Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap20 Fixed-axis rotation in 3D Back tables

Chapter 21 Elementary introduction to 3D rigid-object dynamics

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21.0Introductionstudy PDF
21.1 ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ of points on a 3D rigid objectstudy PDF
21.23D dynamics of a rigid bodystudy PDF

We begin more advanced 3D dynamics here. First we discuss general motion of a rigid body in 3D. Then we discuss a special simple class of problems, instantaneous dynamics in 3D.

Three-dimensional rigid body dynamics is difficult. Here we give a taste of some of the issues.

21.1 Velocity and acceleration of points on a rigid object in 3-D: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$

Since the assumption that objects move as if they were rigid is so common in mechanics, it is important to know how points on a rigid body move. Understanding the motion of points on a rigid body is useful in two different ways. First, one needs to know something about the motion in order to apply the momentum balance equations. Second, formulas involving the motion of points on a rigid body are useful to understand mechanisms, machines where rigid bodies are attached in various ways to each other.

Angular velocity

When any rigid body moves in any way in two or three dimensions it always has an angular velocity. If we call the body by the name $\cal B$ (script B), then we call the angular velocity of the body ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ . The angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ is a vector that may change with time. But what is the angular velocity of a rigid body?

Angular velocity in two dimensions

Let’s first review ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ carefully for the case of a two-dimensional body moving in the plane (see chapter 4). We study 2-D rotation by keeping track of straight lines that are drawn on the body. For each line that we draw on the body we keep track of the angle that the line makes with either the positive $x$ or $y$ axis. Let’s assume that all angles are measured as positive in the counter-clockwise direction. The $x$ and $y$ axes are fixed but the lines on the body rotate with the body. These angles $\theta_{1}$ , $\theta_{2}$ , … all change with time. But, though each angle is different from the other, all the angles change at the same rate. That is:

\dot{\theta}_{1}=\dot{\theta}_{2}=\dot{\theta}_{3}=\dots

Since all lines on a body rotate at the same rate as each other (at a given instant in time) the rotation rate is a single number for the body. We call this number $\omega_{\cal B}$ for body $\cal B$ . In order to make various formulas work out we define a vector which is perpendicular to the $x y$ plane. We make the magnitude of this vector $\omega_{\cal B}$ . So in two dimensions the angular velocity of a rigid body is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}=\underbrace{\omega% _{\cal B}}_{\dot{\theta}}\hat{\boldsymbol{k}}

where $\dot{\theta}$ is the rate of change of the angle of any line marked on the body.

Angular velocity in three dimensions

In three dimensions there is not such a simple geometric description of the angular velocity vector of a rigid body. At any instant in time, the relative velocities of points on a rigid body can be described by thinking of the body as spinning about an axis in space, although this axis may change with time. The direction of this axis is the instantaneous direction of the angular velocity vector. The spinning rate is the instantaneous magnitude of the angular velocity vector.

Although it is a little unpalatable for those who need initial motivation, one way of defining the angular velocity vector is as follows. The angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ of a body $\cal B$ is that vector which makes the formula 17.11 below for relative velocity true. It is a fact that for every rigid body there is, at any instant in time, a unique vector that serves this purpose. That is, for the rigid body $\cal B$ there is one angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ that can be used correctly to describe the relative velocities of all pairs of points on the body.

Relative velocity of two points on a rigid body

For any two points A and B glued to a rigid body $\cal B$ the relative velocity of the points (‘the velocity of B relative to A’) is

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

(21.1)

where ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ is the angular velocity of body $\cal B$ . In words, the relative velocity of two points on a rigid body is given by the cross product of the angular velocity of the body with the relative position of the two points. This expression should not look mysterious. It says that the relative velocity of two points on a rigid body is the same as would be predicted for one of the points if the other were stationary.

Absolute velocity of a point on a rigid body

If one knows the velocity of one point on a rigid body and one also knows the angular velocity of the body, then one can find the velocity of any other point. How?

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+({\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol% {v}}}_{A})$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B/A}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}}_{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}}$

That is, the absolute velocity of the point $B$ is the absolute velocity of the point $A$ plus the velocity of the point $B$ relative to the point $A$ . Because $B$ and $A$ are on the same rigid body, their relative velocity is given by the formula 21.1 given above. The equation is a valid relation no matter what values are known and what are not known. It can be used various ways depending on what is known about the motion of the body being studied and what is known about the motion of the chosen points $A$ and $B$ .

Acceleration of points on a rigid body

Angular acceleration

Because both the magnitude and the direction of the angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ can change with time, we can define the angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ of a rigid body as the rate of change of angular velocity, ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$ . The angular acceleration of a body $\cal B$ is called ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}$ . The relative acceleration of two points on a rigid body depends on the angular acceleration.

Angular acceleration in two dimensions

The concept of angular acceleration is most intuitive for two-dimensional bodies moving in the plane. In this case both the angular velocity and the angular acceleration are always perpendicular to the plane. That is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=\alpha\hat{\boldsymbol{k}}$ . In this case the angular acceleration is only due to the speeding up or slowing down of the rotation rate; i.e., $\alpha=\dot{\omega}$ . There is no change in the direction of the angular velocity vector.

Relative acceleration of two points on a rigid body

For any two points A and B glued to a rigid body $\cal B$ the relative acceleration of the points (‘the acceleration of B relative to A’) is

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

(21.2)

If point A has no acceleration, this formula is the same as that for the acceleration of a point going in circles at variable rate in chapter 5. The formula above has a little more generality, however. In three-dimensional motion ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ can change in both magnitude and direction so the relative motion of $B$ to $A$ is not circular. Nonetheless, the formula above is correct and valid.

Absolute acceleration of a point on a rigid body

If one knows the acceleration of one point on a rigid body and the angular velocity and acceleration of the body, then one can find the acceleration of any other point. How?

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+({\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol% {a}}}_{A})={\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B/A}$		(21.3)
		$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A})+\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{B/A}$		(21.4)

Again, this equation has various uses depending on what is known and what one is trying to find. Equation 21.4 is often called the three term acceleration formula. The acceleration of a point B on a rigid body is the sum of three terms. The first, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$ , is the acceleration of some point A on the body. The second term, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}$ ), is the centripetal acceleration. Its direction is from B towards the line which goes through A and is parallel to the angular velocity. The third term, $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}$ , is due to the change of angular velocity.

Angular velocity

Filename:tfigure8-rel-ang-vel

Figure 21.1: Bodies

\cal B

and

\cal C

move relative to each other.

We want to determine the absolute angular velocity of a moving body $\cal C$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal F}$ . We will calculate it using data from another moving frame $\cal B$ . For example, imagine you drive a car $\cal B$ and rotate the steering wheel $\cal C$ to make a turn. What is the angular velocity of the steering wheel relative to the ground? Once we have found this angular velocity, we use it as the angular velocity of $\cal C$ for use in any of the formulae earlier in this chapter or in Chapter 7.

Angular velocity relative to a body or frame

We start by considering the angular velocity of a body relative to a body or frame. Not only can an arbitrary point $P$ be viewed from a moving frame $\cal B$ , but so also can an entire rigid body $\cal C$ with all of its points. The velocities of each of the points on $\cal C$ can be calculated relative to the moving frame $\cal B$ . So, referring to fig. 21.1, the relative velocities of two points $A$ and $D$ on the body $\cal C$ as calculated in the moving frame $\cal B$ can also be calculated.

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D/\cal B}-{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{A/\cal B}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal C/\cal B}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{D/A}.

The quantity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}$ is the angular velocity of body $\cal C$ relative to body $\cal B$ . This formula is easy enough to understand for relative rotation about a fixed axis but turns out to be true for the general motion of a rigid body.

Absolute angular velocity using data from a moving frame

The equation

\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}}^{\text% {absolute angular velocity of $\cal C$}}=\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}}_{\text{absolute angular % velocity of $\cal B$}}+\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal C/\cal B}}^{\text{angular velocity of $\cal C$ relative to $\cal B$}}

(21.5)

is not only plausible but also true. It is often referred to with the deceptively simple words ‘angular velocities add.’

Filename:tfigure8-ang-vel-ex

Figure 21.2: Example of adding angular velocities to get the absolute angular velocity: rotor on gimbals. The unit vector

\hat{\boldsymbol{n}}_{2}

rotates with the gimbals.

However plausible, this addition formula for angular velocities is not obvious for three-dimensional angular velocity. It is important, however, and is one of the cornerstones of three-dimensional rigid body kinematics. One might develop a feel for the formula by looking at the boxes on rotations on the following pages. Let’s illustrate the calculation of absolute angular velocity with the crane of fig. 21.4.

Example: Angular velocity (3-D): A crane, again

Reconsider the crane in fig. 21.3. We have the following: the angular velocity of the cab is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}}=\dot{\theta}\hat% {\boldsymbol{\jmath}}^{\prime}$ and the angular velocity of the boom relative to the cab is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal A}/{\cal B}}=\dot{% \phi}\hat{\boldsymbol{k}}^{\prime}$ .

