Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap15 Circular motion of a particle 17 Planar motion of an object

Chapter 16 Circular motion of a rigid object

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16.0Introductionstudy PDF
16.12D rigid-object rotationstudy PDF + solutions
16.22D rigid-object angular velocitystudy PDF + solutions
16.3Polar moment of inertiastudy PDF + solutions
16.4Dynamics of rigid-object planar circular motionstudy PDF + solutions

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Here we extend the idea of circular motion from particles to objects. A particle can go in circles. On a rigid object in 2D, all pairs of particles go in circles around each other. The key theoretical idea is of rigid-object rotation. The primary applications are pendulums, gear trains, and rotationally accelerating motors or brakes.

Filename:tfigure8-rel-ang-vel

Figure 16.1: All the points on a gear move in circles, assuming the axle is not translating.

Refer to caption — Figure 16.2: All points on a flywheel move in circles. The points further from the center move faster, so they move further while the camera shutter is open, so they show a bigger motion blur (the constant-thickness spokes look fatter near the rim).

We covered the special case of circular motion of a particle in the previous chapter. Now we consider the kinematics and mechanics of rotating rigid objects in 2D. The key new conceptual idea is that of rotation of a rigid object.

Mechanics of circular motion

For the systems in this chapter, for every system we show in a free-body diagram we have, as always,

\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}}.

Because you already know how to work with forces and moments (the left sides of the top two equations), the primary new skill in this chapter is the evaluation of $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}},\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{/C}$}}}$ , and ${\dot{E}_{\mathrm{K}}}$ for a rotating rigid object. That is, you need to understand the position, velocity and acceleration of points moving in circles. The rest of the skills used are universal, for example solving algebraic or differential equations, plotting, etc.

16.1 Rotation of a rigid object in planar circular motion

Filename:tfigure8-ang-vel

Figure 16.3: a) An object, b) rotated counterclockwise an angle

\theta

about 0.

Filename:tfigure8-ang-accel

Figure 16.4: Rotation of lines on a rotating rigid object. Some real or imagined lines marked on the rigid object are shown. They make the angles

\theta_{1}

\theta_{2}

\theta_{3}

\dots

with respect to various fixed lines which do not rotate. As the object rotates, each of these angles increases by the same amount.

When two parts are glued together or attached by welding, gluing, several tight screws, bolts, rivets or the like we call the connection a ‘rigid attachment’. And, for the purposes of mechanics analysis, the two connected parts make up one bigger object. But most machines have various parts that are connected to each other, but not welded to each other. The most common such non-rigid attachment in engineering is a hinge. In 2D,

a hinge attachment between two objects keeps two points, one from each object, on top of each other while freely allowing relative rotation of the two objects about the hinge point.

In 3D a hinge keeps two lines, one on each body, coincident and allows relative rotation about that line. The common line, or in 2D the line orthogonal to the plane through the points, is called the hinge, the hinge axis, or the axis of relative rotation.

One example of a hinge is a car axle which allows rotation of a wheel relative to the car suspension. The hinge axis is the wheel axle

^†^†margin: The word axis is obviously related to the word axle. More generally the word axis means ’line’. For example the

x

and

y

axes are generally not axles about which anything rotates, they are just lines.

Physically, hinges are made various ways, sometimes by poking a cylindrical pin through the two objects and sometimes with ball bearings (see box 5.3 on page 5.3). So hinges are also called pin connections or bearings (fig. 16.3). In this chapter we limit our attention to a simple use of a hinge: one rigid part is hinged to a second fixed part that doesn’t move at all. Such a non-moving part can be thought of as connected to, or an extension of, the ground or ‘fixed frame’. In the simple case of interest in this chapter, one point on the moving part does not move and the rest of the part rotates about that point.

For definiteness and simplicity let’s call the hinge location 0 and the hinge axis through 0 the $z$ axis.

One function of the hinge is to make the part’s only possible motion to be rotation about O. Thus to understand the dynamics of a hinged part we need to understand the position, velocity and acceleration of points on a rigid object which rotates. This whole section is about the kinematics (the geometry of motion) for this rotation. We will measure the amount of rotation by the angle $\theta$ , and the rate of change of $\theta$ by the angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ (‘omega’), and the rate of change of this angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ by the angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ (‘alpha’).

As simple as this topic seems at first glance, you should pay close attention to the meanings and uses of these quantities. The rest of the book completely depends on the material in this section.

Box 16.1 Rotation is uniquely defined for a rigid object (2D)

Most people will find it self-evident that, starting with a rigid object at a reference orientation, all lines marked on the object rotate by the same angle $\theta$ . Here, for the doubting, we demonstrate this fact.

A rigid object is defined this way:

For every pair of material points A and B on a rigid object the distance $|$ AB $|$ between them does not change as the object moves.

In particular, when a rigid object rotates all distances between all pairs of points are preserved. Thus, by the “side-side-side” similar triangle theorem of elementary geometry, all relative angles between marked line segments are preserved by the rotation. For example, for a triangle ABC the angle at B is constant as the object rotates. Now consider any pair of line segments on the object.

[By ‘on’ the object we don’t mean a projected image drawn with a light pen that can move around on the object relative to the atoms. Rather, by ‘on the object’ we mean something defined by a particular set of atoms that make up the object.]

If the segments do not cross we can extend them to a point of intersection B. Such a pair of intersecting lines is shown here before and after rotation.Initially BA makes an angle $\theta_{0}$ with a horizontal reference line. BC then makes an angle of $\theta_{\rm ABC}+\theta_{0}$ . After rotation we measure the angle to the line BD. BA now makes an angle of $\theta_{0}+\theta$ . By the addition of angles in the rotated configuration line BC now makes an angle of $\theta_{\rm ABC}+\theta_{0}+\theta$ which makes an increase by $\theta$ of the angle made by BC with the horizontal reference line. So both BA and BC rotate by the same angle $\theta$ .

We could use one of these two lines and compare it with an arbitrary third line through B and show that the third line also has equal rotation, and then a fourth, and so on. So all lines on the object through a point B rotate by the same angle $\theta$ . The demonstration for a pair of parallel lines, one of them through B, is easy, they stay parallel so always make a common angle with any reference line.

Because any line on the object either goes through B or is parallel to a line through B, all lines marked on a rigid object rotate by the same angle $\theta$ .

The rotation of a rigid object in 2D is thus unambiguously defined as the angle through which all lines on the object rotate.

Filename:tfigure-rotationisunique

Rotation of an object counterclockwise by $\theta$

We start by imagining the object in a distinguished configuration which we call the reference configuration, reference state or reference position. For example we could take the left figure in fig. 16.3 as the reference configuration. If possible it’s usually best to pick the reference state to be one in which a prominent feature of the object is aligned with the $x$ or $y$ axes. The reference state may or may not be the start of the motion of interest. Even if not, we measure an object’s rotation by the change, relative to the reference state, in the counterclockwise angle $\theta$ of a reference line marked in the object relative to a fixed line outside. Which reference line? Fortunately,

All real or imagined lines marked on a rotating rigid object rotate by the same angle, the rotation angle, $\theta$ . (See box 16.1).

In three dimensions things are more complicated. General rotation of a rigid object is then represented not with a single angle $\theta$ , but rather with 3 angles, or with a unit vector and an angle, or a 3 $\times$ 3 matrix. So we wait to discuss which of the 2D ideas here generalize to 3D and which do not.

Rotated coordinates and base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$

Filename:tfigure8-ang-accel-ex

Figure 16.5: a) A house is drawn by connecting lines between 6 points, b) the house and coordinate system are rotated, thus its coordinates in the rotating system do not change c) But the coordinates in the original system do change

Filename:tfigure8-ang-accel1

Figure 16.6: A rotating rigid object

\cal C

with rotating coordinates

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

rigidly attached.

We pick two orthogonal lines on the rotating object and give them distinguished status as object-fixed (or body-fixed) rotating coordinate axes $x^{\prime}$ and $y^{\prime}$ . Think of these axes as $x^{\prime}y^{\prime}$ coordinate axes on a piece of graph paper that is glued to the object (see fig. 16.5a&b). It’s easiest if we start by assuming that the $x^{\prime}y^{\prime}$ axes have the same origin 0 as the $x y$ axes and are parallel with the fixed $x y$ axes when the object is in the reference configuration (when $\theta=0$ ).

These rotating coordinate axes, $x^{\prime}$ and $y^{\prime}$ , have associated rotating base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ (fig. 16.6 and 16.7). So $\hat{\boldsymbol{\imath}}^{\prime}$ is always in the $x^{\prime}$ direction and $\hat{\boldsymbol{\jmath}}^{\prime}$ always in the $y^{\prime}$ direction. We will use these rotating coordinates and base vectors to keep track of some particle of interest P that is ‘glued’ to the object. To start, note that particle P which is glued to the object has $x^{\prime}$ and $y^{\prime}$ coordinates that don’t change as the rotation progresses.

Example: A particle on the $x^{\prime}$ axis

If a particle P is fixed on the $x^{\prime}$ -axis at position $x^{\prime}=3\,\mathrm{cm}$ , then we have.

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}

\displaystyle=

\displaystyle 3\,\mathrm{cm}\hat{\boldsymbol{\imath}}^{\prime}

for all time, even as the object rotates.

The position vector of a point P fixed to a rigid object hinged at O remains, as the rotation progresses,

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}

\displaystyle=

\displaystyle x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime},\qquad

(16.1)

with $x^{\prime}$ and $y^{\prime}$ both constant. These rotating coordinate system components, $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}]_{x^{\prime}y^{\prime}}=[x^{% \prime},y^{\prime}]$ , are sometimes written as $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}]_{x^{\prime}y^{\prime}}=\left[% \begin{array}[]{c}x^{\prime}\\ y^{\prime}\end{array}\right]$ .

You will see that much of the math for rotating $x^{\prime}y^{\prime}$ coordinates is reminiscent of that for polar coordinates. However, the spirit is a bit different. In polar coordinates the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ axis was picked to track a particular particle of interest. Here we pick axes that rotate with an extended object and use that one set of axes to track any and all particles of interest.

Filename:sfig8-2-3

Figure 16.7: Fixed coordinate axes and rotating coordinate axes.

Note, even though neither $x^{\prime}$ nor $y^{\prime}$ change as $\theta$ changes, the point P they describe moves, in circles actually. How can the particle’s position change if its coordinates don’t change? Well, in eqn. (16.3) the change in position is represented by the base vectors changing as the object rotates. Thus we could write more explicitly that

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}

\displaystyle=

\displaystyle x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}(\theta)+y^{\prime}% \hat{\boldsymbol{\jmath}}^{\prime}(\theta).

(16.3)

Here we show more explicitly that the base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ depend on $\theta$ .

Filename:sfig8-2-3a

Figure 16.8: The

x^{\prime}

and

y^{\prime}

coordinates of a point fixed on a rotating object stay constant while the base vectors

\hat{\boldsymbol{\imath}}^{\prime}

and

\hat{\boldsymbol{\jmath}}^{\prime}

rotate with the object.

Just like for polar base vectors (see eqn. (15.1) on page 15.1) we can express the rotating base vectors in terms of the fixed base vectors and $\theta$ .

	$\displaystyle\hat{\boldsymbol{\imath}}^{\prime}$	$\displaystyle=$	$\displaystyle\phantom{-}\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}},$		(16.4)
	$\displaystyle\hat{\boldsymbol{\jmath}}^{\prime}$	$\displaystyle=$	$\displaystyle-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{% \jmath}}.$

Also we can express the fixed basis vectors in terms of the rotating vectors like this:

	$\displaystyle\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle\cos\theta\hat{\boldsymbol{\imath}}^{\prime}-\sin\theta\hat{% \boldsymbol{\jmath}}^{\prime}$		(16.5)
	$\displaystyle\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle\sin\theta\hat{\boldsymbol{\imath}}^{\prime}+\cos\theta\hat{% \boldsymbol{\jmath}}^{\prime}.$

Please review the section on dot products, 1.2, to see one derivation of these formulae.

We will use the phrase reference frame or just frame to mean “a coordinate system attached to a rigid object”. One can think of the coordinate grid as like an invisible metal framework (hence the word ‘frame’) that rotates with the object. We refer to a calculation based on the rotating coordinates in fig. 16.6 variously as “in the frame $\mathcal{C}$ ” or “using the $x^{\prime}y^{\prime}$ frame” or “in the $\hat{\boldsymbol{\imath}}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}$ frame

^†^†margin: Advanced aside. Sometimes a reference frame is defined as the set of all coordinate systems that could be attached to a rigid object. Two coordinate systems, even if rotated with respect to each other, then represent the same frame so long as they rotate together. Some of the results we will develop only depend on this more slack definition of frame, that the coordinates are glued to the object with no mind of their orientation in the reference configuration.

In computer calculations we usually manipulate lists and arrays of numbers and not geometric vectors. So on a computer we keep track of vectors by keeping track of their lists of components. Let’s look at a point fixed to the object and whose coordinates we know in the reference configuration:

\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}}^{\rm ref}% \right]_{xy}=\left[\begin{array}[]{c}x^{\rm ref}\\ y^{\rm ref}\end{array}\right].

Assuming the object axes and fixed axes coincide in the reference configuration, the object coordinates of a point $\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}\right]_{x^{% \prime}y^{\prime}}$ are equal to the space fixed coordinates of the point in the reference configuration $\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}}^{\rm ref}% \right]_{xy}$ . We can think of the point as defined either way, so

\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}\right]_{x^{% \prime}y^{\prime}}=\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P}}}^{\rm ref}\right]_{xy}.

The rotation matrix $[R]$

Here is a question we often need to answer: What are the fixed basis coordinates of a point that has the rotating-frame coordinates $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}]_{x^{\prime}y^{\prime}}=\left[% \begin{array}[]{c}x^{\prime}\\ y^{\prime}\end{array}\right]$ ?

Here is one way to find the answer:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$	$\displaystyle=$	$\displaystyle x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime}$	(16.6)
	$\displaystyle=$	$\displaystyle x^{\prime}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}})+y^{\prime}(-\sin\theta\hat{\boldsymbol{\imath}}+\cos% \theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\underbrace{\left((\cos\theta)x^{\prime}-(\sin\theta)y^{\prime}% \right)}_{x}\hat{\boldsymbol{\imath}}+\underbrace{\left((\sin\theta)x^{\prime}% +(\cos\theta)y^{\prime}\right)}_{y}\hat{\boldsymbol{\jmath}}$

so we can pull out the $x$ and $y$ coordinates compactly as,

[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}]_{xy}=\left[% \begin{array}[]{c}x\\ y\end{array}\right]=\left[\begin{array}[]{c}\phantom{-}\cos\theta\ x^{\prime}+% \sin\theta(-y^{\prime})\\ \sin\theta\ x^{\prime}+\cos\theta(y^{\prime})\end{array}\right].

(16.7)

But this can, in turn be written in matrix notation as

$\displaystyle\left[\begin{array}[]{c}x\\ y\end{array}\right]$	$\displaystyle=$	$\displaystyle\underbrace{\left[\begin{array}[]{cc}\cos\theta&-\sin\theta\\ \sin\theta&\phantom{-}\cos\theta\end{array}\right]}_{\left[R\right]}\par\left[% \begin{array}[]{c}x^{\prime}\\ y^{\prime}\end{array}\right],\quad\mbox{or}$	(16.14)
$\displaystyle\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}% }\right]_{xy}$	$\displaystyle=$	$\displaystyle\left[R\right]\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\!\mathrm{P}}\right]_{x^{\prime}y^{\prime}},\quad\mbox{or}$	(16.15)
$\displaystyle\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}% }\right]_{xy}$	$\displaystyle=$	$\displaystyle\left[R\right]\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{P}}}^{\rm ref}\right]_{xy}.\par$

The matrix $[R]$ or $[R(\theta)]$ is the rotation matrix for counterclockwise rotations by $\theta$ . As shown above, if you know the coordinates of a point fixed on an object before rotation, you can find its coordinates after rotation by multiplying the coordinate column vector by the matrix $[R]$ . You can remember what $[R]$ is by remembering its components or by remembering that

the first and second column of $[R]$ are the components of $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ , respectively, in the fixed coordinate system.

For example, the first column of $[R]$ consists of the $x$ and $y$ components of $\hat{\boldsymbol{\imath}}^{\prime}$ .

A feature of eqn. (16.14) is that the same matrix $[R]$ prescribes the coordinate change for every different point on the object. Thus for points called 1, 2 and 3 we have

	$\displaystyle\left[\begin{array}[]{c}x_{1}\\ y_{1}\end{array}\right]$	$\displaystyle=\left[R\right]\left[\begin{array}[]{c}x^{\prime}_{1}\\ y^{\prime}_{1}\end{array}\right]\ ,\ \left[\begin{array}[]{c}x_{2}\\ y_{2}\end{array}\right]=\left[R\right]\left[\begin{array}[]{c}x^{\prime}_{2}\\ y^{\prime}_{2}\end{array}\right]\par$	$\displaystyle\&$
	$\displaystyle\left[\begin{array}[]{c}x_{3}\\ y_{3}\end{array}\right]$	$\displaystyle=\left[R\right]\left[\begin{array}[]{c}x^{\prime}_{3}\\ y^{\prime}_{3}\end{array}\right].$

A more compact way to write a matrix times a list of column vectors is to arrange the column vectors one next to the other in a matrix. By multiplying this matrix by $[R]$ we get a new matrix whose columns are the new coordinates of various points. For example,

\left[\begin{array}[]{ccc}x_{1}&x_{2}&x_{3}\\ y_{1}&y_{2}&y_{3}\end{array}\right]=\left[R\right]\left[\begin{array}[]{ccc}x^% {\prime}_{1}&x^{\prime}_{2}&x^{\prime}_{3}\\ y^{\prime}_{1}&y^{\prime}_{2}&y^{\prime}_{3}\end{array}\right].

(16.16)

Eqn. 16.16 is useful for computer animation of rotating things in video games (and in dynamics simulations too) where points 1,2, and 3 are points on an object.

Example: Rotate a picture

If a simple picture of a house is drawn by connecting the six points (fig. 16.5a) with the first point at $(x,y)=(1,2)$ , the second at $(x,y)=(3,2)$ , etc., and the sixth point on top of the first, we have,

\left[\mbox{$xy$ points BEFORE}\right]\equiv\left[\begin{array}[]{cccccc}1&3&3% &2&1&1\\ 2&2&4&5&4&2\end{array}\right].

After a $30^{{}^{\circ}}$ counter-clockwise rotation about O, the coordinates of the house, in a coordinate system that rotates with the house, are unchanged (fig. 16.5b). But in the fixed (non-rotating, Newtonian) coordinate system the new coordinates of the rotated house points are,

$\displaystyle\left[\mbox{$xy$ points AFTER}\right]$	$\displaystyle=$	$\displaystyle\left[R\right]\left[\mbox{$xy$ points BEFORE}\right]=\left[R% \right]\left[\mbox{$x^{\prime}y^{\prime}$ points}\right]$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{cc}\sqrt{3}/2&-.5\\ .5&\sqrt{3}/2\end{array}\right]\left[\begin{array}[]{cccccc}1&3&3&2&1&1\\ 2&2&4&5&4&2\end{array}\right]$
	$\displaystyle\approx$	$\displaystyle\left[\begin{array}[]{cccccc}-0.1&1.6&0.6&-0.8&-1.1&-0.1\\ \phantom{-}2.2&3.2&5.0&\phantom{-}5.3&\phantom{-}4.0&\phantom{-}2.2\end{array}\right]$

as shown in fig. 16.5c.

SAMPLE 16.1

Filename:sfig8-2-3b

Figure 16.9:

Computing rotated position of an object: A rigid object AOB is pinned at point O and is free to rotate about this point. Using the rotation matrix, find the coordinates of points A and B when $\theta=30^{\circ}$ and $110^{\circ}$ respectively.

^†^†margin:

Solution Let $x^{\prime}y^{\prime}$ be a set of axes glued to the object with origin at O which is also the origin of the space-fixed coordinate axes $x y$ . We first write the coordinates of points A and B in the body-fixed axes.

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}|_{x^{\prime}y^{% \prime}}=\left[\begin{array}[]{c}1\,\mathrm{m}\\ 0\end{array}\right],\qquad{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{B}}|_{x^{\prime}y^{\prime}}=\left[\begin{array}[]{c}1\,\mathrm{m}\\ 1\,\mathrm{m}\end{array}\right].

The rotation matrix for the $x^{\prime}y^{\prime}$ axes rotating counterclockwise along with the object is

R=\left[\begin{array}[]{rr}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right].

We can find the coordinates of A and B in the rotated position by multiplying their $x^{\prime}y^{\prime}$ coordinates with the rotation matrix. We first write the coordinates of both points A and B in a single matrix, one column for each, and multiply with $R$ to get the new coordinates:

[\begin{array}[]{cc}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A% }}&{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}\end{array}]_{% xy}=\left[\begin{array}[]{rr}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right]\left[\begin{array}[]{cc}x^{\prime}_{A}% &x^{\prime}_{B}\\ y^{\prime}_{A}&y^{\prime}_{B}\end{array}\right].

Now, we calculate the new coordinates for the given values of $\theta$ .

•

For $\theta=30^{\circ}$ , we have,

$\displaystyle[\begin{array}[]{cc}{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\!\mathrm{A}}&{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}% \end{array}]_{xy}$ $\displaystyle=$ $\displaystyle\left[\begin{array}[]{rr}0.866&-0.5\\ 0.5&0.866\end{array}\right]\left[\begin{array}[]{rr}1\,\mathrm{m}&1\,\mathrm{m% }\\ 0&1\,\mathrm{m}\end{array}\right]$

$\displaystyle=$ $\displaystyle\left[\begin{array}[]{rr}0.866\,\mathrm{m}&0.366\,\mathrm{m}\\ 0.5\,\mathrm{m}&1.366\,\mathrm{m}\end{array}\right].$

Filename:sfig8-3-1

Figure 16.10: Coordinates of points A and B after a rotation of $30^{\circ}$ .

Thus,

Answer: $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}]_{xy}=\left[% \begin{array}[]{r}0.866\,\mathrm{m}\\ 0.5\,\mathrm{m}\end{array}\right],\qquad[{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B}}]_{xy}=\left[\begin{array}[]{r}0.366\,\mathrm{m% }\\ 1.366\,\mathrm{m}\end{array}\right].$
•

Similarly, for $\theta=110^{\circ}$ , we get,

$\displaystyle[\begin{array}[]{cc}{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\!\mathrm{A}}&{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}% \end{array}]_{xy}$ $\displaystyle=$ $\displaystyle\left[\begin{array}[]{rr}-0.342&-0.94\\ 0.94&-0.342\end{array}\right]\left[\begin{array}[]{rr}1\,\mathrm{m}&1\,\mathrm% {m}\\ 0&1\,\mathrm{m}\end{array}\right]$

$\displaystyle=$ $\displaystyle\left[\begin{array}[]{rr}-0.342\,\mathrm{m}&-1.282\,\mathrm{m}\\ 0.94\,\mathrm{m}&0.598\,\mathrm{m}\end{array}\right].$

Filename:sfig8-3-1a

Figure 16.11: Coordinates of points A and B after a rotation of $110^{\circ}$ .

Answer: $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}]_{xy}=\left[% \begin{array}[]{r}-0.342\,\mathrm{m}\\ 0.94\,\mathrm{m}\end{array}\right],\qquad[{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B}}]_{xy}=\left[\begin{array}[]{r}-1.282\,\mathrm{% m}\\ 0.598\,\mathrm{m}\end{array}\right].$

SAMPLE 16.2

Filename:sfig8-5-wiper

Figure 16.12:

Computer program for object rotation: The object shown in the figure rotates counterclockwise with its angular position $\theta$ given by $\theta(t)=\dot{\theta}_{0}t$ where $\dot{\theta}_{0}=10^{\circ}/s$ . Write a computer program to animate the motion of the object by plotting its position at specified time instants. Use your program to plot the position of the object every two seconds for a total of 18 seconds.

Solution

In order to draw the object on the computer screen, we need to define it by taking a sufficient number of points on its boundary so that plotting all those points with connected line segments represents the object as closely as possible. In this case, we can select all the corner points on its boundary and define the object with the coordinates of these points. With the given geometry, it is fairly easy to find these coordinates. Denoting the coordinates of the $k$ th point by $(x_{k},y_{k})$ , we can define this object by the following set of coordinates:

\mbox{object}=\left[\begin{array}[]{ccccc}x_{1}&x_{2}&\ldots&x_{6}&x_{1}\\ y_{1}&y_{2}&\ldots&y_{6}&y_{1}\end{array}\right]=\left[\begin{array}[]{rrrrrrr% }0&2&4&4&2&0&0\\ -1&-1&-2&2&1&1&-1\end{array}\right].

Filename:sfig8-5-wiper-a

Figure 16.13: Coordinates of the six corner points are sufficient to define this object.

Note that the last column is a repeat of the first column. This is essential so that when we plot the object using the $x$ and $y$ coordinates listed here, we get a closed boundary.

Animation of the angular motion of this object basically requires determining the rotated position of the object and plotting it at sufficiently small time intervals. To do this, we need to define the initial orientation (position), define the time increment, find the angle of rotation $\theta$ corresponding to the new time, compute the corresponding rotation matrix, multiply the object coordinates with the rotation matrix, separate out the $x$ and $y$ coordinates, and plot $x$ vs $y$ . We then repeat the whole sequence until we reach the final time.

The following pseudocode can be easily adapted to a computer program to do the required animation.

object = [ 0  2  4  4  2  0  0
          -1 -1 -2  2  1  1 -1]     % coordinates of points
t = 0Ψ                              % specify initial time
t_final = 36                        % specify final time
delta_t = 2                         % specify time increment
thetadot0 = 10*pi/180               % specify thetadot0 in rad/s

while t < t_final
      t = t + delta_t               % increment the time
      theta = thetadot0*t           % compute current theta
      R = [ cos(theta) -sin(theta)  % find the rotation matrix
            sin(theta)  cos(theta)]
      new_position = R*object       % find new coordinates
      x = new_position(1st row)     % extract all x-coordinates
      y = new_position(2nd row)     % extract all y-coordinates
      plot y vs x                   % plot the new position
end

The plot obtained by implementing this code is shown in fig. 16.14.

Filename:sfig8-5-2disks

Figure 16.14: Rotated position of the object plotted at 2 second intervals.

Problems for 16.1 Rotation of a rigid body

Preparatory Problems

16.1.1 Give the definition of each term below in words and, if possible, with an equation.

(a)

rotation angle
(b)

$(x^{\prime},y^{\prime})$ Answer: Rotating frame coordinates of a point.
(c)

$(x,y)$ Answer: Fixed basis coordinates of a point.
(d)

rotation matrix Answer: A matrix that when operated on a column vector of fixed basis coordinates gives the coordinates in a rotated frame.
$\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{bmatrix}$

16.1.2 Assume that in the reference configuration, when $\theta=0$ , that the $x^{\prime}y^{\prime}$ axes attached to an object are aligned with the $x y$ axes. Consider a point P attached to the moving frame (object) that has coordinates $(x^{\prime},y^{\prime})$ . Find each of the quantities below in terms of some or all of $x^{\prime},y^{\prime},\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},% \theta,\dot{\theta}$ and $\ddot{\theta}$ .

(a)

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$ Answer: $\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{bmatrix}\begin{bmatrix}x^{\prime}\\ y^{\prime}\\ \end{bmatrix}$
(b)

The rotation matrix $[R]$ . Answer: $[R]=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{bmatrix}$

16.1.3 Assume that in the reference configuration, when $\theta=0$ , that the $x^{\prime}y^{\prime}$ axes are aligned with the $x y$ axes. Consider two points ${\rm P}_{1}$ and ${\rm P}_{2}$ attached to the moving frame. ${\rm P}_{1}$ and ${\rm P}_{2}$ have coordinates $(x^{\prime}_{1},y^{\prime}_{1})$ $(x^{\prime}_{2},y^{\prime}_{2})$ . Find each of the quantities below in terms of some or all of $x^{\prime}_{1},y^{\prime}_{1},x^{\prime}_{2},y^{\prime}_{2},\hat{\boldsymbol{% \imath}},\hat{\boldsymbol{\jmath}},\theta,\dot{\theta}$ and $\ddot{\theta}$ .

(a)

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P_{1}/P_{2}}}$ Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P_{1}/P_{2}}}=\begin{% bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{bmatrix}\begin{bmatrix}(x^{\prime}_{1}-x^{\prime}_{2})\\ (y^{\prime}_{1}-y^{\prime}_{2})\\ \end{bmatrix}$
(b)

The rotation matrix $[R]$ . Answer: $[R]=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{bmatrix}$

16.1.4 The square plate OABC shown in the figure measures $\ell=1\,\mathrm{m}$ on a side. The coordinate axes $x^{\prime}y^{\prime}$ are fixed to the plate and the axes $x y$ are fixed in space. The plate is rotated by an angle $\theta=60^{\circ}$ as shown.

(a)

Find the coordinates $x^{\prime}$ and $y^{\prime}$ of the corner point B in the rotated frame.
(b)

Find the coordinates $x$ and $y$ of the same point in the fixed frame without using the rotation matrix.
(c)

Write the rotation matrix $R$ for the given rotation.
(d)

Find the coordinates of point B in the fixed frame using its $(x^{\prime},y^{\prime})$ coordinates and the rotation matrix $R$ .
(e)

Find the coordinates of point B in the rotated frame using its $(x,y)$ coordinates and the rotation matrix $R$ in its appropriate form.

Answer: $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}]_{x^{\prime}y^{% \prime}}=\left[\begin{array}[]{r}1\,\mathrm{m}\\ 1\,\mathrm{m}\end{array}\right],\quad[{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{\!\mathrm{B}}]_{xy}=\left[\begin{array}[]{r}-0.37\,\mathrm{m}\\ 1.37\,\mathrm{m}\end{array}\right]$ , $\quad R=\left[\begin{array}[]{rr}0.5&-0.87\\ 0.87&0.5\end{array}\right]$

Filename:sfig8-4-1

Figure 16.15:

16.1.5 The rod shown in the figure rotates counterclockwise, starting from $\theta=0$ . Find the following quantities.

(a)

Write the position vector of points A and B as a list of numbers in an array, e.g., $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}]_{xy}=\left[% \begin{array}[]{c}x_{A}\\ y_{A}\end{array}\right]$ , at $\theta=0$ . Answer: $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}]_{xy}=\left[% \begin{array}[]{c}-1\,\mathrm{m}\\ 0\,\mathrm{m}\end{array}\right]$ , $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}]_{xy}=\left[% \begin{array}[]{c}2\,\mathrm{m}\\ 0\,\mathrm{m}\end{array}\right]$
(b)

Write the rotation matrix $R$ for $\theta=45^{\circ}$ . Answer: $R=\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}$
(c)

Find the coordinates of points A and B using the rotation matrix R for $\theta=45^{\circ}$ . Answer: $\begin{bmatrix}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}&{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}\end{bmatrix}_{xy}=% \begin{bmatrix}\frac{-1}{\sqrt{2}}&\sqrt{2}\\ \frac{-1}{\sqrt{2}}&\sqrt{2}\end{bmatrix}\,\mathrm{m}$
(d)

If the rod rotates at a constant angular speed $\omega=\pi/3\,\mathrm{rad}/\mathrm{s}$ , find the coordinates of points A and B after 2 seconds, assuming the rod is at $\theta=0$ when $t=0$ .

Filename:sfig8-4-1a

Figure 16.16:

16.1.6 Find the angle of rotation corresponding to the following rotation matrices:

(a)

$\left[\begin{array}[]{rr}0.7071&-0.7071\\ 0.7071&0.7071\end{array}\right]$ . Answer: $\theta=45^{\circ}$
(b)

$\left[\begin{array}[]{rr}-0.7071&-0.7071\\ 0.7071&-0.7071\end{array}\right]$ . Answer: $\theta=135^{\circ}$
(c)

$\left[\begin{array}[]{rr}0.7071&0.7071\\ -0.7071&0.7071\end{array}\right]$ . Answer: $\theta=315^{\circ}$

More-Involved Problems

16.1.7 The rod shown in the figure rotates with constant angular speed $\omega=10\,\mathrm{rad}/\mathrm{s}$ . $\theta=0$ at $t=0\,\mathrm{s}$ .

(a)

Find the position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}|_{t_{1}}$ and the velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}|_{t_{1}}$ of point B at $t_{1}=1\,\mathrm{s}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}|_{t_{1}}=-1.68% \hat{\boldsymbol{\imath}}-1.09\hat{\boldsymbol{\jmath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}|_{t_{1}% }=20\,\mathrm{m}/\mathrm{s}\begin{bmatrix}-0.84&0.54\end{bmatrix}$
(b)

Find the position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}|_{t_{2}}$ of point B at $t_{2}=1.2\,\mathrm{s}$ . What is the net displacement of point B during the time interval $\Delta t=t_{2}-t1$ ? Is this net displacement equal to ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}|_{t_{1}% }\Delta t$ ? Why not? Answer: Net displacement $=(3.366\hat{\boldsymbol{\imath}}+0.016\hat{\boldsymbol{\jmath}})\,\mathrm{m}$ . The net displacement is not equal to ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}|_{t_{1}% }\Delta t$ because the motion is not along a straight line.

Filename:sfig8-6-2

Figure 16.17:

16.1.8 Write a computer program to animate the rotation of an object. Your input should be a set of $x$ and $y$ coordinates defining the object (such that plot y vs x draws the object on the screen) and the rotation angle $\theta$ . The output should be the rotated coordinates of the object.

(a)

From the geometric information given in the figure, generate coordinates of enough points to define the given object. Answer: (0, 0); ( $15\sqrt{3}$ , 15); ( $30\sqrt{3}$ , 0); (0, $30\sqrt{3}$ ); (15, $15\sqrt{3}$ ).
(b)

Using your program, plot the object at $\theta=20^{\circ},$ $60^{\circ},$ $100^{\circ},$ $160^{\circ},$ and $270^{\circ}$ .
(c)

Assume that the object rotates with constant angular speed $\omega=2\,\mathrm{rad}/\mathrm{s}$ . Find and plot the position of the object at $t=1\,\mathrm{s},\;2\,\mathrm{s},\;\mbox{and}3\,\mathrm{s}$ .

Filename:sfig8-6-2a

Figure 16.18:

16.1.9 The arrow shaped object shown in the figure is defined by the five points whose coordinates are given in some normalized units. The position shown is at $t=0$ . The time dependent angular position of the object is given by $\theta(t)=C_{1}t^{2}+C_{2}t$ where $C_{1}=0.25\,\mathrm{rad}/\mathrm{s}^{2}$ and $C_{2}=0.1\,\mathrm{rad}/\mathrm{s}$ . Animate the motion of the object and show its position from $t=0$ to $t=5\,\mathrm{s}$ at every second.

Filename:efig1-2-28

Figure 16.19:

16.2 Angular velocity and acceleration of a rigid object in planar circular motion

Angular velocity of a rigid object: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$

Thus far we have talked about rotation, but not how it varies in time. Dynamics is about motion, velocities and accelerations, so we need to think about rotation rates and rotational accelerations.

A 2D rigid object’s net rotation is measured by the rotation angle $\theta$ . Thus, the simplest measure of rotation rate is $\dot{\theta}\equiv{d\theta\over dt}$ . Because all marked lines rotate the same amount $\theta$ they all have the same rates of change. So $\dot{\theta}_{1}=\dot{\theta}_{2}=\dot{\theta}_{3}=$ etc. and, as for rotation, the concept of rotation rate of a rigid object transcends the concept of rotation rate of this or that particular line. We give this rotation rate of a rigid object a special name, angular velocity, and symbol, $\omega$ (omega).

Repeating, for all lines marked on a rigid object,

\omega\equiv\dot{\theta}_{1}=\dot{\theta}_{2}=\dot{\theta}_{3}=\dots=\dot{% \theta}.

(16.23)

Often we think of angular velocity as a vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . Its direction is the axis of the rotation which, for objects in the $x y$ plane is $\hat{\boldsymbol{k}}$ , pointed in the $+z$ direction normal to the $x y$ plane. The scalar part of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is $\omega$ . So, the angular velocity vector is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\equiv\omega\hat{\boldsymbol% {k}}

(16.24)

with $\omega$ as defined in eqn. (16.23). Note ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is the angular velocity of the object (and of every line on it)

^†^†margin: We could also think of rotation rate as a derivative of the rotation matrix

[R]

. This point of view and its relation to the vector point of view here, is described in Box 16.2 on page 16.2.

The usual sign convention for $\omega$ is the same as that for $\theta$ , positive is counter-clockwise (CCW) around the origin. That’s the direction your fingers wrap if you point your thumb in the $+z$ direction and grab the $z$ axis.

Rate of change of $\hat{\boldsymbol{\imath}}^{\prime}$ , $\hat{\boldsymbol{\jmath}}^{\prime}$

Our first use of the angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is to calculate the rate of change of the rotating unit base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ . We can find the rate of change of, say, $\hat{\boldsymbol{\imath}}^{\prime}$ , by taking the time derivative of the first eqn. (16.4), and using the chain rule while recognizing that $\theta=\theta(t)$ . We can also make an analogy with polar coordinates (page 15.1), where we think of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ as like $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ as like $\hat{\boldsymbol{\jmath}}^{\prime}$ . We found there that $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ and $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}=-\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}$ . Either from differentiating $\hat{\boldsymbol{\imath}}^{\prime}$ or from the correlation with polar coordinates, we get

\begin{array}[]{cccccccc}{\dot{\hat{\boldsymbol{\imath}}}^{\prime}}=\phantom{-% }\dot{\theta}\hat{\boldsymbol{\jmath}}^{\prime}&\mbox{\hbox to36.135pt{\hfil or% \hfil}}&{\dot{\hat{\boldsymbol{\imath}}}^{\prime}}={\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}\times\hat{\boldsymbol{\imath}}^{\prime}&\mbox{and}\\ {\dot{\hat{\boldsymbol{\jmath}}}^{\prime}}=-\dot{\theta}\hat{\boldsymbol{% \imath}}^{\prime}&\mbox{\hbox to36.135pt{\hfil or\hfil}}&{\dot{\hat{% \boldsymbol{\jmath}}}^{\prime}}={\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times\hat{\boldsymbol{\jmath}}^{\prime}\\ \end{array}

(16.25)

because $\hat{\boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{k}}^{\prime}\times\hat{% \boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\imath}}^{\prime}=-\hat{\boldsymbol{k}}^{\prime}\times\hat{% \boldsymbol{\jmath}}^{\prime}$ . Depending on the tastes of your lecturer, you may find Equations 16.25 are the most used equations from this point onward

^†^†margin: Eqn. 16.25 is sometimes considered the definition of

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

. In this view, popularized by Tom Kane,

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

is that vector which determines

{\dot{\hat{\boldsymbol{\imath}}}^{\prime}}

and

{\dot{\hat{\boldsymbol{\jmath}}}^{\prime}}

by the formulas

{\dot{\hat{\boldsymbol{\imath}}}^{\prime}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\imath}}^{\prime}

and

{\dot{\hat{\boldsymbol{\jmath}}}^{\prime}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\jmath}}^{\prime}

. Then one needs to show that such a vector exists and that it is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\hat{\boldsymbol{\imath}}^{% \prime}\times{\dot{\hat{\boldsymbol{\imath}}}^{\prime}}

. Luckily this reasoning leads to the same

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

as our

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

Box 16.2 The fixed Newtonian reference frame $\mathcal{F}$

Now we can reconsider the concept of a Newtonian frame, a concept which we had to assume to write the equations of dynamics in the first place.

