find a counter-example. Answer: $M_{Az}$ is always zero.
(c)

Static Balance:. If $d=0$ is it true that all reaction forces are equal to zero? Prove your result or find a counter example. Answer: No. See problems 20.4 and 20.2 for counter-examples.

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Figure 20.135

20.5.9 A uniform block spins at constant angular velocity. The magnitude of the angular velocity is $\omega$ . The center of mass $G$ is on the axis of rotation. The $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ directions are parallel to the sides with lengths $b$ , $c$ and $d$ , respectively. Answer in terms of some or all of $\omega$ , $m$ , $b$ , $c$ , $d$ , $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ .

(a)

Find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{H}$ .
(b)

Find ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{H}$ .
(c)

What is the sum of all the moments about point G of all the reaction forces at $A$ and $B$ ?
(d)

Are there any values of $b$ , $c$ , and $d$ for which this block is dynamically balanced for rotation about an axis through a pair of corners? Why?

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Figure 20.136

20.5.10 Dynamic Balance. Consider each of the objects shown in turn (so to speak), each spinning about the $z$ axis which is horizontal. In each case the center of mass is half way between the bearings (not shown). Gravity is in the $-\hat{\boldsymbol{\jmath}}$ direction. State whether it is possible to spin the object at constant rate with constant bearing reactions of $mg\hat{\boldsymbol{\jmath}}/2$ and no other applied torques (‘yes’) or might not be (‘no’). Give a brief explanation of your answer (enough to distinguish an informed answer from a guess).

(a)

Axis along centerline of a uniform rectangular plate.
(b)

Axis orthogonal to uniform rectangular plate and through its centroid.
(c)

Axis orthogonal to uniform rectangular plate and not through its centroid.
(d)

Axis orthogonal to uniform but totally irregular shaped plate and through its centroid.
(e)

Axis along the diagonal of a uniform rhombus.
(f)

Axis through the center of a uniform sphere.
(g)

Axis through the diagonal of a uniform cube.
(h)

Axis with two equal masses attached as shown.
(i)

Axis with three equal masses attached as shown, all in a common plane with the axis, with $\phi$ not equal to $30^{\circ}$ .
(j)

An arbitrary rigid body which has center of mass on the axis and two of its moment-of-inertia eigenvectors perpendicular to the axis. (no picture)

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Figure 20.137

20.5.11 Static and Dynamic Balance A series of bodies, each of uniform density and each with total mass $m$ , rotate at a constant angular speed $\omega$ about a fixed horizontal axis. Ignore gravity. For each body state whether the body is (i) statically balanced and whether it is (ii) dynamically balanced. Give clear arguments using words or equations to support your claims. (iii, iv) For each body you must either (a) add one point mass $m$ or (b) add two point masses each of mass $m/2$ (your choice) that maintain static and dynamic balance if they are balanced, or that make the bodies statically and dynamically balanced. Justify your placement with words and/or equations. The masses need not be added to the bodies, but could be attached off the bodies by structures with negligible mass. [Hint: none of the placements are unique. You may draw a side view if that helps clarify your placement.]

(a)

A rectangular plate (height $h$ , length $\ell$ ) mounted with the axle perpendicular to the plate and through its center.
(b)

The same plate as in (a) above but mounted at an angle $\phi\neq\pi/2$ from the shaft.
(c)

The numerals ‘203’ cut out of a plate and connected by massless rods. Each letter has mass $m/3$ and the three center-of-mass points of the individual letters are colinear and equally spaced. The shaft goes through the center of the ‘0’ and is perpendicular to the plane of the letters.
(d)

A sphere with radius $R$ where the shaft passes a distance $d<R$ from the center.

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Figure 20.138

20.5.12 Dynamic tire balancing. Where should you place what weights in order to dynamically balance the tire?

Assume that a car tire can be modeled as a uniform disk of radius one foot and mass $20\,{\mathrm{lbm}}$ . Assume the tire has been mounted crooked by one degree (the plane of the disk makes an $89^{\circ}$ angle with the car axle). This crooked wheel is statically balanced since its center of mass is on the axis of rotation. But it is dynamically unbalanced since it wobbles and will wobble the car. Car shops put little weights on unbalanced wheels to balance them. They figure out where to put the weights by using a little machine that measures the wheel wobble. But you don’t need such a machine for this problem because you have been told where all the mass is located.

There are many approaches to solving this problem and there are also many correct solutions. You should find any correct solution. [one approach: guess a good location to put two masses and then figure out how big they have to be to make the tire dynamically balanced.]