So, equation 21.5 on page 21.5 tells us that the angular velocity of the crane boom relative to the fixed frame is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal A}}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{{\cal A}/{\cal B}}=\dot{\theta}\hat{\boldsymbol{\jmath% }}^{\prime}+\dot{\phi}\hat{\boldsymbol{k}}^{\prime}.

Filename:tfigure8-ang-vel

Figure 21.3:

Box 21.1 Some comments on the motions of rigid bodies

The kinematics theory we develop is based on two key facts.

1.

The absolute rate of change of any vector ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ glued to a rotating body ${\cal C}$ is

$\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{{\cal C}}\times{\accentset{\rightharpoonup}{% \boldsymbol{Q}}}.$ (21.6)

More generally, the rate of change of any vector that is fixed in a frame ${\cal C}$ calculated relative to a frame ${\cal B}$ is given by

${{}^{\raise 2.0pt\hbox{$\cal B$}}\dot{\accentset{\rightharpoonup}{\boldsymbol{% Q}}}}={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal C}/{\cal B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{Q}}}.$ (21.7)

If for frame $\cal B$ we use the fixed frame $\cal F$ , then the formula 21.7 above reduces to formula 21.6.
2.

The second key fact is: if body $\cal C$ has angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal C}/{\cal B}}$ with respect to body $\cal B$ and if body $\cal B$ has the angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}}$ , then the absolute angular velocity of $\cal C$ is

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal C}}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{{\cal C}/{\cal B}}.$ (21.8)

More generally, we can write

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal C}/{\cal D}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}/{\cal D}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal C}/{\cal B}}.$ (21.9)

with no need to refer to any fixed frame. If ${\cal D}$ is the fixed frame, then formula 21.9 reduces to formula 21.8.

Equations 21.6-21.9 seem natural enough once one is used to the terms and notation. However, to derive formulas 21.6-21.9 from more basic notions involves several steps of geometric and/or algebraic reasoning. Here is a list of some of the high points in that reasoning.

I.

The net motion of any rigid body from time $t_{1}$ to time $t_{2}$ can be represented as a displacement of a point $A$ on the body and a rotation $\theta$ about an axis $\hat{\boldsymbol{n}}$ through $A$ . If a different point, say $B$ , is chosen, then the same net motion is a displacement of the point $B$ and a rotation about a fixed axis through $B$ .
II.

For a given net motion, the displacement depends on which point $A$ or $B$ is chosen. The direction of the axis $\hat{\boldsymbol{n}}$ of rotation and the amount of rotation about that axis, $\theta$ , does not depend on which point is chosen.
III.

The net finite rotation after two successive finite rotations of a body depends on the order of the rotations. For example, a $90^{\circ}$ rotation about the $y$ -axis followed by a $90^{\circ}$ rotation about the $x$ -axis is not equivalent to a $90^{\circ}$ rotation about the $x$ -axis followed by a $90^{\circ}$ rotation about the $y$ -axis. For a more detailed explanation of this idea, see the box 21.1 on page 21.1.
IV.

For infinitesimal rotations, however, the net infinitesimal rotation does not depend on the order of successive rotations. In fact, the net rotation is given by vector addition:

$\Delta\theta\hat{\boldsymbol{n}}=(\Delta\theta_{1})\hat{\boldsymbol{n}}_{1}+(% \Delta\theta_{2})\hat{\boldsymbol{n}}_{2}.$

For a more detailed explanation of this fact, see the box 21.1 on page 21.1.

Facts I and II imply that the relative motion of any two points is that of rotation of one point about a fixed axis through the other. For small rotations, this geometric result leads to relation

\Delta{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}=(\Delta\theta\hat{% \boldsymbol{n}})\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}.

Dividing by $\Delta t$ and taking the limit as $\Delta t\rightarrow 0$ gives

{\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}}=(\dot{\theta}\hat{% \boldsymbol{n}})\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}.

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ is defined as the vector $\dot{\theta}\hat{\boldsymbol{n}}$ from this formula. The equation of fact IV, when divided by $\Delta t$ , leads to the formula for addition of angular velocities, equation 21.8:

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal C}}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{{\cal B}}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{{\cal C}/{\cal B}}.

(21.10)

Box 21.2 Finite Rotations

The net finite rotation after two successive finite rotations about a fixed axis depends on the order of the rotations.

For example, a $90^{\circ}$ rotation about the $x$ -axis

Filename:tfigure8-book-rot1

followed by a $90^{\circ}$ rotation about the $y$ -axis

Filename:tfigure8-book-rot2

is equivalent to a single $120^{\circ}$ rotation about an axis in the direction $\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}-\hat{\boldsymbol{k}}$

Filename:tfigure8-book-rot3

which is, agreeably, hard to picture ( this rotation corresponds to a cube rotating about one of its main diagonals).

On the other hand, a $90^{\circ}$ rotation about the $y$ -axis

Filename:tfigure8-book-rot4

followed by a $90^{\circ}$ rotation about the $x$ -axis

Filename:tfigure8-book-rot5

is equivalent to a single $120^{\circ}$ rotation about an axis in the direction $\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}$

Filename:tfigure8-book-rot6

which is also hard to see (this rotation corresponds to a $120^{\circ}$ rotation about a different main diagonal of a cube).

Because the order in which the rotations occur affects the resulting net rotation, the result of successive large rotations about axes fixed in space can’t be found by vector addition.

Is rotation a vector?

Rotation can be represented by an arrow with magnitude and direction. The direction is the axis of rotation. The magnitude is the angle of rotation. The net rotation after two rotations is also a vector in that it can be represented with a magnitude and direction. But the net rotation cannot be found by addition of the previous vectors. Vector addition is commutative (the order of addition doesn’t matter) but the previous example shows that rotation is not commutative. So, in the sense that addition of two finite rotation vectors does not have a physical meaning, one can say ‘rotation is not a vector,’ meaning that the rules of vector arithmetic are not physically interpretable.

Small (infinitesimal) rotations and rotation rates do add, however.

An irony

Rotations are not vectors in that vector addition does not have meaning for the composition of two 3-D rotation ‘vectors’. But an odd twist of history is that the very first use of the word ‘vector’ to represent an object with magnitude and direction was to describe rotations. This first use of the word ‘vector’ was by the mathematician and mechanician William Hamilton in 1846 in the context of his invention/discovery: ‘quaternions’. One and a half centuries later, despite this history, some people are emphatic in saying ‘rotation is not a vector.’

Box 21.3 Small (Infinitesimal) Rotations

Unlike finite rotations, the net infinitesimal rotation does not depend on the order of successive rotations.

For example, let’s now look at successive $5^{\circ}$ rotations about the $x$ and $y$ axes. A $5^{\circ}$ rotation about the $x$ -axis

Filename:tfigure8-smallrot1

followed by a

5^{\circ}

rotation about the

y

-axis

Filename:tfigure8-smallrot2

is nearly equal to a single rotation of $\sqrt{5^{2}+5^{2}}^{\circ}=\sqrt{{50}}^{\circ}=7.0711^{\circ}$ about the $\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}$ axis. More exactly, the net rotation is $7.0699^{\circ}$ about the $\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}-(0.031)\hat{\boldsymbol{k}}$ axis. Thus, the component of the axis of rotation in the $z$ direction is negligible compared to the components in the $x$ and $y$ directions as we have supposed.

Similarly, a rotation of $5^{\circ}$ about the $y$ -axis

Filename:tfigure8-smallrot3

followed by a

5^{\circ}

rotation about the

x

-axis

Filename:tfigure8-smallrot4

is nearly equal to a

\sqrt{{50}}^{\circ}=7.0711^{\circ}

rotation about the

\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}

axis. More exactly, the net rotation is

7.0699^{\circ}

about the

\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+(0.031)\hat{\boldsymbol{k}}

axis. Thus, again, the component of the axis of rotation in the

z

direction is negligible compared to the components in the

x

and

y

directions as we have supposed.

The final configuration after the two different sequences of rotations is nearly the same.

So, the order of small rotations doesn’t matter and the result can be found by vector addition. In the example shown, the small rotation vector is

\overbrace{5^{\circ}\hat{\boldsymbol{\imath}}}^{\text{$5^{\circ}$ rotation % about $x$-axis}}\quad``+"\quad\underbrace{5^{\circ}\hat{\boldsymbol{\jmath}}}_{\text% {$5^{\circ}$ rotation about $y$-axis}}\quad\approx\quad\underbrace{5^{\circ}(\hat{\boldsymbol{\imath}}+% \hat{\boldsymbol{\jmath}})}_{\text{the net rotation}}

This reasoning, mathematized, is what justifies the formula for adding angular velocities

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal C/\cal B}.