All of mechanics depends, of course, on the laws of mechanics. The laws of mechanics are equations which involve, in part, the positions of things as a function of time. But how position is perceived to change in time depends on your reference frame. And some reference frames are better than others. The best, from our point of view, are reference frames in which Newton’s laws are accurate. Such a reference frame is called a Newtonian frame. In engineering practice the frames we use as approximations of a Newtonian frame often seem, loosely speaking, somehow still. So we sometimes call such a frame the fixed frame and label it with a script capital $\cal F$ . When we talk about velocity and acceleration of mass points, for use in the equations of mechanics, we are always talking about the velocity and acceleration relative to a $\cal F\!$ ixed, or equivalently, Newtonian frame.

Assume $x$ and $y$ are the coordinates of a vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$ and $\cal F$ is a fixed frame with fixed axis (with associated constant base vectors $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ ). When we write ${\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}}$ we mean $\dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}}$ . But we could be more explicit (and notationally ornate) and write the velocity of P in the Newtonian frame as

{{}^{\mathcal{F}}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}% \over dt}\equiv{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}}_{\rm P}\quad\mbox{by which we mean}\quad% \dot{x}\hat{\boldsymbol{\imath}}+\dot{y}\hat{\boldsymbol{\jmath}}.

The $\mathcal{F}$ in front of the time derivative (or in front of the dot) means that when we calculate a derivative we hold the base vectors of $\mathcal{F}$ constant. This is no surprise, because for $\mathcal{F}$ the base vectors are constant. In general, however, when taking a derivative in a given frame you

•

write vectors in terms of base vectors stuck to the frame, and
•

only differentiate the components.

But only for the $\cal F$ ixed or $\cal N$ ewtonian frame will accelerations calculated this way be directly applicable to Newton’s laws.

We will avoid the ornate notation of labeling frames when it is not needed. For example, if you don’t see any script capital letters floating around in front of derivatives, you can assume that we are taking derivatives relative to a fixed Newtonian frame.

Velocity of a point fixed on a rigid object

Let’s call some rotating object $\mathcal{B}$ (script capital B) to which is glued a coordinate system $x^{\prime}y^{\prime}$ with base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ .

We now introduce the concept of derivative in a frame which we write, for the frame $\mathcal{B}$ , as

{{}^{\mathcal{B}}d\over dt}

which means, in words, the rate of change of something as viewed in the rotating frame $\mathcal{B}$ . Now consider a point P at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$ that is glued to the object. That is, the $x^{\prime}$ and $y^{\prime}$ coordinates of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$ do not change in time.

Box 16.3 Plato’s discussion of spinning in circles as motion (or not)

Here is how a smart person, Plato, thought about these things two thousand years ago. Here he writes about an imagined discussion between Socrates and Glaucon about how an object can maintain contradictory attributes simultaneously.

“Socrates: Now let’s have a more precise agreement so that we won’t have any grounds for dispute as we proceed. If someone were to say of a human being standing still, but moving his hands and head, that the same man at the same time stands still and moves, I don’t suppose we’d claim that it should be said like that, but rather that one part of him stands still and another moves. Isn’t that so?

Glaucon: Yes it is.

Socrates: Then if the man who says this should become still more charming and make the subtle point that tops as wholes stand still and move at the same time when the peg is fixed in the same place and they spin, or that anything else going around in a circle on the same spot does this too, we wouldn’t accept it because it’s not with respect to the same part of themselves that such things are at the same time both at rest and in motion. But we’d say that they have in them both a straight and a circumference; and with respect to the straight they stand still since they don’t lean in any direction –while with respect to the circumference they move in a circle; and when the straight inclines to the right the left, forward, or backward at the same time that it’s spinning, then in no way does it stand still.

Glaucon: And we’d be right.”

This chapter is about things that are still with respect to their own parts (they do not distort) but in which the points do move in circles.

{{}^{\mathcal{B}}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}% \over dt}\equiv{}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}}_{\rm P}=\quad\dot{x}^{\prime}\hat{% \boldsymbol{\imath}}^{\prime}+\dot{y}^{\prime}\hat{\boldsymbol{\jmath}}^{% \prime}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

That is, relative to a moving frame, the velocity of a point glued to the frame is zero (no surprise).

We would like to know the velocity of such a point in the fixed frame. We just take the derivative, using the product rule and the differentiation rules we have developed for the rotating base vectors:

Filename:pfigure-blue-118-2

Figure 16.20: Velocity and acceleration of two points on a rigid object rotating about 0.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$	$\displaystyle=$	$\displaystyle x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{% \boldsymbol{\jmath}}^{\prime}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}={\dot{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{{}_{\rm P}}}$	$\displaystyle=$	$\displaystyle{d\over dt}\left(x^{\prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{% \prime}\hat{\boldsymbol{\jmath}}^{\prime}\right)=x^{\prime}\dot{\hat{% \boldsymbol{\imath}}}^{\prime}+y^{\prime}\dot{\hat{\boldsymbol{\jmath}}}^{% \prime}=x^{\prime}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times% \hat{\boldsymbol{\imath}}^{\prime})+y^{\prime}({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\jmath}}^{\prime})$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times(x^{% \prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^% {\prime})$

where ${\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}}$ is the simple way to write ${{}^{\mathcal{F}}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}% \over dt}$ .

Box 16.4 Acceleration of a point on a rotating body, using ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$

Leaving off the ornate pre-super-script $\mathcal{F}$ for simplicity, we have

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{{}_{\rm P}}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(\frac{d}{dt}\left(x^{\prime}\hat{\boldsymbol{% \imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}\right)\right)$		(16.26)
		$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(x^{\prime}({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\imath}}^{\prime})+y^{\prime}({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat{\boldsymbol{\jmath% }}^{\prime})\right).$		(16.26)

To continue we need to use the product rule of differentiation for the cross product of two time dependent vectors like this:

$\displaystyle\frac{d}{dt}\left({\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}\times\hat{\boldsymbol{\imath}}^{\prime}\right)$	$\displaystyle=\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat% {\boldsymbol{\imath}}^{\prime}+{\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}\times\dot{\hat{\boldsymbol{\imath}}}^{\prime}$	(16.27)
	$\displaystyle=\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat% {\boldsymbol{\imath}}^{\prime}+{\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat{% \boldsymbol{\imath}}^{\prime}),$	(16.28)
$\displaystyle\frac{d}{dt}\left({\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}\times\hat{\boldsymbol{\jmath}}^{\prime}\right)$	$\displaystyle=\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat% {\boldsymbol{\jmath}}^{\prime}+{\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}\times\dot{\hat{\boldsymbol{\jmath}}}^{\prime}$	(16.29)
	$\displaystyle=\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat% {\boldsymbol{\jmath}}^{\prime}+{\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat{% \boldsymbol{\jmath}}^{\prime}).$	(16.30)

Substituting back into eqn. (16.26) we get

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$	$\displaystyle=x^{\prime}\left(\dot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times\hat{\boldsymbol{\imath}}^{\prime}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}% }\times\hat{\boldsymbol{\imath}}^{\prime})\vphantom{\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}\right)$
		$\displaystyle\qquad+y^{\prime}\left(\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\jmath}}^{\prime}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\jmath}}^{\prime})\right)$
		$\displaystyle=\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times(x^{% \prime}\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^% {\prime})+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\left({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\left(x^{\prime}\hat{% \boldsymbol{\imath}}^{\prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^{\prime})% \right)\right)$
		$\displaystyle=\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times\left({\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P}}\right)$

which derives eqn. (16.32).

Thus,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}

(16.31)

We can rewrite eqn. (16.31) in a minimalist or elaborate notation as

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}\qquad\mbox{or}$
	$\displaystyle{{}^{\mathcal{F}}d{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{P}}\over dt}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{}_{\mathcal{% B}/\mathcal{F}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm% {P/O}}.$

Both are correct. In the first case you have to use common sense to know what point you are talking about and that it is on an object rotating with absolute angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . In the second case everything is laid out perfectly clearly (which is why it looks perfectly confusing). On the left side of the equation it says that we are interested in how point P moves relative to, not just any frame, but the fixed frame $\mathcal{F}$ . On the right side we make clear that the rotation rate we are looking at is that of object $\mathcal{B}$ relative to $\mathcal{F}$ and not some other relative rotation. We further make clear that the formula only makes sense if the position of the point P is measured relative to a point which doesn’t move, namely 0.

What we have just found largely duplicates what we already learned in section 7.1 for points moving in circles. The slight generalization is that the same angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ can be used to calculate the velocities of multiple points on one rigid object. But the key idea remains: the velocity of a point going in circles is tangent to the circle it is going around and with magnitude proportional both to distance from the center and to the angular rate of rotation (fig. 16.20a).

Acceleration of a point on a rotating rigid object

Let’s again consider a point stuck on a rotating object and with position

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

Relative to the frame $\mathcal{B}$ to which a point is attached, its acceleration is zero (again no surprise). But what is its acceleration in the fixed frame? We find this acceleration by writing the position vector and then differentiating twice, repeatedly using the product rule and eqn. (16.25) we get (see Box 16.2 for the details).

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{P}}+{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times\left({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}\right)

(16.32)

which is hardly intuitive at a glance

^†^†margin: Although the form eqn. (16.32) is not of much immediate use, if you are going to continue on to the mechanics of mechanisms or three-dimensional mechanics, you should follow the derivation of eqn. (16.32) carefully.

Recalling that in 2D ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ we can use either the right hand rule or manipulation of unit vectors to rewrite eqn. (16.32) as

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}=\dot{\omega}\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% P}}-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}

(16.33)

where $\omega=\dot{\theta}$ and $\dot{\omega}=\ddot{\theta}$

^†^†margin: The derivation of eqn. (16.33) can be written more informally and briefly like this (using minimalist notation):

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}

\displaystyle=

\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{v}}}={d\over dt}({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}})

\displaystyle=

\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}

\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{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}.

Thus, as we found in section 15.1 for a particle going in circles, the acceleration can be written as the sum of two terms, a tangential acceleration $\dot{\omega}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{\!\mathrm{P}}$ due to increasing tangential speed, and a centrally directed (centripetal) acceleration $-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$ due to the direction of the velocity continuously changing towards the center (see fig. 16.20b). The generalization we have made in this section is that the same ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ can be used to calculate the acceleration for all the different points on one rotating object.

Relative motion of points on a rigid object

As you well know by now, the position of point B relative to point A is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}\equiv{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}$ . Similarly the relative velocity and acceleration 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{\rm B}}-{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{\rm A}}\hbox to% 36.135pt{\hfil and\hfil}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm% {\rm B/A}}\equiv{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{\rm B}}% -{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{\rm A}}

(16.34)

So, the relative velocity (as calculated relative to a fixed frame) of two points glued to one spinning rigid object $\mathcal{B}$ is given by

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

where point $O$ is the point in the Newtonian frame on the fixed axis of rotation and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}={\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}_{\mathcal{B}}$ is the angular velocity of $\mathcal{B}$ . Repeating,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{\rm B/A}}.

(16.35)

Because points A and B are fixed on $\mathcal{B}$ their velocities and hence their relative velocity as observed in a reference frame fixed to $\mathcal{C}$ are all ${\accentset{\rightharpoonup}{\boldsymbol{0}}}$ . But, point A has an absolute velocity that is different from that of point B. So they have a relative velocity as seen in the fixed frame. And it is what you would get if B was just going in circles around A.

Filename:pfigure-s95f3a

Figure 16.21: The acceleration of B relative to A if they are both on the same rotating rigid object.

Similarly, the relative acceleration of two points glued to one rigid object spinning at constant rate is

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

(16.36)

Again, the relative acceleration is due to the difference in the points’ positions relative to the point $O$ fixed on the axis. These kinematics results, 16.2 and 16.36, are useful for calculating angular momentum relative to the center-of-mass. They are also sometimes useful for the understanding of the motions of machines with moving connected parts.

The fundamental ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ equation

Equation 16.35 actually applies to any vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ that has the property that it is fixed in the frame which is rotating at ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ (fig. 16.22).

Filename:Danef94s1q2

Figure 16.22: A vector

{\accentset{\rightharpoonup}{\boldsymbol{Q}}}

is fixed to a rotating object. So its rate of change is the velocity of its tip. The magnitude is

\dot{\theta}|{\accentset{\rightharpoonup}{\boldsymbol{Q}}}|

and the direction is orthogonal to

{\accentset{\rightharpoonup}{\boldsymbol{Q}}}

That is, all we needed in the derivation was that ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ ’s components $(Q_{x^{\prime}},Q_{y^{\prime}})$ in the $\hat{\boldsymbol{\imath}}^{\prime}$ - $\hat{\boldsymbol{\jmath}}^{\prime}$ reference frame be constant.

Examples of such ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ would include more than just the relative position of two points fixed on the object, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$ . Both $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ are also are fixed in the rotating frame. Also imagine a bug moving at a constant speed on a straight line marked on a body. That bug’s velocity relative to the rotating frame would also be constant as seen in the frame. For any ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ whose representation as an arrow moves like a line segment drawn on a rigid object rotating at ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ :

\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}.

(16.37)

This is the most fundamental equation of rigid-object kinematics. Know it well. For future reference, equation (16.37) is also valid in three dimensions. Equation 16.37 is the generalization of the three equations

\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}},% \qquad\dot{\hat{\boldsymbol{\imath}}}^{\prime}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\imath}}^{\prime}\qquad\mbox{and}% \qquad\dot{\hat{\boldsymbol{\jmath}}}^{\prime}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times\hat{\boldsymbol{\jmath}}^{\prime}.

Calculating relative velocity directly, using rotating frames

A coordinate system $x^{\prime}y^{\prime}$ attached to a rotating rigid object $\cal C$ , defines a reference frame $\mathcal{C}$ (fig. 16.6 on page 16.6).

Recall, the base vectors in this frame change in time just like any other vector fixed in the rotating frame (eqn. (16.37))

\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}.

Once we know how the base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ change with time, as per the above equations, we can find velocity and acceleration by differentiation of position.

For example, consider two points A and B fixed on a rotating object. The position of B relative to A is

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

where we are using $x^{\prime}$ and $y^{\prime}$ as a shorthand notation for $x^{\prime}_{\textrm{B/A}}$ and $y^{\prime}_{\textrm{B/A}}$ (see fig. 16.23).

Filename:pfigure-blue-123-1

Figure 16.23: Coordinates in a rotating frame.

The coordinates $x^{\prime}$ and $y^{\prime}$ are constant so

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

Now we just differentiate the relative position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}$ with respect to time,

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

We could similarly calculate ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/A}}$ by taking another derivative to get, after a calculation much like that above,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B/A}}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times\left({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}\right)+\dot{\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}_{\cal C}\times{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{B/A}.

For the above calculation the points A and B were fixed on a rotating part. Most machines have many parts, at least some of which move in more complex ways than just circular motion. We will be able to understand such machines by considering points and parts that are moving relative to a part that is itself rotating. Such will be considered in Chapter 19.

Box 16.5 Angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and the rotation matrix $[R]$

The rotation matrix $[R]$ and the angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ are related. Because angular velocity is a rotation rate, in some sense it must be the derivative of the rotation. But the situation is a bit subtle because ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is a vector and $[R]$ is a matrix.

On the one hand we have the rotation equation:

\underbrace{\left[\begin{array}[]{c}x\\ y\end{array}\right]}_{\quad\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{}}}\right]_{xy}}\par=\par\underbrace{\left[\begin{array}[]{cc}% \cos\theta&-\sin\theta\\ \sin\theta&\phantom{-}\cos\theta\end{array}\right]}_{\left[R\right]}\par% \underbrace{\left[\begin{array}[]{c}x^{\prime}\\ y^{\prime}\end{array}\right]}_{\quad\left[{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{}}}^{\ \rm ref}\right]_{xy}}.

This gives the coordinates of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ after rotation in terms of its coordinates before the rotation. Because the coordinates $\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{}}}^{\ \rm ref% }\right]_{xy}$ before rotation are fixed we can take the time derivative of both sides to get

\underbrace{\left[\begin{array}[]{c}\dot{x}\\ \dot{y}\end{array}\right]}_{\quad\left[\dot{\accentset{\rightharpoonup}{% \boldsymbol{r}}}\right]_{xy}}\par=\par\dot{\theta}\underbrace{\left[\begin{% array}[]{cc}-\sin\theta&-\cos\theta\\ \phantom{-}\cos\theta&-\sin\theta\end{array}\right]}_{\left[\dot{R}\right]}% \par\underbrace{\left[\begin{array}[]{c}x^{\prime}\\ y^{\prime}\end{array}\right]}_{\quad\left[{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{}}}^{\ \rm ref}\right]_{xy}}.

On the other hand we have the vector equation

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

where ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . We can relate this vector equation to the matrix component equation above it by expressing the cross product with a matrix multiplication as described in box 1.3 on 1.3. Applying that result to this 2D case where

[{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}]_{xyz}=\left[\begin{array}% []{c}0\\ 0\\ \omega\end{array}\right]\par\qquad\mbox{and}\qquad\par[{\accentset{% \rightharpoonup}{\boldsymbol{r}}}]_{xyz}=\left[\begin{array}[]{c}x\\ y\\ 0\end{array}\right]\par

we get

\par\underbrace{\left[\begin{array}[]{c}-\omega y\\ \phantom{-}\omega x\\ 0\end{array}\right]}_{\quad\left[\dot{\accentset{\rightharpoonup}{\boldsymbol{% r}}}\right]_{xy}}\par\par=\underbrace{\left[\begin{array}[]{rrr}0&-\omega&0\\ \omega&0&0\\ 0&0&0\\ \end{array}\right]}_{[{\cal S}({\accentset{\rightharpoonup}{\boldsymbol{\omega% }}})]}\par\underbrace{\left[\begin{array}[]{c}x\\ y\\ 0\end{array}\right]}_{\quad\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{}}}^{\ \rm ref}\right]_{xy}}.\par

Just keeping the 2D terms we have

	$\displaystyle[{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}]_{xy}$	$\displaystyle=$	$\displaystyle\left[{\cal S}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% )\right]\left[\begin{array}[]{c}x\\ y\end{array}\right]$
		$\displaystyle=$	$\displaystyle\left[\begin{array}[]{cc}0&-\omega\\ \omega&0\end{array}\right]\par\par\left[\begin{array}[]{c}x\\ y\end{array}\right].\par$

But $[R]$ tells us the coordinates in terms of the reference coordinates. So

\par[{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}]_{xy}=\par\left[\begin{array}[]{cc}0&-\omega% \\ \omega&0\end{array}\right]\par\underbrace{[R]\left[\begin{array}[]{c}x^{\prime% }\\ y^{\prime}\end{array}\right]}_{\left[\begin{array}[]{c}x\\ y\end{array}\right]}.\par

Comparing the equations above we see that

\par[\dot{R}]=\left[\begin{array}[]{cc}0&-\omega\\ \omega&0\end{array}\right][R].\par

(16.40)

Using the notation from box 1.3 we can rewrite this as

[\dot{R}]=\left[{\cal S}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}})% \right][R].

(16.41)

Equation 16.41 is the way that angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and the rate of change $[\dot{R}]$ of the rotation matrix are related.

We can also solve for $[{\cal S}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}})]$ as

[{\cal S}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}})]=[\dot{R}][R]^{-1}

from which we can find $\pm\omega$ in the off diagonal elements if we are given $[R]$ and $[\dot{R}]$ .

We can understand eqn. (16.41) better, perhaps, by thinking of the rotation matrix $[R]$ as having two columns which are, respectively, the fixed-coordinate-system components of $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ :

\par[R]=\left[[\hat{\boldsymbol{\imath}}^{\prime}]_{xy}\ |\ [\hat{\boldsymbol{% \jmath}}^{\prime}]_{xy}\right]\par

Thus the columns of $[\dot{R}]$ are the rates of change of these two unit vectors:

\par[\dot{R}]=\left[[\dot{\hat{\boldsymbol{\imath}}}^{\prime}]_{xy}\ |\ [\dot{% \hat{\boldsymbol{\jmath}}}^{\prime}]_{xy}\right]\par=\left[[{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times\hat{\boldsymbol{\imath}}^{\prime}% ]_{xy}\ |\ [{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\hat{% \boldsymbol{\jmath}}^{\prime}]_{xy}\right].

With a total abuse of notation (in general one does not allow cross products of a vectors with matrices) we can write the above equation more memorably as

\textrm{``}\ [\dot{R}]={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times[R].\ \textrm{''}

Thus eqn. (16.41) is just a version of the fundamental kinematic equation 16.37 on page 16.37.

SAMPLE 16.3

Filename:pfigure-blue-119-2

Figure 16.24:

A uniform bar AB of length $\ell=50\,\mathrm{cm}$ rotates counterclockwise about point A with constant angular speed $\omega$ . At the instant shown in fig. 16.24 the linear speed $v_{C}$ of the center-of-mass C is $7.5\,\mathrm{cm}/\,\mathrm{s}$ .

1.

What are the angular speed and angular velocity of the bar?
2.

What is the linear velocity of point B?
3.

By what angles do the angular positions of points C and B change in 2 seconds?

Solution

Filename:pfigure-s95q14

Figure 16.25:

Let the angular velocity of the bar be ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\theta}\hat{% \boldsymbol{k}}$ where $\dot{\theta}$ is the angular speed. We first need to find $\dot{\theta}$ .

1.

The linear speed of point C is given, $v_{C}=7.5\,\mathrm{cm}/\,\mathrm{s}$ . Now,

$\displaystyle v_{C}$ $\displaystyle=$ $\displaystyle\dot{\theta}\,r_{C}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\theta}$ $\displaystyle=$ $\displaystyle\frac{v_{C}}{r_{C}}=\frac{7.5\,\mathrm{cm}/\,\mathrm{s}}{25\,% \mathrm{cm}}=0.3\,\mathrm{rad}/\mathrm{s}.$

Therefore, the angular velocity of the bar is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\theta}\hat{% \boldsymbol{k}}=0.3\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}$ .

Answer: $\dot{\theta}=0.3\,\mathrm{rad}/\mathrm{s},\mbox{ and }{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}=0.3\,\mathrm{rad}/\mathrm{s}\hat{% \boldsymbol{k}}$

Point B is at distance $\ell$ from the pivot point A. Thus it goes around a circle of radius $\ell$ (see fig. 16.26). Therefore,

Filename:Danef94s3q2

Figure 16.26: From the given geometry,

\displaystyle\hat{\boldsymbol{e}}_{r}=\cos\theta\hat{\boldsymbol{\imath}}+\sin% \theta\hat{\boldsymbol{\jmath}}

\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\sin\theta\hat% {\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}}

, and

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=|{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B}|\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta}

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}=\dot{\theta}\hat{\boldsymbol{% k}}\times\ell(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{% \jmath}})$
	$\displaystyle=$	$\displaystyle\dot{\theta}\ell(\cos\theta\hat{\boldsymbol{\jmath}}-\sin\theta% \hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 0.3\,\mathrm{rad}/\mathrm{s}\cdot 50\,\mathrm{cm}\left(\frac{% \sqrt{3}}{2}\hat{\boldsymbol{\jmath}}-\frac{1}{2}\hat{\boldsymbol{\imath}}\right)$
	$\displaystyle=$	$\displaystyle 15\,\mathrm{cm}/\,\mathrm{s}\left(\frac{\sqrt{3}}{2}\hat{% \boldsymbol{\jmath}}-\frac{1}{2}\hat{\boldsymbol{\imath}}\right).$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=15\,\mathrm{cm}/\,\mathrm{s}% \left(\frac{\sqrt{3}}{2}\hat{\boldsymbol{\jmath}}-\frac{1}{2}\hat{\boldsymbol{% \imath}}\right)$

We can also write ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=15\,\mathrm{cm}/\,\mathrm{s}% \,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ where $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=\frac{\sqrt{3}}{2}\hat{% \boldsymbol{\jmath}}-\frac{1}{2}\hat{\boldsymbol{\imath}}.$

Let $\theta_{1}$ be the position of point C at some time $t_{1}$ and $\theta_{2}$ be the position at time $t_{2}$ . We want to find $\Delta\theta=\theta_{2}-\theta_{1}$ for $t_{2}-t_{1}=2\,\mathrm{s}$ .

$\displaystyle\frac{d\theta}{dt}$	$\displaystyle=$	$\displaystyle\dot{\theta}=\rm{constant}=0.3\,\mathrm{rad}/\mathrm{s}.$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}d\theta$	$\displaystyle=$	$\displaystyle(0.3\,\mathrm{rad}/\mathrm{s})dt.$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{\theta_{1}}^{% \theta_{2}}d\theta$	$\displaystyle=$	$\displaystyle\int_{t_{1}}^{t_{2}}(0.3\,\mathrm{rad}/\mathrm{s})dt.$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta_{2}-\theta_{1}$	$\displaystyle=$	$\displaystyle 0.3\,\mathrm{rad}/\mathrm{s}(t_{2}-t_{1})$
$\displaystyle\mbox{or}\quad\Delta\theta$	$\displaystyle=$	$\displaystyle 0.3\frac{\,{\rm rad}}{\cancel{\,\mathrm{s}}}\mathord{\cdot}2% \cancel{\,\mathrm{s}}=0.6\,{\rm rad}.$

The change in angular position of point B is the same as that of point C. In fact, all points on AB undergo the same change in angular position because AB is a rigid body.

Answer: $\Delta\theta_{C}=\Delta\theta_{B}=0.6\,{\rm rad}$

SAMPLE 16.4

Filename:bikefork1-ang-accel

Figure 16.27:

A flywheel of diameter $2\,\mathrm{ft}$ is made of cast iron. To avoid extremely high stresses and cracks it is recommended that the peripheral speed not exceed 6000 to 7000 ft/min. What is the corresponding rpm rating for the wheel?

Solution

	Diameter of the wheel	$\displaystyle=$	$\displaystyle 2\,\mathrm{ft}.$
	$\Rightarrow$ radius of wheel	$\displaystyle=$	$\displaystyle 1\,\mathrm{ft}.$

Now,

$\displaystyle v$	$\displaystyle=$	$\displaystyle\omega r$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega$	$\displaystyle=$	$\displaystyle\frac{v}{r}=\frac{6000\cancel{\,\mathrm{ft}}/\rm{min}}{1\cancel{% \,\mathrm{ft}}}$
	$\displaystyle=$	$\displaystyle 6000\frac{\,{\rm rad}}{\rm{min}}\mathord{\cdot}\frac{1{\rm rev}}% {2\pi\,{\rm rad}}$
	$\displaystyle=$	$\displaystyle 955\,{\rm rpm}.$

Similarly, corresponding to $v=7000\,\mathrm{ft}/\rm{min}$

$\displaystyle\omega$	$\displaystyle=$	$\displaystyle\frac{7000\cancel{\,\mathrm{ft}}/\rm{min}}{1\cancel{\,\mathrm{ft}}}$
	$\displaystyle=$	$\displaystyle 7000\frac{\,{\rm rad}}{\rm{min}}\mathord{\cdot}\frac{1{\rm rev}}% {2\pi\,{\rm rad}}$
	$\displaystyle=$	$\displaystyle 1114\,{\rm rpm}.$

Thus the rpm rating of the wheel should read 955 – 1114 rpm.

Answer: $\omega=955\;\mbox{to}\;1114\,{\rm rpm}.$

SAMPLE 16.5

Filename:bikefork-ang-accel

Figure 16.28:

Two gears A and B have the diameter ratio of 1:2. Gear A drives gear B. If the output at gear B is required to be 150 rpm, what should be the angular speed of the driving gear? Assume no slip at the contact point.

Solution Let C and C^′ be the points of contact on gear A and B respectively at some instant $t$ . Since there is no relative slip between C and C^′, both points must have the same linear velocity at instant $t$ . If the velocities are the same, then the linear speeds must also be the same. Thus

Filename:bikefork1-alt

Figure 16.29:

$\displaystyle v_{C}$	$\displaystyle=$	$\displaystyle v_{C^{\prime}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{A}r_{A}$	$\displaystyle=$	$\displaystyle\omega_{B}r_{B}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{A}$	$\displaystyle=$	$\displaystyle\omega_{B}\mathord{\cdot}\frac{r_{B}}{r_{A}}$
	$\displaystyle=$	$\displaystyle\omega_{B}\mathord{\cdot}\frac{2\cancel{r}}{\cancel{r}}=2\omega_{B}$
	$\displaystyle=$	$\displaystyle(2)\mathord{\cdot}(150\,{\rm rpm})$
	$\displaystyle=$	$\displaystyle 300\,{\rm rpm}.$

Answer: $\omega_{A}=300\,{\rm rpm}$

SAMPLE 16.6

Filename:bikefork-alt

Figure 16.30:

A uniform rigid rod AB of length $\ell=0.6\,\mathrm{m}$ is connected to two rigid links OA and OB. The assembly rotates at a constant rate about point O in the $x y$ plane. At the instant shown, when rod AB is vertical, the velocities of points A and B are ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=-4.64\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\jmath}}-1.87\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=1.87\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}-4.64\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$ . Find the angular velocity of bar AB. What is the length $R$ of the links?

Solution

Let the angular velocity of the rod AB be ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . ^†^†margin: We know that the rod rotates about the $z$ -axis but we do not know the sense of the rotation i.e., $+\hat{\boldsymbol{k}}$ or $-\hat{\boldsymbol{k}}$ . Here we have assumed that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is in the positive $\hat{\boldsymbol{k}}$ direction, although just by sketching ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$ we can easily see that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ must be in the $-\hat{\boldsymbol{k}}$ direction. Since we are given the velocities of two points on the rod we can use the relative velocity formula to find ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ :

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}-{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{A}$
$\displaystyle\mbox{or}\quad\underbrace{\omega\hat{\boldsymbol{k}}}_{{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}}\times\underbrace{\ell\hat{% \boldsymbol{\jmath}}}_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}}$	$\displaystyle=$	$\displaystyle(1.87\hat{\boldsymbol{\imath}}-4.64\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}-(-4.64\hat{\boldsymbol{\jmath}}-1.87\hat{\boldsymbol{% \imath}})\,\mathrm{m}/\mathrm{s}$
$\displaystyle\mbox{or}\quad\omega\ell(-\hat{\boldsymbol{\imath}})$	$\displaystyle=$	$\displaystyle(1.87\hat{\boldsymbol{\imath}}+1.87\hat{\boldsymbol{\imath}})\,% \mathrm{m}/\mathrm{s}-(\cancel{4.64\hat{\boldsymbol{\jmath}}}-\cancel{4.64\hat% {\boldsymbol{\jmath}}})\,\mathrm{m}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle 3.74\hat{\boldsymbol{\imath}}\,\mathrm{m}/\mathrm{s}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega$	$\displaystyle=$	$\displaystyle-\frac{3.74\,\mathrm{m}/\mathrm{s}}{\ell}$	(16.42)
	$\displaystyle=$	$\displaystyle-\frac{3.74}{0.6}\,\mathrm{rad}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle-6.233\,\mathrm{rad}/\mathrm{s}$
Thus,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle-6.233\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}.$	(16.43)

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-6.23\,\mathrm{rad}/\mathrm% {s}\hat{\boldsymbol{k}}$

Let $\theta$ be the angle between link OA and the horizontal axis. Now,

Filename:pfigure-blue-128-1

Figure 16.31: The given velocity vector

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}

. Since

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}

where

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}=R\cos\theta\hat{% \boldsymbol{\imath}}+R\sin\theta\hat{\boldsymbol{\jmath}}

, we can find

R

from the given

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{A}$	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times\underbrace{R(\cos\theta\hat{% \boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{\jmath}})}_{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{A}}$
	$\displaystyle\mbox{or}\quad(-4.64\hat{\boldsymbol{\jmath}}-1.87\hat{% \boldsymbol{\imath}})\,\mathrm{m}/\mathrm{s}$	$\displaystyle=$	$\displaystyle\omega R(\cos\theta\hat{\boldsymbol{\jmath}}+\sin\theta\hat{% \boldsymbol{\imath}})$

Dotting both sides of the equation with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ we get

	$\displaystyle-1.87\,\mathrm{m}/\mathrm{s}$	$\displaystyle=$	$\displaystyle\omega R\sin\theta$		(16.44)
	$\displaystyle-4.64\,\mathrm{m}/\mathrm{s}$	$\displaystyle=$	$\displaystyle\omega R\cos\theta$		(16.45)

Squaring and adding Eqns (16.44) and (16.45) together we get

$\displaystyle\omega^{2}R^{2}$	$\displaystyle=$	$\displaystyle(-4.64\,\mathrm{m}/\mathrm{s})^{2}+(-1.187\,\mathrm{m}/\mathrm{s}% )^{2}$
	$\displaystyle=$	$\displaystyle 25.026\,\mathrm{m}^{2}/\,\mathrm{s}^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad R^{2}$	$\displaystyle=$	$\displaystyle\frac{25.026\,\mathrm{m}^{2}/\,\mathrm{s}^{2}}{(-6.23\,\mathrm{% rad}/\mathrm{s})^{2}}$
	$\displaystyle=$	$\displaystyle 0.645\,\mathrm{m}^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad R$	$\displaystyle=$	$\displaystyle 0.8\,\mathrm{m}$

Answer: $R=0.8\,\mathrm{m}$

SAMPLE 16.7

Filename:tfigure8-syst-bods

Figure 16.32: An ‘L’ shaped bar rotates at speed

\omega

about point O.

Verify the relative velocity formula: The motor at O in Fig. 16.32 rotates the ‘L’ shaped bar OAB in counterclockwise direction at an angular speed which increases at $\dot{\omega}=2.5\,\mathrm{rad}/\mathrm{s}^{2}$ . At the instant shown, the angular speed $\omega=4.5\,\mathrm{rad}/\mathrm{s}$ . Each arm of the bar is of length L = 2 $\,\mathrm{ft}$ .

1.

Find the velocity of point A.
2.

Find the relative velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}\;(={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A})$ and use the result to find the absolute velocity of point B $({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}={\accentset{\rightharpoonup% }{\boldsymbol{v}}}_{A}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A})$ .
3.

Find the velocity of point B directly. Check the answer obtained in part (b) against the new answer.

Solution

1.
Filename:sfig8-7-2

Figure 16.33: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}$ is tangential to the circular path of point A.

As the bar rotates, every point on the bar goes in circles centered at point O. Therefore, we can easily find the velocity of any point on the bar using circular motion formula ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ . Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}=\omega\hat{\boldsymbol{k}}% \times L\hat{\boldsymbol{\imath}}=\omega L\hat{\boldsymbol{\jmath}}$

$\displaystyle=$ $\displaystyle 4.5\,\mathrm{rad}/\mathrm{s}\cdot 2\,\mathrm{ft}\hat{\boldsymbol% {\jmath}}=9\,{\mathrm{ft/s}}\hat{\boldsymbol{\jmath}}.$

The velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$ is shown in Fig. 16.33.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=9\,{\mathrm{ft/s}}\hat{% \boldsymbol{\jmath}}$

Filename:sfig8-7-2a

Figure 16.34:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}

and

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{A}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}

Point B and A are on the same rigid body. Therefore, with respect to point A, point B goes in circles about A. Hence the relative velocity of B with respect to A is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}$
	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\times L\hat{\boldsymbol{\jmath}}=-% \omega L\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-4.5\,\mathrm{rad}/\mathrm{s}\cdot 2\,\mathrm{ft}\hat{\boldsymbol% {\imath}}=-9\,{\mathrm{ft/s}}\hat{\boldsymbol{\imath}}.$
$\displaystyle\hbox to36.135pt{and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B/A}$
	$\displaystyle=$	$\displaystyle 9\,{\mathrm{ft/s}}(-\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}}).$

These velocities are shown in Fig. 16.34.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}=-9\,{\mathrm{ft/s}}\hat{% \boldsymbol{\imath}},\;{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=9\,{% \mathrm{ft/s}}(-\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$

3.
Filename:sfig8-7-2again

Figure 16.35: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}$ .

Since point B goes in circles of radius OB about point O, we can find its velocity directly using circular motion formula:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}$

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

$\displaystyle=$ $\displaystyle 9\,{\mathrm{ft/s}}(-\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}}).$

The velocity vector is shown in Fig. 16.35. Of course this velocity is the same velocity as obtained in part (b) above.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=9\,{\mathrm{ft/s}}(-\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$

Note: Nothing in this sample uses $\dot{\omega}$ !

SAMPLE 16.8

Filename:sfig8-7-2disks

Figure 16.36: The ‘L’ shaped bar rotates counterclockwise while slowing down.

Verify the relative acceleration formula: Consider the ‘L’ shaped bar of Sample 16.31 again. At the instant shown, the bar is rotating at $4\,\mathrm{rad}/\mathrm{s}$ and is slowing down at the rate of $2\,\mathrm{rad}/\mathrm{s}^{2}$ .

(i)

Find the acceleration of point A.
(ii)

Find the relative acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}$ of point B with respect to point A and use the result to find the absolute acceleration of point B ( ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{A}+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}$ ).
(iii)

Find the acceleration of point B directly and verify the result obtained in (ii).

Solution

We are given:

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}=% 4\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}},\hbox to36.135pt{\hfil and\hfil% }\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-\dot{\omega}\hat{% \boldsymbol{k}}=-2\,\mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{k}}.