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Figure 20.139

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle=$	$\displaystyle\,\mathrm{m}_{\rm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{v% }}}_{\rm{cm}}=m({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm{cm/O}})$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{kg}\mathord{\cdot}(1.5\,\mathrm{rad}/\mathrm{s}\hat% {\boldsymbol{k}}\times 1.6\,\mathrm{m}\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 1.2\,{\mathrm{kg}}\mathord{\cdot}{\mathrm{m}}/{\mathrm{s}}\hat{% \boldsymbol{\jmath}}$

$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}I_{z^{\prime}z^{\prime}}^{\rm{O}}\omega^{2}=\frac{1}{2% }(17.68\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2})(1.5\,\mathrm{rad}/\mathrm% {s})^{2}$
	$\displaystyle=$	$\displaystyle 19.89\frac{\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}}{\,% \mathrm{s}^{2}}==19.89\,{\rm N}\mathord{\cdot}{\rm m}$
	$\displaystyle=$	$\displaystyle 19.89\;\rm{J}$

$\displaystyle(m+m^{\prime})r_{1}$	$\displaystyle=$	$\displaystyle mr_{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\quad\quad m^{\prime}$	$\displaystyle=$	$\displaystyle\frac{m^{\prime}(r_{2}-r_{1})}{r_{1}}$
	$\displaystyle=$	$\displaystyle\frac{1\,\mathrm{kg}(0.5\,\mathrm{m}-0.3\,\mathrm{m})}{0.3\,% \mathrm{m}}$
	$\displaystyle=$	$\displaystyle 0.67\,\mathrm{kg}.$

$\displaystyle\sum M_{/AB}$	$\displaystyle=$	$\displaystyle\dot{H}_{/AB}$
$\displaystyle\hbox to36.135pt{or\hfil}M_{z}$	$\displaystyle+$	$\displaystyle\sum M_{/AB\mbox{(due to reaction forces)}}=\dot{H}_{/AB}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad M_{z}$	$\displaystyle=$	$\displaystyle 0.$

	$\displaystyle\overbrace{m^{\prime}\omega^{2}h^{\prime}c}^{\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mbox{\small due to $m^{\prime}$s}}}$	$\displaystyle=$	$\displaystyle\overbrace{m\omega^{2}hb}^{\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\mbox{\small due to $m$s}}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad m^{\prime}\,h^{% \prime}\,c$	$\displaystyle=$	$\displaystyle m\,h\,b$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{\jmath}}\times(L\hat{\boldsymbol{\imath}}+% h\hat{\boldsymbol{\jmath}}+r\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle\omega(-L\hat{\boldsymbol{k}}+r\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 10\,\mathrm{rad}/\mathrm{s}\cdot(-2\,\mathrm{m}\hat{\boldsymbol{% k}}+1\,\mathrm{m}\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle(20\hat{\boldsymbol{\imath}}-10\hat{\boldsymbol{k}})\,\mathrm{m}/% \mathrm{s}.$

Chapter 20 Fixed-axis rotation in 3D

Download this chapter by section

20.1 3-D description of circular motion

Triple cross product

Angular velocity of a rigid body in 3D

Relative motion of points on a rigid body

Relative velocity and acceleration using rotating frames

Mechanics

Problems for 20.1 3-D description of circular motion

20.2 Dynamics of fixed-axis rotation

A stick sweeps out a cone

Evaluation of ∑𝑴⇀O

Evaluation of 𝑯⇀˙/O

The results for the conically swinging stick

The scalar equations governing rotation about an axis

Alternatively,

Problems for 20.2 Dynamics of fixed axis rotation

20.3 Moment of inertia matrices [𝑰cm] and [𝑰O]

The moment of inertias in 3-D: [𝑰cm] and [𝑰O]

The parallel axis theorem for rigid bodies in three dimensions

Matrices and tensors

Eigenvectors and Eigenvalues

Eigenvectors of [𝑰]

Properties of [𝑰]

Derivation 1

Center-of-mass inertia matrix

Comment (aside)

Derivation 2

Relation between 2-D and 3-D parallel axis theorems

Problems for 20.3 Moment of inertia matrices: [𝑰cm] and [𝑰O]

20.4 Mechanics using the moment of inertia matrix

Method 2 of evaluating 𝑯⇀˙/O: using the x⁢y⁢z moment of inertia matrix about point O

Method 3 of calculating 𝑯⇀˙/O: using [𝑰O] and a coordinate system lined up with the rod

The multitude of ways to calculate 𝑯⇀˙/O

Energy of things going in circles at variable rate

The safest bet

The geometry of 𝑯⇀˙/O

Rotation of a rigid body about a fixed axis: the general case

Linear momentum balance

Angular momentum balance

Problems for 20.4 Mechanics using [𝑰cm] and [𝑰O]

20.5 Dynamic balance

Static balance

Dynamic balance

Problems for 20.5 Dynamic balance

Evaluation of $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$

Evaluation of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$

20.3 Moment of inertia matrices $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$

The moment of inertias in 3-D: $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$

Eigenvectors of $[\boldsymbol{I}]$

Properties of $[\boldsymbol{I}]$

Problems for 20.3 Moment of inertia matrices: $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$

The geometry of $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$

Problems for 20.4 Mechanics using $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$