Summary: Rotations and angular velocity

•

Finite rotations ‘don’t add.’
•

Small rotations ‘do add.’
•

Angular velocity ‘does add’,

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal C/\cal B}.$
•

If $D$ and $A$ are fixed on $\cal C$ , then

${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D/\cal B}-{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{A/\cal B}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal C/\cal B}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{D/A}$
•

For any two points $O^{\prime}$ and $D$ fixed in a body or frame $\cal B$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{D/O^{\prime}}=\ ^{\cal F}% \dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{D/O}-\ ^{\cal F}\par\dot{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{\prime}/O}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{D/O^{\prime}}.$

Absolute angular acceleration using data from a moving frame

We want to determine the absolute angular acceleration of a moving body $\cal C$ , ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/\cal F}$ . We will calculate it using data from another moving frame $\cal B$ . See fig. 21.4. For example, imagine again you drive a car $\cal B$ and rotate the steering wheel $\cal C$ to make a turn. What is the absolute angular acceleration of the steering wheel relative to the ground?

Filename:tfigure8-ang-accel

Figure 21.4: Bodies

\cal B

and

\cal C

move relative to each other.

Taking the time derivative of our result for the absolute angular velocity of a rigid body, and making use of the ‘Q-dot’ formula, we get for the bodies $\cal B$ and $\cal C$ in fig. 21.4:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}+% \dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}$	(differentiating)
	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}+% \overbrace{\ {}^{\cal B}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% _{\cal C/\cal B}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}}^{% \dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}}$	( $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula)
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}+{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/\cal B}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}.$	(change of notation)

Thus, the absolute angular acceleration of body $\cal C$ with respect to body $\cal B$ is

{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\alpha}}}_{\cal C/\cal B}+{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \cal C/\cal B}

or, equivalently,

{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\alpha}}}_{\cal C/\cal B}+{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \cal C},

(21.11)

because ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal F}={\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{B}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ .

Note: angular accelerations do not add unless the angular velocities are parallel.

Filename:tfigure8-ang-accel-ex

Figure 21.5: Example of total angular acceleration: rotor on gimbals. The unit vector

\hat{\boldsymbol{n}}_{2}

rotates with the gimbals.

We now have all the kinematics we need to calculate the right hand sides of the momentum balance and energy equations for a variety of complex problems.

Let’s illustrate the calculation of absolute angular acceleration with the crane of fig. 21.4.

Example: Angular acceleration (3-D): A crane, again

Filename:tfigure8-ang-accel1

Figure 21.6:

Reconsider the crane in fig. 21.6. The angular acceleration of the cab is ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{{\cal B}}=\ddot{\theta}% \hat{\boldsymbol{\jmath}}^{\prime}$ , and the angular acceleration of the boom relative to the cab is ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{{\cal A}/{\cal B}}=\ddot{% \phi}\hat{\boldsymbol{k}}^{\prime}$ .

Equation 21.11 on page 21.11 tells us that the angular acceleration of the boom relative to the ground is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{{\cal A}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{{\cal B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{{\cal A}/{\cal B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal A/\cal B}$
	$\displaystyle=$	$\displaystyle\ddot{\theta}\hat{\boldsymbol{\jmath}}^{\prime}+\ddot{\phi}\hat{% \boldsymbol{k}}^{\prime}+\dot{\theta}\hat{\boldsymbol{\jmath}}^{\prime}\times% \dot{\phi}\hat{\boldsymbol{k}}^{\prime}$
	$\displaystyle=$	$\displaystyle\dot{\theta}\dot{\phi}\hat{\boldsymbol{\imath}}^{\prime}+\ddot{% \theta}\hat{\boldsymbol{\jmath}}^{\prime}+\ddot{\phi}\hat{\boldsymbol{k}}^{% \prime}.$

In this example, the absolute acceleration of the crane boom is due to:

1.

the change in magnitude of the angular velocity of the cab,
2.

the change in the magnitude of the angular velocity of the boom relative to the cab,
3.

the change in direction of the angular velocity of the boom relative to the cab caused by the rotation of the cab. This contribution points in the $\hat{\boldsymbol{\imath}}^{\prime}$ direction.

Consider the following special case at the instant of interest. Suppose the crane boom does not have a rotational acceleration relative to the cab and the cab does not have a rotational acceleration relative to the ground; thus, $\ddot{\phi}=0$ and $\ddot{\theta}=0$ . Even in this case, the boom does have angular acceleration relative to the ground, ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A/\cal F}=\dot{\theta% }\dot{\phi}\hat{\boldsymbol{\imath}}^{\prime}+0\hat{\boldsymbol{\jmath}}^{% \prime}+0\hat{\boldsymbol{k}}^{\prime}\neq{\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ . In this special case, the absolute angular acceleration of the boom is due to the changing direction of its angular velocity vector caused by the rotation of the cab at constant rate relative to the ground.

In the 2-D case, if a body $\cal A$ rotates relative to a fixed frame at constant rate and a frame $\cal B$ rotates relative to the $\cal A$ at constant rate, body $\cal B$ does not have an angular acceleration relative to the fixed frame. That compounding angular velocities does not lead to angular acceleration in planar problems can be explained mathematically as follows: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}=\omega_{\cal A}% \hat{\boldsymbol{k}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/A}=\omega_{\cal B/A% }\hat{\boldsymbol{k}}$ so that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}\times{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/A}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ .

Summary of angular acceleration

Relative angular accelerations do not ‘add’ (an extra term is needed).

{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\alpha}}}_{\cal C/\cal B}+\overbrace{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal C/\cal B}}^{\mbox{extra term}}

or, equivalently, because ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal F}={\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{B}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ ,

{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C}={\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\alpha}}}_{\cal C/\cal B}+{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}_{\cal B}\times\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal C}}_{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C% /\cal F}}.

SAMPLE 21.1

Filename:sfig8-2-3

Figure 21.7:

Velocity and acceleration of a point on a spinning and precessing body. A three-mass symmetric system is supported by a fan-like structure (Fig. 21.7). The three masses spin about the horizontal shaft at a constant rate $\omega=10\,\mathrm{rad}/\mathrm{s}$ . The horizontal shaft is rigidly attached to a vertical shaft which rotates about the vertical axis at a constant rate $\Omega=2\,\mathrm{rad}/\mathrm{s}$ . At the instant shown the three masses are in the $x z$ -plane and mass P is aligned with the $x$ -axis.

1.

Find the absolute velocity of mass P.
2.

Find the absolute acceleration of mass P.

Solution Let us attach a frame $\cal B$ to the horizontal shaft so that this frame rotates with the horizontal shaft with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}=\Omega\hat{% \boldsymbol{k}}$ . For calculations in the rotating frame, let us attach a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ to the shaft at point $O^{\prime}$ . Since this coordinate system is attached to the rotating frame, its basis vectors $\hat{\boldsymbol{\imath}}^{\prime},\,\hat{\boldsymbol{\jmath}}^{\prime}\,\hat{% \boldsymbol{k}}^{\prime}$ also rotate with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ , but at the instant of interest

\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}},\quad\hat{% \boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}},\quad\hat{\boldsymbol{% k}}^{\prime}=\hat{\boldsymbol{k}}.

Velocity of P: The absolute velocity of point P is given by

Filename:sfig8-2-3a

Figure 21.8:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{P^{\prime}}+\underbrace{{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\rm rel}}_{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/% {\cal B}}}

where $P^{\prime}$ is a point fixed in the rotating frame and coincides with point P at the instant of interest. To visualize the motion of $P^{\prime}$ , we draw a rigid arm of any shape (usually taking advantage of the given geometry) to the point $P^{\prime}$ (which is at point P). This imaginary rigid arm rotates with the same angular velocity as the rotating frame $\cal B$ , since $P^{\prime}$ is a point fixed in the rotating frame. From Fig. 21.8 we see that point $P^{\prime}$ goes in circles of radius $AP^{\prime}$ at constant rate $\Omega$ . Considering the rigid arm $OAP^{\prime}$ we can find the velocity of $P^{\prime}$ as

Filename:sfig8-2-3b

Figure 21.9:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}}=\Omega\hat{% \boldsymbol{k}}\times(h\hat{\boldsymbol{k}}+L\hat{\boldsymbol{\jmath}}^{\prime% }+r\hat{\boldsymbol{\imath}}^{\prime})$
		$\displaystyle=$	$\displaystyle-\Omega L\hat{\boldsymbol{\imath}}^{\prime}+\Omega r\hat{% \boldsymbol{\jmath}}^{\prime}.$

To find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$ , the velocity of point P relative to the rotating frame, we imagine ourselves sitting anywhere in the rotating frame and observe the motion of point P. That is, we forget about the rotation of the frame (if we are sitting in the rotating frame we cannot see the rotation of the frame) and watch the motion of point P as if the horizontal and the vertical shafts were stationary. It is easy to see that with respect to $\cal B$ the three masses execute circular motion at a constant rate $\omega$ . From Fig. 21.9 we see that