Filename:sfig8-4-4

Figure 16.37:

(i)

Point A is going in circles of radius L. Hence,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A})=\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{A}-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}$
	$\displaystyle=$	$\displaystyle-\dot{\omega}\hat{\boldsymbol{k}}\times L\hat{\boldsymbol{\imath}% }-\omega^{2}L\hat{\boldsymbol{\imath}}=-\dot{\omega}L\hat{\boldsymbol{\jmath}}% -\omega^{2}L\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-2\,\mathrm{rad}/\mathrm{s}\cdot 2\,\mathrm{ft}\hat{\boldsymbol{% \jmath}}-(4\,\mathrm{rad}/\mathrm{s})^{2}\cdot 2\,\mathrm{ft}\hat{\boldsymbol{% \imath}}$
	$\displaystyle=$	$\displaystyle-(4\hat{\boldsymbol{\jmath}}+32\hat{\boldsymbol{\imath}})\,{% \mathrm{ft}}/{\mathrm{s}}^{2}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=-(4\hat{\boldsymbol{\jmath}}% +32\hat{\boldsymbol{\imath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}$

(ii)

Filename:sfig8-4-4a

Figure 16.38:

The relative acceleration of point B with respect to point A is found by considering the motion of B with respect to A. Since both the points are on the same rigid body, point B executes circular motion with respect to point A. Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}+{\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A})=\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}-\omega^{2}L\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-\dot{\omega}\hat{\boldsymbol{k}}\times L\hat{\boldsymbol{\jmath}% }-\omega^{2}L\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle\dot{\omega}L\hat{\boldsymbol{\imath}}-\omega^{2}L\hat{% \boldsymbol{\jmath}}=2\,\mathrm{rad}/\mathrm{s}^{2}\cdot 2\,\mathrm{ft}\hat{% \boldsymbol{\imath}}-(4\,\mathrm{rad}/\mathrm{s})^{2}\cdot 2\,\mathrm{ft}\hat{% \boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle(4\hat{\boldsymbol{\imath}}-32\hat{\boldsymbol{\jmath}})\,{% \mathrm{ft}}/{\mathrm{s}}^{2},$

and

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B/A}=(-28\hat{\boldsymbol{\imath}}-36\hat{% \boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=-(28\hat{\boldsymbol{\imath}% }+36\hat{\boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}$

(iii)

Filename:sfig8-4-4b

Figure 16.39:

Since point B is going in circles of radius OB about point O, we can find the acceleration of B as follows.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B})$
	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B}$
	$\displaystyle=$	$\displaystyle-\dot{\omega}\hat{\boldsymbol{k}}\times(L\hat{\boldsymbol{\imath}% }+L\hat{\boldsymbol{\jmath}})-\omega^{2}(L\hat{\boldsymbol{\imath}}+L\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(-\dot{\omega}L-\omega^{2}L)\hat{\boldsymbol{\jmath}}+(\dot{% \omega}L-\omega^{2}L)\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(-4-32)\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}+% (4-32)\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(-36\hat{\boldsymbol{\jmath}}-28\hat{\boldsymbol{\imath}})\,{% \mathrm{ft}}/{\mathrm{s}}^{2}.$

This acceleration is, naturally again, the same acceleration as found in (ii) above.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=-(28\hat{\boldsymbol{\imath}% }+36\hat{\boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}$

Problems for 16.2 Angular velocity

Preparatory Problems

16.2.1 Give the definition of each term below in words and, but for the first, with an equation.

(a)

angular velocity Answer: The rate of change of any marked or imaginary line on a rigid object.
(b)

angular acceleration Answer: The rate of change of the angular velocity.

16.2.2 Assume that in the reference configuration, when $\theta=0$ , that the $x^{\prime}y^{\prime}$ axes attached to an object are aligned with the $x y$ axes. Consider a point P attached to the moving frame (object) that has coordinates $(x^{\prime},y^{\prime})$ . Find each of the quantities below in terms of some or all of $x^{\prime},y^{\prime},\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},% \theta,\dot{\theta}$ and $\ddot{\theta}$ .

(a)

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$ Answer: $\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{bmatrix}\begin{bmatrix}x^{\prime}\\ y^{\prime}\\ \end{bmatrix}$
(b)

${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}$ Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}$
(c)

${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}$ Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{P}}+{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}})$

16.2.3 Assume that in the reference configuration, when $\theta=0$ , the $x^{\prime}y^{\prime}$ axes are aligned with the $x y$ axes. Consider two points ${\rm P}_{1}$ and ${\rm P}_{2}$ attached to the moving frame. ${\rm P}_{1}$ and ${\rm P}_{2}$ have coordinates $(x^{\prime}_{1},y^{\prime}_{1})$ $(x^{\prime}_{2},y^{\prime}_{2})$ . Find each of the quantities below in terms of some or all of $x^{\prime}_{1},y^{\prime}_{1},x^{\prime}_{2},y^{\prime}_{2},\hat{\boldsymbol{% \imath}},\hat{\boldsymbol{\jmath}},\theta,\dot{\theta}$ and $\ddot{\theta}$ .

(a)

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P_{1}/P_{2}}}$ Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P_{1}/P_{2}}}=(x^{% \prime}_{2}-x^{\prime}_{1})\hat{\boldsymbol{\imath}}+(y^{\prime}_{2}-y^{\prime% }_{1})\hat{\boldsymbol{\jmath}}$
(b)

${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P_{1}/P_{2}}}$ Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P_{1}/P_{2}% }}=\dot{\theta}\left((x^{\prime}_{2}-x^{\prime}_{1})\hat{\boldsymbol{\jmath}}-% (y^{\prime}_{2}-y^{\prime}_{1})\hat{\boldsymbol{\imath}}\right)$
(c)

${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P_{1}/P_{2}}}$ Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{P_{1}/P_{2}}}=\ddot{% \theta}\left((x^{\prime}_{2}-x^{\prime}_{1})\hat{\boldsymbol{\jmath}}-(y^{% \prime}_{2}-y^{\prime}_{1})\hat{\boldsymbol{\imath}}\right)-\dot{\theta}^{2}% \left((x^{\prime}_{2}-x^{\prime}_{1})\hat{\boldsymbol{\imath}}+(y^{\prime}_{2}% -y^{\prime}_{1})\hat{\boldsymbol{\jmath}}\right)$

16.2.4 Find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , if ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=1.5\,\mathrm{rad}/\mathrm{s% }\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=2\,\mathrm{m}\hat{\boldsymbol{% \imath}}-3\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ . Answer: $(5.4\hat{\boldsymbol{\imath}}+3.0\hat{\boldsymbol{\jmath}})\,\mathrm{m}/% \mathrm{s}$

16.2.5 A rod OB rotates with its end O fixed as shown in the figure with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=5\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ and angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=2\,\mathrm{rad}/\mathrm{s}^% {2}\hat{\boldsymbol{k}}$ at the instant of interest. Find, draw, and label the tangential and normal acceleration of point B at $\theta=60^{{}^{\circ}}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{t}}=20\hat{\boldsymbol{% \jmath}}\,{\mathrm{in}}/{\mathrm{s}}^{2}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{n}}=-250\hat{% \boldsymbol{\imath}}\,{\mathrm{in}}/{\mathrm{s}}^{2}$ .

Filename:sfig8-4-4c

Figure 16.40:

16.2.6 A disc rotates at 15 rpm. How many seconds does it take to rotate by 180 degrees? What is the angular speed of the disc in rad/s? Answer: $\,\mathrm{rad}/\mathrm{s}=\frac{\pi}{2}\,\mathrm{rad}/\mathrm{s}$ .

16.2.7 A motor turns a uniform disc of radius $R$ counter-clockwise about its mass center at a constant rate $\omega$ . The disc lies in the $x y$ -plane and its angular displacement $\theta$ is measured (positive counter-clockwise) from the $x$ -axis. What is the angular displacement $\theta(t)$ of the disc if it starts at $\theta(0)=\theta_{0}$ and $\dot{\theta}(0)=\omega$ ? What are the velocity and acceleration of a point $P$ at position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}$ ? Answer: $\theta(t)=\theta_{0}+\omega t$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\omega(x\hat{\boldsymbol{\jmath}% }-y\hat{\boldsymbol{\imath}})$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\omega(x\hat{\boldsymbol{\imath% }}+y\hat{\boldsymbol{\jmath}})$ .

16.2.8 Find the angular velocities of the second, minute, and hour hands of a clock. Answer: $\omega_{\rm{sec}}=\frac{\pi}{30}\,\mathrm{rad}/\mathrm{s}$ , $\omega_{\rm{min}}=\frac{2\pi}{3600}\,\mathrm{rad}/\mathrm{s}$ , and $\omega_{\rm{hour}}=\frac{2\pi}{43200}\,\mathrm{rad}/\mathrm{s}$ .

16.2.9 A disc $\cal C$ spins at a constant rate of two revolutions per second counter-clockwise about its geometric center, G, which is fixed. A point P is marked on the disk at a radius of one meter. At the instant of interest, point P is on the $x$ -axis of an $x y$ -coordinate system centered at point G.

(a)

Draw a neat diagram showing the disk, the particle, and the coordinate axes.
(b)

What is the angular velocity of the disk, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{C}}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{C}}=4\pi\,\mathrm% {rad}/\mathrm{s}$ .
(c)

What is the angular acceleration of the disk, $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm C}$ ? Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm C}=0\,\mathrm{rad}% /\mathrm{s}^{2}$ .
(d)

What is the velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ of point P? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{p}=4\pi\hat{\boldsymbol{\jmath}% }\,\mathrm{m}/\mathrm{s}$ .
(e)

What is the acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ of point P? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{p}=-16\pi^{2}\hat{\boldsymbol{% \imath}}\,\mathrm{m}/\mathrm{s}^{2}$ .

16.2.10 A uniform rigid rod rotates at constant speed in the $x y$ -plane about a peg at point $O$ . The center of mass of the rod may not exceed a specified acceleration $a_{max}=0.5\,\mathrm{m}/\mathrm{s}^{2}$ . Find the maximum angular velocity of the rod. Answer: $\omega_{\rm{max}}=1\,\mathrm{rad}/\mathrm{s}$

Filename:pfigure-s94h13p2

Figure 16.41:

16.2.11 A dumbbell AB is welded to a rigid arm OC such that OC is perpendicular to AB. Arm OC rotates about O at a constant angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=10\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ . At the instant when $\theta=60^{{}^{\circ}}$ ,

(a)

Find the relative velocity of B with respect to A. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}=4\,\mathrm{m}/\mathrm{s}(% \hat{\boldsymbol{\imath}}+\sqrt{3}\hat{\boldsymbol{\jmath}})$
(b)

Find the acceleration of point B relative to point A. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}=-40\,\mathrm{m}/\mathrm{s}% ^{2}(\sqrt{3}\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}})$

Filename:pfigure-f93f5

Figure 16.42:

More-Involved Problems

16.2.12 Two discs $\cal A$ and $\cal B$ rotate at constant speeds about their centers. Disc $\cal A$ rotates at $100\,{\rm rpm}$ and disc $\cal B$ rotates at $10\,\mathrm{rad}/\mathrm{s}$ . Which is rotating faster? Answer: Disc $\cal A$ is faster.

16.2.13 A motor turns a uniform disc of radius $R$ counter-clockwise about its mass center at a constant rate $\omega$ . The disc lies in the $x y$ -plane and its angular displacement $\theta$ is measured (positive counter-clockwise) from the $x$ -axis. What are the velocity and acceleration of a point $P$ at position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}=c\hat{\boldsymbol{\imath}}+d% \hat{\boldsymbol{\jmath}}$ relative to the velocity and acceleration of a point $Q$ at position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{Q}=0.5(-d\hat{\boldsymbol{% \imath}}+c\hat{\boldsymbol{\jmath}}$ ) on the disk? ( $c^{2}+d^{2}<R^{2}$ .)

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/Q}=\omega\left((c+\frac{d}{2}% )\hat{\boldsymbol{\jmath}}-(d-\frac{c}{2})\hat{\boldsymbol{\imath}}\right)$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P/Q}=-\omega^{2}\left((c+\frac{% d}{2})\hat{\boldsymbol{\imath}}+(d-\frac{c}{2})\hat{\boldsymbol{\jmath}}\right)$

16.2.14 A $0.4\,\mathrm{m}$ long rod $A B$ has many holes along its length such that it can be pegged at any of the various locations. It rotates counter-clockwise at a constant angular speed about a peg whose location is not known. At some instant $t$ , the velocity of end $B$ is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=-3\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\jmath}}$ . After $\frac{\pi}{20}\,{\rm s}$ , the velocity of end $B$ is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=-3\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}$ . If the rod has not completed one revolution during this period,

(a)

find the angular velocity of the rod, and Answer: $\omega=30\,\mathrm{rad}/\mathrm{s}$ .
(b)

find the location of the peg along the length of the rod. Answer: $\frac{1}{10}\,\mathrm{m}$ from $B$ .

Filename:pfigure-s94h13p3

Figure 16.43:

16.2.15 A circular disc of radius $r=250\,{\rm mm}$ rotates in the $x y$ -plane about a point which is at a distance $d=2r$ away from the center of the disk. At the instant of interest, the linear speed of the center $C$ is $0.60\,\mathrm{m}/\mathrm{s}$ and the magnitude of its centripetal acceleration is $0.72\,\mathrm{m}/\mathrm{s}^{2}$ .

(a)

Find the rotational speed of the disk. Answer: $\dot{\theta}=1.2\,\mathrm{rad}/\mathrm{s}$ .
(b)

Is the given information enough to locate the center of rotation of the disk? Answer: No.
(c)

If the acceleration of the center has no component in the $\hat{\boldsymbol{\jmath}}$ direction at the instant of interest, can you locate the center of rotation? If yes, is the point you locate unique? If not, what other information is required to make the point unique? Answer: If the acceleration of the center has no component in the $\hat{\boldsymbol{\jmath}}$ direction the center of rotation can lie anywhere on the x-axis.

Filename:p-s96-p3-3

Figure 16.44:

16.2.16 A uniform disc of radius $r=200\,{\rm mm}$ is mounted eccentrically on a motor shaft at point $O$ . The motor rotates the disc at a constant angular speed. At the instant shown, the velocity of the center of mass is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}=-1.5\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\jmath}}$ .

(a)

Find the angular velocity of the disc. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-11.25\,\mathrm{rad}/% \mathrm{s}\hat{\boldsymbol{k}}$
(b)

Find the point with the highest linear speed on the disc. What is its velocity? Answer: Speed is maximum at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=\frac{5r}{3}\hat{\boldsymbol{% \imath}}$ ; ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm{max}}=-3.75\,\mathrm{m}/% \mathrm{s}\hat{\boldsymbol{\jmath}}$

Filename:bikefork1-ang-mom

Figure 16.45:

16.2.17 The circular disc of radius $R=100\,{\rm mm}$ rotates about its center $O$ . At a given instant, point $A$ on the disk has a velocity $v_{A}=0.8\,\mathrm{m}/\mathrm{s}$ in the direction shown. At the same instant, the tangent of the angle $\theta$ made by the total acceleration vector of any point B with its radial line to $O$ is 0.6. Compute the angular acceleration $\alpha$ of the disc. Answer: $\alpha=38.4\,\mathrm{rad}/\mathrm{s}^{2}$

Filename:bikefork-ang-mom

Figure 16.46:

16.2.18 Show that, for non-constant rate circular motion, the acceleration of all points in a given radial line are parallel.

Filename:summer95f-5-a

Figure 16.47:

16.2.19 A motor turns a uniform disc of radius $R$ about its mass center at a variable angular rate $\omega$ with rate of change $\dot{\omega}$ , counter-clockwise. The disc lies in the $x y$ -plane and its angular displacement $\theta$ is measured from the $x$ -axis, positive counter-clockwise. What are the velocity and acceleration of a point $P$ at position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}=c\hat{\boldsymbol{\imath}}+d% \hat{\boldsymbol{\jmath}}$ relative to the velocity and acceleration of a point $Q$ at position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{Q}=0.5(-d\hat{\boldsymbol{% \imath}}+c\hat{\boldsymbol{\jmath}}$ ) on the disk? ( $c^{2}+d^{2}<R^{2}$ .) Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/Q}=\omega\left((c+\frac{d}{2}% )\hat{\boldsymbol{\jmath}}-(d-\frac{c}{2})\hat{\boldsymbol{\imath}}\right)$ , and
${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P/Q}=-\left(\dot{\omega}(d-% \frac{c}{2})+\omega^{2}(c+\frac{d}{2})\right)\hat{\boldsymbol{\imath}}+\left(% \dot{\omega}(c+\frac{d}{2})-\omega^{2}(d-\frac{c}{2})\right)\hat{\boldsymbol{% \jmath}}$

16.2.20 The dumbbell AB shown in the figure rotates counterclockwise about point O with angular acceleration $3\,\mathrm{rad}/\mathrm{s}^{2}$ . Bar AB is perpendicular to bar OC. At the instant of interest, $\theta=45^{{}^{\circ}}$ and the angular speed is 2 rad/s.

(a)

Find the velocity of point B relative to point A. Will this relative velocity be different if the dumbbell were rotating at a constant rate of 2 rad/s? Answer: $\frac{1}{\sqrt{2}}\,\mathrm{m}/\mathrm{s}(\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}})$
(b)

Without calculations, draw a vector approximately representing the acceleration of B relative to A.
(c)

Find the acceleration of point B relative to A. What can you say about the direction of this vector as the motion progresses in time? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}=(-0.353\hat{\boldsymbol{% \imath}}+2.474\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}$

Filename:pfigure4-2-rp10

Figure 16.48:

16.2.21 Bit-stream kinematics of a CD. A Compact Disk (CD) has bits of data etched on concentric circular tracks. The data from a track is read by a beam of light from a head that is positioned under the track. The angular speed of the disk remains constant as long as the head is positioned over a particular track. As the head moves to the next track, the angular speed of the disk changes, so that the linear speed at any track is always the same. The data stream comes out at a constant rate $4.32\times 10^{6}$ bits/second. When the head is positioned on the outermost track, for which $r=56\,{\rm mm}$ , the disk rotates at $200\,{\rm rpm}$ .

(a)

What is the number of bits of data on the outermost track? Answer: $1.296$ Megabits.
(b)

find the angular speed of the disk when the head is on the innermost track ( $r=22\,{\rm mm}$ ), and Answer: 509.09 rpm.
(c)

find the number of bits on the innermost track. Answer: $0.509$ Megabits

16.2.22 2-D constant rate gear train. The angular velocity of the input shaft (driven by a motor not shown) is a constant, $\omega_{\rm{input}}=\omega_{A}$ . What is the angular velocity $\omega_{\rm{output}}=\omega_{C}$ of the output shaft and the speed of a point on the outer edge of disc $C$ , in terms of $R_{A}$ , $R_{B}$ , $R_{C}$ , and $\omega_{A}$ ? Answer: $\omega_{C}=\frac{\omega_{A}R_{A}}{R_{B}}$ and $v_{C}=\frac{\omega_{A}R_{A}R_{C}}{R_{B}}$

Filename:pfigure-blue-125-2

Figure 16.49:

Gear B is welded to C and engages with A.

16.2.23 A horizontal disk $\cal D$ of diameter $d=500\,{\rm mm}$ is driven at a constant speed of $100\,{\rm rpm}$ . A small disk $\cal C$ can be positioned anywhere between $r=10\,{\rm mm}$ and $r=240\,{\rm mm}$ on disk $\cal D$ by sliding it along the overhead shaft and then fixing it at the desired position with a set screw (see the figure). Disk $\cal C$ rolls without slip on disk $\cal D$ . The overhead shaft rotates with disk $\cal C$ and, therefore, its rotational speed can be varied by varying the position of disk $\cal C$ . This gear system is called brush gearing. Find the maximum and minimum rotational speeds of the overhead shaft. Answer: $\omega_{\rm{min}}=10\,{\rm rpm}$ and $\omega_{\rm{max}}=240\,{\rm rpm}$

Filename:pfigure-blue-68-1

Figure 16.50:

16.2.24 Two points A and B are on the same machine part that is hinged at an as yet unknown location C. Assume you are given that points at positions ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}$ are supposed to move in given directions, indicated by unit vectors $\hat{\boldsymbol{\lambda}}_{A}$ and $\hat{\boldsymbol{\lambda}}_{B}$ . For each of the problem parts below, illustrate your results with two numerical examples (in consistent units): i) ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}=1\hat{\boldsymbol% {\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}=1\hat{\boldsymbol% {\jmath}}$ , $\hat{\boldsymbol{\lambda}}_{A}=1\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{\lambda}}_{B}=-1\hat{\boldsymbol{\imath}}$ (thus ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ ), and ii) a more complex example of your choosing.

(a)

Describe in detail what equations must be satisfied by the point ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}$ .
(b)

Write a computer program that takes as input the 4 pairs of numbers $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}]$ , $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}]$ , $[\hat{\boldsymbol{\lambda}}_{A}]$ and $[\hat{\boldsymbol{\lambda}}_{B}]$ and gives as output the pair of numbers $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}]$ .
(c)

Find a formula of the form ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}=\dots$ that explicitly gives the position vector for point C in terms of the 4 given vectors. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}+\frac{({\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}-{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B}})\cdot\hat{\boldsymbol{\lambda}}_{A}}{\left(({% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}\times\hat{% \boldsymbol{\lambda}}_{B})\times\hat{\boldsymbol{\lambda}}_{B}\right)\cdot\hat% {\boldsymbol{\lambda}}_{A}}\left(({\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{C}}\times\hat{\boldsymbol{\lambda}}_{B})\times\hat{\boldsymbol{% \lambda}}_{B}\right)$

16.3 Polar moment of inertia

Filename:pfigure-blue-58-1

Figure 16.51: A bit of mass

d m

on a general planar object.

All of the basic equations of Dynamics concern the motion of mass. In particular, linear momentum, angular momentum, and energy are all expressed as sums over all the bits of mass in a system, with each bit of mass multiplied by some terms concerning position, velocity and acceleration.

From the earlier sections in this chapter we know how to find the velocity and acceleration of every bit of mass on a 2-D rigid object as it spins about a fixed axis. So it is just a matter of doing integrals or sums to calculate the various momentum and energy quantities of interest. As an object moves and rotates the region of integration and the values of the integrands change. So, in principle, in order to analyze a rigid object one has to evaluate a different integral or sum at every different configuration. But there is a shortcut: for a rotating rigid object a sum (over all atoms, say), or a difficult integral (for example, over the complex region representing a machine part) is reduced to simple multiplication.

The moment of inertia

$I^{\textrm{cm}}$ ^†^†margin: The COM is the most common reference point. The moment of inertia of a given object depends on what reference point is used. Most often when people say ‘the’ moment of inertia they mean the moment of inertia with respect to the center-of-mass. For clarity, this moment of inertia is often notated as $I^{\textrm{cm}}$ in this book. If a different reference point, say point $O$ is used, the moment of inertia is notated as $I^{0}$ .

simplifies the expressions for the angular momentum, the rate of change of angular momentum, and the energy of a rigid object. For more general motions the shortcuts need a $3\times 3$ matrix $[\boldsymbol{I}^{\mathrm{cm}}]$ But for 2D mechanics only one component of the matrix $[\boldsymbol{I}^{\mathrm{cm}}]$ is relevant, it is $I_{zz}^{\textrm{cm}}$ , called just $I$ or $J$ for short.

Here are the main results. A flat object spinning with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ in the $x y$ plane has a mass distribution which gives, by means of a calculation which we will discuss shortly, a moment of inertia $I^{\textrm{cm}}$ so that:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle I^{\textrm{cm}}\omega\hat{\boldsymbol{k}}$	(16.46)
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle I^{\textrm{cm}}\dot{\omega}\hat{\boldsymbol{k}}$	(16.47)
$\displaystyle{E_{\mathrm{K}}}_{/cm}$	$\displaystyle=$	$\displaystyle{1\over 2}\omega^{2}I^{\textrm{cm}}.$	(16.48)

There are two main skills you need to develop associated with $I$ .

•

Using the formulas above properly in angular momentum and energy balance. This is covered in sec. 16.4.
•

Finding I of an object, as is covered in most freshman calculus texts and is reviewed in this section.

Some facts about moment of inertia are summarized on the inside back cover.

The moments of inertia in 2-D : $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ .

The definition of moment of inertia

^†^†margin: What do the words ‘moment of inertia’ mean? The word inertia, in this case, can be translated to mean mass. So it’s ‘moment of mass’. The word ‘moment’ is similar to the word ‘moment’ as in torque, or (force)

\cdot

(distance), but slightly more general. The more general word moment is ‘the product of a quantity and its distance, raised to a power, from a reference point,’ the n’th moment of a quantity rel to 0

\displaystyle=\int x^{n}(\text{quantity at }x)\,dx.

So, what we used to call ‘moment’ (meaning torque) we can think of as short for ‘the first moment of perpendicular force’. And the moment of inertia is ‘the second moment of mass’. The words ‘first’ and ‘second’ refer to the exponent

n

$I^{\textrm{cm}}$ is

	$\displaystyle I^{cm}$	$\displaystyle\equiv\int(\underbrace{x^{2}+y^{2}}_{r^{2}})\,dm$
		$\displaystyle=\int\int r^{2}\overbrace{\underbrace{\left(m_{\mathrm{tot}}\over A% \right)}_{\text{The mass per unit area.}}\,dA}^{dm}\qquad\hbox{for a uniform % planar object}$

where $x=x_{\rm/cm}$ , $y=y_{\rm/cm}$ and $r=r_{\rm/cm}$ are the distances in the $x$ and $y$ directions of the bits of mass ( $d m$ ) from the center of mass and $r=r_{/cm}$ is the total distance. Similarly if all distances are measured relative to a point C or O (instead of relative to the center of mass) then similar formulas as above calculate $I^{C}$ or $I^{O}$ . See box 16.3 on page 16.3 for the calculation of the moments of inertia of some simple objects.

The term $I^{\textrm{cm}}=I^{\textrm{cm}}_{zz}$ is sometimes called the polar moment of inertia, or polar mass moment of inertia to distinguish it from the $I_{xx}$ and $I_{yy}$ , terms which have little utility in planar dynamics ( $I_{xx}$ and $I_{yy}$ , as area inertias rather than mass inertias, are all-important when calculating the bending stiffness or the stress in elastic beams, however).

What, physically, is the polar moment of inertia? It is a measure of the extent to which mass is far from the given reference point. Every bit of mass contributes to $I$ in proportion to the square of its distance from the reference point. As we will show in sec. 16.4 on page 16.4 (see eqn. (16.51) on page 16.51) $I^{\textrm{cm}}$ is just the quantity we need to do mechanics problems.

Radius of gyration

Another measure of the extent to which mass is spread from the reference point, besides the moment of inertia $I^{\textrm{cm}}$ , is the radius of gyration, $r_{\textrm{gyr}}$

^†^†margin:

k

for stiffness. The radius of gyration

r_{\textrm{gyr}}

is sometimes called

k

but we save

k

for stiffness in this book.

The radius of gyration is defined as:

r_{\textrm{gyr}}\equiv\sqrt{I/m}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}r^{2}% _{\textrm{gyr}}m=I.

That is, the radius of gyration of an object is the radius of an equivalent ring of mass that has the same $I$ and the same mass as the given object. It is an equivalent single radius for the whole mass distribution

^†^†margin: Radius of gyration and average distance of mass. The radius of gyration is an equivalent radius of the collection of mass. But the radius of gyration is not exactly the average radius of the distributed mass. Rather the radius of gyration is the square root of the mean square radius:

r_{\textrm{gyr}}\equiv\sqrt{\frac{\int r^{2}\,dm}{m_{\mathrm{tot}}}}.

For the special cases where all the mass is at the same radius

R

(as for a circular hoop or for masses distributed on the vertices of, say, a regular pentagon) the radius of gyration is

R

. For all other cases the radius of gyration is larger than the average radius of the mass.

Example: Radius of gyration of a hoop

The radius of gyration about its center of a circular hoop of mass $m$ and radius $R$ about its center is $r_{\textrm{gyr}}=R$ .

Other reference points

For the most part it is $I^{cm}$ which is of primary interest. Other reference points are useful

^†^†margin: Notation: simple vs precise. When just the symbol

I

is used one assumes

I=I^{\mathrm{cm}}_{zz}

, and if just

I^{\textrm{cm}}

is used, one assumes

I^{\textrm{cm}}=I^{\mathrm{cm}}_{zz}

. Similarly

I^{0}

and

I^{\textrm{C}}

are assumed to mean

I^{0}_{zz}

and

I^{\textrm{C}}_{zz}

, respectively.

1.

If the rigid object is hinged at a fixed point $O$ then $I^{0}$ can be used in a slight short cut in calculation of angular momentum and energy; and
2.

If one wants to calculate the moment of inertia of a composite object about its center-of-mass then it is useful to first find the moment of inertia of each of its parts about that system’s center of mass (which is generally not the center-of-mass of any of the separate parts).
3.

Sometimes it is easiest to set up the integral for moment of inertia about a special point C that is not the center of mass and then use the parallel axis theorem (eqn. (16.49)) to find the moment of inertia about the center of mass (see, for example the semi-circle example on 16.3).

The parallel axis theorem

The planar parallel axis theorem is the equation

I^{C}_{zz}=I^{cm}_{zz}+m_{\mathrm{tot}}\underbrace{r^{2}_{cm/C}}_{d^{2}}.

(16.49)

In this equation $d=r_{cm/C}$ is the distance from the center-of-mass to a line parallel to the $z$ -axis which passes through point $C$ . See box 16.3 on page 16.3 for a derivation of the parallel axis theorem for planar objects.

Note that $I^{C}_{zz}\geq I^{cm}_{zz}$ , always.

Moment of inertia of complex objects. One can calculate the moment of inertia of a composite object about its center of mass, in terms of the masses and moments of inertia of the separate parts. Say the position of the center of mass of $m_{i}$ is $(x_{i},y_{i})$ relative to a fixed origin, and the moment of inertia of that part about its center of mass is $I_{i}$ . We can then find the moment of inertia of the composite $I_{tot}$ about its center-of-mass $(x_{cm},y_{cm})$ by the following sequence of calculations, in order:

\begin{array}[]{l lcl}(1)&m_{\mathrm{tot}}&=&\sum m_{i}\\[3.0pt] (2)&x_{cm}&=&\left[\sum x_{i}m_{i}\right]/m_{\mathrm{tot}}\\[3.0pt] &y_{cm}&=&\left[\sum y_{i}m_{i}\right]/m_{\mathrm{tot}}\\[3.0pt] (3)&d^{2}_{i}&=&(x_{i}-x_{cm})^{2}+(y_{i}-y_{cm})^{2}\\[3.0pt] (4)&I_{tot}&=&\sum\left[I_{i}^{cm}+m_{i}d^{2}_{i}\right].\end{array}

You can reduce this recipe to one grand formula with lots of summation signs. But you would end up doing the calculations in about the order prescribed above in any case. This sequence of steps lends itself naturally to a computer spread sheet or to any program that deals easily with arrays of numbers. This is laid out for the similar center-of-mass calculation on page 2.2

^†^†margin: The tidy recipe just presented is actually more commonly used, with slight modification, in strength of materials than in dynamics. The need for finding area moments of inertia of strange beam cross sections arises more frequently than the need to find the polar mass moment of inertia of a strange cutout shape.

Example: Inertia of a semicircle about its COM.

The moment of inertia of a semicircle with mass $m$ and radius $r$ about O (see fig. 16.52) is

Filename:pfigure-blue-157-1

Figure 16.52: The moment of inertia of a semi-circle about the COM is calculated by calculating about point O and then using the parallel axis theorem.

I^{0}=\int r_{/0}^{2}\,dm=mr^{2}.

From the example on page 2.2 the center of mass is at

$\displaystyle y_{\textrm{G}}=\frac{2}{\pi}r.$

By the parallel axis theorem

I^{0}=I^{\textrm{cm}}+my_{\textrm{G}}^{2}

which we can solve for $I^{\textrm{cm}}$ to get

I^{\textrm{cm}}=I^{0}-my_{\textrm{G}}^{2}=mr^{2}-\frac{4}{\pi^{2}}mr^{2}=\left% (1-\frac{4}{\pi^{2}}\right)mr^{2}\approx 0.6\,mr^{2}\ .

The radius of gyration is that radius $\rho_{\textrm{gyr}}$ so that $m\rho_{\textrm{gyr}}^{2}=I^{\textrm{cm}}$ and is

\rho_{\textrm{gyr}}=\sqrt{1-\frac{4}{\pi^{2}}}\ \ r.

The perpendicular axis theorem for planar rigid bodies

The perpendicular axis theorem for planar objects is the equation

I_{zz}=I_{xx}+I_{yy}

which is derived in box 16.3 on page 16.3. It gives the ‘polar’ inertia $I_{zz}$ in terms of the inertias $I_{xx}$ and $I_{yy}$ . Unlike the parallel axis theorem, the perpendicular axis theorem does not have a three-dimensional counterpart. The theorem is of greatest utility when one wants to study the three-dimensional mechanics of a flat object and thus are in need of its full moment of inertia matrix.

Box 16.6 The 2-D parallel axis theorem and the perpendicular axis theorem

Sometimes one wants to know the moment of inertia relative to the center of mass. And, sometimes, if the object is held at a hinge joint at $O$ , relative to that hinge point $O$ There is a simple relation between these two moments of inertia known as the parallel axis theorem.

2-D parallel axis theorem

For the two-dimensional mechanics of two-dimensional objects, our only concern is $I^{o}_{zz}$ and $I^{cm}_{zz}$ and not the full moment of inertia matrix. In this case, $I^{o}_{zz}=\int r_{/o}^{2}\,dm$ and $I^{cm}_{zz}=\int r_{/cm}^{2}\,dm$ . Now, let’s prove the theorem in two dimensions referring to the figure.

Filename:tfigure4-para-axis

$\displaystyle I^{O}_{zz}$	$\displaystyle=$	$\displaystyle\int r_{/O}^{2}\,dm$
	$\displaystyle=$	$\displaystyle\int(x_{/O}^{2}+y_{/O}^{2})dm$
	$\displaystyle=$	$\displaystyle\int[{\underbrace{(x_{cm/O}+x_{/cm})}_{x_{/O}}}^{2}+{\underbrace{% (y_{cm/O}+y_{/cm})}_{y_{/O}}}^{2}]dm$
	$\displaystyle=$	$\displaystyle\int[(x_{cm/O}^{2}+2x_{cm/O}x_{/cm}+x_{/cm}^{2})+$
		$\displaystyle\hskip 50.58878pt(y_{cm/O}^{2}+2y_{cm/O}y_{/cm}+y_{/cm}^{2})]dm$
	$\displaystyle=$	$\displaystyle(x_{cm/O}^{2}+y_{cm/O}^{2})\underbrace{\int dm}_{m}+2x_{cm/O}% \underbrace{\int x_{/cm}dm}_{0}+$
		$\displaystyle\hskip 21.68121pt2y_{cm/O}\underbrace{\int y_{/cm}dm}_{0}+\int(x_% {/cm}^{2}+y_{/cm}^{2})dm$
	$\displaystyle=$	$\displaystyle r_{cm/O}^{2}m+\underbrace{\int(x_{/cm}^{2}+y_{/cm}^{2})dm}_{I^{% cm}_{zz}}$
	$\displaystyle=$	$\displaystyle I^{cm}_{zz}+\underbrace{r_{cm/O}^{2}}_{d^{2}}m$

The cancellation $\int y_{\rm/cm}\,dm=\int x_{\rm/cm}\,dm=0$ comes from the definition of center of mass.

Sometimes, people write the parallel axis theorem more simply as

I^{0}=I^{cm}+md^{2}\qquad\mbox{or}\qquad J_{O}=J_{cm}+md^{2}

using the symbol $J$ to mean $I_{zz}$ . One thing to note about the parallel axis theorem is that the moment of inertia about any point $O$ is always greater than the moment of inertia about the center of mass. For a given object, the minimum moment of inertia is about the center-of-mass.

Why the name parallel axis theorem? We use the name because the two $I$ ’s calculated are the moments of inertia about two parallel axes (both in the $z$ direction) through the two points $c m$ and $O$ .

One way to think about the theorem is the following. The moment of inertia of an object about a point $O$ not at the center-of-mass is the same as that of the object about the $c m$ plus that of a point mass located at the center-of-mass. If the distance from $O$ to the $c m$ is larger than the outer radius of the object, then the $d^{2}m$ term is larger than $I_{zz}^{cm}$ . The distance of equality of the two terms is the radius of gyration, $r_{\textrm{gyr}}$ .

Perpendicular axis theorem (applies to planar objects only)

For planar objects,

$\displaystyle I^{O}_{zz}$	$\displaystyle=$	$\displaystyle\int\|{\accentset{\rightharpoonup}{\boldsymbol{r}}}\|^{2}\,dm$
	$\displaystyle=$	$\displaystyle\int(x_{/O}^{2}+y_{/O}^{2})dm$
	$\displaystyle=$	$\displaystyle\int x_{/O}^{2}dm+\int y_{/O}^{2}dm$
	$\displaystyle=$	$\displaystyle I^{O}_{yy}+I^{O}_{xx}$

Similarly,

I^{cm}_{zz}=I^{cm}_{xx}+I^{cm}_{yy}.

That is, the moment of inertia about the $z$ -axis is the sum of the inertias about the two perpendicular axes $x$ and $y$ . Note that the objects must be planar ( $z=0$ everywhere) or the theorem would not be true. For example, $I^{o}_{xx}=\int(y_{/O}^{2}+z_{/O}^{2})dm\neq\int y_{/O}^{2}dm$ for a three-dimensional object.

Filename:tfigure4-perp-axis-thm

Box 16.7 Some examples of 2-D Moment of Inertia

Here, we illustrate some simple moment of inertia calculations for two-dimensional objects. The needed formulas are summarized, in part, by the lower right corner components (that is, the elements in the third column and third row (3,3)) of the matrices in the table on the inside back cover.

One point mass

Filename:tfigure4-pointmassIo

If we assume that all mass is concentrated at one or more points, then the integral

I^{o}_{zz}=\int r_{/o}^{2}\,dm

reduces to the sum

I^{o}_{zz}=\sum r_{i/o}^{2}m_{i}

which reduces to one term if there is only one mass,

I^{o}_{zz}=r^{2}m=(x^{2}+y^{2})m.

So, if $x=3\in$ , $y=4\in$ , and $m=0.1\,{\mathrm{lbm}}$ , then $I^{o}_{zz}=2.5\,{\mathrm{lbm}}\in^{2}$ . Note that, in this case, $I^{cm}_{zz}=0$ since the radius from the center-of-mass to the center-of-mass is zero.

Two point masses

Filename:tfigure4-inertia-fig0

In this case, the sum that defines $I^{o}_{zz}$ reduces to two terms, so

I^{o}_{zz}=\sum r_{i/o}^{2}m_{i}=m_{1}r_{1}^{2}+m_{2}r_{2}^{2}.

Note that, if $r_{1}=r_{2}=r$ , then $I^{o}_{zz}=m_{tot}r^{2}$ .

A thin uniform rod

Filename:tfigure4-2DrodIo

Consider a thin rod with uniform mass density, $\rho$ , per unit length, and length $\ell$ . We calculate $I^{o}_{zz}$ as

$\displaystyle I^{o}_{zz}$	$\displaystyle=$	$\displaystyle\int r^{2}\overbrace{dm}^{\rho ds}$
	$\displaystyle=$	$\displaystyle\int_{-\ell_{1}}^{\ell_{2}}s^{2}\rho ds\,\,\,\,\,\,\,\,(s=r)$
	$\displaystyle=$	$\displaystyle\left.\frac{1}{3}\rho s^{3}\right\|_{-\ell_{1}}^{\,\,\,\,\ell_{2}}% \,\,\,({\rm since}\,\,\,\,\,\,\,\,\rho\equiv const.)$
	$\displaystyle=$	$\displaystyle\frac{1}{3}\rho(\ell_{1}^{3}+\ell_{2}^{3}).$

If either $\ell_{1}=0$ or $\ell_{2}=0$ , then this expression reduces to $I^{o}_{zz}=\frac{1}{3}m\ell^{2}$ . If $\ell_{1}=\ell_{2}$ , then $O$ is at the center-of-mass and

I^{o}_{zz}=I^{cm}_{zz}=\frac{1}{3}\rho\left(\left(\frac{\ell}{2}\right)^{3}+% \left(\frac{\ell}{2}\right)^{3}\right)=\frac{ml^{2}}{12}.

We can illustrate one last point. With a little bit of algebraic histrionics of the type that only hindsight can inspire, you can verify that the expression for $I^{0}_{zz}$ can be arranged as follows:

$\displaystyle I^{0}_{zz}$	$\displaystyle=$	$\displaystyle\frac{1}{3}\rho(\ell_{1}^{3}+\ell_{2}^{3})$
	$\displaystyle=$	$\displaystyle\underbrace{\rho(\ell_{1}+\ell_{2})}_{m}{\left(\underbrace{\frac{% \ell_{2}-\ell_{1}}{2}}_{d}\right)}^{2}+\underbrace{\rho\frac{(\ell_{1}+\ell_{2% })^{3}}{12}}_{m\ell^{2}/12}$
	$\displaystyle=$	$\displaystyle md^{2}+m\frac{\ell^{2}}{12}$
	$\displaystyle=$	$\displaystyle md^{2}+I^{cm}_{zz}$

That is, the moment of inertia about point $O$ is greater than that about the center of mass by an amount equal to the mass times the distance from the center-of-mass to point $O$ squared. This derivation of the parallel axis theorem is for one special case, that of a uniform thin rod.

A uniform hoop

Filename:tfigure4-2DhoopIo

For a hoop of uniform mass density, $\rho$ , per unit length, we might consider all of the points to have the same radius $R$ . So,

I^{o}_{zz}=\int r^{2}dm=\int R^{2}dm=R^{2}\int dm=R^{2}m.

Or, a little more tediously,

$\displaystyle I^{o}_{zz}$	$\displaystyle=$	$\displaystyle\int r^{2}dm$
	$\displaystyle=$	$\displaystyle\int_{0}^{2\pi}R^{2}\rho Rd\theta$
	$\displaystyle=$	$\displaystyle\rho R^{3}\int_{0}^{2\pi}d\theta$
	$\displaystyle=$	$\displaystyle 2\pi\rho R^{3}=\underbrace{(2\pi\rho R)}_{m}R^{2}=mR^{2}.$

This $I^{o}_{zz}$ is the same as for a single point mass $m$ at a distance $R$ from the origin $O$ . It is also the same as for two point masses if they both are a distance $R$ from the origin. For the hoop, however, $O$ is at the center-of-mass so $I^{o}_{zz}=I^{cm}_{zz}$ which is not the case for a single point mass.

A uniform disk

Filename:tfigure4-2DdiskIo

Assume the disk has uniform mass density, $\rho$ , per unit area. For a uniform disk centered at the origin, the center-of-mass is at the origin so

$\displaystyle I^{o}_{zz}=I^{cm}_{zz}$	$\displaystyle=$	$\displaystyle\int r^{2}dm$
	$\displaystyle=$	$\displaystyle\int_{0}^{R}\int_{0}^{2\pi}r^{2}\rho rd\theta dr$
	$\displaystyle=$	$\displaystyle\int_{0}^{R}2\pi\rho r^{3}dr$
	$\displaystyle=$	$\displaystyle\left.2\pi\rho\frac{r^{4}}{4}\right\|_{0}^{R}=\pi\rho\frac{R^{4}}{% 2}=(\pi\rho R^{2})\frac{R^{2}}{2}$
	$\displaystyle=$	$\displaystyle m\frac{R^{2}}{2}.$

For example, a $1\,\mathrm{kg}$ plate of $1\,\mathrm{m}$ radius has the same moment of inertia as a $1\,\mathrm{kg}$ hoop with a $70.7\,\mathrm{cm}$ radius.

Uniform rectangular plate

Filename:tfigure4-2DplateIo

For the special case that the center of the plate is at point $O$ , the center-of-mass is also at $O$ and $I^{o}_{zz}=I^{cm}_{zz}$ .

$\displaystyle I^{o}_{zz}=I^{cm}_{zz}$	$\displaystyle=$	$\displaystyle\int r^{2}dm$
	$\displaystyle=$	$\displaystyle\int_{-\frac{b}{2}}^{\frac{b}{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}% }(x^{2}+y^{2})\overbrace{\rho dxdy}^{dm}$
	$\displaystyle=$	$\displaystyle\left.\int_{-\frac{b}{2}}^{\frac{b}{2}}\rho\left(\frac{x^{3}}{3}+% xy^{2}\right)\right\|^{x=\frac{a}{2}}_{x=-\frac{a}{2}}dy$
	$\displaystyle=$	$\displaystyle\left.\left.\rho\left(\frac{x^{3}y}{3}+\frac{xy^{3}}{3}\right)% \right\|^{x=\frac{a}{2}}_{x=-\frac{a}{2}}\right\|^{y=\frac{b}{2}}_{y=-\frac{b}{2}}$
	$\displaystyle=$	$\displaystyle\rho\left(\frac{a^{3}b}{12}+\frac{ab^{3}}{12}\right)$
	$\displaystyle=$	$\displaystyle\frac{m}{12}(a^{2}+b^{2}).$

Note that $\int r^{2}dm=\int x^{2}dm+\int y^{2}dm$ for all planar objects (the perpendicular axis theorem). For a uniform rectangle, $\int y^{2}dm=\rho\int y^{2}dA$ . But the integral $y^{2}dA$ is just the term often used for $I$ , the area moment of inertia, in strength of materials calculations for the stresses and stiffnesses of beams in bending. You may recall that $\int y^{2}dA=\frac{ab^{3}}{12}=\frac{Ab^{2}}{12}$ for a rectangle. Similarly, $\int x^{2}dA=\frac{Aa^{2}}{12}$ . So, the polar moment of inertia $J=I^{o}_{zz}=m\frac{1}{12}(a^{2}+b^{2})$ can be recalled by remembering the area moment of inertia of a rectangle combined with the perpendicular axis theorem.

SAMPLE 16.9

Filename:summer95f-5

Figure 16.53:

Moment of inertia of point masses: A pendulum is made up of two unequal point masses $m$ and $2m$ connected by a massless rigid rod of length $4r$ . The pendulum is pivoted at distance $r$ along the rod from the small mass.

1.

Find the moment of inertia $I^{\mathrm{cm}}_{zz}$ of the pendulum.
2.

Find the moment of inertia $I_{zz}^{O}$ of the pendulum.
3.

Find the radius of gyration of the pendulum.

Solution

1.

First we need to find the center of mass of the system. Let the center of mass C be located at distance $r_{cm}$ from the origin O. An easy way to find the location of C will be to consider the pendulum to lie along the x-axis with the origin at O (see fig. 16.54). Then

Filename:pfigure-blue-127-2

Figure 16.54:

$\displaystyle x_{cm}$ $\displaystyle=$ $\displaystyle\frac{\sum m_{i}x_{i}}{\sum m_{i}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}r_{cm}$ $\displaystyle=$ $\displaystyle\frac{m(-r)+2m(3r)}{m+2m}=\frac{5}{3}r.$

So, the moment of inertia about the center of mass is

$I^{\mathrm{cm}}_{zz}=\sum m_{i}r_{i/cm}^{2}=m\left(r+\frac{5}{3}r\right)^{2}+2% m\left(3r-\frac{5}{3}r\right)^{2}=\frac{32}{3}mr^{2}.$

Answer: $I^{\mathrm{cm}}_{zz}=10.67mr^{2}$
2.

We can calculate $I^{\scriptscriptstyle\mathrm{O}}_{zz}$ in two different ways, one directly by summing the contributions of each mass about point O, and the other by using $I^{\mathrm{cm}}_{zz}$ and the parallel axis theorem.

$\displaystyle I^{\scriptscriptstyle\mathrm{O}}_{zz}$ $\displaystyle=$ $\displaystyle mr^{2}+2m(3r)^{2}=19mr^{2}$

$\displaystyle\hbox to36.135pt{or\hfil}I^{\scriptscriptstyle\mathrm{O}}_{zz}$ $\displaystyle=$ $\displaystyle I^{\mathrm{cm}}_{zz}+m_{tot}r_{cm/O}^{2}$

$\displaystyle=$ $\displaystyle\frac{32}{3}mr^{2}+3m\left(\frac{5}{3}r\right)^{2}=19mr^{2}.$

Answer: $I_{zz}^{O}=19mr^{2}$
3.

The radius of gyration, by definition, is the distance that gives the desired moment of inertia if the entire mass of the system is concentrated there. Thus, if $r_{gyr}$ is the radius of gyration for the moment of inertia about point O, then

Filename:pfigure-s94h13p4

Figure 16.55:

$\displaystyle I^{\scriptscriptstyle\mathrm{O}}_{zz}$ $\displaystyle=$ $\displaystyle(3m)\mathord{\cdot}r_{gyr}^{2}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad 19\cancel{m}r^{2}$ $\displaystyle=$ $\displaystyle 3\cancel{m}r_{gyr}^{2}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad r_{gyr}$ $\displaystyle=$ $\displaystyle\sqrt{\frac{19}{3}}r=2.52r.$

Thus the radius of gyration $r_{gyr}$ of the given pendulum is $r_{gyr}=2.52\,r$ .

Answer: $r_{gyr}=2.52\,r$

SAMPLE 16.10

Filename:pfigure-blue-90-2

Figure 16.56:

Moment of inertia of a rod: A uniform rigid rod AB of mass $M=2\,\mathrm{kg}$ and length $3\ell=1.5\,\mathrm{m}$ swings about the $z$ -axis passing through the pivot point O.

1.

Find the moment of inertia $I_{zz}^{O}$ of the bar using the fundamental definition $I_{zz}^{O}=\int_{m}r_{/O}^{2}dm$ .
2.

Find $I_{zz}^{O}$ using the parallel axis theorem given that $I_{zz}^{\rm{cm}}=\frac{1}{12}m\ell^{2}$ where $m=$ total mass, and $\ell=$ total length of the rod. (You can find $I_{zz}^{\rm{cm}}$ for many commonly encountered objects in the table on the inside backcover of the text).

Solution

Filename:s92f1p7

Figure 16.57:

Since we need to carry out the integral, $I_{zz}^{O}=\int_{m}r_{/O}^{2}dm$ , to find $I_{zz}^{O}$ , let us consider an infinitesimal length segment $d\ell^{\prime}$ of the bar at distance $\ell^{\prime}$ from the pivot point O. (see Figure 16.57). Let the mass of the infinitesimal segment be $d m$ .
Now the mass of the segment may be written as

	$\displaystyle dm$	$\displaystyle=$	(mass per unit length of the bar) $\mathord{\cdot}$ (length of the segment)
		$\displaystyle=$	$\displaystyle\frac{M}{3\ell}\,d\ell^{\prime}\quad\quad\left(\mbox{Note:}\;% \frac{\rm{mass}}{\rm{unit\;length}}=\frac{\rm{total\;mass}}{\rm{total\;length}% }\right).$

We also note that the distance of the segment from point O, $r_{/O}=\ell$ . Substituting the values found above for $r_{/O}$ and $d m$ in the formula we get

$\displaystyle I_{zz}^{O}$	$\displaystyle=$	$\displaystyle\int_{-\ell}^{2\ell}\underbrace{(\ell^{\prime})^{2}}_{r_{/O}^{2}}% \underbrace{\frac{M}{3\ell}d\ell^{\prime}}_{dm}$
	$\displaystyle=$	$\displaystyle\frac{M}{3\ell}\int_{-\ell}^{2\ell}(\ell^{\prime})^{2}d\ell^{% \prime}=\frac{M}{3\ell}\left[\frac{\ell^{\prime 3}}{3}\right]_{-\ell}^{2\ell}$
	$\displaystyle=$	$\displaystyle\frac{M}{3\ell}\left[\frac{8\ell^{3}}{3}-\left(-\frac{\ell^{3}}{3% }\right)\right]=M\ell^{2}$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{kg}\mathord{\cdot}(0.5\,\mathrm{m})^{2}=0.5\,\mathrm{% kg}\mathord{\cdot}\,\mathrm{m}^{2}.$

Answer: $I_{zz}^{O}=0.5\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$

2.

The parallel axis theorem states that

Filename:twodisks-ang-mom

Figure 16.58:

$I^{O}_{zz}=I_{zz}^{\rm{cm}}+Mr_{O/\rm{cm}}^{2}.$

Since the rod is uniform, its center-of-mass is at its geometric center, i.e., at distance $\frac{3\ell}{2}$ from either end. From the Fig 16.58 we can see that

$r_{O/\rm{cm}}=AG-AO=\frac{3\ell}{2}-\ell=\frac{\ell}{2}$

$\displaystyle\mbox{Therefore,}\quad I_{zz}^{O}$ $\displaystyle=$ $\displaystyle\underbrace{\frac{1}{12}M(3\ell)^{2}}_{I_{zz}^{\rm{cm}}}+M(\frac{% \ell}{2})^{2}$

$\displaystyle=$ $\displaystyle\frac{9}{12}M\ell^{2}+M\frac{\ell^{2}}{4}=M\ell^{2}$

$\displaystyle=$ $\displaystyle 0.5\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}\quad\mbox{(same % as in (a), of course)}$

Answer: $I_{zz}^{O}=0.5\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$

SAMPLE 16.11

Filename:pfigure-blue-90-1

Figure 16.59:

Moment of inertia of a wheel with a cut-out: A uniform rigid wheel of radius $r=1\,\mathrm{ft}$ is made eccentric by cutting out a portion of the wheel. The center-of-mass of the eccentric wheel is at C, a distance $e=\frac{r}{3}$ from the geometric center O. The mass of the wheel (after deducting the cut-out) is 3.2 lbm. The moment of inertia of the wheel about point O, $I_{zz}^{\rm{O}}$ , is 1.8 $\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2}$ . We are interested in the moment of inertia $I_{zz}$ of the wheel about points A and B on the perimeter.

1.

Without any calculations, guess which point, A or B, gives a higher moment of inertia. Why?
2.

Calculate $I_{zz}^{\rm{C}}$ , $I_{zz}^{\rm{A}}$ and $I_{zz}^{\rm{B}}$ and compare with the guess in (a).

Solution

1.

The moment of inertia $I_{zz}^{\rm{B}}$ should be higher. Moment of inertia $I_{zz}$ measures the geometric distribution of mass about the $z$ -axis. But the distance of the mass from the axis counts more than the mass itself ( $I_{zz}^{O}=\int_{m}r_{/O}^{2}dm$ ). The distance $r_{/O}$ of the mass appears as a quadratic term in $I_{zz}^{\rm{O}}$ . The total mass is the same whether we take the moment of inertia about point A or about point B. However, the distribution of mass is not the same about the two points. Due to the cut-out being closer to point B there are more “ $d m$ ’s” at greater distances from point B than from point A. So, we guess that

Answer: $I_{zz}^{\rm{B}}>I_{zz}^{\rm{A}}$

If we know the moment of inertia $I_{zz}^{\rm{C}}$ (about the center-of-mass) of the wheel, we can use the parallel axis theorem to find $I_{zz}^{\rm{A}}$ and $I_{zz}^{\rm{B}}$ . In the problem, we are given $I_{zz}^{\rm{O}}$ . But,

Filename:pfigure-blue-110-1

Figure 16.60:

$\displaystyle I_{zz}^{\rm{O}}$	$\displaystyle=$	$\displaystyle I_{zz}^{\rm{C}}+Mr_{O/C}^{2}\quad\mbox{(parallel axis theorem)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad I_{zz}^{\rm{C}}$	$\displaystyle=$	$\displaystyle I_{zz}^{\rm{O}}-Mr_{O/C}^{2}$
	$\displaystyle=$	$\displaystyle 1.8\,{\mathrm{lbm}}\,\mathrm{ft}^{2}-3.2\,{\mathrm{lbm}}% \underbrace{\left(\frac{1\,\mathrm{ft}}{3}\right)^{2}}_{r_{O/C}=e=\frac{r}{3}}$
	$\displaystyle=$	$\displaystyle 1.44\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2}$
$\displaystyle\mbox{Now,}\quad I_{zz}^{\rm{A}}$	$\displaystyle=$	$\displaystyle I_{zz}^{\rm{C}}+Mr_{A/C}^{2}=I_{zz}^{\rm{C}}+M\left(\frac{2r}{3}% \right)^{2}$
	$\displaystyle=$	$\displaystyle 1.44\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2}+3.2\,{% \mathrm{lbm}}\left(\frac{2\,\mathrm{ft}}{3}\right)^{2}$
	$\displaystyle=$	$\displaystyle 2.86\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2}$
$\displaystyle\mbox{and }\quad I_{zz}^{\rm{B}}$	$\displaystyle=$	$\displaystyle I_{zz}^{\rm{C}}+Mr_{B/C}^{2}=I_{zz}^{\rm{C}}+M\left(r+\frac{r}{3% }\right)^{2}$
	$\displaystyle=$	$\displaystyle 1.44\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2}+3.2\,{% \mathrm{lbm}}\left(1\,\mathrm{ft}+\frac{1\,\mathrm{ft}}{3}\right)^{2}$
	$\displaystyle=$	$\displaystyle 7.13\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2}$

Answer: $I_{zz}^{\rm{C}}=1.44\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2},\;I_{zz}^% {\rm{A}}=2.86\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2},\;I_{zz}^{\rm{B}% }=7.13\,{\mathrm{lbm}}\mathord{\cdot}\,\mathrm{ft}^{2}$

Clearly, $I_{zz}^{\rm{B}}>I_{zz}^{\rm{A}}$ , as guessed in (a).

SAMPLE 16.12

Filename:pfigure-blue-107-1

Figure 16.61:

Moment of inertia: modeling a sphere as a point mass: A uniform solid sphere of mass $m$ and radius $r$ is attached to a massless rigid rod of length $\ell$ . The sphere swings in the $x y$ plane. Find the error in calculating $I_{zz}^{O}$ as a function of $r/\ell$ if the sphere is treated as a point mass concentrated at the center-of-mass of the sphere.

Solution

The exact moment of inertia of the sphere about point O can be calculated using parallel axis theorem:

	$\displaystyle I_{zz}^{O}$	$\displaystyle=$	$\displaystyle I_{zz}^{cm}+m\ell^{2}$
		$\displaystyle=$	$\displaystyle\frac{2}{5}mr^{2}+m\ell^{2}.\quad\mbox{({\footnotesize See Table IV on inside cover})}$

If we treat the sphere as a point mass, the moment of inertia $I_{zz}^{O}$ is

\tilde{I}_{zz}^{O}=m\ell^{2}.

Therefore, the relative error in $I_{zz}^{O}$ is

error	$\displaystyle=$	$\displaystyle\frac{I_{zz}^{O}-\tilde{I}_{zz}^{O}}{I_{zz}^{O}}$
	$\displaystyle=$	$\displaystyle\frac{\frac{2}{5}mr^{2}+m\ell^{2}-m\ell^{2}}{\frac{2}{5}mr^{2}+m% \ell^{2}}$
	$\displaystyle=$	$\displaystyle\frac{\frac{2}{5}\frac{r^{2}}{\ell^{2}}}{\frac{2}{5}\frac{r^{2}}{% \ell^{2}}+1}$

From the above expression we see that for $r\ll\ell$ the error is very small. From the graph of error in Fig. 16.62 we see that even for $r=\ell/5$ , the error in $I_{zz}^{O}$ due to approximating the sphere as a point mass is less than 2%.

Filename:pfigure-s94h14p4

Figure 16.62: Relative error in

I_{zz}^{O}

of the sphere as a function of

r/\ell

Problems for 16.3 Polar moment of inertia

Preparatory Problems

16.3.1 Find an expression for the polar moment of inertia for each of the following systems about the specified points. Answer in terms of some or all of the variables given. If a position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is given then you can assume the coordinates $(x,y)$ are given so that ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}$ .

(a)
A particle with mass $m$ at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ about
1. (a)
  
  Its center of mass
2. (b)
  
  The origin
Answer: $I^{cm}_{zz}=0$ , $I^{O}_{zz}=mr^{2}$ .
(b)

Two particles with masses $m_{1}$ and $m_{2}$ a distance $d$ apart located at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{2}$ , respectively, about
1. i)
  
  Mass 1
2. ii)
  
  Mass 2
3. i)
  
  The center of mass of the system
4. iv)
  
  The origin
Answer: $I^{m1}_{zz}=m_{2}\left|r_{21}^{2}\right|$ , $I^{m2}_{zz}=m_{1}\left|r_{12}^{2}\right|$ , $I^{cm}_{zz}=\left(\frac{m_{1}m_{2}}{m_{1}+m_{2}}\right)d^{2}$ , and $I^{cm}_{zz}=m1\left|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}\right|^{% 2}+m2\left|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{2}\right|^{2}$
(c)
A collection of $n$ particles $m_{i}$ at locations ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/0}}$ about
1. i)
  
  The origin
2. ii)
  
  The center of mass G of the system
Answer: $I^{O}_{zz}=\sum\limits_{i=1}^{n}m_{i}\left|{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{i}\right|^{2}$ and
$I^{cm}_{zz}=\sum\limits_{i=1}^{n}m_{i}\left|{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{i}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}\right|% ^{2}$
(d)

A uniform line segment with length $\ell$ and mass $m$ with center of mass at location ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/0}}$ about
1. i)
  
  Its center of mass G
2. ii)
  
  One end
3. iii)
  
  The origin
Answer: $I^{G}_{zz}=\frac{ml^{2}}{12}$ , $I^{A}_{zz}=\frac{ml^{2}}{3}$ , and
$I^{O}_{zz}=\frac{ml^{2}}{12}+m\left|{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{G/O}\right|^{2}$ .
(e)
A uniform hoop with mass $m$ and radius $R$ and center at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/0}}$ about
1. i)
  
  Its center of mass G
2. ii)
  
  A point on the hoop
3. iiii)
  
  The origin
Answer: $I^{G}_{zz}=mR^{2}$ , $I^{P}_{zz}=2mR^{2}$ , and
$I^{O}_{zz}=I^{G}_{zz}+m\left|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/% O}\right|^{2}$ .
(f)

A uniform disk with mass $m$ and radius $R$ and center at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/0}}$ about
1. i)
  
  Its center of mass G
2. ii)
  
  A point on the perimeter
3. iii)
  
  The origin
Answer: $I^{G}_{zz}=\frac{mR^{2}}{2}$ , $I^{P}_{zz}=\frac{3}{2}mR^{2}$ , and
$I^{O}_{zz}=I^{G}_{zz}+m\left|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/% O}\right|^{2}$ .
(g)

A uniform rectangle with mass $m$ and sides $\ell$ and $w$ and with center at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/0}}$ about
1. i)
  
  Its center of mass G
2. ii)
  
  A corner
3. iii)
  
  The origin
Answer: $I^{G}_{zz}=\frac{m}{12}\left(l^{2}+w^{2}\right)$ , $I^{P}_{zz}=\frac{m}{3}\left(l^{2}+w^{2}\right)$ , and
$I^{O}_{zz}=I^{G}_{zz}+m\left|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/% O}\right|^{2}$ .

16.3.2 Two objects have masses $m_{1}$ and $m_{2}$ and polar moments of inertia $I_{1}$ and $I_{2}$ about their respective centers of mass G₁ and G₂ which are a distance $d$ apart. What is the moment of inertia of the system about its center of mass? Answer: $I^{G}_{zz}=I_{1}+I_{2}+\left(\frac{m_{1}m_{2}}{m_{1}+m_{2}}\right)d^{2}$ .

16.3.3 A point mass $m=0.5\,\mathrm{kg}$ is located at $x=0.3\,\mathrm{m}$ and $y=0.4\,\mathrm{m}$ in the $x y$ -plane. Find the moment of inertia of the mass about the $z$ -axis. Answer: $I_{zz}=0.125\,\mathrm{kg}\cdot{\,\mathrm{m}}^{2}$ .

16.3.4 A small ball of mass $0.2\,\mathrm{kg}$ is attached to a $1\,\mathrm{m}$ massless rod.

(a)

What is the value of $I_{zz}$ of the ball about the other end of the rod? Answer: $0.2\,\mathrm{kg}\cdot{\,\mathrm{m}}^{2}$ .
(b)

How much must you shorten the rod to reduce the moment of inertia of the ball by half? Answer: $0.29\,\mathrm{m}$ .

More-Involved Problems

16.3.5 Two identical point masses are attached to the two ends of a rigid massless bar of length $\ell$ (one mass at each end). Locate a point along the length of the bar about which the polar moment of inertia of the system is $20\%$ more than that calculated about the mid point of the bar. Answer: At $0.72\ell$ from either end

16.3.6 A dumbbell consists of a rigid massless bar of length $\ell$ and two identical point masses $m$ and $m$ , one at each end of the bar.

(a)

About which point on the dumbbell is its polar moment of inertia $I_{zz}$ a minimum and what is this minimum value? Answer: $(I_{zz})_{min}=m\ell^{2}/2$ , about the midpoint.
(b)

About which point on the dumbbell is its polar moment of inertia $I_{zz}$ a maximum and what is this maximum value? Answer: $(I_{zz})_{max}=m\ell^{2}$ , about either end

16.3.7 A light rigid rod $A B$ of length $3\ell$ has a point mass $m$ at end $A$ and a point mass $2m$ at end $B$ . Point $C$ is the center of mass of the system. First, answer the following questions without any calculations and then do calculations to verify your guesses.

(a)

About which point $A$ , $B$ , or $C$ , is the polar moment of inertia $I_{zz}$ of the system a minimum? Answer: $C$
(b)

About which point is $I_{zz}$ a maximum? Answer: $A$
(c)

What is the ratio of $I_{zz}^{A}$ and $I_{zz}^{B}$ ? Answer: $I_{zz}^{A}/I_{zz}^{B}=2$
(d)

Is the radius of gyration of the system greater, smaller, or equal to the length of the rod? Answer: smaller, $r_{gyr}=\sqrt{I_{zz}^{C}/(3m)}=\sqrt{2}\ell$

Filename:pfigure-blue-112-1

Figure 16.63:

16.3.8 Three identical particles of mass $m$ are connected to three identical massless rods of length $\ell$ and welded together at point $O$ as shown in the figure.

(a)

Guess (no calculations) which of the three moment of inertia terms $I^{O}_{xx}$ , $I^{O}_{yy}$ , $I^{O}_{zz}$ is the smallest and which is the biggest. Answer: Biggest: $I^{O}_{zz}$ ; smallest: $I^{O}_{yy}=I^{O}_{xx}$ .
(b)

Calculate the three moments of inertia to check your guess. Answer: $\displaystyle I^{O}_{xx}=\frac{3}{2}m\ell^{2}=I^{O}_{yy}$ . $I^{O}_{zz}=3m\ell^{2}$ .
(c)

If the orientation of the system is changed, so that one mass is along the $x$ -axis, will your answer to part (a) change? Answer: Answer does not change.
(d)

Find the radius of gyration of the system for the polar moment of inertia. Answer: $r_{gyr}=\ell$ .

Filename:tfigure8-alt-app2c

Figure 16.64:

16.3.9 Show that the polar moment of inertia $I^{O}_{zz}$ of the uniform bar of length $\ell$ and mass $m$ , shown in the figure, is $\frac{1}{3}m\ell^{2}$ , in two different ways:

(a)

by using the basic definition of polar moment of inertia $\displaystyle I^{O}_{zz}=\int r^{2}\ dm$ , and
(b)

by computing $I^{cm}_{zz}$ first and then using the parallel axis theorem.

Filename:sfig4-6-4a

Figure 16.65:

16.3.10 Approximately locate the center of mass of the tapered rod shown in the figure and compute the polar moment of inertia $I^{cm}_{zz}$ . [Hint: use the variable thickness of the rod to define an approximate equivalent variable mass density per unit length.] Answer: $x_{cm}=40\,{\rm mm}$ and $I^{cm}_{zz}=733.3\,\mathrm{kg}\cdot\,{\rm mm}^{2}$ .

Filename:sfig4-6-3

Figure 16.66:

16.3.11 A short rod of mass $m$ and length $h$ hangs from an inextensible and massless rod of length $\ell$ .

(a)

Find the moment of inertia $I^{O}_{zz}$ of the rod. Answer: $I^{O}_{zz}=\frac{mh^{2}}{12}+m\left(\ell+\frac{h}{2}\right)^{2}$ .
(b)

Find the moment of inertia of the rod $I^{O}_{zz}$ by considering it as a point mass located at its center of mass. Answer: $I^{O}_{zz}=m\left(\ell+\frac{h}{2}\right)^{2}$ .
(c)

Find the percent error in $I^{O}_{zz}$ in treating the bar as a point mass by comparing the expressions in parts (a) and (b). Plot the percent error versus $h/\ell$ . For what values of $h/\ell$ is the percentage error less than $5\%$ ? Answer: Percent error in $I^{O}_{zz}$ is $100\left(1+12(\frac{\ell}{h}+0.5)^{2}\right)^{-1}$ . The percentage error less than $5\%$ if $h/\ell<1.32$ .

Filename:sfig4-6-3a

Figure 16.67:

16.3.12 Parallel axis theorem? A massless square plate $A B C D$ has four identical point masses located at its corners.

(a)

Find the polar moment of inertia $I^{cm}_{zz}$ . Answer: $I^{cm}_{zz}=2m\ell^{2}$
(b)

Find a point $P$ on the plate about which the system’s moment of inertia $I_{zz}$ is maximum. Answer: $P\equiv A,\;B,\;C,$ or $D$ , often, but not always, diagonally opposite the corner with the largest mass
(c)

Find the radius of gyration of the system about the rectangle center. Answer: $r_{gyr}=\ell/\sqrt{2}$

Filename:pfigure4-4-rp12

Figure 16.68:

16.3.13 For the massless square plate with four point masses on the corners, the polar moment of inertia $I^{cm}_{zz}=0.6\,\mathrm{kg}\cdot\,\mathrm{m}^{2}$ . Find $I^{cm}_{xx}$ of the system. Answer: $I^{O}_{xx}=I^{O}_{yy}=0.3\,\mathrm{kg}\cdot\,\mathrm{m}^{2}$ .

Filename:pfigure-blue-38-2

Figure 16.69:

16.3.14 A uniform square plate ( $2\,\mathrm{m}$ on edge) has a corner cut out. The total mass of the remaining plate is 3 $\,\mathrm{kg}$ .

(a)

Where is the center of mass of the plate at the instant shown? Answer: $x_{cm}=1.67\,\mathrm{m},y_{cm}=0.83\,\mathrm{m}$ .
(b)

What is the moment of inertia of the plate about point O? Answer: $I^{O}_{zz}=8\,\mathrm{kg}\cdot\,\mathrm{m}^{2}$ .
(c)

What is the moment of inertia of the plate about its center of mass? Answer: $I^{cm}_{zz}=1.83\,\mathrm{kg}\cdot\,\mathrm{m}^{2}$ .

Filename:pfigure-blue-49-2

Figure 16.70:

16.3.15 A uniform thin triangular plate of mass $m$ , height $h$ , and base $b$ lies in the $x y$ -plane.

(a)

Set up the integral to find the polar moment of inertia $I^{O}_{zz}$ of the plate. Answer: $\displaystyle I^{O}_{zz}=\frac{2m}{bh}\int_{0}^{b}\int_{0}^{hx/b}(x^{2}+y^{2})% \ dy\ dx$ .
(b)

Show that $\displaystyle I^{O}_{zz}=\frac{m}{6}(h^{2}+3b^{2})$ by evaluating the integral in part (a).
(c)

Locate the center of mass of the plate and calculate $I_{zz}^{cm}$ . Answer: $x_{cm}=\frac{2b}{3},y_{cm}=\frac{h}{3}$ .

Filename:pfigure4-4-rp13

Figure 16.71:

16.3.16 A uniform thin plate of mass $m$ is cast in the shape of a semi-circular disk of radius $R$ as shown in the figure.

(a)

Find the location of the center of mass of the plate Answer: $x_{cm}=0,y_{cm}=\frac{4R}{3\pi}$ .
(b)

Find the polar moment of inertia of the plate, $I^{cm}_{zz}$ . [Hint: It may be easier to set up and evaluate the integral for $I^{O}_{zz}$ and then use the parallel axis theorem to calculate $I^{cm}_{zz}$ .] Answer: $I^{cm}_{zz}=\frac{(9\pi^{2}-31)mR^{2}}{18\pi^{2}}$ .

Filename:f92h7p1

Figure 16.72:

16.3.17 A uniform square plate of side $\ell=250\,{\rm mm}$ has a circular cut-out of radius $r=50\,{\rm mm}$ . The mass of the plate is $\displaystyle m=\frac{1}{2}\,\mathrm{kg}$ .

(a)

Find the polar moment of inertia of the plate about its center. Answer: $I^{cm}_{zz}=\frac{m\ell^{2}}{6}\left(\frac{1-3\pi(r/\ell)^{4})}{1-\pi(r/\ell)^% {2}}\right)$ .
(b)

Plot $I^{cm}_{zz}$ versus $r/\ell$ .
(c)

Find the limiting values of $I_{zz}^{cm}$ for $r=0$ and $r=\ell$ . Answer: $\displaystyle\lim_{r\to 0}I_{zz}^{cm}=\frac{m\ell^{2}}{6}$ and
$\displaystyle\lim_{r\to\ell}I_{zz}^{cm}=\frac{m\ell^{2}}{6}\left(\frac{3\pi-1}% {\pi-1}\right)$ .

Filename:pfigure-blue-52-2

Figure 16.73:

16.3.18 A uniform thin circular disk of radius $r=100\,{\rm mm}$ and mass $m=2\,\mathrm{kg}$ has a rectangular slot of width $w=10\,{\rm mm}$ cut into it as shown in the figure. You can approximate the right edge of the cutout as a straight line (thus your calculation would be exact for a system that has a very thin sliver of material closing the gap at the right).

(a)

Find the polar moment of inertia $I^{O}_{zz}$ of the disk. Answer: $\displaystyle I^{O}_{zz}=0.01010904\,\mathrm{kg}\cdot\,\mathrm{m}^{2}$ .
(b)

Locate the center of mass of the disk and calculate $I^{cm}_{zz}$ . Answer: $x_{cm}=-1.64\,{\rm mm}$ , $y_{cm}=0\,{\rm mm}$ , and $I^{cm}_{zz}=0.01010364\,\mathrm{kg}\cdot\,\mathrm{m}^{2}$ .
(c)

estimate the error in your calculation of the mass, center of mass location and moment of inertia due to not subtracting out the sliver of material at the right edge of the slot. Answer: Not subtracting out the sliver of material at the right edge of the slot leads to an error of $0.053\%$ .

Filename:pfigure4-4-rp16

Figure 16.74:

16.3.19 Calculate the polar moment of inertia of a uniform square plate with mass $m$ and side $\ell$ various ways. Choose numerical values for $m$ and $\ell$ if you want to address this as a numerical calculation rather than a theoretical one.

(a)

As a single uniform plate
(b)

As a composite of 4 plates, each with sides $\ell/2$
(c)

As a composite of 100 plates, each with sides $\ell/10$ (These first three methods should give exactly the same answer).
(d)

For the composite of 100 plates now neglect the terms which concern the moments of inertia of each small square about its center of mass. What is the error in making this neglect? Answer: Exactly 1% .

16.4 Dynamics of a rigid object in planar circular motion

We now want to find what we can about the forces on and motions of a single planar object that is hinged at one point. The most famous examples are pendula and gears. But many things move like pendula, like people or towers falling over or boats rocking in the water. And many machine parts besides gears rotate about a bearing including propellers, lawn mower blades, and printing-press rollers. None of these things are strictly planar objects, but, for many analyses, the planar approximation is more or less accurate

^†^†margin: Accuracy of planar analysis. The planar analysis might be accurate for one of two different reasons: 1) the objects are nearly planar, or 2) the projection of the 3D equations to 2D gives the planar equations.

The basic method, as always, is to use a free-body diagram and kinematics results to evaluate terms in the linear momentum, angular momentum and energy balance equations. Then these equations are used to calculate unknown forces or to find differential equations of motion. The terms involving force and moment are found from a free-body diagram exactly as in statics. The terms involving motion, namely momentum, angular momentum and energy, are evaluated using the various tools developed earlier in this chapter.

Mechanics and the motion quantities

We can evaluate all the momentum and energy terms in the equations of motion (inside cover), namely: ${\accentset{\rightharpoonup}{\boldsymbol{L}}},\dot{\accentset{\rightharpoonup}% {\boldsymbol{L}}},{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/C}},% \dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{}_{\rm/C}},{E_{\mathrm{K}}}$ and ${\dot{E}_{\mathrm{K}}}$ (for any reference point C of our choosing) if we can calculate the velocity and acceleration of every point in the system. For circular motion of a rigid object about point C, we know from sec. 16.2 that the velocities and accelerations of a point at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{/C}}$ are

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}),$
	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$

where ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is the angular velocity of the object relative to a fixed frame. This situation is a little more complex than the special case of straight-line motion in chapter 14, where all points in a system had the same acceleration as each other, but is still quite manageable.

The most general case of 2D rigid-object circular motion is an arbitrarily shaped 2D rigid object with arbitrary $\omega$ and $\dot{\omega}$ . The situation shown in fig. 16.75 shows the general case (but in that example exactly two forces are applied, as opposed to an arbitrary number).