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}=\omega\hat{% \boldsymbol{\jmath}}^{\prime}\times r\hat{\boldsymbol{\imath}}^{\prime}$
		$\displaystyle=$	$\displaystyle-\omega r\hat{\boldsymbol{k}}^{\prime}=-\omega r\hat{\boldsymbol{% k}}.\qquad\mbox{(since $\hat{\boldsymbol{k}}^{\prime}=\hat{\boldsymbol{k}}$)}$

Thus, the absolute velocity of point P is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}=-\Omega L\hat{% \boldsymbol{\imath}}+\Omega r\hat{\boldsymbol{\jmath}}-\omega r\hat{% \boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle-2\,\mathrm{rad}/\mathrm{s}\cdot 2\,\mathrm{m}\hat{\boldsymbol{% \imath}}+2\,\mathrm{rad}/\mathrm{s}\cdot 1\,\mathrm{m}\hat{\boldsymbol{\jmath}% }-10\,\mathrm{rad}/\mathrm{s}\cdot 1\,\mathrm{m}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(-4\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}-10\hat{% \boldsymbol{k}})\,\mathrm{m}/\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=(-4\hat{\boldsymbol{\imath}}% +2\hat{\boldsymbol{\jmath}}-10\hat{\boldsymbol{k}})\,\mathrm{m}/\mathrm{s}$

Acceleration of point P: The absolute acceleration of point P is given by

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{O^{\prime}/O}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{B}}\times({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{P/O^{\prime}})+\dot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/% O^{\prime}}}_{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}}+% \underbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P/\mathcal{B}}}_{{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm rel}}+\underbrace{2\omega_{% \mathcal{D}}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/\mathcal{B}% }}_{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm cor}}

where ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$ is the acceleration of point $P^{\prime}$ discussed above, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm cor}$ is the Coriolis acceleration, and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm rel}\equiv{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{P/{\cal B}}$ is the relative acceleration of P with respect to the rotating frame. Now we calculate each term separately. Considering the motion of point $P^{\prime}$ in Fig. 21.8, we see that

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}$	$\displaystyle=$	$\displaystyle\underbrace{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}% }_{\cal B}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P^{\prime}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times\overbrace{({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{P^{\prime}})}^{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P^{\prime}}}$
	$\displaystyle=$	$\displaystyle\Omega\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{P^{\prime}}$
	$\displaystyle=$	$\displaystyle\Omega\hat{\boldsymbol{k}}\times(-\Omega L\hat{\boldsymbol{\imath% }}+\Omega r\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-\Omega^{2}(L\hat{\boldsymbol{\jmath}}+r\hat{\boldsymbol{\imath}}).$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm cor}$	$\displaystyle=$	$\displaystyle 2{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm rel}$
	$\displaystyle=$	$\displaystyle 2\Omega\hat{\boldsymbol{k}}\times(-\omega r\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm rel}$	$\displaystyle=$	$\displaystyle\underbrace{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}% }}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}+\overbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^% {\prime}})}^{-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{% \prime}}}$
	$\displaystyle=$	$\displaystyle-\omega^{2}r\hat{\boldsymbol{\imath}}^{\prime}$
	$\displaystyle=$	$\displaystyle-\omega^{2}r\hat{\boldsymbol{\imath}}\qquad\mbox{(since $\hat{% \boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}}$)}.$

Thus, the absolute acceleration of point P is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P^{\prime}}+{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm cor}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\rm rel}$
	$\displaystyle=$	$\displaystyle-\Omega^{2}(L\hat{\boldsymbol{\jmath}}+r\hat{\boldsymbol{\imath}}% )+{\accentset{\rightharpoonup}{\boldsymbol{0}}}-\omega^{2}r\hat{\boldsymbol{% \imath}}$
	$\displaystyle=$	$\displaystyle-(\Omega^{2}+\omega^{2})r\hat{\boldsymbol{\imath}}-\Omega^{2}L% \hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-(4+100)(\,\mathrm{rad}/\mathrm{s})^{2}\cdot 1\,\mathrm{m}\hat{% \boldsymbol{\imath}}-4(\,\mathrm{rad}/\mathrm{s})^{2}\cdot 2\,\mathrm{m}\hat{% \boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-(104\hat{\boldsymbol{\imath}}+8\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}^{2}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-(104\hat{\boldsymbol{\imath% }}+8\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}$

SAMPLE 21.2

Filename:sfig8-3-1

Figure 21.10:

A student stands on a turntable that rotates about the $z$ -axis at a constant rate $\omega_{1}=2\,\mathrm{rad}/\mathrm{s}$ . He holds a wheel that rotates with respect to his arm at a constant rate $\omega_{2}=6\,\mathrm{rad}/\mathrm{s}$ (see Figure 21.10. At the instant when the student’s arm is parallel to the $y$ -axis, find

1.

the absolute angular velocity of the wheel,
2.

the magnitude of the angular velocity of the wheel, and
3.

the axis of instantaneous rotation of the wheel.

Solution Let a coordinate system with basis vectors ( $\hat{\boldsymbol{e}}_{1},\hat{\boldsymbol{e}}_{2},\hat{\boldsymbol{e}}_{3}$ ) be attached to the hand of the student. At the instant of interest, these basis vectors are parallel to the fixed basis vectors ( $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},\hat{\boldsymbol{k}}$ ), respectively, i.e.,

\hat{\boldsymbol{e}}_{1}=\hat{\boldsymbol{\imath}},\quad\hat{\boldsymbol{e}}_{% 2}=\hat{\boldsymbol{\jmath}},\hbox to36.135pt{\hfil and\hfil}\hat{\boldsymbol{% e}}_{3}=\hat{\boldsymbol{k}}.

1.

The absolute angular velocity of the wheel is:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm arm}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm wheel/arm}$

$\displaystyle=$ $\displaystyle\omega_{1}\hat{\boldsymbol{k}}+\omega_{2}(-\hat{\boldsymbol{e}}_{% 2})$

$\displaystyle=$ $\displaystyle\omega_{1}\hat{\boldsymbol{k}}-\omega_{2}\hat{\boldsymbol{\jmath}}$

$\displaystyle=$ $\displaystyle(2\hat{\boldsymbol{k}}-6\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/% \mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=(2\hat{\boldsymbol{k}}-6% \hat{\boldsymbol{\jmath}})\,\mathrm{rad}/\mathrm{s}$
2.

The magnitude of the absolute angular velocity of the wheel is:

$\omega=|{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}|=\sqrt{4+36}\,% \mathrm{rad}/\mathrm{s}=6.32\,\mathrm{rad}/\mathrm{s}.$

Answer: $\omega=6.32\,\mathrm{rad}/\mathrm{s}$
3.
Filename:sfig8-3-1a

Figure 21.11: The angular velocity of the wheel and the instantaneous axis of rotation.

The instantaneous axis of rotation is the direction of the angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . Let $\hat{\boldsymbol{\lambda}}$ , a unit vector in the direction of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , represent the axis of rotation. Then,

$\displaystyle\hat{\boldsymbol{\lambda}}$ $\displaystyle=$ $\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}{\omega}% =\frac{(2\hat{\boldsymbol{k}}-6\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/% \mathrm{s}}{6.32\,\mathrm{rad}/\mathrm{s}}$

$\displaystyle=$ $\displaystyle 0.32\hat{\boldsymbol{k}}-0.95\hat{\boldsymbol{\jmath}}.$

Answer: $\hat{\boldsymbol{\lambda}}=0.32\hat{\boldsymbol{k}}-0.95\hat{\boldsymbol{% \jmath}}$

Comments:

1.

After computing any unit vector, you should check that it has a magnitude of $1$ .
2.

$\hat{\boldsymbol{\lambda}}\equiv 0.32\hat{\boldsymbol{k}}-0.95\hat{\boldsymbol% {\jmath}}$ is a unit vector along the axis of instantaneous rotation. Therefore, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ can now also be written as

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{% \lambda}}=6.37\,\mathrm{rad}/\mathrm{s}(0.32\hat{\boldsymbol{k}}-0.95\hat{% \boldsymbol{\jmath}})$

.

SAMPLE 21.3

Filename:sfig8-5-wiper

Figure 21.12:

Windshield wiper of a bus. The windshield wipers of big vehicles such as buses and trucks are usually designed such that the blades of the wipers always keep a fixed angle during the motion of the connecting link. This angle is maintained by making the blade rotate with respect to the link about pin B. At any instant, let the angular velocity of the link be ${\accentset{\rightharpoonup}{\boldsymbol{\Omega}}}=\omega_{1}\hat{\boldsymbol{% k}}.$ What must be the angular velocity of the blade with respect to the link so that the blade stays horizontal?

Solution Intuitively, the answer should be more or less obvious. If the link rotates by $\Delta\theta$ in some time interval $\Delta t$ , then the blade must rotate, relative to the link, by the same amount $\Delta\theta$ in the opposite direction so that its net rotation is zero and it stays horizontal. Now let us see how we can get the same answer using angular velocities and rotating frames.