Filename:pfigure4-4-rp14

Figure 16.75: a) A rigid object rotates about 0. It has center of mass at G. Forces

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}

and

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}

are applied. b) The free-body diagram shows all the forces acting on the object, including the reaction force (

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{0}=F_{0x}\hat{\boldsymbol{% \imath}}+F_{0y}\hat{\boldsymbol{\jmath}}

) at 0 and the gravity force

m g

at G.

Linear momentum: ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$

For any system in any motion we have the general result that

{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}\hbox to36.135pt{\hfil and\hfil% }\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset% {\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}.

For a rigid object, the center-of-mass is a particular point G that is fixed relative to the object. So the velocity and acceleration of that point can be expressed the same way as for any other point. So, for an object in planar rotational motion about 0

{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{G/0}}

and

\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}\underbrace{% \left(\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/0}}-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/0}}\right)}_{{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\textrm{G}}}.

If the center-of-mass is at 0 the momentum and its rate of change are both zero. But if the center-of-mass is off the axis of rotation, there must be a net force on the object with a component parallel to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0/G}}$ (if $\omega\neq 0$ ) and also a component orthogonal to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0/G}}$ (if $\dot{\omega}\neq 0$ ). This net force might be applied at 0 or G or any other place(s) on the object.

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}$}}}$ 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}$}}}$

The angular momentum itself is easy enough to calculate, using the short hand notation that ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is the position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/0}}$ of a point relative to hinge point O.

\begin{array}[]{rclrc}\displaystyle\par{\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\int_{\rm all\ mass}{{\accentset{\rightharpoonup}{\boldsymbol% {r}}}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}}\,dm&\mbox{(a)}\\ &=&\displaystyle\int{{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times\left(% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}\right)}\,dm&\mbox{(b)}\\ &=&\displaystyle\omega\hat{\boldsymbol{k}}\int r^{2}\,dm&\mbox{(c)}\\[10.0pt] &=&\underbrace{I^{O}_{zz}}_{\text{$I^{O}_{zz}$ is the `polar' moment of % inertia.}}\omega\hat{\boldsymbol{k}}\\[55.0pt] \hbox to36.135pt{\hfil$\Rightarrow$\hfil}H_{0}&=&\displaystyle\omega I^{O}_{zz% }&\mbox{(d)}\\ \end{array}

(16.50)

Here eqn. (16.50)c is a vector equation. But since both sides are in the $\hat{\boldsymbol{k}}$ direction we can dot both sides with $\hat{\boldsymbol{k}}$ to get the scalar moment equation eqn. (16.50)d, taking both $M_{\rm net}$ and $\omega$ as positive when counterclockwise.

The moment of inertia shortcut. Note that an integral over all the mass of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\times{\accentset{\rightharpoonup% }{\boldsymbol{v}}}$ has been reduced to a simple multiplication by using that

\int r^{2}\,dm=I^{0}_{zz}.

This shortcut is especially useful because $\int r^{2}\,dm$ does not change as the object rotates about 0.

However, the moment of inertia is never needed for problem solving. One can just evaluate, for example, ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\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 evaluating the integral $\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}\,dm$ for the specific problem at hand.

To get the all important term for angular momentum balance, namely $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\textrm{/0}}$ , for this system we can differentiate eqn. (16.50). We can also use the general expression 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}$}}}$ to write the angular momentum balance equation as follows.

\begin{array}[]{rclrc}\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{$\scriptscriptstyle\mathrm{/O}$}}% }&=&\mbox{rate of change of angular momentum}_{/0}&\mbox{(a)}\\ &=&\displaystyle\int_{\rm all\ mass}{{\accentset{\rightharpoonup}{\boldsymbol{% r}}}\times{\accentset{\rightharpoonup}{\boldsymbol{a}}}}\,dm&\mbox{(b)}\\[10.0% pt] &=&\displaystyle\int{{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times\left(% \underbrace{-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}+\dot{% \omega}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}% }_{{\accentset{\rightharpoonup}{\boldsymbol{a}}}}\right)}\,dm&\mbox{(c)}\\ &=&\displaystyle\int{{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times\left(% \dot{\omega}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol% {r}}}\right)}\,dm\quad(\mbox{because }{\accentset{\rightharpoonup}{\boldsymbol% {r}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}})&\mbox{(d)}\\ &=&\displaystyle\int{{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times\left(% \dot{\omega}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol% {r}}}\right)}\,dm&\mbox{(e)}\\ \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\dot{\omega}\hat{% \boldsymbol{k}}\int r^{2}\,dm\ =\ \dot{\omega}I^{0}_{zz}\hat{\boldsymbol{k}}&% \mbox{(f)}\\[9.0pt] \hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{H}_{/0}&=&\displaystyle\dot{% \omega}\int r^{2}\,dm\ =\ \dot{\omega}I^{0}_{zz}&\mbox{(g)}\\ \end{array}

(16.51)

We get from eqn. (16.51)e to eqn. (16.51)f by noting that ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is perpendicular to $\hat{\boldsymbol{k}}$ . Thus, using the right hand rule twice we get ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\times(\hat{\boldsymbol{k}}\times% {\accentset{\rightharpoonup}{\boldsymbol{r}}})=r^{2}\hat{\boldsymbol{k}}$ .

Eqn. 16.51f and eqn. (16.51)g are the vector and scalar versions for the rate of change of angular momentum with respect to point 0 for rotation of a planar object about 0. Repeating,

	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{k}}\int r^{2}\,dm\ =\ \dot{\omega}I^% {0}_{zz}\hat{\boldsymbol{k}}$
	$\displaystyle\qquad\mbox{and}\qquad\dot{H}_{/0}$	$\displaystyle=$	$\displaystyle\dot{\omega}\int r^{2}\,dm\ =\ \dot{\omega}I^{0}_{zz}.$		(16.52)

Again, $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\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 always be evaluated either of two ways: 1) as a sum or integral of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/0}}\times{\accentset% {\rightharpoonup}{\boldsymbol{a}}}$ over all the mass or 2) using the shortcut with $I^{0}_{zz}$ .

Power and Energy

Let’s assume that there are a set of point forces applied to an object

^†^†margin: Forces distributed on a curve, surface or throughout a volume could be treated similarly with sums replaced by line, surface, or volume integrals.

And, to be contrary, let’s assume the mass is continuously distributed (the derivation for rigidly connected point masses would be similar). The power balance equation for one rotating rigid object is (discussed below):

\begin{array}[]{rclrcccccc}\displaystyle\mbox{Net power in}&=&\mbox{rate of % change of kinetic energy}&\mbox{(a)}\\ P&=&{\dot{E}_{\mathrm{K}}}&\mbox{(b)}\\ \displaystyle\sum_{\rm all\ applied\ forces}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{i}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}&=&% \displaystyle{d\over dt}\int_{\rm all\ mass}{{1\over 2}v^{2}}\,dm&\mbox{(c)}\\ \displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}\cdot\left({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{i}\right)&=&\displaystyle{d\over dt}\int{{1% \over 2}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}})\cdot({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}})}\,dm% &\mbox{(d)}\\ \displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot({% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i})&=&\displaystyle{d\over dt}\int{1\over 2% }\par\omega^{2}r^{2}\,dm&\mbox{(e)}\\ \displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot\sum({% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i})&=&\displaystyle{d\over dt}\left({1\over 2% }\omega^{2}\right)\int\par r^{2}\,dm&\mbox{(f)}\\ \displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot\sum{% \accentset{\rightharpoonup}{\boldsymbol{M}}}_{i}&=&\displaystyle\dot{\omega}% \omega\int r^{2}\,dm&\mbox{(g)}\\ \displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{M}}}_{\rm tot}&=&\displaystyle\dot{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot\underbrace{\left({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\int r^{2}\,dm\right)}_{{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/0}}}&\mbox{(h)}\\ \end{array}

(16.53)

When not directly labeled, positions and moments are assumed to be relative to the hinge at 0. Derivation 16.53 is two derivations in one. The left side about power and the right side about kinetic energy. Let’s discuss one at a time.

Power. On the left side of eqn. (16.53) we note in (c) that the power of each force is the dot product of the force with the velocity of the point it touches. In (d) we use what we know about the velocities of points on rotating rigid bodies. In (e) we use the vector identity ${\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{\boldsymbol{C}}}={% \accentset{\rightharpoonup}{\boldsymbol{B}}}\cdot{\accentset{\rightharpoonup}{% \boldsymbol{C}}}\times{\accentset{\rightharpoonup}{\boldsymbol{A}}}$ (see sec. 1.3, 1.3). In (f) we note that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is common to all points so factors out of the sum. In (g) we note that ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\times{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{i}$ is the moment of the force about pt O. And in (g) we sum the moments of the forces. So the power of a set of forces acting on a rigid object is the product of their net moment (about 0) and the object angular velocity,

P={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\rm tot}.

(16.54)

Kinetic energy. On the right side of eqn. (16.53) we note in (c) that the kinetic energy is the sum of the kinetic energy of the mass increments. In (d) we use what we know about the velocities of these bits of mass, given that they are on a common rotating object. In (e) we use that the magnitude of the cross product of orthogonal vectors is the product of the magnitudes ( $|{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{B}}}|=AB$ ) and that the dot product of a vector with itself is its magnitude squared ( ${\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{A}}}=A^{2}$ ). In (f) we factor out $\omega^{2}$ because it is common to all the mass increments and note that the remaining integral is constant in time for a rigid object. In (g) we carry out the derivative. In (h) we de-simplify the result from (g) in order to show a more general form that we will find later in 3D mechanics. Eqn. (h) follows from (g) because ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is parallel to $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ for 2D rotations.

Note that we started here with the basic power balance equation from the front inside cover. Instead, we could have derived power balance from our angular momentum balance expression (see box 16.4 on 16.4).

Box 16.8 The relation between angular momentum balance and power balance

For this system, angular momentum balance can be derived from power balance and vice versa. Thus neither is essentially more fundamental than the other and both are reliable. First we can derive power balance from angular momentum balance as follows:

\begin{array}[]{rclccc}{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{% net}}&=&\dot{\omega}\hat{\boldsymbol{k}}\int r^{2}\,dm\\ {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{net}}&=&{\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}\cdot\left(\dot{\omega}\hat{\boldsymbol{k}}\int r^{2}\,% dm\right).\\ P&=&{\dot{E}_{\mathrm{K}}}\end{array}

(16.55)

That is, when we dot both sides of the angular momentum equation with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ we get on the left side a term which we recognize as the power of the forces and on the right side a term which is the rate of change of kinetic energy.

The opposite derivation starts with the power balance fig. 16.53(g)

\begin{array}[]{rclr}\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\cdot\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{i}&=&% \displaystyle\dot{\omega}\omega\int r^{2}\,dm&\mbox{(g)}\\ \hbox to36.135pt{\hfil$\Rightarrow$\hfil}\displaystyle\omega\left(\hat{% \boldsymbol{k}}\cdot\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{i}% \right)&=&\displaystyle\dot{\omega}\omega\int r^{2}\,dm\\ \hbox to36.135pt{\hfil$\Rightarrow$\hfil}\displaystyle\left(\hat{\boldsymbol{k% }}\cdot\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{i}\right)&=&% \displaystyle\dot{\omega}\int r^{2}\,dm\\ \end{array}

(16.56)

and, assuming $\omega\neq 0$ , divide by $\omega$ to get the angular momentum equation for planar rotational motion.

Example: Spinning disk

The round flat uniform disk in fig. 16.76 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:

Filename:pfigure4-4-rp15

Figure 16.76: A uniform disk turned by a motor at a constant rate. The free-body diagram shows a force and moment equivalent to the force system that acts including gravity, bearing forces, etc. A bit of mass

d m

occupies the space between the radial lines at

\theta

and

\theta+d\theta

and between the circles at radii

R

and

R+dR

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

So the net force and moment needed are ${\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ . Like a particle that moves at constant velocity with no force, a uniform disk rotates at constant rate with no torque (at least in 2D).

Using moment-of-inertia in 2-D circular motion dynamics

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 earlier chapters. 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. As you recall from earlier chapters, the 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 wishes for something so simple for the calculation of angular momentum.

It turns out that we are in luck if we are only interested in the two-dimensional motion of two-dimensional rigid bodies. The luck is not so great for 3-D rigid bodies but still there is some simplification. For general motion of non-rigid bodies there is no simplification to be had.

The simplification is to use the moment of inertia for the bodies rather than evaluating the momenta and energy quantities as integrals and sums. Of course one may have to do a sum or integral to evaluate $I\equiv I^{cm}_{zz}$ or $[\boldsymbol{I}^{\mathrm{cm}}]$ but once this calculation is done, one need not work with the integrals while worrying about the dynamics. At this point we will assume that you are comfortable calculating and looking-up moments of inertia. We proceed to use it for the purposes of studying mechanics.

For constant rate rotation, we can calculate the velocity and acceleration of various points on a rigid body using ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}})$ . So we can calculate the various motion quantities of interest: linear momentum ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ , rate of change of linear momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ , angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}$ , rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ , and kinetic energy ${E_{\mathrm{K}}}$ .

Consider a two-dimensional rigid body like that shown in fig. 16.77.

Filename:tfigure4-spherical-rotaxis

Figure 16.77: A two-dimensional body is rotating around the point O at constant rate

\omega

. A differential bit of mass

d m

is shown. The center-of-mass is also shown.

Now let us consider the various motion quantities in turn. First the linear momentum ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ . The linear momentum of any system in any motion is ${\accentset{\rightharpoonup}{\boldsymbol{L}}}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{cm}m_{tot}$ . So, for a rigid body spinning at constant rate $\omega$ about point O (using ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ ):

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

Similarly, for any system, we can calculate the rate of change of linear momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ as $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{cm}m_{tot}$ . So, for a rigid body spinning at constant rate,

\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{cm}m_{tot}={\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{cm/o})m_{tot}.

That is, the linear momentum is correctly calculated for this special motion, as it is for all motions, by thinking of the body as a point mass at the center-of-mass.

Unlike the calculation of linear momentum, the angular momentum turns out to be something different than would be calculated by using a point mass at the center of mass. You can remember this important fact by looking at the case when the rotation is about the center-of-mass (point O coincides with the center-of-mass). In this case one can intuitively see that the angular momentum of a rigid body is not zero even though the center-of-mass is not moving. Here’s the calculation just to be sure:

\begin{array}[]{rclp{20 pt}}{\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&\hbox{(by 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{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{/O})\,dm&\hbox{(using ${\accentset{\rightharpoonup}{% \boldsymbol{v}}}={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}$)}\\ &=&\int(x_{/O}\hat{\boldsymbol{\imath}}+y_{/O}\hat{\boldsymbol{\jmath}})\times% \left[(\omega\hat{\boldsymbol{k}})\times(x_{/O}\hat{\boldsymbol{\imath}}+y_{/O% }\hat{\boldsymbol{\jmath}})\right]\,dm&\mbox{(substituting ${\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{/O}$ and ${\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}$)}\\ &=&\{\int(x_{/O}^{2}+y_{/O}^{2})\,dm\}\omega\hat{\boldsymbol{k}}&\hbox{(doing % cross products)}\\ &=&\{\int r_{/O}^{2}\,dm\}\omega\hat{\boldsymbol{k}}\\[10.0pt] &=&\underbrace{I^{O}_{zz}}_{\text{$I^{O}_{zz}$ is the `polar' moment of % inertia.}}\omega\hat{\boldsymbol{k}}\end{array}

We have defined the ‘polar’ moment of inertia as $I^{o}_{zz}=\int r_{/o}^{2}\,dm$ . In order to calculate $I^{o}_{zz}$ for a specific body, assuming uniform mass distribution for example, one must convert the differential quantity of mass $d m$ into a differential of geometric quantities. For a line or curve, $dm=\rho d\ell$ ; for a plate or surface, $dm=\rho dA$ , and for a 3-D region, $dm=\rho dV$ . $d\ell$ , $d A$ , and $d V$ are differential line, area, and volume elements, respectively. In each case, $\rho$ is the mass density per unit length, per unit area, or per unit volume, respectively. To avoid clutter, we do not define a different symbol for the density in each geometric case. The differential elements must be further defined depending on the coordinate systems chosen for the calculation; e.g., for rectangular coordinates, $dA=dxdy$ or, for polar coordinates, $dA=rdrd\theta$ .

Since ${\accentset{\rightharpoonup}{\boldsymbol{H}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ always point in the $\hat{\boldsymbol{k}}$ direction for two dimensional problems people often just think of angular momentum as a scalar and write the equation above simply as ‘ $H=I\omega$ ,’ the form usually seen in elementary physics courses.

The derivation above has a feature that one might not notice at first sight. The quantity called $I^{\scriptscriptstyle\mathrm{O}}_{zz}$ does not depend on the rotation of the body. That is, the value of the integral does not change with time, so $I^{\scriptscriptstyle\mathrm{O}}_{zz}$ is a constant. So, perhaps unsurprisingly, a two-dimensional body spinning about the $z$ -axis through $O$ has constant angular momentum about $O$ if it spins at a constant rate. ^†^†margin: Note that the angular momentum about some other point than O will not be constant unless the center-of-mass does not accelerate (i.e., is at point 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{0% }}}.

Now, of course we could find this result about constant rate motion of 2-D bodies somewhat more cumbersomely by plugging in the general formula for rate of change of angular momentum as follows:

\begin{array}[]{lll}\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{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{/O}\times{\accentset{\rightharpoonup}{% \boldsymbol{a}}}\,dm\\ &=\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O})% )\,dm\\ &=\int(x_{/O}\hat{\boldsymbol{\imath}}+y_{/O}\hat{\boldsymbol{\jmath}})\times% \left[\omega\hat{\boldsymbol{k}}\times(\omega\hat{\boldsymbol{k}}\times(x_{/O}% \hat{\boldsymbol{\imath}}+y_{/O}\hat{\boldsymbol{\jmath}}))\right]\,dm\\ &={\accentset{\rightharpoonup}{\boldsymbol{0}}}.\\ \end{array}

(16.57)

Using moment of inertia about the center of mass. Often it is easier to think of the motion as composed of two parts, motion of the center of mass and motion relative to the center of mass, as explained in box 16.4 on page 16.4. Thus we have two terms for 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}$}}}$ 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}$}}}$ :

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times m{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}+I_{zz}^{cm}\omega\hat{% \boldsymbol{k}}$		(16.58)
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}+I_{zz}^{cm}\alpha\hat{% \boldsymbol{k}}$		(16.59)

Kinetic energy. Finally, we can calculate the kinetic energy by adding up $\frac{1}{2}m_{i}v_{i}^{2}$ for all the bits of mass on a 2-D body spinning about the $z$ -axis:

{E_{\mathrm{K}}}=\int{1\over 2}v^{2}\,dm=\int{1\over 2}(\omega r)^{2}\,dm={1% \over 2}\omega^{2}\int r^{2}\,dm={1\over 2}I^{o}_{zz}\omega^{2}\quad.

(16.60)

The kinetic energy can also be written as a sum of contributions of motion of the center of mass and motion relative to the center of mass,

{E_{\mathrm{K}}}={1\over 2}mv_{cm}^{2}+{1\over 2}I^{cm}_{zz}\omega^{2}\quad.

(16.61)

Example: Pendulum disk

Filename:tfigure5-7

Figure 16.78:

For the disk shown in fig. 16.78, we can calculate the rate of change of angular momentum about point $O$ 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}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}+I_{zz}^{cm}\alpha\hat{% \boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle R^{2}m\ddot{\theta}\hat{\boldsymbol{k}}+I_{zz}^{cm}\ddot{\theta}% \hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(I_{zz}^{cm}+R^{2}m)\ddot{\theta}\hat{\boldsymbol{k}}.$

Alternatively, we could calculate directly

	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle I^{O}_{zz}\alpha\hat{\boldsymbol{k}}$
		$\displaystyle=$	$\displaystyle\underbrace{(I_{zz}^{cm}+R^{2}m)}_{\text{by the parallel axis % theorem}}\ddot{\theta}\hat{\boldsymbol{k}}.$

Note that we are using the planarity of the objects and of their motion for our calculations

^†^†margin: 2D vs 3D. Beware of falling into the common misconception that the formula

M=I\alpha

applies in three dimensions by just thinking of the scalars as vectors and matrices. In 3D the 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{\alpha}}}}

is only correct when

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

is zero or when

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

is an eigen vector of

[I_{/O}]

. That is, the vector equation

\sum\mbox{Moments about O}=[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]% \cdot{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}

is generally wrong in 3D.

The equation for linear momentum balance is the same as always, we just need to calculate the acceleration of the center-of-mass of the spinning body.

\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{tot}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{cm}=m_{tot}\left[{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/O% })+\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{cm/O}\right]

(16.62)

Finally, the kinetic energy for a planar rigid body rotating in the plane is:

{E_{\mathrm{K}}}={1\over 2}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \cdot([\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}})+{1\over 2}m\underbrace{v_{\rm cm}^{2}}_{\text{${% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{{% \rm cm/O}}$}}.

Box 16.9 Simplifying ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}$ using the center of mass

The definition of angular momentum relative to a point C is

{\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{$% \scriptscriptstyle\mathrm{/C}$}}}=\sum{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{i/C}\times m_{i}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}.

If we rewrite ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}$ as

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}=({\accentset{\rightharpoonup% }{\boldsymbol{v}}}_{i}-{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm})+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{i/cm}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}

and

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}=({\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{i}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm})+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}={\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{i/cm}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}

then

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle=$	$\displaystyle\sum\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/cm}\right)\times\left[{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}+{\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{i/cm}\right]m_{i}.$
	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}m_{i}+\sum{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{i/cm}\times{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{i/cm}m_{i}$
		$\displaystyle+\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{i/cm}m_{i}+\sum{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{i/cm}\times{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{cm}m_{i}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}m_{\mathrm{tot}}+\sum{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/cm}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{i/cm}\ m_{i}$
		$\displaystyle+\ {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}\times% \underbrace{\left[\sum{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i/cm}\,% \mathrm{m}_{i}\right]}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+% \underbrace{\left[\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/cm}\ m_% {i}\right]}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{cm}$

So,

{\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{$% \scriptscriptstyle\mathrm{/C}$}}}=\underbrace{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{cm}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}m% _{\mathrm{tot}}}_{\text{contribution of center of mass motion}}\ +\ \underbrace{\sum{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{i\mathrm{/cm}}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i/cm}% \ m_{i}}_{\text{contribution of motion relative to center-of-mass}}.

The reason $\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/cm}\ m_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ is somewhat intuitive. It is what you would calculate if you were looking for the center-of-mass relative to the center of mass. More formally,

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/cm}\ m_{i}$	$\displaystyle=$	$\displaystyle\sum({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}-{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm})\ m_{i}$
	$\displaystyle=$	$\displaystyle\underbrace{\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}% \ m_{i}}_{m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}}-% m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Similarly, $\sum{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i/cm}\ m_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ because it is what you would calculate if you were looking for the velocity of the center-of-mass relative to the center of mass.

The central result of this box is that

angular momentum of any system is that due to motion of the center-of-mass plus motion relative to the center-of-mass.

SAMPLE 16.13

Filename:tfigure5-gen-rigid-body

Figure 16.79: A rod goes in circles at a constant rate.

A rod in a constant rate circular motion: A uniform rod of mass $m$ and length $\ell$ is connected to a motor at end O. A ball of mass $m$ is attached to the rod at end B. The motor turns the rod in the counterclockwise direction at a constant angular speed $\omega$ . There is gravity pointing in the $-\hat{\boldsymbol{\jmath}}$ direction. Find the torque applied by the motor (i) at the instant shown and (ii) when $\theta=0^{\circ},\;90^{\circ},\;180^{\circ}$ . How does the torque change if the angular speed is doubled?

Solution

The FBD of the rod and ball system is shown in Fig. 16.80(a). Since the system is undergoing circular motion at a constant speed, the acceleration of the ball as well as every point on the rod is just radial (pointing towards the center of rotation O) and is given by ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\omega^{2}r\hat{\boldsymbol{% \lambda}}$ where $r$ is the radial distance from the center O to the point of interest and $\hat{\boldsymbol{\lambda}}$ is a unit vector along OB pointing away from O (Fig. 16.80(b)).

Filename:tfigure5-term1-a

Figure 16.80: A rod goes in circles at a constant rate.

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{$\scriptscriptstyle\mathrm{/O}$}}}

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times(-mg\hat% {\boldsymbol{\jmath}})+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/O}% \times(-mg\hat{\boldsymbol{\jmath}})+M\hat{\boldsymbol{k}}$	(16.63)
	$\displaystyle=$	$\displaystyle-\frac{\ell}{2}\cos\theta mg\hat{\boldsymbol{k}}-\ell\cos\theta mg% \hat{\boldsymbol{k}}+M\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(M-\frac{3\ell}{2}mg\cos\theta)\hat{\boldsymbol{k}}$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/O}% \times m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}}^{({\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{/O}$}}}})_{\rm ball}}+\overbrace{\int_{m}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{dm/O}\times{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{dm% }\,dm}^{({\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}})_{\rm rod}}$	(16.64)
	$\displaystyle=$	$\displaystyle\ell\hat{\boldsymbol{\lambda}}\times(-m\omega^{2}\ell\hat{% \boldsymbol{\lambda}})+\int_{m}\overbrace{s\hat{\boldsymbol{\lambda}}}^{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{dm/O}}\times\overbrace{(-\omega^% {2}s\hat{\boldsymbol{\lambda}})}^{{\accentset{\rightharpoonup}{\boldsymbol{a}}% }_{dm}}dm$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}\makebox[144.54pt][r% ]{(since $\hat{\boldsymbol{\lambda}}\times\hat{\boldsymbol{\lambda}}={{% \accentset{\rightharpoonup}{\boldsymbol{0}}}}$)}$

(i) Equating (16.63) and (16.64) we get

M=\frac{3}{2}mg\ell\cos\theta.

Answer: $M=\frac{3}{2}mg\ell\cos\theta$

(ii) Substituting the given values of $\theta$ in the above expression we get

M(\theta=0^{\circ})=\frac{3}{2}mg\ell,\quad M(\theta=90^{\circ})=0\quad M(% \theta=180^{\circ})=-\frac{3}{2}mg\ell

Answer: $M(0^{\circ})=\frac{3}{2}mg\ell,\quad M(90^{\circ})=0\quad M(180^{\circ})=-% \frac{3}{2}mg\ell$

^†^†margin: The values obtained above make sense (at least qualitatively). To make the rod and the ball go up from the

0^{\circ}

position, the motor has to apply some torque in the counterclockwise direction. In the

90^{\circ}

position no torque is required for the dynamic balance. In the

180^{\circ}

position the system is accelerating downwards under gravity; therefore, the motor has to apply a clockwise torque to make the system maintain a uniform speed.

It is clear from the expression of the torque that it does not depend on the value of the angular speed $\omega$ ! Therefore, the torque will not change if the speed is doubled. In fact, as long as the speed remains constant at any value, the only torque required to maintain the motion is the torque to counteract the moments due to gravity at O.

SAMPLE 16.14

Filename:tfigure5-term1-b

Figure 16.81: A rectangular plate is released from rest from the position shown.

At the onset of circular motion: A $2^{\prime}\times 4^{\prime}$ rectangular plate of mass $20\,{\mathrm{lbm}}$ is pivoted at one of its corners as shown in the figure. The plate is released from rest in the position shown. Find the force on the support immediately after release.

Solution The free-body diagram of the plate is shown in Fig. 16.82. The force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ applied on the plate by the support is unknown.

Filename:sfig1-2-12

Figure 16.82: (a) The free-body diagram of the plate. (b) Computation of the integral in

\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{\omega}\hat{\boldsymbol{k}}\int_{m}r^% {2}dm

. (c) The geometry of motion. From the given dimensions,

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\frac{a\hat{\boldsymbol{\imath}}-% b\hat{\boldsymbol{\jmath}}}{\sqrt{(a^{2}+b^{2})}}

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=\frac{b\hat{\boldsymbol{% \imath}}+a\hat{\boldsymbol{\jmath}}}{\sqrt{(a^{2}+b^{2})}}

, and

r_{G/O}=\frac{\sqrt{a^{2}+b^{2}}}{2}

The linear momentum balance for the plate gives

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}-mg\hat{\boldsymbol{% \jmath}}$	$\displaystyle=$	$\displaystyle m(\ddot{\theta}\,r_{G/O}\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta}-\dot{\theta}^{2}r_{G/O}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})$	(16.65)
	$\displaystyle=$	$\displaystyle m\,\ddot{\theta}\,r_{G/O}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\quad\mbox{(since $\dot{\theta}=0$ at $t=0$)}.$	(16.65)

Thus to find ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ we need to find $\ddot{\theta}$ .

The angular momentum balance for the plate about the fixed support point O gives

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{O}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm O}$
where
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{O}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times mg(-% \hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(\underbrace{\frac{a}{2}\hat{\boldsymbol{\imath}}-\frac{b}{2}\hat% {\boldsymbol{\jmath}}}_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}})% \times mg(-\hat{\boldsymbol{\jmath}})=-mg\frac{a}{2}\hat{\boldsymbol{k}},$
and
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{k}}\int_{m}r^{2}dm=\ddot{\theta}\hat% {\boldsymbol{k}}\int_{0}^{b}\int_{0}^{a}\overbrace{(x^{2}+y^{2})}^{r^{2}}% \overbrace{\frac{m}{ab}dxdy}^{dm}$
	$\displaystyle=$	$\displaystyle\frac{m(a^{2}+b^{2})}{3}\ddot{\theta}\hat{\boldsymbol{k}}.$
Thus,
$\displaystyle-mg\frac{a}{2}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\frac{m(a^{2}+b^{2})}{3}\ddot{\theta}\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{3ga}{2(a^{2}+b^{2})}$
	$\displaystyle=$	$\displaystyle-\frac{3\cdot 32.2\,{\mathrm{ft}}/{\mathrm{s}}^{2}\cdot 4\,% \mathrm{ft}}{2(16+4)\,\mathrm{ft}^{2}}=-9.66\,\mathrm{rad}/\mathrm{s}^{2}.$

From eqn. (16.65), the support force is now readily calculated:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle mg\hat{\boldsymbol{\jmath}}+m\ddot{\theta}\,r_{G/O}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle mg\hat{\boldsymbol{\jmath}}+m\ddot{\theta}\frac{\sqrt{a^{2}+b^{2% }}}{2}\frac{b\hat{\boldsymbol{\imath}}+a\hat{\boldsymbol{\jmath}}}{\sqrt{(a^{2% }+b^{2})}}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}m\ddot{\theta}b\hat{\boldsymbol{\imath}}+(mg+\frac{1}{% 2}m\ddot{\theta}a)\hat{\boldsymbol{\jmath}}$

Using the given numerical values of $m, a$ , and $b$ , $\ddot{\theta}=-9.66\,\mathrm{rad}/\mathrm{s}^{2}$ , and $g=32.2\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ , we get

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=(-6\hat{\boldsymbol{\imath}}+8% \hat{\boldsymbol{\jmath}})\,{\rm lbf}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=(-6\hat{\boldsymbol{\imath}}+8% \hat{\boldsymbol{\jmath}})\,{\rm lbf}$

SAMPLE 16.15

Filename:sfig4-6-5

Figure 16.83: A compound gear train.

A compound gear train. When the gear of an input shaft, often called the driver or the pinion, is directly meshed in with the gear of an output shaft, the motion of the output shaft is opposite to that of the input shaft. To get the output motion in the same direction as that of the input motion, an idler gear is used. If the idler shaft has more than one gear in mesh, then the gear train is called a compound gear train.

In the gear train shown in Fig. 16.83, the input shaft is rotating at 2000 rpm and the input torque is 200 N-m. The efficiency (defined as the ratio of output power to input power) of the train is 0.96 and the various radii of the gears are: $R_{A}=5\,\mathrm{cm},\,R_{B}=8\,\mathrm{cm},\,R_{C}=4\,\mathrm{cm},$ and $R_{D}=10\,\mathrm{cm}$ . Find

1.

the input power $P_{\rm in}$ and the output power $P_{\rm out}$ ,
2.

the output speed $\omega_{\rm out}$ , and
3.

the output torque.

Solution

The power:

$\displaystyle P_{\rm in}$	$\displaystyle=$	$\displaystyle M_{\rm in}\omega_{\rm in}=200\,{\rm N}\mathord{\cdot}{\rm m}% \cdot 2000\,{\rm rpm}$
	$\displaystyle=$	$\displaystyle 400000\,{\rm N}\mathord{\cdot}{\rm m}\cdot\frac{\cancel{\,\mbox{% rev}}}{\cancel{\,{\rm min}}}\cdot\frac{2\pi}{1\cancel{\,\mbox{rev}}}\cdot\frac% {1\cancel{\,{\rm min}}}{60\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 41887.9\,{\mbox{N}}\,\mathrm{m}/\mathrm{s}\approx 42\mbox{ kW}.$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}p_{\rm out}$	$\displaystyle=$	$\displaystyle\mbox{efficiency}\cdot P_{\rm in}=0.96\cdot 42\mbox{ kW}\approx 4% 0\mbox{ kW}$

Answer: $P_{\rm in}=42\mbox{ kW},\quad p_{\rm out}=40\mbox{ kW}$

Filename:sfig4-6-5a

Figure 16.84:

The angular speed of meshing gears can be easily calculated by realizing that the linear speed of the point of contact has to be the same irrespective of which gear’s speed and geometry is used to calculate it. Thus,

$\displaystyle v_{P}$	$\displaystyle=$	$\displaystyle\omega_{\rm in}\,R_{A}=\omega_{B}\,R_{B}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega_{B}$	$\displaystyle=$	$\displaystyle\omega_{\rm in}\cdot\frac{R_{A}}{R_{B}}$
$\displaystyle\hbox to36.135pt{and\hfil}v_{R}$	$\displaystyle=$	$\displaystyle\omega_{C}\,R_{C}=\omega_{\rm out}\,R_{D}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega_{\rm out}$	$\displaystyle=$	$\displaystyle\omega_{C}\cdot\frac{R_{C}}{R_{D}}$
$\displaystyle\mbox{But}\qquad\omega_{C}$	$\displaystyle=$	$\displaystyle\omega_{B}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega_{\rm out}$	$\displaystyle=$	$\displaystyle\omega_{\rm in}\cdot\frac{R_{A}}{R_{B}}\cdot\frac{R_{C}}{R_{D}}$
	$\displaystyle=$	$\displaystyle 2000\,{\rm rpm}\cdot\frac{5}{8}\cdot\frac{4}{10}=500\,{\rm rpm}.$

Answer: $\omega_{\rm out}=500\,{\rm rpm}$

3.

The output torque,

$\displaystyle M_{\rm out}=\frac{P_{\rm out}}{\omega_{\rm out}}$ $\displaystyle=$ $\displaystyle\frac{40\mbox{ kW}}{500\,{\rm rpm}}=\frac{40}{500}\cdot 1000\frac{\,{\rm N}\mathord{\cdot}{% \rm m}}{\cancel{\,\mathrm{s}}}\cdot\frac{\cancel{\,{\rm min}}}{\cancel{\,\mbox% {rev}}}\cdot\frac{1\cancel{\,\mbox{rev}}}{2\pi}\cdot\frac{60\cancel{\,\mathrm{% s}}}{1\cancel{\,{\rm min}}}$

$\displaystyle=$ $\displaystyle 764\,{\rm N}\mathord{\cdot}{\rm m}.$

Answer: $M_{\rm out}=764\,{\rm N}\mathord{\cdot}{\rm m}$

SAMPLE 16.16

Filename:sfig4-6-5b

Figure 16.85: An accelerating compound gear train.

An accelerating gear train. In the gear train shown in Fig. 16.85, the torque at the input shaft is $M_{\rm in}=200\,{\rm N}\mathord{\cdot}{\rm m}$ and the angular acceleration is $\alpha_{\rm in}=50\,\mathrm{rad}/\mathrm{s}^{2}$ . The radii of the various gears are: $R_{A}=5\,\mathrm{cm},\,R_{B}=8\,\mathrm{cm},\,R_{C}=4\,\mathrm{cm},$ and $R_{D}=10\,\mathrm{cm}$ and the moments of inertia about the shaft axis passing through their respective centers are: $I_{A}=0.1\,\mathrm{kg}\,\,\mathrm{m}^{2},\;I_{BC}=5I_{A},\;I_{D}=4I_{A}$ . Find the output torque $M_{\rm out}$ of the gear train.

Solution

Since the difference between the input power and the output power is used in accelerating the gears, we may write

P_{\rm in}-P_{\rm out}=\dot{{E_{\mathrm{K}}}}

Let $M_{\rm out}$ be the output torque of the gear train. Then,

P_{\rm in}-P_{\rm out}=M_{\rm in}\,\omega_{\rm in}-M_{\rm out}\,\omega_{\rm out}.

(16.66)

Now,

$\displaystyle\dot{{E_{\mathrm{K}}}}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}({E_{\mathrm{K}}})$
	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(\frac{1}{2}I_{A}\,\omega_{\rm in}^{2}+\frac{1}{% 2}I_{BC}\,\omega_{BC}^{2}+\frac{1}{2}I_{D}\,\omega_{\rm out}^{2}\right)$
	$\displaystyle=$	$\displaystyle I_{A}\,\omega_{\rm in}\,\dot{\omega}_{\rm in}+I_{BC}\,\omega_{BC% }\,\dot{\omega}_{BC}+I_{D}\,\omega_{\rm out}\,\dot{\omega}_{\rm out}$
	$\displaystyle=$	$\displaystyle I_{A}\,\omega_{\rm in}\,\alpha_{\rm in}+5I_{A}\,\omega_{BC}\,% \alpha_{BC}+4I_{A}\,\omega_{\rm out}\,\alpha_{\rm out}.$	(16.68)

Filename:sfig4-6-5c

Figure 16.86:

The velocity or acceleration of the point of contact between two meshing gears has to be the same irrespective of which meshing gear’s geometry and motion is used to compute them.

The different $\omega$ ’s and the $\alpha$ ’s can be related by realizing that the linear speed or the tangential acceleration of the point of contact between any two meshing gears has to be the same irrespective of which gear’s speed and geometry is used to calculate it. Thus, using the linear speed and tangential acceleration calculations for points P and R in Fig. 16.86, we can find $\omega_{B}$ , $\alpha_{B}$ and $\omega_{\rm out}$ , $\alpha_{\rm out}$ . Considering the linear speed of point P, we find

	$\displaystyle v_{P}$	$\displaystyle=$	$\displaystyle\omega_{\rm in}\,R_{A}=\omega_{B}\,R_{B}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{B}$	$\displaystyle=$	$\displaystyle\omega_{\rm in}\cdot\frac{R_{A}}{R_{B}},$

and considering the tangential acceleration of point P, $(a_{P})_{\theta}$ , we find

	$\displaystyle(a_{P})_{\theta}$	$\displaystyle=$	$\displaystyle\alpha_{\rm in}\,R_{A}=\alpha_{B}\,R_{B}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\alpha_{B}$	$\displaystyle=$	$\displaystyle\alpha_{\rm in}\cdot\frac{R_{A}}{R_{B}}.$

Similarly,

	$\displaystyle v_{R}$	$\displaystyle=$	$\displaystyle\omega_{C}\,R_{C}=\omega_{\rm out}\,R_{D}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{\rm out}$	$\displaystyle=$	$\displaystyle\omega_{C}\cdot\frac{R_{C}}{R_{D}},$

and

	$\displaystyle(a_{R})_{\theta}$	$\displaystyle=$	$\displaystyle\alpha_{C}\,R_{C}=\alpha_{\rm out}\,R_{D}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\alpha_{\rm out}$	$\displaystyle=$	$\displaystyle\alpha_{C}\cdot\frac{R_{C}}{R_{D}}.$

But

	$\displaystyle\omega_{C}$	$\displaystyle=$	$\displaystyle\omega_{B}=\omega_{BC}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{\rm out}$	$\displaystyle=$	$\displaystyle\omega_{\rm in}\cdot\frac{R_{A}}{R_{B}}\cdot\frac{R_{C}}{R_{D}}$

and

	$\displaystyle\alpha_{C}$	$\displaystyle=$	$\displaystyle\alpha_{B}=\alpha_{BC}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\alpha_{\rm out}$	$\displaystyle=$	$\displaystyle\alpha_{\rm in}\cdot\frac{R_{A}}{R_{B}}\cdot\frac{R_{C}}{R_{D}}.$

Substituting these expressions for $\omega_{\rm out},\;\alpha_{\rm out},\;\omega_{BC}$ and $\alpha_{BC}$ in equations (16.66) and (16.68), we get

$\displaystyle P_{\rm in}-P_{\rm out}$	$\displaystyle=$	$\displaystyle M_{\rm in}\,\omega_{\rm in}-M_{\rm out}\omega_{\rm in}\cdot\frac% {R_{A}}{R_{B}}\cdot\frac{R_{C}}{R_{D}}$
	$\displaystyle=$	$\displaystyle\omega_{\rm in}\left(M_{\rm in}-M_{\rm out}\cdot\frac{R_{A}}{R_{B% }}\cdot\frac{R_{C}}{R_{D}}\right).$
$\displaystyle\dot{{E_{\mathrm{K}}}}$	$\displaystyle=$	$\displaystyle I_{A}\left[\omega_{\rm in}\alpha_{\rm in}+5\omega_{\rm in}\alpha% _{\rm in}\left(\frac{R_{A}}{R_{B}}\right)^{2}+4\omega_{\rm in}\alpha_{\rm in}% \left(\frac{R_{A}}{R_{B}}\cdot\frac{R_{C}}{R_{D}}\right)^{2}\right]$
	$\displaystyle=$	$\displaystyle I_{A}\omega_{\rm in}\left[\alpha_{\rm in}+5\alpha_{\rm in}\left(% \frac{R_{A}}{R_{B}}\right)^{2}+4\alpha_{\rm in}\left(\frac{R_{A}}{R_{B}}\cdot% \frac{R_{C}}{R_{D}}\right)^{2}\right].$

Now equating the two quantities, $P_{\rm in}-P_{\rm out}$ and $\dot{{E_{\mathrm{K}}}}$ , and canceling $\omega_{\rm in}$ from both sides, we obtain

$\displaystyle M_{\rm out}\frac{R_{A}}{R_{B}}\cdot\frac{R_{C}}{R_{D}}$	$\displaystyle=$	$\displaystyle M_{\rm in}-I_{A}\alpha_{\rm in}\left[1+5\left(\frac{R_{A}}{R_{B}% }\right)^{2}+4\left(\frac{R_{A}}{R_{B}}\cdot\frac{R_{C}}{R_{D}}\right)^{2}\right]$
$\displaystyle M_{\rm out}\frac{5}{8}\cdot\frac{4}{10}$	$\displaystyle=$	$\displaystyle 200\,{\rm N}\mathord{\cdot}{\rm m}-5\,\mathrm{kg}\,\,\mathrm{m}^% {2}\cdot\,\mathrm{rad}/\mathrm{s}^{2}\left[1+5\left(\frac{5}{8}\right)^{2}+4% \left(\frac{5}{8}\cdot\frac{4}{10}\right)^{2}\right]$
$\displaystyle M_{\rm out}$	$\displaystyle=$	$\displaystyle 735.94\,{\rm N}\mathord{\cdot}{\rm m}$
	$\displaystyle\approx$	$\displaystyle 736\,{\rm N}\mathord{\cdot}{\rm m}.$

Answer: $M_{\rm out}=736\,{\rm N}\mathord{\cdot}{\rm m}$

SAMPLE 16.17

Filename:sfig4-6-5d

Figure 16.87: Two drums with strings wrapped around are used to pull up a mass

m

Drums used as pulleys. Two drums, A and B of radii $R_{o}=200\,{\rm mm}$ and $R_{i}=100\,{\rm mm}$ are welded together. The combined mass of the drums is $m_{D}=20\,\mathrm{kg}$ and the combined moment of inertia about the $z$ -axis passing through their common center O is $I_{zz/O}=1.6\,\mathrm{kg}\,\,\mathrm{m}^{2}$ . A string attached to and wrapped around drum B supports a mass $m=2\,\mathrm{kg}$ . The string wrapped around drum A is pulled with a force $F=20\,{\mbox{N}}$ as shown in Fig. 16.87. Assume there is no slip between the strings and the drums. Find

1.

the angular acceleration of the drums,
2.

the tension in the string supporting mass $m$ , and
3.

the acceleration of mass $m$ .

^†^†margin:

Solution

The free-body diagram of the drums and the mass are shown in Fig. 16.88 separately where $T$ is the tension in the string supporting mass $m$ and $O_{x}$ and $O_{y}$ are the support reactions at O. Since the drums can only rotate about the $z$ -axis, let

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}% \hbox to36.135pt{\hfil and\hfil}\dot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}=\dot{\omega}\hat{\boldsymbol{k}}.

Filename:sfig4-5-6

Figure 16.88: Free-body diagram of the drums and the mass

m

T

is the tension in the string supporting mass

m

and

O_{x}

and

O_{y}

are the reactions of the support at O.

Now, let us do angular momentum balance about the center of rotation 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{$\scriptscriptstyle\mathrm{/O}$}}}

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle TR_{i}\hat{\boldsymbol{k}}-FR_{o}\hat{\boldsymbol{k}}$
		$\displaystyle=$	$\displaystyle(TR_{i}-FR_{o})\hat{\boldsymbol{k}}.$

Since the motion is restricted to the $x y$ -plane (i.e., 2-D motion), 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}$}}}$ may be computed 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}$}}}$	$\displaystyle=$	$\displaystyle I_{zz/cm}\;\dot{\omega}\hat{\boldsymbol{k}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{cm/O}\times{{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{cm}}\;\;m_{D}$
	$\displaystyle=$	$\displaystyle I_{zz/O}\;\dot{\omega}\hat{\boldsymbol{k}}+\underbrace{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{O/O}}_{0}\;\times\underbrace{{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}}_{0}\;\;m_{D}$
	$\displaystyle=$	$\displaystyle I_{zz/O}\;\dot{\omega}\hat{\boldsymbol{k}}.$

Setting $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\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

TR_{i}-FR_{o}=I_{zz/O}\,\dot{\omega}.

(16.69)

Now, let us write linear momentum balance, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ , for mass $m$ :

\underbrace{(T-mg)\hat{\boldsymbol{\jmath}}}_{\sum{\accentset{\rightharpoonup}% {\boldsymbol{F}}}}=m{\accentset{\rightharpoonup}{\boldsymbol{a}}}.

Do we know anything about acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ of the mass? Yes, we know its direction ( $\pm\hat{\boldsymbol{\jmath}}$ ) and we also know that it has to be the same as the tangential acceleration $({\accentset{\rightharpoonup}{\boldsymbol{a}}}_{D})_{\theta}$ of point D on drum B (why?). Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{a}}}_{D})_{\theta}$	(16.70)
	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{k}}\times(-R_{i}\hat{\boldsymbol{% \imath}})$
	$\displaystyle=$	$\displaystyle-\dot{\omega}R_{i}\hat{\boldsymbol{\jmath}}.$

Therefore,

T-mg=-m\dot{\omega}R_{i}.

(16.71)

Calculation of $\dot{\boldsymbol{\omega}}$ : We now have two equations, (16.69) and (16.71), and two unknowns, $\dot{\omega}$ and $T$ . Subtracting $R_{i}$ times Eqn.(16.71) from Eqn. (16.69) we get

$\displaystyle-FR_{o}+mgR_{i}$	$\displaystyle=$	$\displaystyle(I_{zz/O}+mR_{i}^{2})\dot{\omega}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\dot{\omega}$	$\displaystyle=$	$\displaystyle\frac{-FR_{o}+mgR_{i}}{(I_{zz/O}+mR_{i}^{2})}$
	$\displaystyle=$	$\displaystyle\frac{-20\,{\mbox{N}}\cdot 0.2\,\mathrm{m}+2\,\mathrm{kg}\cdot 9.% 81\,\mathrm{m}/\mathrm{s}^{2}\cdot 0.1m}{1.6\,\mathrm{kg}\,\,\mathrm{m}^{2}+2% \,\mathrm{kg}\cdot(0.1\,\mathrm{m})^{2}}$
	$\displaystyle=$	$\displaystyle\frac{-2.038\,\mathrm{kg}\,\,\mathrm{m}^{2}/\,\mathrm{s}^{2}}{1.6% 2\,\mathrm{kg}\,\,\mathrm{m}^{2}}$
	$\displaystyle=$	$\displaystyle-1.258\frac{1}{\,\mathrm{s}^{2}}$

Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-1.26\,\mathrm{rad}/% \mathrm{s}^{2}\hat{\boldsymbol{k}}$

2.

Calculation of tension $\bf T$ : From equation (16.71):

$\displaystyle T$ $\displaystyle=$ $\displaystyle mg-m\dot{\omega}R_{i}$

$\displaystyle=$ $\displaystyle 2\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}-2\,\mathrm{% kg}\cdot(-1.26\,\mathrm{s}^{-2})\cdot 0.1\,\mathrm{m}$

$\displaystyle=$ $\displaystyle 19.87\,{\mbox{N}}$

Answer: $T=19.87\,{\mbox{N}}$
3.

Calculation of acceleration of the mass: Since the acceleration of the mass is the same as the tangential acceleration of point D on the drum, we get (from eqn. (16.70))

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ $\displaystyle=$ $\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{a}}}_{D})_{\theta}=-% \dot{\omega}R_{i}\hat{\boldsymbol{\jmath}}$

$\displaystyle=$ $\displaystyle-(-1.26\,\mathrm{s}^{-2})\cdot 0.1\,\mathrm{m}$

$\displaystyle=$ $\displaystyle 0.126\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\jmath}}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=0.13\,\mathrm{m}/\mathrm{s}^{2}% \hat{\boldsymbol{\jmath}}$

Comments: It is important to understand why the acceleration of the mass is the same as the tangential acceleration of point D on the drum. We have assumed (as is common practice) that the string is massless and inextensible. Therefore each point of the string supporting the mass must have the same linear displacement, velocity, and acceleration as the mass. Now think about the point on the string which is momentarily in contact with point D of the drum. Since there is no relative slip between the drum and the string, the two points must have the same vertical acceleration. This vertical acceleration for point D on the drum is the tangential acceleration $({\accentset{\rightharpoonup}{\boldsymbol{a}}}_{D})_{\theta}$ .

SAMPLE 16.18 Energy method for pulley dynamics: Consider the pulley problem of Sample 16.86 again. Use energy method to

1.

find the angular acceleration of the pulley, and
2.

the acceleration of the mass.

Solution

In energy method we use speeds, not velocities. Therefore, we have to be careful in our thinking about the direction of motion. In the present problem, let us assume that the pulley rotates and accelerates clockwise. Consequently, the mass moves up against gravity.

The energy equation we want to use is

P=\dot{E}_{K}.

The power $P$ is given by $P=\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{i}$ where the sum is carried out over all external forces. For the mass and pulley system the external forces that do work are $F$ and $m g$ .

The other external forces on the system—the reaction force of the support point O and the weight of the pulley—are acting at point O (see fig. 16.89). But, since point O is stationary, these forces do no work.

Filename:sfig4-5-6a

Figure 16.89: Free-body-diagram of the pulley-mass system together. Note that the forces acting at the support point O do no work since point O is stationary.

Therefore,

$\displaystyle P$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{A}+m{\accentset{\rightharpoonup}{% \boldsymbol{g}}}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{m}$	(16.72)
	$\displaystyle=$	$\displaystyle F\hat{\boldsymbol{\imath}}\cdot v_{A}\hat{\boldsymbol{\imath}}+(% -mg\hat{\boldsymbol{\jmath}})\cdot\underbrace{v_{D}\hat{\boldsymbol{\jmath}}}_% {{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{m}}$
	$\displaystyle=$	$\displaystyle Fv_{A}-mgv_{D}.$

Now we need to calculate the rate of change of kinetic energy $\dot{E}_{K}$ . There are two objects here that have kinetic energy—the hanging mass and the pulley. Hence,

\dot{E}_{K}=(\dot{E}_{K})_{m}+(\dot{E}_{K})_{\rm pulley}.

The hanging mass has pure translational motion and hence its kinetic energy is

({E}_{K})_{m}=\frac{1}{2}mv^{2}

where $v$ is the linear speed of the mass. If we assume the string to be inextensible, then the linear speed $v$ of the mass has to be the same as the tangential speed of point D of the pulley. Thus $v=v_{D}$ and

({E}_{K})_{m}=\frac{1}{2}mv_{D}^{2}.

The pulley, on the other hand, has pure rotational motion about point O, and hence its kinetic energy is given by

({E}_{K})_{\rm pulley}=\frac{1}{2}I^{\scriptscriptstyle\mathrm{O}}_{zz}\omega^% {2}.

Summing the two kinetic energies and differentiating with respect to time $t$ , we get

	$\displaystyle\dot{E}_{K}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(\frac{1}{2}\,m\,v_{D}^{2}+\frac{1}{2}I^{% \scriptscriptstyle\mathrm{O}}_{zz}\omega^{2}\right)$		(16.73)
		$\displaystyle=$	$\displaystyle m\,v_{D}\,\dot{v}_{D}+I^{\scriptscriptstyle\mathrm{O}}_{zz}\,% \omega\,\dot{\omega}.$		(16.74)

Now equating the power and the rate of change of kinetic energy from eqns. eqn. (16.72) and eqn. (16.74), we get

F\,v_{A}-mg\,v_{D}=m\,v_{D}\,\dot{v}_{D}+I^{\scriptscriptstyle\mathrm{O}}_{zz}% \,\omega\,\dot{\omega}.

From kinematics of circular motion,

$\displaystyle v_{A}$	$\displaystyle=$	$\displaystyle\omega R_{o},$
$\displaystyle v_{D}$	$\displaystyle=$	$\displaystyle\omega R_{i}$
$\displaystyle\hbox to36.135pt{and\hfil}\dot{v}_{D}\equiv(a_{D})_{\theta}$	$\displaystyle=$	$\displaystyle\dot{\omega}R_{i}.$

Substituting these values in the power balance equation above, we get

$\displaystyle{\omega}(FR_{o}-mgR_{i})$	$\displaystyle=$	$\displaystyle{\omega}\,\dot{\omega}(mR_{i}^{2}+I^{\scriptscriptstyle\mathrm{O}% }_{zz})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\omega}$	$\displaystyle=$	$\displaystyle\frac{FR_{o}-mgR_{i}}{(I^{\scriptscriptstyle\mathrm{O}}_{zz}+mR_{% i}^{2})}$
	$\displaystyle=$	$\displaystyle\frac{20\,{\mbox{N}}\cdot 0.2\,\mathrm{m}-2\,\mathrm{kg}\cdot 9.8% 1\,\mathrm{m}/\mathrm{s}^{2}\cdot 0.1m}{1.6\,\mathrm{kg}\,\,\mathrm{m}^{2}+2\,% \mathrm{kg}\cdot(0.1\,\mathrm{m})^{2}}$
	$\displaystyle=$	$\displaystyle 1.258\frac{1}{\,\mathrm{s}^{2}}.\qquad\mbox{(same as the answer % before.)}$

Since the sign of $\dot{\omega}$ is positive, our initial assumption of clockwise acceleration of the pulley is correct.

Answer: $\dot{\omega}=1.26\,\mathrm{rad}/\mathrm{s}^{2}$

2.

From kinematics of circular motion,

$\displaystyle a_{m}$ $\displaystyle=$ $\displaystyle(a_{D})_{\theta}$

$\displaystyle=$ $\displaystyle\dot{\omega}R_{i}$

$\displaystyle=$ $\displaystyle 0.126\,\mathrm{m}/\mathrm{s}^{2}.$

Answer: $a_{m}=0.13\,\mathrm{m}/\mathrm{s}^{2}$

SAMPLE 16.19 Energy Accounting: Consider the pulley problem of Sample 16.86 again.

1.

What percentage of the input energy (work done by the applied force $F$ ) is used in raising the mass by 1 $\,\mathrm{m}$ ?
2.

Where does the rest of the energy go? Provide an energy-balance sheet.

Solution

1.

Let $W_{i}$ and $W_{h}$ be the input energy and the energy used in raising the mass by 1 $\,\mathrm{m}$ , respectively. Then the percentage of energy used in raising the mass is

$\mbox{\% of input energy used}=\frac{W_{h}}{W_{i}}\times 100.$

Thus we need to calculate $W_{i}$ and $W_{h}$ to find the answer. $W_{i}$ is the work done by the force $F$ on the system during the interval in which the mass moves up by 1 $\,\mathrm{m}$ . Let $s$ be the displacement of the force $F$ during this interval. Since the displacement is in the same direction as the force (we know it is from Sample 16.86), the input-energy is

$W_{i}=F\,s.$

So to find $W_{i}$ we need to find $s$ .

For the mass to move up by 1 $\,\mathrm{m}$ the inner drum B must rotate by an angle $\theta$ where

$1\,\mathrm{m}=\theta\,R_{i}\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad% \theta=\frac{1\,\mathrm{m}}{0.1\,\mathrm{m}}=10\,{\rm rad}.$

Since the two drums, A and B, are welded together, drum A must rotate by $\theta$ as well. Therefore the displacement of force $F$ is

$s=\theta\,R_{o}=10\,{\rm rad}\cdot 0.2\,\mathrm{m}=2\,\mathrm{m},$

and the energy input is

$W_{i}=F\,s=20\,{\mbox{N}}\cdot 2\,\mathrm{m}=40{\rm J}.$

Now, the work done in raising the mass by $1\,\mathrm{m}$ is

$W_{h}=mgh=2\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot 1\,\mathrm{% m}=19.62{\rm J}.$

Therefore, the percentage of input-energy used in raising the mass

$=\frac{19.62\,{\rm N}\mathord{\cdot}{\rm m}}{40}\times 100=49.05\%\approx 49\%.$

The rest of the energy ( $=51\%$ ) goes in accelerating the mass and the pulley. Let us find out how much energy goes into each of these activities. Since the initial state of the system from which we begin energy accounting is not prescribed (that is, we are not given the height of the mass from which it is to be raised 1 $\,\mathrm{m}$ , nor do we know the velocities of the mass or the pulley at that initial height), let us assume that at the initial state, the angular speed of the pulley is $\omega_{o}$ and the linear speed of the mass is $v_{o}$ . At the end of raising the mass by 1 $\,\mathrm{m}$ from this state, let the angular speed of the pulley be $\omega_{f}$ and the linear speed of the mass be $v_{f}$ . Then, the energy used in accelerating the pulley is

	$\displaystyle(\Delta{E_{\mathrm{K}}})_{\rm pulley}$	$\displaystyle=\mbox{final kinetic energy}-\mbox{initial kinetic energy}$
		$\displaystyle=\frac{1}{2}I\omega_{f}^{2}-\frac{1}{2}I\omega_{o}^{2}$
		$\displaystyle=\frac{1}{2}I\underbrace{(\omega_{f}^{2}-\omega_{o}^{2})}_{\text{% assuming constant acceleration, $\omega_{f}^{2}=\omega_{o}^{2}+2\alpha\theta$,% or $\omega_{f}^{2}-\omega_{o}^{2}=2\alpha\theta$.}}$
		$\displaystyle=I\alpha\,\theta\qquad\mbox{(from Sample~{}\ref{sample5.6.pulley2}, $\alpha=1.258\,\mathrm{rad}/\mathrm{s}^{2}.$% ) }$
		$\displaystyle=1.6\,\mathrm{kg}\,\mathrm{m}^{2}\cdot 1.258\,\mathrm{rad}/% \mathrm{s}^{2}\cdot 10\,{\rm rad}$
		$\displaystyle=20.13\,{\rm N}\mathord{\cdot}{\rm m}=20.13\;{\rm J}.$

Similarly, the energy used in accelerating the mass is

$\displaystyle(\Delta{E_{\mathrm{K}}})_{\rm mass}$	$\displaystyle=$	$\displaystyle\mbox{final kinetic energy}-\mbox{initial kinetic energy}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{o}^{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}m(\underbrace{v_{f}^{2}-v_{o}^{2}}_{2ah})$
	$\displaystyle=$	$\displaystyle mah$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{kg}\cdot 0.126\,\mathrm{m}/\mathrm{s}^{2}\cdot 1\,% \mathrm{m}$
	$\displaystyle=$	$\displaystyle 0.25\;{\rm J}.$

We can calculate the percentage of input energy used in these activities to get a better idea of energy allocation. Here is the summary table:

Activities	Energy Spent
	in Joule	as % of input energy
In raising the mass by 1 $\,\mathrm{m}$	19.62	49.05%
In accelerating the mass	0.25	0.62 %
In accelerating the pulley	20.13	50.33 %
Total	40.00	100 %

So, what would you change in the set-up so that more of the input energy is used in raising the mass? Think about what aspects of the motion would change due to your proposed design.

SAMPLE 16.20

Filename:sfig4-6-8

Figure 16.90: A uniform rod swings in the plane about its pinned end O.

Equation of motion of a swinging stick: A uniform bar of mass $m$ and length $\ell$ is pinned at one of its ends O. The bar is displaced from its vertical position by an angle $\theta$ and released (Fig. 16.90).

1.

Find the equation of motion using momentum balance.
2.

Find the reaction at O as a function of $(\theta,\,\dot{\theta},\,g,\,m,\,\ell)$ .

Solution First we draw a simple sketch of the given problem showing relevant geometry (Fig. 16.90(a)), and then a free-body diagram of the bar (Fig. 16.90(b)).

Filename:sfig4-6-8a

Figure 16.91:

(a) A line sketch of the swinging rod and (b) free-body diagram of the rod.

We should note for future reference that

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\omega\hat{\boldsymbol{k}}\equiv\dot{\theta}\hat{\boldsymbol{k}}$
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{k}}\equiv\ddot{\theta}\hat{% \boldsymbol{k}}$

Equation of motion using momentum balance: We can write angular momentum balance about point O as

\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 now calculate both sides of this equation:

Filename:sfig4-6-8b

Figure 16.92: Computation 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% {$\scriptscriptstyle\mathrm{/O}$}}}

by integration over the rod.

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times mg(-% \hat{\boldsymbol{\jmath}})$	(16.75)
	$\displaystyle=$	$\displaystyle\frac{\ell}{2}(\sin\theta\hat{\boldsymbol{\imath}}-\cos\theta\hat% {\boldsymbol{\jmath}})\times mg(-\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-\frac{\ell}{2}mg\,\sin\theta\hat{\boldsymbol{k}}.$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{k}}\int_{m}r^{2}dm$	(16.76)
	$\displaystyle=$	$\displaystyle\ddot{\theta}\hat{\boldsymbol{k}}\int_{0}^{\ell}s^{2}\frac{m}{% \ell}ds\quad(dm=m/\ell\cdot ds)$
	$\displaystyle=$	$\displaystyle\frac{m\ddot{\theta}}{\ell}\left[\left.\frac{s^{3}}{3}\right\|_{0}% ^{\ell}\right]=\frac{m\ell^{2}}{3}\ddot{\theta}\hat{\boldsymbol{k}}$

Equating (16.75) and (16.76) we get

$\displaystyle-\frac{\ell}{2}\cancel{m}\,g\,\sin\theta$	$\displaystyle=$	$\displaystyle\cancel{m}\frac{\ell^{2}}{3}\ddot{\theta}$
$\displaystyle\hbox to36.135pt{or\hfil}\dot{\omega}+\frac{3g}{2\ell}\sin\theta$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{\theta}+\frac{3g}{2\ell}\sin\theta$	$\displaystyle=$	$\displaystyle 0.$	(16.77)

Answer: $\ddot{\theta}+\frac{3g}{2\ell}\sin\theta=0$

Reaction at O: Using linear momentum balance

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m\,{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G},$
$\displaystyle\hbox to36.135pt{where\hfil}\sum{\accentset{\rightharpoonup}{% \boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle R_{x}\hat{\boldsymbol{\imath}}+(R_{y}-mg)\hat{\boldsymbol{\jmath% }},$
$\displaystyle\hbox to36.135pt{and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{G}$	$\displaystyle=$	$\displaystyle\frac{\ell}{2}\dot{\omega}(\cos\theta\hat{\boldsymbol{\imath}}+% \sin\theta\hat{\boldsymbol{\jmath}})+\frac{\ell}{2}\omega^{2}(-\sin\theta\hat{% \boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\frac{\ell}{2}[(\dot{\omega}\cos\theta-\omega^{2}\sin\theta)\hat{% \boldsymbol{\imath}}+(\dot{\omega}\sin\theta+\omega^{2}\cos\theta)\hat{% \boldsymbol{\jmath}}].$

Dotting both sides of $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m\,{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G}$ with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ and rearranging, we get

$\displaystyle R_{x}$	$\displaystyle=$	$\displaystyle m\frac{\ell}{2}(\dot{\omega}\cos\theta-\omega^{2}\sin\theta)$
	$\displaystyle\equiv$	$\displaystyle m\frac{\ell}{2}(\ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin% \theta),$
$\displaystyle R_{y}$	$\displaystyle=$	$\displaystyle mg+m\frac{\ell}{2}(\dot{\omega}\sin\theta+\omega^{2}\cos\theta)$
	$\displaystyle\equiv$	$\displaystyle mg+m\frac{\ell}{2}(\ddot{\theta}\sin\theta+\dot{\theta}^{2}\cos% \theta).$

Now substituting the expression for $\ddot{\theta}$ from (16.77) in $R_{x}$ and $R_{y}$ , we get

	$\displaystyle R_{x}$	$\displaystyle=$	$\displaystyle-m\sin\theta\left(\frac{3}{4}g\cos\theta+\frac{\ell}{2}\dot{% \theta}^{2}\right),$		(16.78)
	$\displaystyle R_{y}$	$\displaystyle=$	$\displaystyle mg\left(1-\frac{3}{4}\sin^{2}\theta\right)+m\frac{\ell}{2}\dot{% \theta}^{2}\cos\theta.$		(16.79)

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=-m(\frac{3}{4}g\cos\theta+\frac{% \ell}{2}\dot{\theta}^{2})\sin\theta\hat{\boldsymbol{\imath}}+[mg(1-\frac{3}{4}% \sin^{2}\theta)+m\frac{\ell}{2}\dot{\theta}^{2}\cos\theta]\hat{\boldsymbol{% \jmath}}$

Check: We can check the reaction force in the special case when the rod does not swing but just hangs from point O. The forces on the bar in this case have to satisfy static equilibrium. Therefore, the reaction at O must be equal to $m g$ and directed vertically upwards. Plugging $\theta=0$ and $\dot{\theta}=0$ (no motion) in Eqn. (16.78) and (16.79) we get $R_{x}=0$ and $R_{y}=mg$ , the values we expect.

SAMPLE 16.21

Filename:sfig5-5-2

Figure 16.93: A uniform rod swings in the plane about its pinned end O.

Swinging stick dynamics using moment of inertia: A uniform bar of mass $m$ and length $\ell$ is pinned at one of its ends O. The bar is displaced from its vertical position by an angle $\theta$ and released (Fig. 16.93). Find the equation of motion of the stick.

^†^†margin:

Solution

We repeat the problem solved in Sample 16.88 here with just one different step of finding the rate of change of angular momentum with the help of moment of inertia formula. As usual, we first draw a free-body diagram of the bar (Fig. 16.94). We assume, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}% \equiv\dot{\theta}\hat{\boldsymbol{k}}$ , and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\omega}\hat{% \boldsymbol{k}}\equiv\ddot{\theta}\hat{\boldsymbol{k}}$

Filename:sfig5-5-2a

Figure 16.94: The free-body diagram of the rod.

We can write angular momentum balance about point O as

\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 now calculate both sides of this equation:

Filename:sfig5-5-2b

Figure 16.95: Radial and tangential components of

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}

. Since the radial component is parallel to

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}\times{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G}=\frac{\ell^{2}}{4}\dot{\omega}\hat{% \boldsymbol{k}}

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times mg(-% \hat{\boldsymbol{\jmath}})$	(16.80)
	$\displaystyle=$	$\displaystyle\frac{\ell}{2}(\sin\theta\hat{\boldsymbol{\imath}}-\cos\theta\hat% {\boldsymbol{\jmath}})\times mg(-\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-\frac{\ell}{2}mg\,\sin\theta\hat{\boldsymbol{k}}.$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle I_{zz/G}\,\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% +{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}\times m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G}$	(16.81)
	$\displaystyle=$	$\displaystyle\frac{m\ell^{2}}{12}\dot{\omega}\hat{\boldsymbol{k}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{G}\times m(\overbrace{\dot{\omega}\hat{% \boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}-\omega^% {2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}}^{{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G}})$
	$\displaystyle=$	$\displaystyle\frac{m\ell^{2}}{12}\dot{\omega}\hat{\boldsymbol{k}}+\frac{m\ell^% {2}}{4}\dot{\omega}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle\frac{m\ell^{2}}{3}\dot{\omega}\hat{\boldsymbol{k}}$	(16.82)

where the last step, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}\times m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G}=\frac{m\ell^{2}}{4}\dot{\omega}\hat{% \boldsymbol{k}}$ , should be clear from Fig. 16.95. Equating (16.75) and (16.76) we get

$\displaystyle-\frac{\ell}{2}\cancel{m}\,g\,\sin\theta$	$\displaystyle=$	$\displaystyle\cancel{m}\frac{\ell^{2}}{3}\dot{\omega}$
$\displaystyle\hbox to36.135pt{or\hfil}\dot{\omega}+\frac{3g}{2\ell}\sin\theta$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{\theta}+\frac{3g}{2\ell}\sin\theta$	$\displaystyle=$	$\displaystyle 0.$	(16.83)

Answer: $\ddot{\theta}+\frac{3g}{2\ell}\sin\theta=0$

SAMPLE 16.22 Numerical solution of the swinging stick motion: For the swinging stick considered in Samples 16.88 or 16.97, find the time that the rod takes to fall from $\theta=\pi/2$ to $\theta=0$ if it is released from rest at $\theta=\pi/2$ ?

Solution

The given initial angle $\pi/2$ is a big value of $\theta$ – big in that we cannot assume $\sin\theta\approx\theta$ (obviously $1\neq 1.5708$ ). Therefore we may not use the linearized equation (16.86) to solve for $t$ explicitly. We have to solve the full nonlinear equation (16.83) to find the required time. Unfortunately, we cannot get a closed form solution of this equation using mathematical skills you have at this level. Therefore, we resort to numerical integration of this equation.

For numerical integration, we need to first write the given differential equation as a set of first order ordinary differential equations. To do so, we introduce $\omega$ as a new variable and rewrite eqn. (16.83) as

	$\displaystyle\dot{\theta}$	$\displaystyle=$	$\displaystyle\omega$		(16.84)
	$\displaystyle\dot{\omega}$	$\displaystyle=$	$\displaystyle-\frac{3g}{2\ell}\sin\theta$		(16.85)

Now we need to specify the initial conditions and the time duration for integration, and solve the equations using some ODE solver program. Here is a pseudo-code that lists the steps:

   g = 9.81,   L = 1      % define constants
   ODES = { thetadot = omega
            omegadot = -3*g/(2*L) * sin(theta) }
    ICs = { theta_0 = pi/2
            omega_0 = 0 }
   solve ODES with ICs for t = 0 to 4 s
   plot theta vs t and plot omega vs t

The results obtained from the numerical solution are shown in Fig. 16.96.

Filename:sfig7-4-2

Figure 16.96: Numerical solution is shown by plotting

\theta

and

\omega

against time.

The problem of finding the time taken by the bar to fall from $\theta=\pi/2$ to $\theta=0$ numerically is nontrivial. It is called a boundary value problem. We have only illustrated how to solve initial value problems. However, we can get a fairly good estimate of the time from the solution obtained.

We first plot $\theta$ against time $t$ as shown in fig. 16.97. We find the values of $t$ and the corresponding values of $\theta$ that bracket $\theta=0$ . Now, we can use linear interpolation to find the value of $t$ at $\theta=0$ . Proceeding this way, we get $t=0.48$ (seconds), a little more than we get from the linear ODE in sample 16.97 ( $t=0.41$ ).

Comments: Additionally, we can also get the value of $\dot{\theta}\equiv\omega$ when $\theta=0$ using similar interpolation. In fact, from the $\omega$ vs $t$ plot, we find that at $t=0.48\,\mathrm{s}$ , $\omega=-5.42\,\mathrm{rad}/\mathrm{s}$ . How does this result compare with the analytical value of $\omega$ from sample 16.97 (which did not depend on the small angle approximation)? Well, we found that

\omega=-\sqrt{\frac{3g}{\ell}}=-\sqrt{\frac{3\cdot 9.81\,\mathrm{m}/\mathrm{s}% ^{2}}{1\,\mathrm{m}}}=-5.4249\,\mathrm{s}^{-1}.

Thus, we get a fairly accurate value from numerical integration.

Filename:pfigure-blue-99-1

Figure 16.97: Time

t

corresponding to

\theta=0

SAMPLE 16.23

Filename:pfigure-blue-47-2

Figure 16.98: Work done by the force of gravity in moving from G^′ to G

\int{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot d{\accentset{% \rightharpoonup}{\boldsymbol{r}}}=-mg\hat{\boldsymbol{\jmath}}\cdot h\hat{% \boldsymbol{\jmath}}=-mgh

Filename:pfigure-blue-89-1

Figure 16.99: The infinitesimal mass

d m

considered in the calculation of

{E_{\mathrm{K}}}

The swinging stick dynamics with energy balance: Consider the same swinging stick as in Sample 16.88. The stick is, again, displaced from its vertical position by an angle $\theta$ and released (See Fig. 16.90).

1.

Find the equation of motion using energy balance.
2.

What is $\dot{\theta}$ at $\theta=0$ if $\theta(t=0)=\pi/2$ ?
3.

Find the period of small oscillations about $\theta=0$ .

Solution

1.

Equation of motion using energy balance: We use the power equation, $\dot{{E_{\mathrm{K}}}}~{}=~{}P$ , to derive the equation of motion of the bar. Now, the kinetic energy is given by

${E_{\mathrm{K}}}=\frac{1}{2}\int_{m}v^{2}dm$

where $v$ is the speed of the infinitesimal mass element $d m$ . Referring to fig. 16.99, we can write, $dm=(m/\ell)ds$ , and $v=\omega s\equiv\dot{\theta}s$ . Thus,

$\displaystyle{E_{\mathrm{K}}}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\int_{0}^{\ell}\dot{\theta}^{2}s^{2}\frac{m}{\ell}ds$

$\displaystyle=$ $\displaystyle\frac{m\dot{\theta}^{2}}{2\ell}\int_{0}^{\ell}s^{2}ds$

$\displaystyle=$ $\displaystyle\frac{1}{6}m\ell^{2}\dot{\theta}^{2}$

and, therefore,

$\dot{{E_{\mathrm{K}}}}=\frac{d}{dt}(\frac{1}{6}m\ell^{2}\omega^{2})=\frac{1}{3% }m\ell^{2}\omega\,\dot{\omega}=\frac{1}{3}m\ell^{2}\dot{\theta}\ddot{\theta}.$

Calculation of power ( $P$ ): There are only two forces acting on the bar, the reaction force, ${\accentset{\rightharpoonup}{\boldsymbol{R}}}(=R_{x}\hat{\boldsymbol{\imath}}+% R_{y}\hat{\boldsymbol{\jmath}})$ and the force due to gravity, $-mg\hat{\boldsymbol{\jmath}}$ . Since the support point O does not move, no work is done by ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ . Therefore,

$\displaystyle W$ $\displaystyle=$ $\displaystyle\mbox{Work done by gravity force in moving from G${}^{\prime}$ to% G.}=-mgh$

Note that the negative sign stands for the work done against gravity.

Now,

$h=OG^{\prime}-OG^{\prime\prime}=\frac{\ell}{2}-\frac{\ell}{2}\cos\theta=\frac{% \ell}{2}(1-\cos\theta).$

Therefore,

$\displaystyle W$ $\displaystyle=$ $\displaystyle-mg\frac{\ell}{2}(1-\cos\theta)$

$\displaystyle\hbox to36.135pt{and\hfil}P$ $\displaystyle=$ $\displaystyle\dot{W}=\frac{dW}{dt}=-mg\frac{\ell}{2}\sin\theta\dot{\theta}.$

Equating $\dot{{E_{\mathrm{K}}}}$ and $P$ we get

$\displaystyle-\cancel{m}\,g\frac{\ell}{2}\sin\theta\cancel{\dot{\theta}}$ $\displaystyle=$ $\displaystyle\frac{1}{3}\cancel{m}\,\ell^{2}\cancel{\dot{\theta}}\ddot{\theta}$

$\displaystyle\hbox to36.135pt{or\hfil}\ddot{\theta}+\frac{3g}{2\ell}\sin\theta$ $\displaystyle=$ $\displaystyle 0.$

Answer: $\ddot{\theta}+\frac{3g}{2\ell}\sin\theta=0$

This equation is, of course, the same as we obtained using balance of angular momentum in Sample 16.88.
2.