Let ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{2}=\omega_{2}\hat{% \boldsymbol{k}}$ be the angular velocity of the blade with respect to the link. Let us attach a frame $\mathcal{B}$ to the blades and fix a coordinate axes $x^{\prime}y^{\prime}z^{\prime}$ in this frame. (see Figure 21.13). Then

Filename:sfig8-5-wiper-a

Figure 21.13:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm{blade}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm{blade/link}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm{link}}$
		$\displaystyle=$	$\displaystyle\omega_{2}\hat{\boldsymbol{k}}+\omega_{1}\hat{\boldsymbol{k}}=(% \omega_{2}+\omega_{1})\hat{\boldsymbol{k}}.$

Since, in our example, the blade remains horizontal, the unit vector $\hat{\boldsymbol{\imath}}^{\prime}$ does not change its direction, i.e.,

$\displaystyle\dot{\hat{\boldsymbol{\imath}}}^{\prime}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$
$\displaystyle\rm{But}\quad\dot{\hat{\boldsymbol{\imath}}}^{\prime}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times\hat{\boldsymbol{\imath}}^{\prime}$
	$\displaystyle=$	$\displaystyle(\omega_{1}+\omega_{2})\hat{\boldsymbol{k}}\times\hat{\boldsymbol% {\imath}}^{\prime}$
	$\displaystyle=$	$\displaystyle(\omega_{1}+\omega_{2})\hat{\boldsymbol{\jmath}}^{\prime}.$

Therefore,

$\displaystyle(\omega_{1}+\omega_{2})\hat{\boldsymbol{\jmath}}^{\prime}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{1}+\omega_{2}$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{2}$	$\displaystyle=$	$\displaystyle-\omega_{1}.$

Answer: $\omega_{2}=-\omega_{1}$

Comments: In general, justification of such simple ideas in such detail would preclude efficient solving of more complex problems. We present it just to show how the formulae do ultimately agree with common sense.

SAMPLE 21.4 Relative rotations in 2-D and 3-D. In the figures shown below, the rigid arm AB rotates with constant angular speed $\Omega=5\,\mathrm{rad}/\mathrm{s}$ and the disk $\mathcal{D}$ rotates with respect to the arm at constant angular speed $\omega=10\,\mathrm{rad}/\mathrm{s}$ . Is the angular acceleration of the disk the same in each case?

Filename:sfig8-5-2disks

Figure 21.14:

Solution In each case, let us attach a frame $\mathcal{B}$ to the rod and fix coordinate axes $x^{\prime}y^{\prime}z^{\prime}$ in $\mathcal{B}$ with the origin at the center of the disk. Let the primed coordinate axes be parallel to the inertial coordinate axes at the moment of interest.

In case (a):

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\Omega\hat{% \boldsymbol{k}}\quad\rm{and}\quad{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\mathcal{D}/\mathcal{B}}=\omega\hat{\boldsymbol{k}}^{\prime}=\omega% \hat{\boldsymbol{k}}.

Therefore, the angular acceleration of the disk is

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{D}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{% \mathcal{D}/\mathcal{B}}}^{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal{B}}$
		$\displaystyle=$	$\displaystyle\Omega\hat{\boldsymbol{k}}\times\omega\hat{\boldsymbol{k}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}.$

In case (b):

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\Omega\hat{% \boldsymbol{k}}\quad\rm{and}\quad{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\mathcal{D}/\mathcal{B}}=\omega\hat{\boldsymbol{\jmath}}^{\prime}=% \omega\hat{\boldsymbol{\jmath}}.

Therefore, the angular acceleration of the disk is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{D}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{% \mathcal{D}/\mathcal{B}}}^{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal{B}}$
	$\displaystyle=$	$\displaystyle\Omega\hat{\boldsymbol{k}}\times\omega\hat{\boldsymbol{\jmath}}=-% \Omega\omega\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-50\,\mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}.$

Thus, the angular acceleration of the disk is not the same in each case.

SAMPLE 21.5

Filename:sfig8-4-1

Figure 21.15:

Relative and absolute angular velocity and acceleration. A three-mass body is made up of three point masses P, Q, and R, connected by three identical rigid rods at $120^{o}$ . The three-mass system is attached to shaft CDE and spins with respect to the shaft at a non-constant rate $\omega_{2}$ . Shaft CDE rotates with a constant rate $\omega_{1}$ with respect to the base sleeve which, in turn, rotates with the base shaft AB with constant angular speed $\Omega$ . At the instant shown, $\omega_{2}=10\,\mathrm{rad}/\mathrm{s}$ and is changing at the rate of $\dot{\omega}_{2}=2\,\mathrm{rad}/\mathrm{s}^{2}$ , $\omega_{1}=5\,\mathrm{rad}/\mathrm{s}$ (constant), and $\Omega=2\,\mathrm{rad}/\mathrm{s}$ (constant).

1.

Find the angular velocity of the three-mass body PQR with respect to the base shaft AB.
2.

Find the absolute angular velocity of the body PQR.
3.

Find the angular acceleration of the body PQR relative to the base shaft AB.
4.

Find the absolute angular acceleration of the body PQR.

Solution

Filename:sfig8-4-1a

Figure 21.16: Frame

\cal B

is attached to the shaft CDE and frame

\cal C

to the base shaft AB.

\cal D

represents the three mass system.

\cal D

rotates with respect to

\cal B

\cal B

rotates with respect to

\cal C

, and

\cal C

rotates with respect to the fixed frame.

Let us attach a frame $\cal B$ to the shaft CDE and a frame $\cal C$ to the base shaft AB. Thus frame $\cal B$ rotates with respect to frame $\cal C$ with constant speed $\omega_{1}$ or at the instant shown,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/% \mathcal{C}}$	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}\quad\mbox{ and }$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{C}}$	$\displaystyle=$	$\displaystyle\Omega\hat{\boldsymbol{\imath}}.$

Let the symbol $\cal D$ denote the three-mass rigid body PQR. Also, let the coordinate axes $x^{\prime}y^{\prime}z^{\prime}$ be attached to the frame $\mathcal{D}$ at point E.

1.

The angular velocity of $\cal D$ with respect to frame $\cal C$ is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D/C}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/% \mathcal{B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/% \mathcal{C}}$

$\displaystyle=$ $\displaystyle\omega_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{1}\hat{% \boldsymbol{k}}$

$\displaystyle=$ $\displaystyle(10\hat{\boldsymbol{\jmath}}+5\hat{\boldsymbol{k}})\,\mathrm{rad}% /\mathrm{s}$

where the last line follows from the fact that at the instant of interest the primed coordinate axes $x^{\prime},\,y^{\prime}$ , and $z^{\prime}$ , glued to the frame $\cal B$ , are parallel to the fixed coordinates $x,\,y$ , and $z$ .

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D/C}=(10\hat{% \boldsymbol{\jmath}}+5\hat{\boldsymbol{k}})\,\mathrm{rad}/\mathrm{s}$
2.

The absolute angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}\equiv{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal{F}}$ where $\cal F$ denotes the fixed frame, can be written as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D/C}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{C}}$

$\displaystyle=$ $\displaystyle\omega_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{1}\hat{% \boldsymbol{k}}+\Omega\hat{\boldsymbol{\imath}}$

$\displaystyle=$ $\displaystyle(10\hat{\boldsymbol{\jmath}}+5\hat{\boldsymbol{k}}+2\hat{% \boldsymbol{\imath}})\,\mathrm{rad}/\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}=(2\hat{% \boldsymbol{\imath}}+10\hat{\boldsymbol{\jmath}}+5\hat{\boldsymbol{k}})\,% \mathrm{rad}/\mathrm{s}$

Now we calculate the angular acceleration. We have to be extra careful in calculating angular accelerations relative to the intermediate frames. Explicit notations for the time derivatives taken in different frames usually help.

The angular acceleration of $\cal D$ relative to $\cal C$ may be calculated as follows.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal D/C}$	$\displaystyle=$	$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{C}}$
	$\displaystyle=$	$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{B}}+{}^{% \lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{B}/\mathcal{C}}$
	$\displaystyle=$	$\displaystyle\underbrace{{}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/\mathcal{C}}% \times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal% {B}}}_{{}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{B}}}\quad+\quad{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}/\mathcal{C}}$
	$\displaystyle=$	$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{B}}+\omega_{1}% \hat{\boldsymbol{k}}\times\omega_{2}\hat{\boldsymbol{\jmath}}+{\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{B}/\mathcal{C}}$
	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}-\omega_{1}\omega_{2}% \hat{\boldsymbol{\imath}}+\underbrace{\dot{\omega}_{1}}_{0}\;\;\hat{% \boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{\jmath}}-10\,% \mathrm{rad}/\mathrm{s}\cdot 5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(-50\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}})\,% \mathrm{rad}/\mathrm{s}^{2}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal D/C}=(-50\hat{% \boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/\mathrm{s}^{2}$

The absolute angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{D}}$ of the body can be found in a similar way by carrying the time derivative of the absolute angular velocity in the fixed frame. In the following calculations, all angular quantities without explicit reference to a frame stand for quantities with respect to the fixed frame $\cal F$ .

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{D}}$	$\displaystyle=$	$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}}$
	$\displaystyle=$	$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{B}}+{}^{% \lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{B}/\mathcal{C}}+{}^{\lxWithClass{rp-bodyframe% }{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{% \mathcal{C}}$
	$\displaystyle=$	$\displaystyle({}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{B}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal{B}})+({}^{% \lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{B}/\mathcal{C}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{C}}\times{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{B}/\mathcal{C}})+\underbrace{\dot{\Omega}}_{0}% \;\;\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}+(\omega_{1}\hat{% \boldsymbol{k}}+\Omega\hat{\boldsymbol{\imath}})\times\omega_{2}\hat{% \boldsymbol{\jmath}})+(\underbrace{\dot{\omega}_{1}}_{0}\;\;\hat{\boldsymbol{k% }}+\Omega\hat{\boldsymbol{\imath}}\times\omega_{1}\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle(\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}-\omega_{1}\omega_{2}% \hat{\boldsymbol{\imath}}+\Omega\omega_{2}\hat{\boldsymbol{k}})+(-\Omega\omega% _{1}\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(2\,\mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{\jmath}}-50\,% \mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}+20\,\mathrm{rad}/\mathrm{% s}^{2}\hat{\boldsymbol{k}})-10\,\mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{% \jmath}}$
	$\displaystyle=$	$\displaystyle(-50\hat{\boldsymbol{\imath}}-8\hat{\boldsymbol{\jmath}}+20\hat{% \boldsymbol{k}})\,\mathrm{rad}/\mathrm{s}^{2}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\mathcal{D}}=(-50\hat{% \boldsymbol{\imath}}-8\hat{\boldsymbol{\jmath}}+20\hat{\boldsymbol{k}})\,% \mathrm{rad}/\mathrm{s}^{2}$

SAMPLE 21.6

Filename:sfig8-6-2

Figure 21.17:

Relative and absolute angular velocity and acceleration. Consider Sample 21.14 again: A three-mass body consists of three point masses P, Q, and R, connected by three identical rigid rods $120^{\rm o}$ apart. The three-mass system is attached to shaft CDE and spins with respect to the shaft at a non-constant rate $\omega_{2}$ . Shaft CDE rotates with a constant rate $\omega_{1}$ with respect to the base sleeve which, in turn, rotates with the base shaft AB with constant angular speed $\Omega$ . At the instant shown, $\omega_{2}=10\,\mathrm{rad}/\mathrm{s}$ and is changing at the rate of $\dot{\omega}_{2}=2\,\mathrm{rad}/\mathrm{s}^{2}$ ; $\omega_{1}=5\,\mathrm{rad}/\mathrm{s}$ (constant), and $\Omega=2\,\mathrm{rad}/\mathrm{s}$ (constant).

1.

Find the angular acceleration of the body relative to the base shaft AB.
2.

Find the absolute angular acceleration of the body.

Solution Let ${\cal D}$ represent the rotating mass system. Let us attach frame $\mathcal{B}$ to the shaft CDE and frame $\mathcal{C}$ to the shaft AB. Coordinate axes $x^{\prime}y^{\prime}z^{\prime}$ are fixed in $\mathcal{B}$ and rotate with shaft CDE.

The angular velocity of ${\cal D}$ with respect to shaft AB (or frame $\mathcal{C}$ ) is:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D/C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/% \mathcal{B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/% \mathcal{C}}$
		$\displaystyle=$	$\displaystyle\omega_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{1}\hat{% \boldsymbol{k}}^{\prime}.$

Therefore, the angular acceleration of ${\cal D}$ with respect to shaft AB is:

Filename:sfig8-6-2a

Figure 21.18:

$\displaystyle{}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{C}}$	$\displaystyle=$	$\displaystyle\rm{time\;derivative\;of\;}{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal D/C}\rm{\;in\;frame\;}\mathcal{C}.$
	$\displaystyle=$	$\displaystyle\frac{{}^{\mathcal{C}}d}{dt}[\omega_{2}\hat{\boldsymbol{\jmath}}^% {\prime}+\omega_{1}\hat{\boldsymbol{k}}^{\prime}]$
	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{2}\dot% {\hat{\boldsymbol{\jmath}}}^{\prime}+\underbrace{\dot{\omega}_{1}}_{0}\hat{% \boldsymbol{k}}^{\prime}+\omega_{1}\dot{\hat{\boldsymbol{k}}^{\prime}}$
	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{2}({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/\mathcal{C}}% \times\hat{\boldsymbol{\jmath}}^{\prime})+\omega_{1}({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/\mathcal{C}}\times\hat{% \boldsymbol{k}}^{\prime})$
	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{2}(% \omega_{1}\underbrace{\hat{\boldsymbol{k}}\times\hat{\boldsymbol{\jmath}}^{% \prime}}_{-\hat{\boldsymbol{\imath}}^{\prime}})+\omega_{1}(\omega_{1}% \underbrace{\hat{\boldsymbol{k}}\times\hat{\boldsymbol{k}}^{\prime}}_{0})$
	$\displaystyle=$	$\displaystyle-\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}^{\prime}+\dot{% \omega}_{2}\hat{\boldsymbol{\jmath}}^{\prime}$
	$\displaystyle=$	$\displaystyle(-50\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}})\,% \mathrm{rad}/\mathrm{s}^{2}$

since $\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}}$ at the instant of interest.

Answer: ${}^{\lxWithClass{rp-bodyframe}{\mathcal{C}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{D}/\mathcal{C}}=(-50\hat{\boldsymbol{\imath}}% +2\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/\mathrm{s}^{2}$

The absolute angular velocity of ${\cal D}$ is

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/% \mathcal{F}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D/C}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{C}/\mathcal{F}}$
		$\displaystyle=$	$\displaystyle\underbrace{\omega_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{% 1}\hat{\boldsymbol{k}}^{\prime}}_{{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal D/C}}+\Omega\hat{\boldsymbol{\imath}}.$

Therefore, the absolute angular acceleration of ${\cal D}$ is

${}^{\mathcal{F}}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D% /\mathcal{F}}$	$\displaystyle=$	${}^{\mathcal{F}}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D% /\mathcal{C}}+^{\mathcal{F}}\dot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\mathcal{C}/\mathcal{F}}$
	$\displaystyle=$	$\displaystyle\frac{{}^{\mathcal{F}}d}{dt}[\omega_{2}\hat{\boldsymbol{\jmath}}^% {\prime}+\omega_{1}\hat{\boldsymbol{k}}^{\prime}]+\frac{{}^{\mathcal{F}}d}{dt}% [\Omega\hat{\boldsymbol{\imath}}]$
	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{2}\dot% {\hat{\boldsymbol{\jmath}}}^{\prime}+\underbrace{\dot{\omega}_{1}}_{0}\hat{% \boldsymbol{k}}^{\prime}+\omega_{1}\dot{\hat{\boldsymbol{k}}}^{\prime}+% \underbrace{\dot{\Omega}}_{0}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{2}({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/\mathcal{F}}% \times\hat{\boldsymbol{\jmath}}^{\prime})+\omega_{1}({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/\mathcal{F}}\times\hat{% \boldsymbol{k}}^{\prime}).$

But

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/% \mathcal{F}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{C}/% \mathcal{F}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}/% \mathcal{C}}$
		$\displaystyle=$	$\displaystyle\Omega\hat{\boldsymbol{\imath}}+\omega_{1}\hat{\boldsymbol{k}}.$

Therefore,

${}^{\mathcal{F}}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D% /\mathcal{F}}$	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}^{\prime}+\omega_{2}([% \Omega\hat{\boldsymbol{\imath}}+\omega_{1}\hat{\boldsymbol{k}}]\times\hat{% \boldsymbol{\jmath}}^{\prime})+\omega_{1}([\Omega_{1}+\omega_{1}\hat{% \boldsymbol{k}}]\times\hat{\boldsymbol{k}}^{\prime})$
	$\displaystyle=$	$\displaystyle\dot{\omega}_{2}\hat{\boldsymbol{\jmath}}+\omega_{2}\Omega\hat{% \boldsymbol{k}}-\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}-\omega_{1}\Omega% \hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}+(\dot{\omega}_{2}-% \omega_{1}\Omega)\hat{\boldsymbol{\jmath}}+\omega_{2}\Omega\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(-50\hat{\boldsymbol{\imath}}-8\hat{\boldsymbol{\jmath}}+20\hat{% \boldsymbol{k}})\,\mathrm{rad}/\mathrm{s}^{2}\par$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D/\mathcal{F}}=(-50% \hat{\boldsymbol{\imath}}-8\hat{\boldsymbol{\jmath}}+20\hat{\boldsymbol{k}})\,% \mathrm{rad}/\mathrm{s}^{2}$

Comments: Again, this approach based on direct differentiation, agrees with the result obtained by quoting the formula for general motion.