Find $\omega$ at $\theta=0$ : We are given that at $t=0,\;\theta=\pi/2$ and $\dot{\theta}\equiv\omega=0$ (released from rest). This position is (1) shown in Fig. 16.100. In position (2) $\theta=0$ , i.e., the rod is vertical. Since there are no dissipative forces, the total energy of the system remains constant. Therefore, taking datum for potential energy as shown in Fig. 16.100, we may write

$\displaystyle\underbrace{{{E_{\mathrm{K}}}}_{1}}_{0}+V_{1}$ $\displaystyle=$ $\displaystyle{{E_{\mathrm{K}}}}_{2}+\underbrace{V_{2}}_{0}$

$\displaystyle\hbox to36.135pt{or\hfil}mg\frac{\ell}{2}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\int_{m}v^{2}dm$

$\displaystyle=$ $\displaystyle\frac{1}{6}m\ell^{2}\omega^{2}\quad(\mbox{see part (a)})$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega$ $\displaystyle=$ $\displaystyle\pm\sqrt{\frac{3g}{\ell}}$

Filename:pfigure-blue-35-2

Figure 16.100: The total energy between positions (1) and (2) is constant.

Answer: $\omega=\pm\sqrt{\frac{3g}{\ell}}$
3.

Period of small oscillations: The equation of motion is

$\ddot{\theta}+\frac{3g}{2\ell}\sin\theta=0.$

For small $\theta$ , $\sin\theta\approx\theta$

$\Rightarrow$ $\displaystyle\ddot{\theta}+\frac{3g}{2\ell}\theta=0$ (16.86)

or $\displaystyle\ddot{\theta}+\lambda^{2}\theta=0$

where $\displaystyle\lambda^{2}=\frac{3g}{2\ell}.$

Therefore,

the circular frequency $\displaystyle=$ $\displaystyle{\lambda}=\sqrt{\frac{3g}{2\ell}},$

and the time period T $\displaystyle=$ $\displaystyle\frac{2\pi}{\lambda}=2\pi\sqrt{\frac{2\ell}{3g}}.$

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

[Say for $g=9.81\,\mathrm{m}/\mathrm{s}^{2},\ell=1\,\mathrm{m}\quad\mbox{we get}\quad% \frac{T}{4}=\frac{\pi}{2}\sqrt{\frac{2}{3}\frac{1}{9.81}\,\mathrm{s}}=0.4097\,% \mathrm{s}$ ]

SAMPLE 16.24 The swinging stick with a destabilizing torque. Consider the swinging stick of Sample 16.88 once again.

1.

Find the equation of motion of the stick, if a torque ${\accentset{\rightharpoonup}{\boldsymbol{M}}}=M\hat{\boldsymbol{k}}$ is applied at end O and a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\imath}}$ is applied at the other end A.
2.

Take $F=0$ and $M=C\theta$ . For $C=0$ you get the equation of free oscillations obtained in Sample 16.88 or 16.97. For small $C$ , does the period of the pendulum increase or decrease?
3.

What happens if $C$ is big?

Solution

A free-body diagram of the bar is shown in Fig. 16.101. Once again, we can use $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\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}$}}}$ to derive the equation of motion as in Sample 16.88. We calculated $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\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 $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\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 Sample 16.88. 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}$}}}$ remains the same in the present problem. We only need to recalculate $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\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}$}}}$	$\displaystyle=$	$\displaystyle M\hat{\boldsymbol{k}}+{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{G/O}\times mg(-\hat{\boldsymbol{\jmath}})+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{A/O}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}$
	$\displaystyle=$	$\displaystyle M\hat{\boldsymbol{k}}-\frac{\ell}{2}mg\sin\theta\hat{\boldsymbol% {k}}+F\ell\cos\theta\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(M+F\ell\cos\theta-\frac{\ell}{2}mg\sin\theta)\hat{\boldsymbol{k}}$
and
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle m\ddot{\theta}\frac{\ell^{2}}{3}\hat{\boldsymbol{k}}\quad\mbox{(% see Sample~{}\ref{ch5.4.swingstick})}$

Filename:pfigure4-3Dpend

Figure 16.101: Free-body diagram of the bar with applied torque

{\accentset{\rightharpoonup}{\boldsymbol{M}}}

and force

{\accentset{\rightharpoonup}{\boldsymbol{F}}}

Therefore, from $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\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}$}}}$

	$\displaystyle M+F\ell\cos\theta-\frac{\ell}{2}mg\sin\theta$	$\displaystyle=$	$\displaystyle m\ddot{\theta}\frac{\ell^{2}}{3}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\ddot{\theta}+\frac% {3g}{2\ell}\sin\theta-\frac{3F}{m\ell}\cos\theta-\frac{3M}{m\ell^{2}}$	$\displaystyle=$	$\displaystyle 0.$

Answer: $\ddot{\theta}+\frac{3g}{2\ell}\sin\theta-\frac{3F}{m\ell}\cos\theta-\frac{3M}{% m\ell^{2}}=0$

2.

Now, setting $F=0$ and $M=C\,\theta$ we get

$\ddot{\theta}+\frac{3g}{2\ell}\sin\theta-\frac{3C\theta}{m\ell^{2}}=0$ (16.87)

Numerical Solution: We can numerically integrate (16.87) just as in the previous Sample to find $\theta(t)$ . Here is the pseudo-code that can be used for this purpose.
```
   g = 9.81, L = 1      % specify parameters
   m = 1, C = 4
   ODES = { thetadot = omega
            omegadot = -(3*g/(2*L)) * sin(theta)
                      + 3*C/(m*L^2) * theta      }
   ICs =  { thetazero = pi/20
            omegazero = 0  }
   solve ODES with ICs until t = 10
```
Using this pseudo-code, we find the response of the pendulum. Figure 16.102 shows different responses for various values of $C$ . Note that for $C=0$ , it is the same case as unforced bar pendulum considered above.

From Fig. 16.102 it is clear that the bar has periodic motion for small $C$ , with the period of motion increasing with increasing values of $C$ . It makes sense if you look at Eqn. (16.87) carefully. Gravity acts as a restoring force while the applied torque acts as a destabilizing force. Thus, with the resistance of the applied torque, the stick swings more sluggishly making its period of oscillation bigger.

Filename:pfigure-s94h6p3

Figure 16.102: $\theta(t)$ with applied torque $M=C\,\theta$ for $C=0,\;1,\;2,\;4,\;4.905,\;5.$ Note that for small $C$ the motion is periodic but for large $C$ ( $C\geq 4.4$ ) the motion becomes aperiodic.
3.

From Fig. 16.102, we see that at about $C\approx 4.9$ the stability of the system changes completely. $\theta(t)$ is not periodic anymore. It keeps on increasing at a faster and faster rate, that is, the bar makes complete loops about point O with ever increasing speed. Does it make physical sense? Yes, it does. As the value of $C$ is increased beyond a certain value (can you guess the value?), the applied torque overcomes any restoring torque due to gravity. Consequently, the bar is forced to rotate continuously in the direction of the applied force.

SAMPLE 16.25

Filename:pfigure-s94h6p4

Figure 16.103:

A torsional pendulum with linear springs: A uniform rigid bar of mass $m=2\,\mathrm{kg}$ and length $\ell=1\,\mathrm{m}$ is pinned at one end and connected to two springs, each with spring constant $k$ , at the other end. The bar is tweaked slightly from its vertical position. It then oscillates about its original position. The bar is timed for 20 full oscillations which take 12.5 seconds. Ignore gravity.

1.

Find the equation of motion of the rod.
2.

Find the spring constant $k$ .
3.

What should be the spring constant of a torsional spring if the bar is attached to one at the bottom and has the same oscillating motion characteristics?

Solution

Filename:pg84-3

Figure 16.104:

Refer to the free-body diagram in figure 16.104. Angular momentum balance for the rod 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{$\scriptscriptstyle\mathrm{/O}$}}}

$\displaystyle\mbox{where}\quad{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.% 0pt\hbox{$\scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle-2k\overbrace{x}^{\ell\sin\theta}\mathord{\cdot}\ell\cos\theta% \hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle-2k\ell^{2}\sin\theta\,\cos\theta\hat{\boldsymbol{k}},$
$\displaystyle\mbox{and}\quad\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_% {\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle I_{zz}^{O}\ddot{\theta}\hat{\boldsymbol{k}}=\underbrace{\frac{1}% {3}m\ell^{2}}_{I_{zz}^{O}}\ \ddot{\theta}\hat{\boldsymbol{k}}.$

Thus

\frac{1}{3}m\ell^{2}\,\ddot{\theta}=-2k\ell^{2}\sin\theta\cos\theta.

However, for small $\theta,\;\;\cos\theta\approx 1$ and $\sin\theta\approx\theta$ ,

\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\theta}+\frac{6k\cancel{\ell^{2% }}}{m\cancel{\ell^{2}}}\theta=0.

(16.88)

Answer: $\ddot{\theta}+\frac{6k}{m}\theta=0$

2.

Comparing Eqn. (16.88) with the standard harmonic oscillator equation $\ddot{x}+\lambda^{2}x=0$ , we get

$\displaystyle\mbox{angular frequency}\quad\lambda$ $\displaystyle=$ $\displaystyle\sqrt{\frac{6k}{m}},$

$\displaystyle\mbox{and the time period}\quad T$ $\displaystyle=$ $\displaystyle\frac{2\pi}{\lambda}$

$\displaystyle=$ $\displaystyle 2\pi\sqrt{\frac{m}{6k}}.$

From the measured time for 20 oscillations, the time period (time for one oscillation) is

$T=\frac{12.5}{20}\,\mathrm{s}=0.625\,\mathrm{s}$

Now equating the measured $T$ with the derived expression for $T$ we get

$\displaystyle 2\pi\sqrt{\frac{m}{6k}}$ $\displaystyle=$ $\displaystyle 0.625\,\mathrm{s}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad k$ $\displaystyle=$ $\displaystyle 4\pi^{2}\mathord{\cdot}\frac{m}{6(0.625\,\mathrm{s})^{2}}$

$\displaystyle=$ $\displaystyle\frac{4\pi^{2}\mathord{\cdot}2\,\mathrm{kg}}{6(0.625\,\mathrm{s})% ^{2}}$

$\displaystyle=$ $\displaystyle 33.7\,{\mbox{N}}/\,\mathrm{m}.$

Answer: $k=33.7\,{\mbox{N}}/\,\mathrm{m}$
3.

If the two linear springs are to be replaced by a torsional spring at the bottom, we can find the spring constant of the torsional spring by comparison. Let $k_{\rm{tor}}$ be the spring constant of the torsional spring. Then, as shown in the free-body diagram (see figure 16.105), the restoring torque applied by the spring at an angular displacement $\theta$ is $k_{\rm{tor}}\theta$ . Now, writing the angular momentum balance about point O, we get

Filename:pfigure-blue-36-1

Figure 16.105:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\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}$}}}$

$\displaystyle-k_{\rm{tor}}\theta\hat{\boldsymbol{k}}$ $\displaystyle=$ $\displaystyle I_{zz}^{O}(\ddot{\theta}\hat{\boldsymbol{k}})$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\ddot{\theta}+\frac% {k_{\rm{tor}}}{I_{zz}^{O}}\,\theta$ $\displaystyle=$ $\displaystyle 0.$

Comparing with the standard harmonic equation, we find the angular frequency

$\lambda=\sqrt{\frac{k_{\rm{tor}}}{I_{zz}^{O}}}=\sqrt{\frac{k_{\rm{tor}}}{\frac% {1}{3}m\ell^{2}}}.$

If this system has to have the same period of oscillation as the first system, the two angular frequencies must be equal, i.e.,

$\displaystyle\sqrt{\frac{k_{\rm{tor}}}{\frac{1}{3}m\ell^{2}}}$ $\displaystyle=$ $\displaystyle\sqrt{\frac{6k}{m}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad k_{\rm{tor}}$ $\displaystyle=$ $\displaystyle 6k\mathord{\cdot}\frac{1}{3}\ell^{2}=2k\ell^{2}$

$\displaystyle=$ $\displaystyle 2\mathord{\cdot}(33.7\,{\mbox{N}}/\,\mathrm{m})\mathord{\cdot}(1% \,\mathrm{m})^{2}$

$\displaystyle=$ $\displaystyle 67.4\,{\rm N}\mathord{\cdot}{\rm m}.$

Answer: $k_{\rm{tor}}=67.4\,{\rm N}\mathord{\cdot}{\rm m}$

SAMPLE 16.26

Filename:pfigure-s95q8

Figure 16.106:

A spring loaded seesaw: A kid, modelled as a point mass with $m=10\,\mathrm{kg}$ , is sitting at end B of a rigid rod AB of negligible mass. The rod is supported by a spring at end A and a pin at point O. The system is in static equilibrium when the rod is horizontal. Someone pushes the kid vertically downwards by a small distance $y$ and lets go. Given that $AB=3\,\mathrm{m},\;AC=0.5\,\mathrm{m},\;k=1\,{\rm kN}/\!\,\mathrm{m}$ ; find

1.

the unstretched (relaxed) length of the spring,
2.

the equation of motion (a differential equation relating the position of the mass to its acceleration) of the system, and
3.

the natural frequency of the system.

If the rod is pinned at the midpoint instead of at O, what is the natural frequency of the system? How does the new natural frequency compare with that of a mass $m$ simply suspended by a spring with the same spring constant?

Solution

Filename:pfigure4-rpi

Figure 16.107: Free-body diagrams

Static Equilibrium: The FBD of the (rod + mass) system is shown in Fig. 16.107. Let the stretch in the spring in this position be $y_{st}$ and the relaxed length of the spring be $\ell_{0}$ . The balance of angular momentum about point O gives:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/{\rm o}}=\dot% {\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/{\rm o}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}\qquad\mbox{(no % motion)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad(ky_{st})d_{1}-(mg)% d_{2}$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad y_{st}$	$\displaystyle=$	$\displaystyle\frac{mg}{k}\cdot\frac{d_{2}}{d_{1}}$
	$\displaystyle=$	$\displaystyle\frac{10\,\mathrm{kg}\cdot 9.8\,\mathrm{m}/\mathrm{s}^{2}\cdot 2% \ell}{1000\,{\mbox{N}}/\,\mathrm{m}\cdot\ell}=0.196\,\mathrm{m}$
$\displaystyle\hbox to57.81621pt{Therefore,\hfil}\ell_{0}$	$\displaystyle=$	$\displaystyle AC-y_{st}$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{m}-0.196\,\mathrm{m}=0.304\,\mathrm{m}.$

Answer: $\ell_{0}=30.4\,\mathrm{cm}$

Equation of motion: As point B gets displaced downwards by a distance $y$ , point A moves up by a proportionate distance $y_{a}$ . From geometry, ^†^†margin: Here, we are considering a very small $y$ so that we can ignore the arc the point mass B moves on and take its motion to be just vertical (i.e., $\sin\theta\approx\theta$ for small $\theta$ ).

	$\displaystyle y$	$\displaystyle\approx$	$\displaystyle d_{2}\theta\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad% \theta=\frac{y}{d_{2}}$
	$\displaystyle y_{a}$	$\displaystyle\approx$	$\displaystyle d_{1}\theta=\frac{d_{1}}{d_{2}}y$

Therefore, the total stretch in the spring, in this position,

\Delta y=y_{a}+y_{st}=\frac{d_{1}}{d_{2}}y+\frac{d_{2}}{d_{1}}\frac{mg}{k}

Now, Angular Momentum Balance about point O gives:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/{\rm o}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/{\rm o}}$
$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/{\rm o}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}\times mg\hat{% \boldsymbol{\jmath}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}\times k% \Delta y\hat{\boldsymbol{\jmath}}$	(16.89)
	$\displaystyle=$	$\displaystyle(d_{2}mg-d_{1}k\Delta y)\hat{\boldsymbol{k}}$	(16.89)
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/{\rm o}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B}\times m\ddot{y}\hat{\boldsymbol{\jmath}}$	(16.90)
	$\displaystyle=$	$\displaystyle d_{2}m\ddot{y}\hat{\boldsymbol{k}}$	(16.91)

Equating (16.89) and (16.91) we get

$\displaystyle d_{2}mg-d_{1}k\Delta y$	$\displaystyle=$	$\displaystyle d_{2}m\ddot{y}$
$\displaystyle\hbox to36.135pt{or\hfil}d_{2}mg-d_{1}k\left(\frac{d_{1}}{d_{2}}y% +\frac{d_{2}mg}{d_{1}k}\right)$	$\displaystyle=$	$\displaystyle d_{2}m\ddot{y}$
$\displaystyle\hbox to36.135pt{or\hfil}\cancel{d_{2}mg}-k\frac{d_{1}^{2}}{d_{2}% }y-\cancel{d_{2}mg}$	$\displaystyle=$	$\displaystyle d_{2}m\ddot{y}$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{y}+\frac{k}{m}\left(\frac{d_{1}}{d% _{2}}\right)^{2}y$	$\displaystyle=$	$\displaystyle 0$

Answer: $\ddot{y}+\frac{k}{m}\left(\frac{d_{1}}{d_{2}}\right)^{2}y=0$

3.

The natural frequency of the system: We may also write the previous equation as

$\ddot{y}+\lambda^{2}y=0\makebox[72.26999pt]{where}\lambda^{2}=\frac{k}{m}\frac% {d_{1}^{2}}{d_{2}^{2}}.$ (16.92)

Substituting $d_{1}=\ell$ and $d_{2}=2\ell$ in the expression for $\lambda$ we get the natural frequency of the system

$\lambda=\frac{1}{2}\sqrt{\frac{k}{m}}=\frac{1}{2}\sqrt{\frac{1000\,{\mbox{N}}/% \,\mathrm{m}}{10\,\mathrm{kg}}}=5\,\mathrm{s}^{-1}$

Answer: $\lambda=5\,\mathrm{s}^{-1}$
4.

Comparison with a simple spring mass system:

Filename:pfigure-blue-86-1

Figure 16.108:

When $d_{1}=d_{2}$ , the equation of motion (16.92) becomes

$\ddot{y}+\frac{k}{m}y=0$

and the natural frequency of the system is simply

$\lambda=\sqrt{\frac{k}{m}}$

which corresponds to the natural frequency of a simple spring mass system shown in Fig. 16.108.

In our system (with $d_{1}=d_{2}$ ) any vertical displacement of the mass at B induces an equal amount of stretch or compression in the spring which is exactly the case in the simple spring-mass system. Therefore, the two systems are mechanically equivalent. Such equivalences are widely used in modeling complex physical systems with simpler mechanical models.

Problems for 16.4 Dynamics

Preparatory Problems

16.4.1 An object consists of a massless bar with two attached masses $m_{1}$ and $m_{2}$ . The object is hinged at $O$ .

(a)

What is the moment of inertia of the object about point $O$ ( $I^{O}_{zz}$ )? Answer: $I^{cm}_{zz}=m_{1}\ell_{1}^{2}+m_{2}\ell^{2}$ .
(b)

Given $\theta$ , $\dot{\theta}$ , and $\ddot{\theta}$ , 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}$}}}$ , the angular momentum 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}$}}}=\dot{\theta}(m_{1}\ell_{1}^{2}+m_{2}\ell^{2}% )\hat{\boldsymbol{k}}$ .
(c)

Given $\theta$ , $\dot{\theta}$ , and $\ddot{\theta}$ , what is $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\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 angular momentum about point $O$ ? 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}$}}}=\ddot{\theta}(m_{1}\ell_{1}^{2}+m_{2}\ell^% {2})\hat{\boldsymbol{k}}$ .
(d)

Given $\theta$ , $\dot{\theta}$ , and $\ddot{\theta}$ , what is $T$ , the total kinetic energy? Answer: $E_{\rm{K}}=\frac{\dot{\theta}^{2}}{2}(m_{1}\ell_{1}^{2}+m_{2}\ell^{2})\hat{% \boldsymbol{k}}$ .
(e)

Assume that you don’t know $\theta$ , $\dot{\theta}$ or $\ddot{\theta}$ but you do know that $F_{1}$ is applied to the rod, perpendicular to the rod at $m_{1}$ . What is $\ddot{\theta}$ ? (Neglect gravity.) Answer: $\ddot{\theta}=\frac{F_{1}\ell_{1}}{m_{1}\ell_{1}^{2}+m_{2}\ell^{2}}$
(f)

If $F_{1}$ were applied to $m_{2}$ instead of $m_{1}$ , would $\ddot{\theta}$ be bigger or smaller? Answer: $\ddot{\theta}$ will be bigger if $F_{1}$ is applied to $m_{2}$ instead of $m_{1}$ .

Filename:pfigure-blue-88-1

Figure 16.109:

16.4.2 A uniform circular disc rotates at constant angular speed $\omega$ about the origin, which is also the center of the disc. It’s radius is $R$ . It’s total mass is $M$ .

(a)

What is the total force and moment required to hold it in place (use the origin as the reference point of angular momentum and torque). Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=0\,{\mbox{N}}$ .
(b)

What is the total kinetic energy of the disk? Answer: $E_{\rm{K}}=\frac{1}{4}MR^{2}\omega^{2}$ .

Filename:pfigure4-2-rp3

Figure 16.110:

16.4.3 The hinged disk of mass $m$ (uniformly distributed) is acted upon by a force $P$ shown in the figure. Determine the initial angular acceleration and the reaction forces at the pin $O$ . Answer: $\ddot{\theta}_{0}=\frac{P}{mR}$ and ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{1}=P(\sin{30}\hat{\boldsymbol{% \imath}}-\cos{30}\hat{\boldsymbol{\jmath}})$ .

Filename:pfigure-s94q6p1

Figure 16.111:

16.4.4 A motor turns a bar. A uniform bar of length $\ell$ and mass $m$ is turned by a motor whose shaft is attached to the end of the bar at O. The angle that the bar makes (measured counter-clockwise) from the positive $x$ axis is $\theta=2\pi\,\,t^{2}/\hbox{s}^{2}$ . Neglect gravity.

(a)

Draw a free-body diagram of the bar.
(b)

Find the force acting on the bar from the motor and hinge at $t=1\,$ s. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{\rm{BM}}=-\frac{4\pi m\ell}{2}% \left(4\pi\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}\right)$ .
(c)

Find the torque applied to the bar from the motor at $t=1\,$ s. Answer: $M=\frac{4\pi\,\mathrm{rad}/\mathrm{s}^{2}}{3}m\ell^{2}$ .
(d)

What is the power produced by the motor at $t=1\,$ s? Answer: $P=\frac{16\pi^{2}\,\mathrm{s}^{-3}}{3}m\ell^{2}$ .

Filename:pfigure-spinningbrick

Figure 16.112:

16.4.5 A physical pendulum. A swinging stick is sometimes called a ‘physical’ pendulum. Take the ‘body’, the system of interest, to be the whole stick.

(a)

Draw a free-body diagram of the system.
(b)

Write the equation of angular momentum balance for this system about point $O$ .
(c)

Evaluate the left-hand-side as explicitly as possible in terms of the forces showing on your Free-Body Diagram.
(d)

Evaluate the right hand side as completely as possible. You may use the following facts:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ $\displaystyle=$ $\displaystyle l\dot{\theta}\cos\theta\hat{\boldsymbol{\jmath}}+-l\dot{\theta}% \sin\theta\hat{\boldsymbol{\imath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ $\displaystyle=$ $\displaystyle-l\dot{\theta}^{2}\left[\cos\theta\hat{\boldsymbol{\imath}}+\sin% \theta\hat{\boldsymbol{\jmath}}\right]$

$\displaystyle\qquad+l\ddot{\theta}\left[\cos\theta\hat{\boldsymbol{\jmath}}-% \sin\theta\hat{\boldsymbol{\imath}}\right]$

where $\ell$ is the distance along the pendulum from the top, $\theta$ is the angle by which the pendulum is displaced counter-clockwise from the vertically down position, $\hat{\boldsymbol{\imath}}$ is vertically down, and $\hat{\boldsymbol{\jmath}}$ is to the right. You will have to set up and evaluate an integral. Answer: $\sum{{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{O}}=-\frac{\ell}{2}mg\sin{% \theta}\hat{\boldsymbol{k}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_% {\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}=\frac{m\ell^{2}}{3}% \ddot{\theta}\hat{\boldsymbol{k}}$ .

Filename:pfigure-s94f1p2

Figure 16.113:

16.4.6 A uniform one meter bar is hung from a hinge that is at the end. It is allowed to swing freely. $g=10\,\mathrm{m}/\mathrm{s}^{2}$ .

(a)

What is the period of small oscillations for this pendulum? Answer: $T=1.62\,\mathrm{s}$ .
(b)

Suppose the rod is hung $0.4\,\mathrm{m}$ from one end. What is the period of small oscillations for this pendulum? Can you explain why it is longer or shorter than when it is hung by its end? Answer: The period of oscillations is longer if the rod is hung at $0.4\,\mathrm{m}$ because the torque due to weight of the portion above the hinge negates the torque due to the portion below the hinge. One way to see it is that if the bar is hinged at mid, net torque about the hinge becomes zero leading to infinitely large period of oscillation.

More-Involved Problems

16.4.7 Motor turns a dumbbell. Two uniform bars of length $\ell$ and mass $m$ are welded at right angles. At the ends of the horizontal bar are two more masses $m$ . The bottom end of the vertical rod is attached to a hinge at O where a motor keeps the structure rotating at constant rate $\omega$ (counter-clockwise). What is the net force and moment that the motor and hinge cause on the structure at the instant shown? Answer: $F_{x}=0,F_{y}=\frac{7\ell m\omega^{2}}{2}$ , ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{O}=0\hat{\boldsymbol{k}}$ .

Filename:s97f2

Figure 16.114:

16.4.8 The structure shown in the figure consists of two point masses connected by three rigid, massless rods such that the whole structure behaves like a rigid body. The structure rotates counterclockwise at a constant rate of $60\,{\rm rpm}$ . At the instant shown, find the force in each rod. Answer: $F_{1}=0\,{\mbox{N}}$ , $F_{2}=2\pi^{2}\,{\mbox{N}}$ , and $F_{3}=2\sqrt{2}\pi^{2}\,{\mbox{N}}$ .

Filename:pg92-2

Figure 16.115:

16.4.9 Balancing a system of rotating particles. A wire frame structure is made of four concentric loops of massless and rigid wires, connected to each other by four rigid wires presently coincident with the $x$ and $y$ axes. Three masses, $m_{1}=200\,{\mathrm{grams}}$ , $m_{2}=150\,{\mathrm{grams}}$ and $m_{3}=100\,{\mathrm{grams}}$ , are glued to the structure as shown in the figure. The structure rotates counter-clockwise at a constant rate $\dot{\theta}=5\,\mathrm{rad}/\mathrm{s}$ . There is no gravity.

(a)

Find the net force exerted by the structure on the support at the instant shown. Answer: (a) ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=0.33\,{\mbox{N}}\hat{\boldsymbol% {\imath}}-0.54\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$ .
(b)

You are to put a mass $m$ at an appropriate location on the third loop so that the net force on the support is zero. Find the appropriate mass and the location on the loop. Answer: The net force on the support becomes zero if we put $m=0.84\,\mathrm{kg}$ on $0.3\,\mathrm{m}$ radius loop at $\theta=302.33^{\circ}$ .

Filename:summer95p2-3

Figure 16.116:

16.4.10 Motor turns a bent bar. Two uniform bars of length $\ell$ and uniform mass $m$ are welded at right angles. One end is attached to a hinge at O where a motor keeps the structure rotating at a constant rate $\omega$ (counterclockwise). What is the net force and moment that the motor and hinge cause on the structure at the instant shown.

(a)

neglecting gravity Answer: Net force: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{net}=-(\frac{3m\omega^{2}L}{2})% \hat{\boldsymbol{\imath}}-(\frac{m\omega^{2}L}{2})\hat{\boldsymbol{\jmath}}$ , Net moment: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ .
(b)

including gravity. Answer: Net force: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{net}=-(\frac{3m\omega^{2}L}{2})% \hat{\boldsymbol{\imath}}+(2mg-\frac{m\omega^{2}L}{2})\hat{\boldsymbol{\jmath}}$ , Net moment: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{net}=\frac{3mgL}{2}\hat{% \boldsymbol{k}}$ .

Filename:p-f96-p3-3

Figure 16.117:

A bent bar is rotated by a motor.

16.4.11 A uniform disk of mass $M$ and radius $R$ rotates about a hinge $O$ in the $x y$ -plane. A point mass $m$ is fixed to the disk at a distance $R/2$ from the hinge. A motor at the hinge drives the disk/point mass assembly with constant angular acceleration $\alpha$ . What torque at the hinge does the motor supply to the system? Answer: $T=\frac{\alpha R^{2}}{4}(2M+m)$ .

16.4.12 A rigid rod of length $\ell$ and total mass $m$ is held fixed at one end and whirled around in circular motion at a constant rate $\omega$ in the horizontal plane. Ignore gravity.

(a)

Find the tension in the rod as a function of $r$ , the radial distance from the center of rotation to any desired location on the rod. Answer: $T(r)=\frac{m\omega^{2}}{2L}(L^{2}-r^{2})$
(b)

Where does the maximum tension occur in the rod? Answer: at $r=0$ ; i.e., at the center of rotation.
(c)

At what distance from the center of rotation does the tension drop to half its maximum value? Answer: $r=L/\sqrt{2}$ .

16.4.13 A thin uniform circular disc of mass $M$ and radius $R$ rotates in the $x y$ plane about its center of mass point $O$ . Driven by a motor, it has rate of change of angular speed proportional to angular position, $\alpha=k\theta^{5/2}$ . The disc starts from rest at $\theta=0$ .

(a)

What is the rate of change of angular momentum about the origin at $\theta=\frac{\pi}{3}\,{\rm rad}$ ? 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}$}}}=\left(k\frac{\pi}{3}\right)^{2.5}\left(% \frac{MR^{2}}{2}\right)\hat{\boldsymbol{k}}$ .
(b)

What is the torque of the motor at $\theta=\frac{\pi}{3}\,{\rm rad}$ ? Answer: $T=k\left(\frac{\pi}{3}\right)^{2.5}\left(\frac{MR^{2}}{2}\right)$ .
(c)

What is the total kinetic energy of the disk at $\theta=\frac{\pi}{3}\,{\rm rad}$ ? Answer: $E_{\rm{K}}=k^{2}\left(\frac{\pi}{3}\right)^{3.5}\left(\frac{MR^{2}}{2}\right)$ .

16.4.14 Neglecting gravity, calculate $\alpha=\dot{\omega}=\ddot{\theta}$ at the instant shown for the system in the figure. Answer: $\ddot{\theta}=100\,{{\rm rev}/{\rm sec}}$ .

Filename:s97p3-3

Figure 16.118:

16.4.15 The uniform square shown is released from rest at $t=0$ . What is $\alpha={\dot{\omega}}=\ddot{\theta}$ immediately after release? Answer: $\ddot{\theta}=-\frac{3g}{2L}\sin\theta$

Filename:pfigure-s94h8p1

Figure 16.119:

16.4.16 Acceleration of a trap door. A uniform bar AB of mass $m$ and a ball of the same mass are released from rest from the same horizontal position. The bar is hinged at end A. There is gravity.

(a)

Which point on the rod has the same acceleration as the ball, immediately after release. Answer: Point at $2L/3$ from A
(b)

What is the reaction force on the bar at end A just after release? Answer: $mg/4$ directed upwards.

Filename:sfig4-7-DH1

Figure 16.120:

16.4.17 A disk with radius $R$ has a string wrapped around it which is pulled with a force $F$ . The disk is free to rotate about the axis through $O$ normal to the page. The moment of inertia of the disk about $O$ is $I_{o}$ . A point $A$ is marked on the string. Given that $x_{A}$ (0) = 0 and that ${\dot{x}}_{A}$ (0) = 0, what is $x_{A}(t)$ ? Answer: $x_{A}(t)=\frac{Ft^{2}}{2\left(m+\frac{I}{R^{2}}\right)}$

Filename:sfig4-7-DH2

Figure 16.121:

16.4.18 A uniform stick with length $\ell$ and mass $M_{o}$ is welded to a pulley hinged at the center $O$ . The pulley has negligible mass and radius $R_{p}$ . A string is wrapped many times around the pulley. At time $t=0$ , the pulley, stick, and string are at rest and a force $F$ is suddenly applied to the string. How long does it take for the pulley to make one full revolution? Ignore gravity. Answer: $T_{rev}=\sqrt{\frac{2M_{O}\ell^{2}\pi}{3FR_{p}}}$ .

Filename:sfig4-7-DH3

Figure 16.122:

String wraps around a pulley with a stick glued to it.

Gears, Belt Drives, and Gear Trains

16.4.19 Constant speed gear train. Gear A is connected to a motor (not shown) and gear B, which is welded to gear C, is connected to a taffy-pulling mechanism. Assume you know the torque $M_{\rm input}=M_{A}$ and angular velocity $\omega_{\rm{input}}=\omega_{A}$ of the input shaft. Assume the bearings and contacts are frictionless.

(a)

What is the input power? Answer: $P_{\rm in}=M_{A}\omega_{A}$ .
(b)

What is the output power? Answer: $P_{\rm out}=M_{A}\omega_{A}$ .
(c)

What is the output torque $M_{\rm{output}}=M_{C}$ , the torque that gear C applies to its surroundings in the clockwise direction? Answer: $M_{C}=M_{A}\frac{R_{B}}{R_{A}}$

Filename:sfig4-7-1

Figure 16.123:

16.4.20 At the input to a gear box a $100\,{\rm lbf}$ force is applied to gear A. At the output, the machinery (not shown) applies a force of $F_{B}$ to the output gear. Gear $A$ rotates at constant angular rate $\omega=2\,\mathrm{rad}/\mathrm{s}$ , clockwise.

(a)

What is the angular speed of the right gear? Answer: $\omega_{B}=\omega\frac{R_{A}}{R_{C}}$
(b)

What is the velocity of point $P$ ? Answer: $V_{P}=\frac{\omega R_{A}R_{B}}{R_{C}}$
(c)

What is $F_{B}$ ? Answer: $F_{B}=\frac{F_{A}R_{C}}{R_{B}}$
(d)

If the gear bearings had friction, would $F_{B}$ have to be larger or smaller in order to achieve the same constant velocity? Answer: $F_{B}$ will have to be smaller.
(e)

If instead of applying a $100\,{\rm lbf}$ to the left gear it is driven by a motor (not shown) at constant angular speed $\omega$ , what is the angular speed of the right gear? Answer: The left gear is constrained to rotate at $\omega_{B}=\frac{\omega R_{A}}{R_{C}}$ .

Filename:sfig4-7-1a

Figure 16.124:

Two gears with end loads.

16.4.21 A bevel gear system. A bevel-type gear system, shown in the figure, is used to transmit power between two shafts that are perpendicular to each other. The driving gear has a mean radius of $50\,{\rm mm}$ and rotates at a constant speed $\omega=150\,{\rm rpm}$ . The mean radius of the driven gear is $80\,{\rm mm}$ and the driven shaft is expected to deliver a torque of $M_{out}=25\,{\rm N}\mathord{\cdot}{\rm m}$ . Assuming no power loss, find the input torque supplied by the driving shaft. Answer: $M_{\rm in}=15.625\,{\mbox{N}}\cdot\,\mathrm{m}$ .

Filename:sfig4-7-1b

Figure 16.125:

A bevel gear

16.4.22 Belt drives are used to transmit power between parallel shafts. Two parallel shafts, $3\,\mathrm{m}$ apart, are connected by a belt passing over the pulleys $A$ and $B$ fixed to the two shafts. The driver pulley $A$ rotates at a constant $200\,{\rm rpm}$ . The speed ratio between the pulleys $A$ and $B$ is 1:2.5. The input torque is $350\,{\rm N}\mathord{\cdot}{\rm m}$ . Assume no loss of power between the two shafts.

(a)

Find the input power. Answer: $P_{in}=7.33\mbox{kilo-watts}$
(b)

Find the rotational speed of the driven pulley $B$ Answer: $500\,{\rm rpm}$ .
(c)

Find the output torque at $B$ . Answer: $M_{out}=140\,{\rm N}\mathord{\cdot}{\rm m}$

Filename:sfig4-7-2

Figure 16.126:

16.4.23 In the belt drive system shown, assume that the driver pulley rotates at a constant angular speed $\omega$ . If the motor applies a constant torque $M_{O}$ on the driver pulley, show that the tensions in the two parts, $A B$ and $C D$ , of the belt must be different. Which part has a greater tension? Does your conclusion about unequal tension depend on whether the pulley is massless or not? Assume any dimensions you need. Answer: $M_{out}=140\,{\rm N}\mathord{\cdot}{\rm m}$

Filename:sfig4-7-2a

Figure 16.127:

16.4.24 Two racks connected by three constant rate gears. A $100\,{\rm lbf}$ force is applied to one rack. At the output, the machinery (not shown) applies a force of $F_{B}$ to the other rack.

(a)

Assume the gear-train is spinning at constant rate and is frictionless. What is $F_{B}$ ? Answer: $F_{B}=100\,{\rm lbf}$ .
(b)

If the gear bearings had friction would that increase or decrease $F_{B}$ to achieve the same constant rate?
(c)

If instead of applying a $100\,{\rm lbf}$ to the left rack it is driven by a motor (not shown) at constant speed $v$ , what is the speed of the right rack? Answer: $v_{right}=v$ .
(d)

If the angular velocity of the gear is increasing at rate $\alpha$ does this increase or decrease $F_{B}$ at the given $\omega$ .

Filename:sfig4-7-2b

Figure 16.128:

Two racks connected by three gears.

16.4.25 -3-D accelerating gear train. This is really a 2-D problem; each gear turns in a different parallel plane. Shaft B is rigidly connected to gears $G_{4}$ and $G_{5}$ . $G_{3}$ meshes with gear $G_{6}$ . Gears $G_{6}$ and $G_{5}$ are both rigidly attached to shaft AD. Gear $G_{5}$ meshes with $G_{2}$ which is welded to shaft A. Shaft A and shaft B spin independently. Assume you know the torque $M_{\mbox{input}}$ , angular velocity $\omega_{\mbox{input}}$ and the angular acceleration $\alpha_{\mbox{input}}$ of the input shaft. Assume the bearings and contacts are frictionless.

(a)

What is the input power? Answer: $P_{\rm in}=2500\pi\,{\mbox{N}}\cdot\,\mathrm{m}/\mathrm{s}$ .
(b)

What is the output power? Answer: $P_{\rm out}=2500\pi\,{\mbox{N}}\cdot\,\mathrm{m}/\mathrm{s}$ .
(c)

What is the angular velocity $\omega_{\mbox{output}}$ of the output shaft? Answer: $\omega_{\mbox{output}}=25\pi\,\mathrm{rad}/\mathrm{s}$ .
(d)

What is the output torque $M_{\mbox{output}}$ ? Answer: $M_{\mbox{output}}=100\,{\mbox{N}}\cdot\,\mathrm{m}$ .