Problems for 21.1 Velocity and acceleration of points on a rigid body: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$

21.1.1 A particle moves on a helix. Say you know that a particle moves according to the equation

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=R\cos\theta\hat{\boldsymbol{% \imath}}+R\sin\theta\hat{\boldsymbol{\jmath}}+z\hat{\boldsymbol{k}}

where $R$ is a constant, $\theta=c_{1}t$ , and $z=c_{2}t$ .

(a)

Find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ at general time $t$ .
(b)

Find ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ at general time $t$ .
(c)

Find $\hat{\boldsymbol{e}}_{t}$ at general time $t$ .
(d)

Find $\hat{\boldsymbol{e}}_{n}$ at general time $t$ .
(e)

Write ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ in terms of $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ .
(f)

What is the radius of the osculating circle? Check your answer to see if it makes sense in the special case when $c_{2}=0$ .

21.1.2 A curious chef tosses a potato in the air. Being a part-time dynamicist, she has just the right instruments in her kitchen and measures the absolute angular velocity of the potato to be ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=1\hat{\boldsymbol{\imath}}+% 2\hat{\boldsymbol{\jmath}}+3\hat{\boldsymbol{k}}(\,\mathrm{rad}/\mathrm{s})$ and the absolute center of mass velocity to be ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}=3\hat{\boldsymbol{\imath}}+4% \hat{\boldsymbol{\jmath}}+5\hat{\boldsymbol{k}}(\,{\mathrm{in/s}})$ , at a particular instant in time. At the same instant in time, she notes the position of an eye on the potato relative to its center of mass, point $G$ , is ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{E/G}=1\in\hat{\boldsymbol{% \imath}}$ . (The high quality potato came from the market with the center of mass already marked inside it. ) $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ are basis vectors in the kitchen frame.

(a)

What is the velocity of the potato eye relative to $G$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{E/G}$ ?
(b)

What is the absolute velocity of the potato eye, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{E}$ ?

21.1.3 Find the rotation matrix [R] for $\theta=30^{{}^{\circ}}$ such that $\{\hat{\boldsymbol{e}}^{\prime}\}=\mbox{[}{\bf R}]\{\hat{\boldsymbol{e}}\}$ , where $\hat{\boldsymbol{e}}=\{\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},% \hat{\boldsymbol{k}}\}$ . Using the rotation matrix, find the components of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=2\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}-3\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}+1\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{k}}$ in the rotated coordinate system.

Filename:efig1-2-28

Figure 21.19

21.1.4 A circular plate rotates at a constant angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=5\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ . A set of coordinate axes $x^{\prime}y^{\prime}z^{\prime}$ is glued to the plate at its center and thus rotates with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . Find the rate of change of basis vectors $\hat{\boldsymbol{\imath}}^{\prime}$ , $\hat{\boldsymbol{\jmath}}^{\prime}$ , and $\hat{\boldsymbol{k}}^{\prime}$ using the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula.

21.1.5 A rigid body $\mathcal{B}$ rotates in space with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=(2\hat{\boldsymbol{\imath}}% +3\hat{\boldsymbol{\jmath}}-\hat{\boldsymbol{k}})\,\mathrm{rad}/\mathrm{s}$ . A set of local coordinate axes $x^{\prime}y^{\prime}z^{\prime}$ is fixed to the body at its center of mass. The position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$ of a point P is given in local coordinates: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}=(5\hat{\boldsymbol{\imath}}^% {\prime}+\hat{\boldsymbol{\jmath}}^{\prime})\,\mathrm{m}$ . Find the velocity ( $\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$ ) of point P using the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula.

21.1.6 The rotation matrix between two coordinate systems with basis vectors $(\hat{\boldsymbol{e}}_{1},\hat{\boldsymbol{e}}_{2},\hat{\boldsymbol{e}}_{3})$ and $(\hat{\boldsymbol{E}}_{1},\hat{\boldsymbol{E}}_{2},\hat{\boldsymbol{E}}_{3})$ is Q such that $\{{\hat{\boldsymbol{e}}}\}=[\mbox{\bf Q}]\{{\boldsymbol{E}}\}$ . Find the components of ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=(-2\hat{\boldsymbol{e}}_{1}+3% \hat{\boldsymbol{e}}_{3})\,\mathrm{m}/\mathrm{s}^{2}$ in the $(\hat{\boldsymbol{E}}_{1},\hat{\boldsymbol{E}}_{2},\hat{\boldsymbol{E}}_{3})$ basis if ⇀ Q = $\left[\begin{array}[]{rrr}1&0&0\\ 0&\frac{1}{2}&-\frac{\sqrt{3}}{2}\\ 0&\frac{\sqrt{3}}{2}&\frac{1}{2}\end{array}\right]$ .

21.1.7 Let $\cal B$ and $\cal F$ be two moving rigid bodies. Show that ${}^{\lxWithClass{rp-bodyframe}{\mathcal{B}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{B}/\mathcal{F}}={}^{\lxWithClass{rp-bodyframe% }{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}_{% \mathcal{B}/\mathcal{F}}$ .

21.1.8 George is driving his car on the highway on a rainy day. His windshield wipers are on. He notices that a drop of water on the wiper blade is moving along the blade at approximately $3\,{\mathrm{in/s}}$ . Although the wipers rotate at variable angular speed, he approximates that when the wiper is parallel to his nose, its angular speed is $\pi/2\,\mathrm{rad}/\mathrm{s}$ . In George’s coordinate system (attached to his frame of reference and moving with him), the wiper is rotating in the positive $\hat{\boldsymbol{\jmath}}$ direction and the water-drop is moving in the positive $\hat{\boldsymbol{k}}$ direction at the instant of interest. Find the velocity of the water-drop in George’s frame of reference.

21.1.9 A person sits on a bar stool which spins at angular speed $\Omega$ . Simultaneously, (s)he hoists a beer glass of mass $m$ up at angular speed $\omega$ relative to the bar stool with a straight, rigid arm of length $L$ . $\theta$ measures the angle of her/his arm with respect to the $X Y$ plane. $X Y Z$ is an inertial coordinate system; $Y Z$ lies in the plane of the paper. If you use moving axes to do this problem, specify clearly his/her orientation and motion.

(a)

What is the angular velocity of the person’s arm?
(b)

What is the angular acceleration of the person’s arm?
(c)

What is the linear velocity of the beer glass relative to inertial space?
(d)

What is the linear acceleration of the beer glass relative to inertial space?
(e)

Describe how you would point an absolutely full glass such that the beer will not spill out.

Filename:pfigure-blue-118-2

Figure 21.20

21.1.10 Kinematics of a disk spinning on a spinning rod. The rigid rod $\cal R$ rotates at constant rate $\Omega$ about the $z$ axis. Attached to this rod, a distance $R$ from the origin, is the center $C$ of a disk $\cal C$ with diameter $D$ . The disk is rotating at constant rate $\omega$ about an axis $y^{\prime}$ that is attached to, and rotates with, the rod. A point $B$ is on the outer edge of the disk. At the instant of interest, the system is in the configuration shown with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{CB}=(D/2)\hat{\boldsymbol{k}}$ .

(a)

What is the absolute angular velocity of the disk, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D/\cal F}=\Omega\hat{% \boldsymbol{k}}+\omega\hat{\boldsymbol{\jmath}}$ .
(b)

What is the absolute angular acceleration of the disk ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal D}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal D/\cal F}=-\Omega% \omega\hat{\boldsymbol{\imath}}$ .
(c)

What is the absolute velocity of point B, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=\Omega R\hat{\boldsymbol{% \jmath}}+\frac{1}{2}\omega D\hat{\boldsymbol{\imath}}$ .
(d)

What is the absolute acceleration of point B, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=-\Omega^{2}R\hat{\boldsymbol% {\imath}}+\Omega\omega D\hat{\boldsymbol{\jmath}}-\frac{1}{2}\omega^{2}D\hat{% \boldsymbol{k}}$ .

Filename:pfigure-s95f3a

Figure 21.21

21.1.11 See also problems 21.27, 21.29, and 21.2. Body $\cal A$ rotates with respect to a Newtonian frame $\cal F$ at a constant angular rate $\omega_{0}$ about axis $OO^{\prime}$ . Body $\cal B$ , a ‘fork’, rotates with respect to body $\cal A$ at a constant angular rate $\omega_{1}$ about axis $PP^{\prime}$ . Bodies $\cal A$ and $\cal B$ are of negligible mass. Body $\cal D$ , a uniform disc of mass $m$ , rotates with respect to body $\cal B$ , at a constant angular rate $\omega_{2}$ about axis $CC^{\prime}$ . Each of the bodies is turned at constant rate by motors(not shown.) At the instant shown, find the absolute angular velocity of body $\cal D$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal{F}}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal D}/{\cal F}}=\omega_% {2}\hat{\boldsymbol{\imath}}+\omega_{1}\hat{\boldsymbol{\jmath}}+\omega_{0}% \hat{\boldsymbol{k}}$ .