Filename:sfig4-7-2c

Figure 16.129:

A 3-D gear train.

16.4.26 Gear A with radius $R_{A}=400\,{\rm mm}$ is rigidly connected to a drum B with radius $R_{B}=200\,{\rm mm}$ . The combined moment of inertia of the gear and the drum about the axis of rotation is $I_{zz}~{}=~{}0.5~{}\,\mathrm{kg}~{}\cdot~{}\,\mathrm{m}^{2}$ . Gear A is driven by gear C which has radius $R_{C}=300\,{\rm mm}$ . As the drum rotates, a $5\,\mathrm{kg}$ mass $m$ is pulled up by a string wrapped around the drum. At the instant of interest, The angular speed and angular acceleration of the driving gear are $60\,{\rm rpm}$ and $12\,{\rm rpm}/{\rm s}$ , respectively. Find the acceleration of the mass $m$ . Answer: $a_{m}=0.188\,\mathrm{m}/\mathrm{s}^{2}$

Filename:pfigure-s94q7p1

Figure 16.130:

16.4.27 A spindle and pulley arrangement is used to hoist a $50\,\mathrm{kg}$ mass as shown in the figure. Assume that the pulley is to be of negligible mass. When the motor is running at a constant $100\,{\rm rpm}$ ,

(a)

Find the velocity of the mass at $B$ . Answer: $V_{B}=3.77\,\mathrm{m}/\mathrm{s}$ .
(b)

Find the tension in strings $A B$ and $C D$ . Answer: $T_{\rm AB}=500\,{\mbox{N}}$ , $T_{\rm CD}=1000\,{\mbox{N}}$ .

Filename:pfigure-s94h7p2

Figure 16.131:

16.4.28 Accelerating rack and pinion. The two gears shown are welded together and spin on a frictionless bearing. The inner gear has radius $0.5\,\mathrm{m}$ and negligible mass. The outer disk has $1\,\mathrm{m}$ radius and a uniformly distributed mass of $0.2\,\mathrm{kg}$ . They are loaded as shown with the force $F=20\,{\mbox{N}}$ on the massless rack which is held in place by massless frictionless rollers. At the time of interest the angular velocity is $\omega=2\,\mathrm{rad}/\mathrm{s}$ (though $\omega$ is not constant). The point P is on the disk a distance $1\,\mathrm{m}$ from the center. At the time of interest, point P is on the positive $y$ axis.

(a)

What is the speed of point P? Answer: $\left|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{p}\right|=2\,\mathrm{m}/% \mathrm{s}$ .
(b)

What is the velocity of point P? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{p}=-2\hat{\boldsymbol{\imath}}% \,\mathrm{m}/\mathrm{s}$ .
(c)

What is the angular acceleration $\alpha$ of the gear? Answer: $\alpha=100\,\mathrm{rad}/\mathrm{s}^{2}$ .
(d)

What is the acceleration of point P? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{p}=(100\hat{\boldsymbol{\imath}% }-4\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}$ .
(e)

What is the magnitude of the acceleration of point P? Answer: $\left|{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{p}\right|=100.08\,\mathrm% {m}/\mathrm{s}^{2}$ .
(f)

What is the rate of increase of the speed of point P? Answer: $\frac{{\rm d}\left|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{a}\right|}{{% \rm d}t}=-100\,\mathrm{m}/\mathrm{s}^{2}$ .

Filename:ch4-7-c

Figure 16.132:

Accelerating rack and pinion

16.4.29 Two racks connected by a gear. A $100\,{\rm lbf}$ force is applied to one rack. At the output the machinery (not shown) applies a constant force $F_{B}$ to the other rack.

(a)

Assume the gear is spinning at constant rate and is frictionless. What is $F_{B}$ ? Answer: $F_{B}=100\,{\rm lbf}$ .
(b)

If the gear bearing had friction, would that increase or decrease $F_{B}$ to achieve the same constant rate? Answer: If gear bearing has friction, $F_{B}$ has to increase.
(c)

If the angular velocity of the gear is increasing at rate $\alpha$ , does this increase or decrease $F_{B}$ at the given $\omega$ . Answer: $F_{B}$ will decrease if the angular velocity of the gear is increasing.
(d)

If the output load $F_{B}$ is given then the motion of the machine can be found, assuming friction is negligible. Assume that the machine starts from rest with a given output load. So long as rack B moves down and the output force $F_{B}$ is positive, the output power is positive. The machine starts from rest and runs for a fixed amount of time $T$ . For what value of $F_{B}$ is the output work maximum? Once you find an answer, can you find an intuitive explanation for the answer? Answer: For work to be maximum, $F_{B}=50\,{\rm lbf}$ . Initially work increases with $F_{B}$ but too large $F_{B}$ leads to small $\omega_{0}$ thus giving smaller net work.

Filename:pfigure-blue-87d-1

Figure 16.133:

Two racks connected by a gear.

16.4.30 A rack and pinion constrained by a linear spring. Neglect gravity. The spring is relaxed when the angle $\theta=0$ . Assume the system is released from rest at $\theta=\theta_{0}$ . What is the acceleration of the point $P$ at the end of the stick when $\theta=0$ ? Answer in terms of any or all of $m$ , $R$ , $\ell$ , $\theta_{0}$ , $k$ , $\hat{\boldsymbol{\imath}}$ , and $\hat{\boldsymbol{\jmath}}$ . [Hint: There are several steps of reasoning required. You might want to draw FBD(s), use angular momentum balance, set up a differential equation, solve it, plug values into this solution, and use the result to find the quantities of interest. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-\frac{3\theta_{0}^{2}R^{2}k% }{\ell m}\hat{\boldsymbol{\jmath}}$

Filename:Mikef91p3

Figure 16.134:

Pendulum Problems

16.4.31 A small particle of mass $m$ is attached to the end of a thin rod of mass M (uniformly distributed), which is pinned at hinge O, as depicted in the figure.

(a)

Obtain the equation of motion governing the rotation $\theta$ of the rod. Answer: $\left(\frac{(M+4m)\ell^{2}}{9}\right)\ddot{\theta}+\left(\frac{(M+4m)g\ell}{6}% \right)\sin\theta=0$ .
(b)

What is the natural frequency of the system for small oscillations $\theta$ ? Answer: $f=\frac{1}{2\pi}\sqrt{\frac{3g(M+4m)}{2\ell(M+4m)}}$

Filename:pfigure-s94h7p3

Figure 16.135:

16.4.32 A rigid massless rod has two equal masses $m_{B}$ and $m_{C}$ ( $m_{B}=m_{C}=m$ ) attached to it at distances $2\ell$ and $3\ell$ , respectively, measured along the rod from a frictionless hinge located at a point $A$ . The rod swings freely from the hinge. There is gravity. Let $\phi$ denote the angle of the rod measured from the vertical. Assume that $\phi$ and $\dot{\phi}$ are known at the instant of interest.

(a)

What is $\ddot{\phi}$ ? Find $\ddot{\phi}$ in terms of $m$ , $\ell$ , $g$ , $\phi$ and $\dot{\phi}$ . Answer: $\ddot{\phi}=-\frac{5g}{13\ell}\sin{\phi}$ .
(b)

What is the force of the hinge on the rod? Solve in terms of $m$ , $\ell$ , $g$ , $\phi$ , $\dot{\phi}$ , $\ddot{\phi}$ and any unit vectors you may need to define. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=-5m\ell(\dot{\phi}^{2}cos\phi+% \ddot{\phi}\sin\phi)\hat{\boldsymbol{\imath}}-2mg\hat{\boldsymbol{\imath}}+5ml% (\ddot{\phi}\cos\phi-\dot{\phi}^{2}\sin\phi)\hat{\boldsymbol{\jmath}}$ .
(c)

Would you get the same answers if you put a mass $2m$ at $2.5\ell$ ? Why or why not? Answer: Answer will change because although total moment of force about the hinge stays same but moment of inertia reduces. In this case we get $\ddot{\phi}=-\frac{5g}{12.5\ell}\sin\phi$ .

Filename:p-f96-p2-3

Figure 16.136:

16.4.33 For the pendula in the figure :

(a)

Without doing any calculations, try to figure out the relative durations of the periods of oscillation for the five pendula (i.e. the order, slowest to fastest) Assume small angles of oscillation.
(b)
Calculate the period of small oscillations. [Hint: use balance of angular momentum about the point 0]. Answer:
1. (b)
  
  $T=2.29\,\mathrm{s}$
2. (e)
  
  $T=1.99\,\mathrm{s}$
(b) has a longer period than (e) does since in (b) the moment of inertia about the center of mass (located at the same position as the mass in (e)) is non-zero.
(c)

Rank the relative duration of oscillations and compare to your intuitive solution in part (a), and explain in words why things work the way they do.

Filename:s97p3-2

Figure 16.137:

16.4.34 A pegged compound pendulum. A uniform bar of mass $m$ and length $\ell$ hangs from a peg at point C and swings in the vertical plane about an axis passing through the peg. The distance $d$ from the center of mass of the rod to the peg can be changed by putting the peg at some other point along the length of the rod.

(a)

Find the angular momentum of the rod about point C. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{c}=\left(\frac{m\ell^{2}}{12}+% md^{2}\right)\dot{\theta}\hat{\boldsymbol{k}}$ .
(b)

Find the rate of change of angular momentum of the rod about C. Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{c}=\left(\frac{m\ell^{2}}{1% 2}+md^{2}\right)\ddot{\theta}\hat{\boldsymbol{k}}$ .
(c)

How does the period of the pendulum vary with $d$ ? Show the variation by plotting the period against $d\over\ell$ . [Hint, you must first find the equations of motion, linearize for small $\theta$ , and then solve.] Answer: $T={2\pi\over\sqrt{g/\ell}}\sqrt{{1\over 12(d/\ell)}+{d\over\ell}}$
(d)

Find the total energy of the rod (using point C as a datum for potential energy). Answer: $E_{\rm T}=\frac{1}{2}\left(\frac{m\ell^{2}}{12}+md^{2}\right)\dot{\theta}^{2}-% mgd\cos\theta$ .
(e)

Find $\ddot{\theta}$ when $\theta=\pi/6$ . Answer: $\ddot{\theta}=-\frac{6gd}{12d^{2}+\ell^{2}}$ .
(f)

Find the reaction force on the rod at C, as a function of $m$ , $d$ , $\ell$ , $\theta$ , and $\dot{\theta}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{c}=-m\left(\frac{12gd}{12d^{2}+% \ell^{2}}\cos\theta\sin\theta+\dot{\theta}^{2}d\sin\theta\right)\hat{% \boldsymbol{\imath}}+\left(mg-\frac{12mgd}{12d^{2}+\ell^{2}}\sin{\theta}^{2}+m% \dot{\theta}^{2}d\cos\theta\right)\hat{\boldsymbol{\jmath}}$ .
(g)

For the given rod, what should be the value of $d$ (in terms of $\ell$ ) in order to have the fastest pendulum? Answer: $d=0.29\,\ell$
(h)

Test of Schuler’s pendulum. The pendulum with the value of $d$ obtained in (g) is called the Schuler’s pendulum. It is not only the fastest pendulum but also the “most accurate pendulum”. The claim is that even if $d$ changes slightly over time due to wear at the support point, the period of the pendulum does not change much. Verify this claim by calculating the percent error in the time period of a pendulum of length $\ell=1\,\mathrm{m}$ under the following three conditions: (i) initial $d=0.15\,\mathrm{m}$ and after some wear $d=0.16\,\mathrm{m}$ , (ii) initial $d=0.29\,\mathrm{m}$ and after some wear $d=0.30\,\mathrm{m}$ , and (iii) initial $d=0.45\,\mathrm{m}$ and after some wear $d=0.46\,\mathrm{m}$ . Which pendulum shows the least error in its time period? What is the connection between this result and the plot obtained in (c)? Answer: The error in the time period of the pendulum for the three cases is (i) $1.767\%$ , (ii) $0.034\%$ , and (iii) $0.470\%$ respectively.

Filename:pfigure-dynbalancers

Figure 16.138:

16.4.35 A uniform stick of length $\ell$ and mass $m$ is a hair away from vertically up position when it is released with no angular velocity (a ‘hair’ is a technical word that means ‘very small amount, zero for some purposes’). It falls to the right. What is the force on the stick at point $O$ when the stick is horizontal. Solve in terms of $\ell$ , $m$ , $g$ , $\hat{\boldsymbol{\imath}}$ , and $\hat{\boldsymbol{\jmath}}$ . Carefully define any coordinates, base vectors, or angles that you use. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=-\frac{m\ell}{2}\dot{\theta}^{2}% \hat{\boldsymbol{\imath}}+\frac{7mg}{4}\hat{\boldsymbol{\jmath}}$ .

Filename:pfigure-s94h7p5

Figure 16.139:

16.4.36 A massless 10 meter long bar is supported by a frictionless hinge at one end and has a $3.759\,$ kg point mass at the other end. It is released at $t=0$ from a tip angle of $\phi=.02\,$ radians measured from vertically upright position (hinge at the bottom). Use $g=10{\rm m}\,\mathrm{s}^{-2}$ .

(a)

Using a small angle approximation and the solution to the resulting linear differential equation, find the angle of tip at $t=1$ s and $t=7$ s. Use a calculator, not a numerical integrator. Answer: Angle of tip is $0.031\,{\rm rad}$ at $t=1\,\mathrm{s}$ and $10.967\,{\rm rad}$ at $t=7\,\mathrm{s}$ .
(b)

Using numerical integration of the non-linear differential equation for an inverted pendulum find $\phi$ at $t=1$ s and $t=7$ s.
(c)

Make a plot of the angle versus time for your numerical solution. Include on the same plot the angle versus time from the approximate linear solution from part (a).
(d)

Comment on the similarities and differences in your plots.

16.4.37 A zero length spring (relaxed length $\ell_{0}=0$ ) with stiffness $k=5\,{\rm N}/{\rm m}$ supports the pendulum shown.

(a)

Find $\ddot{\theta}$ assuming $\dot{\theta}=2\,\mathrm{rad}/\mathrm{s}$ , $\theta=\pi/2$ . Answer: $\ddot{\theta}=0\ \,\mathrm{rad}/\mathrm{s}^{2}$ .
(b)

Find $\ddot{\theta}$ as a function of $\dot{\theta}$ and $\theta$ (and $k$ , $\ell$ , $m$ , and $g$ .) Answer: $\ddot{\theta}=\frac{\sin\theta}{m\ell}(Dk-mg)$ .

[Hint: use vectors (otherwise it’s hard)]

[Hint: For the special case, $kD=mg$ , the solution simplifies greatly.]

Filename:pfigure-blue-81-1

Figure 16.140:

16.4.38 A spring-mass-damper system is depicted in the figure. The horizontal damping force applied at B is given by $F_{D}=-c\dot{y}_{B}$
The dimensions and parameters are as follows:

$\displaystyle r_{B/0}$	$\displaystyle=$	$\displaystyle 2ft$
$\displaystyle r_{A/0}$	$\displaystyle=$	$\displaystyle\ell=3ft$
$\displaystyle k$	$\displaystyle=$	$\displaystyle 2\,{\rm lbf}/\,\mathrm{ft}$
$\displaystyle c$	$\displaystyle=$	$\displaystyle 0.3\,{\rm lbf}\cdot\,\mathrm{s}/\,\mathrm{ft}$

For small $\theta$ , assume that $\sin(\theta)\approx\theta$ and $\cos(\theta)\approx 1$ .

(a)

Determine the natural circular frequency of small oscillations about equilibrium for the pendulum shown. The static equilibrium position is $\theta=0$ (pendulum hanging vertically), so the spring is at its rest point in this position.

Idealize the pendulum as a point mass attached to a rigid massless rod of length $1$ , so $I^{O}=m\ell^{2}$ . Also use the “small angle approximation” where appropriate. Answer: $\omega_{n}=\sqrt{\frac{k\ell_{B}^{2}+\ell_{A}mg}{m\ell_{A}^{2}}}$ .
(b)

Sketch a graph of $\theta$ as a function of $t(t\geq 0)$ if the pendulum is released from rest at position $\theta=0.2\,{\rm rad}$ when $t=0$ . Your graph should show the correct qualitative behavior, but calculations are not necessary.

Filename:pfigure-blue-137-2

Figure 16.141:

16.4.39 The asymmetric dumbbell shown in the figure is pivoted in the center and also attached to a spring at one quarter of its length from the bigger mass. When the bar is horizontal, the compression in the spring is $y_{s}$ . At the instant of interest, the bar is at an angle $\theta$ from the horizontal; $\theta$ is small enough so that $y\approx\frac{L}{2}\theta$ . If, at this position, the velocity of mass ‘ $m$ ’ is $v\hat{\boldsymbol{\jmath}}$ and that of mass $3m$ is $-v\hat{\boldsymbol{\jmath}}$ , evaluate the power term ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}$ ) in the energy balance equation. Answer: $P=2mg-\frac{kv}{2}\left(\frac{L\theta}{4}+y_{s}\right)$ .

Filename:pfigure-blue-78-2

Figure 16.142:

16.4.40 The dumbbell shown in the figure has a torsional spring with spring constant $k$ (torsional stiffness units are $\frac{\,{\rm kN}\cdot\,\mathrm{m}}{\,{\rm rad}}$ ). The dumbbell oscillates about the horizontal position with small amplitude $\theta$ . At an instant when the angular velocity of the bar is $\dot{\theta}\hat{\boldsymbol{k}}$ , the velocity of the left mass is $-L\dot{\theta}\hat{\boldsymbol{\jmath}}$ and that of the right mass is $L\dot{\theta}\hat{\boldsymbol{\jmath}}$ . Find the expression for the power $P$ of the spring on the dumbbell at the instant of interest. Answer: $P=mgL\dot{\theta}-k\theta\dot{\theta}$ .

Filename:pfigure-blue-130-2

Figure 16.143:

16.4.41 A square plate with side $\ell$ and mass $m$ is hinged at one corner in a gravitational field $g$ . Find the period of small oscillation. Answer: $T=2\pi\sqrt{\frac{2\sqrt{2}\ell}{3g}}$ .

Filename:pfigure-s94h10p4

Figure 16.144:

16.4.42 A thin hoop of radius $R$ and mass $M$ is hung from a point on its edge and swings in its plane. Assuming it swings near to the position where its center of mass $G$ is below the hinge:

(a)

What is the period of its swinging oscillations? Answer: $T=2\pi\sqrt{\frac{2R}{g}}$ .
(b)

If, instead, the hoop was set to swinging in and out of the plane would the period of oscillations be greater or less? Answer: The period will reduce to $T=2\pi\sqrt{\frac{3R}{2g}}$ .
(c)

What length $\ell$ of simple pendulum (a massless rod with a point mass at the end) has the same period of oscillation of this hoop, swinging in its plane? Answer: $\ell=2R$

Filename:pfigure-blue-81-2

Figure 16.145:

16.4.43 Oscillating disk. A uniform disk with mass $m$ and radius $R$ pivots around a frictionless hinge at its center. It is attached to a massless spring which is horizontal and relaxed when the attachment point is directly above the center of the disk. Assume small rotations and the consequent geometrical simplifications. Assume the spring can carry compression. What is the period of oscillation of the disk if it is disturbed from its equilibrium configuration? [You may use the fact that, for the disk shown, $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}={1\over 2}mR^{2}\ddot{\theta}\hat{% \boldsymbol{k}}$ , where $\theta$ is the angle of rotation of the disk.]. Answer: period = $\pi\sqrt{\frac{2m}{k}}$ .

Filename:pg147-2

Figure 16.146:

16.4.44

Robotics problem: Simplest balancing of an inverted pendulum. You are holding a stick upside down, one end is in your hand, the other end sticking up. To simplify things, think of the stick as massless but with a point mass at the upper end. Also, imagine that it is only a two-dimensional problem (either you can ignore one direction of falling for simplicity or imagine wire guides that keep the stick from moving in and out of the plane of the paper on which you draw the problem).

You note that if you model your holding the stick as just having a stationary hinge then you get $\ddot{\phi}={g\over l}\sin\phi$ . Assuming small angles, this hinge leads to exponentially growing solutions. Upside-down sticks fall over. How can you prevent this falling?

One way to do keep the stick from falling over is to firmly grab it with your hand, and if the stick tips, apply a torque in order to right it. This corrective torque is (roughly) how your ankles keep you balanced when you stand upright. Your task in this assignment is to design a robot that keeps an inverted pendulum balanced by applying appropriate torque.

Your model is: Inverted pendulum, length $\ell$ , point mass $m$ , and a hinge at the bottom with a motor that can apply a torque $T_{m}$ . The stick might be tipped an angle $\phi$ from the vertical. A horizontal disturbing force $F(t)$ is applied to the mass (representing wind, annoying friends, etc).

(a)

Draw a picture and a FBD
(b)

Write the equation for angular momentum balance about the hinge point. Answer: $-F(t)\ell\cos\phi-mg\ell\sin\phi+T_{m}=-m\ell^{2}\ddot{\phi}$ .
(c)

Imagine that your robot can sense the angle of tip $\phi$ and its rate of change $\dot{\phi}$ and can apply a torque in response to that sensing. That is you can make $T_{m}$ any function of $\phi$ and $\dot{\phi}$ that you want. Can you find a function that will make the pendulum stay upright? Make a guess (you will test it below).
(d)

Test your guess the following way: plug it into the equation of motion from part (b), linearize the equation, assume the disturbing force is zero, and see if the solution of the differential equation has exponentially growing (i.e. unstable) solutions. Go back to (c) if it does and find a control strategy that works.
(e)

Pick numbers and model your system on a computer using the full non-linear equations. Use initial conditions both close to and far from the upright position and plot $\phi$ versus time.
(f)

If you are ambitious, pick a non-zero forcing function $F(t)$ (say a sine wave of some frequency and amplitude) and see how that affects the stability of the solution in your simulations.

Filename:pfigure-s94h9p1

Figure 16.147:

16.4.45 A thin rod of mass $m$ and length $\ell$ is hinged with a torsional spring of stiffness $K$ at A, and is connected to a thin disk of mass $M$ and radius $R$ at B. The spring is uncoiled when $\theta=0$ . Determine the natural frequency $\omega_{n}$ of the system for small oscillations $\theta$ , assuming that the disk is:

(a)

welded to the rod, and Answer: $\omega_{n}=\sqrt{\frac{gL\left(M+\frac{m}{2}\right)+K}{\left(M+\frac{m}{3}% \right)L^{2}+M\frac{R^{2}}{2}}}$ .
(b)

pinned frictionlessly to the rod. Answer: $\omega_{n}=\sqrt{\frac{gL\left(M+\frac{m}{2}\right)+K}{\left(M+\frac{m}{3}% \right)L^{2}}}$ . Frequency higher than in (a)

Filename:pfigure-blue-151-4

Figure 16.148:

Mixed Problems

16.4.46 Assume that the pulley shown in figure(a) rotates at a constant speed $\omega$ . Let the angle of contact between the belt and pulley surface be $\theta$ . Assume that the belt is massless and that the condition of impending slip exists between the pulley and the belt. The free-body diagram of an infinitesimal section $a b$ of the belt is shown in figure(b).

(a)

Write the equations of linear momentum balance for section $a b$ of the belt in the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ directions.
(b)

Eliminate the normal force $N$ from the two equations in part (a) and get a differential equation for the tension $T$ in terms of the coefficient of friction $\mu$ and The contact angle $\theta$ . Answer: $\frac{T_{1}}{T_{2}}=\frac{R_{o}+R_{i}\sin\theta}{R_{o}-R_{i}\sin\theta}$ , where $\theta=\tan^{-1}{\mu}$ .
(c)

Show that the solution to the equation in part (b) satisfies $\frac{T_{1}}{T_{2}}=e^{\mu\theta}$ , where $T_{1}$ and $T_{2}$ are the tensions in the lower and the upper segments of the belt, respectively.

Filename:pfigure4-rpf

Figure 16.149:

16.4.47 A belt drive is required to transmit $15\mbox{ kW}$ power from a $750\,{\rm mm}$ diameter pulley rotating at a constant $300\,{\rm rpm}$ to a $500\,{\rm mm}$ diameter pulley. The centers of the pulleys are located $2.5\,\mathrm{m}$ apart. The coefficient of friction between the belt and pulleys is $\mu=0.2$ .

(a)

(See problem 16.4.) Draw a neat diagram of the pulleys and the belt-drive system and find the angle of lap, the contact angle $\theta$ , of the belt on the driver pulley.
(b)

Find the rotational speed of the driven pulley. Answer: $N_{2}=450\,{\rm rpm}$ .
(c)

(See the figure in problem 16.4.) The power transmitted by the belt is given by power = net tension $\times$ belt speed, i.e., $P=(T_{1}-T_{2})v$ , where $v$ is the linear speed of the belt. Find the maximum tension in the belt. [Hint: $\frac{T_{1}}{T_{2}}=e^{\mu\theta}$ (see problem 16.4).] Answer: $T_{1_{\rm max}}=2793.8\,{\mbox{N}}$ .
(d)

The belt in use has a $15\,{\rm mm}\times 5\,{\rm mm}$ rectangular cross-section. Find the maximum tensile stress in the belt. Answer: $\sigma_{\rm max}=37.25$ MPa.

16.4.48 Slippery money A round uniform flat horizontal platform with radius $R$ and mass $m$ is mounted on frictionless bearings with a vertical axis at 0. At the instant of interest it is rotating counter clockwise (looking down) with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . A force in the $x y$ plane with magnitude $F$ is applied at the perimeter at an angle of $30^{\circ}$ from the radial direction. The force is applied at a location that is $\phi$ from the fixed positive $x$ axis. At the instant of interest a small coin sits on a radial line that is an angle $\theta$ from the fixed positive $x$ axis (with mass much much smaller than $m$ ). Gravity presses it down, the platform holds it up, and friction (coefficient= $\mu$ ) keeps it from sliding.

Find the biggest value of $d$ for which the coin does not slide in terms of some or all of $F,m,g,R,\omega,\theta,\phi$ , and $\mu$ . Answer: $d_{\rm max}=\frac{\mu mgR}{\sqrt{F^{2}+(mR\omega)^{2}}}$ .

Filename:pfigure-slipperymoney

Figure 16.150:

16.4.49 Frequently parents will build a tower of blocks for their children. Just as frequently, kids knock them down. In falling (even when they start to topple aligned), these towers invariably break in two (or more) pieces at some point along their length. Why does this breaking occur? What condition is satisfied at the point of the break? Will the stack bend towards or away from the floor after the break? Answer: The stack will bend away from the floor after the break. This breaking occurs because gravity is not able to provide equal angular momentum to all the falling blocks.

Filename:pfigure-blue-129-1

Figure 16.151:

	$\displaystyle[\begin{array}[]{cc}{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\!\mathrm{A}}&{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}% \end{array}]_{xy}$	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}0.866&-0.5\\ 0.5&0.866\end{array}\right]\left[\begin{array}[]{rr}1\,\mathrm{m}&1\,\mathrm{m% }\\ 0&1\,\mathrm{m}\end{array}\right]$
		$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}0.866\,\mathrm{m}&0.366\,\mathrm{m}\\ 0.5\,\mathrm{m}&1.366\,\mathrm{m}\end{array}\right].$

	$\displaystyle[\begin{array}[]{cc}{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\!\mathrm{A}}&{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}% \end{array}]_{xy}$	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}-0.342&-0.94\\ 0.94&-0.342\end{array}\right]\left[\begin{array}[]{rr}1\,\mathrm{m}&1\,\mathrm% {m}\\ 0&1\,\mathrm{m}\end{array}\right]$
		$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}-0.342\,\mathrm{m}&-1.282\,\mathrm{m}\\ 0.94\,\mathrm{m}&0.598\,\mathrm{m}\end{array}\right].$

	$\displaystyle v_{C}$	$\displaystyle=$	$\displaystyle\dot{\theta}\,r_{C}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\theta}$	$\displaystyle=$	$\displaystyle\frac{v_{C}}{r_{C}}=\frac{7.5\,\mathrm{cm}/\,\mathrm{s}}{25\,% \mathrm{cm}}=0.3\,\mathrm{rad}/\mathrm{s}.$

	$\displaystyle x_{cm}$	$\displaystyle=$	$\displaystyle\frac{\sum m_{i}x_{i}}{\sum m_{i}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}r_{cm}$	$\displaystyle=$	$\displaystyle\frac{m(-r)+2m(3r)}{m+2m}=\frac{5}{3}r.$

$\displaystyle I^{\scriptscriptstyle\mathrm{O}}_{zz}$	$\displaystyle=$	$\displaystyle mr^{2}+2m(3r)^{2}=19mr^{2}$
$\displaystyle\hbox to36.135pt{or\hfil}I^{\scriptscriptstyle\mathrm{O}}_{zz}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}+m_{tot}r_{cm/O}^{2}$
	$\displaystyle=$	$\displaystyle\frac{32}{3}mr^{2}+3m\left(\frac{5}{3}r\right)^{2}=19mr^{2}.$

$\displaystyle I^{\scriptscriptstyle\mathrm{O}}_{zz}$	$\displaystyle=$	$\displaystyle(3m)\mathord{\cdot}r_{gyr}^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad 19\cancel{m}r^{2}$	$\displaystyle=$	$\displaystyle 3\cancel{m}r_{gyr}^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad r_{gyr}$	$\displaystyle=$	$\displaystyle\sqrt{\frac{19}{3}}r=2.52r.$

$\displaystyle\mbox{Therefore,}\quad I_{zz}^{O}$	$\displaystyle=$	$\displaystyle\underbrace{\frac{1}{12}M(3\ell)^{2}}_{I_{zz}^{\rm{cm}}}+M(\frac{% \ell}{2})^{2}$
	$\displaystyle=$	$\displaystyle\frac{9}{12}M\ell^{2}+M\frac{\ell^{2}}{4}=M\ell^{2}$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}\quad\mbox{(same % as in (a), of course)}$

	$\displaystyle M_{\rm out}=\frac{P_{\rm out}}{\omega_{\rm out}}$	$\displaystyle=$	$\displaystyle\frac{40\mbox{ kW}}{500\,{\rm rpm}}=\frac{40}{500}\cdot 1000\frac{\,{\rm N}\mathord{\cdot}{% \rm m}}{\cancel{\,\mathrm{s}}}\cdot\frac{\cancel{\,{\rm min}}}{\cancel{\,\mbox% {rev}}}\cdot\frac{1\cancel{\,\mbox{rev}}}{2\pi}\cdot\frac{60\cancel{\,\mathrm{% s}}}{1\cancel{\,{\rm min}}}$
		$\displaystyle=$	$\displaystyle 764\,{\rm N}\mathord{\cdot}{\rm m}.$

$\displaystyle T$	$\displaystyle=$	$\displaystyle mg-m\dot{\omega}R_{i}$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}-2\,\mathrm{% kg}\cdot(-1.26\,\mathrm{s}^{-2})\cdot 0.1\,\mathrm{m}$
	$\displaystyle=$	$\displaystyle 19.87\,{\mbox{N}}$

$\displaystyle a_{m}$	$\displaystyle=$	$\displaystyle(a_{D})_{\theta}$
	$\displaystyle=$	$\displaystyle\dot{\omega}R_{i}$
	$\displaystyle=$	$\displaystyle 0.126\,\mathrm{m}/\mathrm{s}^{2}.$

$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\int_{0}^{\ell}\dot{\theta}^{2}s^{2}\frac{m}{\ell}ds$
	$\displaystyle=$	$\displaystyle\frac{m\dot{\theta}^{2}}{2\ell}\int_{0}^{\ell}s^{2}ds$
	$\displaystyle=$	$\displaystyle\frac{1}{6}m\ell^{2}\dot{\theta}^{2}$

	$\displaystyle W$	$\displaystyle=$	$\displaystyle-mg\frac{\ell}{2}(1-\cos\theta)$
	$\displaystyle\hbox to36.135pt{and\hfil}P$	$\displaystyle=$	$\displaystyle\dot{W}=\frac{dW}{dt}=-mg\frac{\ell}{2}\sin\theta\dot{\theta}.$

	$\displaystyle-\cancel{m}\,g\frac{\ell}{2}\sin\theta\cancel{\dot{\theta}}$	$\displaystyle=$	$\displaystyle\frac{1}{3}\cancel{m}\,\ell^{2}\cancel{\dot{\theta}}\ddot{\theta}$
	$\displaystyle\hbox to36.135pt{or\hfil}\ddot{\theta}+\frac{3g}{2\ell}\sin\theta$	$\displaystyle=$	$\displaystyle 0.$

$\displaystyle\underbrace{{{E_{\mathrm{K}}}}_{1}}_{0}+V_{1}$	$\displaystyle=$	$\displaystyle{{E_{\mathrm{K}}}}_{2}+\underbrace{V_{2}}_{0}$
$\displaystyle\hbox to36.135pt{or\hfil}mg\frac{\ell}{2}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\int_{m}v^{2}dm$
	$\displaystyle=$	$\displaystyle\frac{1}{6}m\ell^{2}\omega^{2}\quad(\mbox{see part (a)})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega$	$\displaystyle=$	$\displaystyle\pm\sqrt{\frac{3g}{\ell}}$

$\Rightarrow$	$\displaystyle\ddot{\theta}+\frac{3g}{2\ell}\theta=0$	(16.86)
or	$\displaystyle\ddot{\theta}+\lambda^{2}\theta=0$
where	$\displaystyle\lambda^{2}=\frac{3g}{2\ell}.$

	the circular frequency	$\displaystyle=$	$\displaystyle{\lambda}=\sqrt{\frac{3g}{2\ell}},$
	and the time period T	$\displaystyle=$	$\displaystyle\frac{2\pi}{\lambda}=2\pi\sqrt{\frac{2\ell}{3g}}.$

$\displaystyle\mbox{angular frequency}\quad\lambda$	$\displaystyle=$	$\displaystyle\sqrt{\frac{6k}{m}},$
$\displaystyle\mbox{and the time period}\quad T$	$\displaystyle=$	$\displaystyle\frac{2\pi}{\lambda}$
	$\displaystyle=$	$\displaystyle 2\pi\sqrt{\frac{m}{6k}}.$

$\displaystyle 2\pi\sqrt{\frac{m}{6k}}$	$\displaystyle=$	$\displaystyle 0.625\,\mathrm{s}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad k$	$\displaystyle=$	$\displaystyle 4\pi^{2}\mathord{\cdot}\frac{m}{6(0.625\,\mathrm{s})^{2}}$
	$\displaystyle=$	$\displaystyle\frac{4\pi^{2}\mathord{\cdot}2\,\mathrm{kg}}{6(0.625\,\mathrm{s})% ^{2}}$
	$\displaystyle=$	$\displaystyle 33.7\,{\mbox{N}}/\,\mathrm{m}.$

$\displaystyle\sqrt{\frac{k_{\rm{tor}}}{\frac{1}{3}m\ell^{2}}}$	$\displaystyle=$	$\displaystyle\sqrt{\frac{6k}{m}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad k_{\rm{tor}}$	$\displaystyle=$	$\displaystyle 6k\mathord{\cdot}\frac{1}{3}\ell^{2}=2k\ell^{2}$
	$\displaystyle=$	$\displaystyle 2\mathord{\cdot}(33.7\,{\mbox{N}}/\,\mathrm{m})\mathord{\cdot}(1% \,\mathrm{m})^{2}$
	$\displaystyle=$	$\displaystyle 67.4\,{\rm N}\mathord{\cdot}{\rm m}.$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle l\dot{\theta}\cos\theta\hat{\boldsymbol{\jmath}}+-l\dot{\theta}% \sin\theta\hat{\boldsymbol{\imath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle-l\dot{\theta}^{2}\left[\cos\theta\hat{\boldsymbol{\imath}}+\sin% \theta\hat{\boldsymbol{\jmath}}\right]$
		$\displaystyle\qquad+l\ddot{\theta}\left[\cos\theta\hat{\boldsymbol{\jmath}}-% \sin\theta\hat{\boldsymbol{\imath}}\right]$

Chapter 16 Circular motion of a rigid object

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Mechanics of circular motion

16.1 Rotation of a rigid object in planar circular motion

Rotation of an object counterclockwise by θ

Rotated coordinates and base vectors ıˆ′ and ȷˆ′

The rotation matrix [R]

Problems for 16.1 Rotation of a rigid body

Preparatory Problems

More-Involved Problems

16.2 Angular velocity and acceleration of a rigid object in planar circular motion

Angular velocity of a rigid object: 𝝎⇀

Rate of change of ıˆ′, ȷˆ′

Velocity of a point fixed on a rigid object

Acceleration of a point on a rotating rigid object

Relative motion of points on a rigid object

The fundamental 𝝎⇀ equation

Calculating relative velocity directly, using rotating frames

Problems for 16.2 Angular velocity

Preparatory Problems

More-Involved Problems

16.3 Polar moment of inertia

The moments of inertia in 2-D : [𝑰cm] and [𝑰O].

Radius of gyration

Other reference points

The parallel axis theorem

The perpendicular axis theorem for planar rigid bodies

2-D parallel axis theorem

Perpendicular axis theorem (applies to planar objects only)

One point mass

Two point masses

A thin uniform rod

A uniform hoop

A uniform disk

Uniform rectangular plate

Problems for 16.3 Polar moment of inertia

Preparatory Problems

More-Involved Problems

16.4 Dynamics of a rigid object in planar circular motion

Mechanics and the motion quantities

Linear momentum: 𝑳⇀ and 𝑳⇀˙

Angular momentum: 𝑯⇀/O and 𝑯⇀˙/O

Power and Energy

Using moment-of-inertia in 2-D circular motion dynamics

Problems for 16.4 Dynamics

Preparatory Problems

More-Involved Problems

Gears, Belt Drives, and Gear Trains

Pendulum Problems

Mixed Problems

Rotation of an object counterclockwise by $\theta$

Rotated coordinates and base vectors $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$

The rotation matrix $[R]$

Angular velocity of a rigid object: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$

Rate of change of $\hat{\boldsymbol{\imath}}^{\prime}$ , $\hat{\boldsymbol{\jmath}}^{\prime}$

The fundamental ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ equation

The moments of inertia in 2-D : $[\boldsymbol{I}^{\mathrm{cm}}]$ and $[\boldsymbol{I}^{\scriptscriptstyle\mathrm{O}}]$ .

Linear momentum: ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$