Filename:Danef94s1q2

Figure 21.22

21.1.12 A particle $P$ of mass $m$ is attached to the edge of the disk of radius $R$ , as shown in the figure. The disk spins about its center line with constant angular speed $\omega\,\mathrm{rad}/\mathrm{s}$ relative to the shaft $O B$ . The shaft $O B$ rotates with angular speed $\Omega\,\mathrm{rad}/\mathrm{s}$ . For the instant when the disk lies in the $Y Z$ plane, and the particle is on the $Y$ axis, as shown:

(a)

Find the particle’s velocity relative to $X Y Z$ .
(b)

Calculate the particle’s acceleration relative to $X Y Z$ .
(c)

Calculate the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ acting on the particle.

If you choose to use another coordinate system, define it explicitly.

Filename:pfigure-blue-123-1

Figure 21.23

21.1.13 See also problem 21.57. Two identical thin disks, bodies $\mathcal{C}$ and $\mathcal{D}$ , of radius $r$ are connected (perpendicularly) to opposite ends of a thin axle, as depicted below. The axle rotates about the fixed $Z$ axis at a constant angular speed $\Omega$ , and the disks roll without slipping. Determine:

1.

The total angular velocity of each disk, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{C}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ , and
2.

The total angular acceleration of each disk, $\boldsymbol{\alpha}_{\mathcal{C}}$ and $\boldsymbol{\alpha}_{\mathcal{D}}$ .

(Express your answers relative to the rotating basis { $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},\hat{\boldsymbol{k}}$ }.)

Filename:pfigure-blue-119-2

Figure 21.24

21.1.14 Kinematics of a rolling cone. See also problem 20.1. A solid cone with circular-base radius $r$ and height $h$ moves as follows. Point O is fixed in space. The cone surface always contacts the plane but may roll and/or slide. The axis of the cone $O B$ rotates about the vertical axis with constant angular velocity $-\omega_{1}\hat{\boldsymbol{\imath}}$ . The cone spins about this moving axis $\hat{\boldsymbol{\jmath}}^{\prime}$ with constant angular rate $\omega_{2}$ .

(a)

At the instant shown what is the velocity of point B? (in terms of some or all of $\omega_{1}$ , $\omega_{2}$ , $r$ , $h$ , $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ ) Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=-\frac{\omega_{1}h^{2}}{% \sqrt{h^{2}+r^{2}}}\hat{\boldsymbol{k}}$ .
(b)

At the instant shown what is the angular acceleration of the cone? (in terms of some or all of $\omega_{1}$ , $\omega_{2}$ , $r$ , $h$ , $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , and $\hat{\boldsymbol{k}}$ ) Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/\cal F}=-\frac{% \omega_{1}\omega_{2}}{\sqrt{1+\left(\frac{r}{h}\right)^{2}}}\hat{\boldsymbol{k}}$ .
(c)

If the disk rolls with no slip, find $\omega_{1}$ in terms of $\omega_{2}$ , $r$ and $h$ . Answer: $\omega_{1}=\frac{\omega_{2}}{\sqrt{h^{2}+r^{2}}}$ .
(d)

If the disk rolls with no slip and point B has constant speed $v$ find the magnitude of the angular acceleration of the cone in terms of $v$ , $r$ and $h$ . Answer: $\left|{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/\cal F}\right% |=\frac{v^{2}}{hr}\left(1+\left(\frac{r}{h}\right)^{2}\right)$ .

Filename:pfigure-s95q14

Figure 21.25

Cone rolling on a plane

21.1.15 See also problems 21.26, 21.28, 21.44, and 21.44. Body $\cal A$ rotates with respect to a Newtonian frame $\cal F$ at a constant angular rate $\omega_{0}$ about axis $OO^{\prime}$ . Body $\cal B$ , a ‘fork’, rotates with respect to body $\cal A$ at a constant angular rate $\omega_{1}$ about axis $PP^{\prime}$ . Bodies $\cal A$ and $\cal B$ are of negligible mass. Body $\cal D$ , a uniform disc of mass $m$ , rotates with respect to body $\cal B$ , at a constant angular rate $\omega_{2}$ about axis $CC^{\prime}$ . Each of the bodies is turned at constant rate by motors(not shown.) At the instant shown, find the absolute velocity and acceleration of point $Q$ on the disk.

Filename:Danef94s3q2

Figure 21.26

21.1.16 (See also problems 21.28, 21.44, and 21.44.) For the system in problem 21.25, find the absolute angular acceleration of frame $\cal D$ , ${}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{D}}$ , at the instant shown.

Filename:bikefork1-ang-accel

Figure 21.27

21.1.17 (See also problems 21.29, 21.2, and 21.2.) For the system in problem 21.21, find the absolute angular acceleration of frame $\cal D$ , ${}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{D}}$ , at the instant shown. Answer: ${}^{\lxWithClass{rp-bodyframe}{\mathcal{F}}}\dot{{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}_{\mathcal{D}}=-\omega_{0}\omega_{1}\hat{\boldsymbol{% \imath}}+\omega_{0}\omega_{2}\hat{\boldsymbol{\jmath}}-\omega_{1}\omega_{2}% \hat{\boldsymbol{k}}$ .

Filename:bikefork-ang-accel

Figure 21.28

21.1.18 (See also problems 21.26, 21.44, and 21.44.) For the configuration in problem 21.25, find the velocity and acceleration of point Q on the disk using the alternative method.

Filename:bikefork1-alt

Figure 21.29

21.1.19 (See also problems 21.27, 21.2, and 21.2.) For the configuration in problem 21.21, find the velocity and acceleration of the point on the edge of the disk directly above its center point $E$ using the alternative method.

Filename:bikefork-alt

Figure 21.30

21.1.20 To simulate the flight conditions of a space vehicle, engineers have developed the centrifuge, shown diagrammatically in the figure. A main truss arm of length $\ell=15\,\mathrm{m}$ , rotates about the AA axis. The pilot sits in the capsule which may rotate about axis CC. The seat for the pilot may rotate inside the capsule about an axis perpendicular to the page and going through the point B. These rotations are controlled by a computer that is set to simulate certain maneuvers corresponding to the entry and exit from the earth’s atmosphere, malfunctions of the control system, etc. When a pilot sits in the capsule, his head, particularly his ears, has the position shown in the enlarged figure, a distance $r=1\,\mathrm{m}$ from point B.

The main arm is rotating at $\Omega=10\,{\rm rpm}$ and accelerating at ${\dot{\Omega}}=5\,{{\rm rev}/{\rm min}^{2}}$ , the capsule is rotating at a constant speed about CC at $\omega_{c}=10\,{\rm rpm}$ , and the seat rotates at a speed of $\omega_{s}=5\,{\rm rpm}$ inside the capsule.

(a)

Determine the acceleration, in terms of how many $g$ ’s, that the pilot’s head is subjected to by choosing a moving coordinates system $x y z$ centered at B and fixed to the capsule.
(b)

Repeat the calculations of (A) by letting $x y z$ be fixed to the seat. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-(\ell{\dot{\Omega}}+2r% \Omega\omega_{s})\hat{\boldsymbol{\imath}}+(2r\Omega\omega_{c}-\ell\Omega^{2})% \hat{\boldsymbol{\jmath}}-r(\omega_{c}^{2}+\omega_{s}^{2})\hat{\boldsymbol{k}}$ .
(c)

Choose another moving coordinates system to solve the same problem.
(d)
A pilot has different tolerances for acceleration components. They are roughly:
1. (a)
  
  In a vertical direction, i.e., toe to head – 5 $g$ ’s. (Here the pilot experiences blackout or may pass out completely.)
2. (b)
  
  Front to rear – 1.5 $g$ ’s. (Here vision becomes greatly distorted.)

Arrange a test program on the centrifuge so that each acceleration is reached separately while the other accelerations are kept below one- half their tolerance levels.

Filename:pfigure-blue-128-1

Figure 21.31

21.2 3D dynamics of a rigid body

Momenta and energy of rigid bodies

The kinematics formulas we have developed allow us to calculate velocities and accelerations of points on a rigid body. We can therefore calculate the motion quantities: the momenta, their rates of change, and kinetic energy (and its rate of change).

The three basic laws of mechanics that are at our disposal are:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}};$	Linear momentum balance
$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}};$	Angular momentum balance
$\displaystyle\sum(\mbox{Power in})$	$\displaystyle=$	$\displaystyle\frac{d}{dt}({E_{\mathrm{K}}})+\frac{d}{dt}({E_{\mathrm{P}}})$
		$\displaystyle+\mbox{Dissipation rate};$	Energy/power balance

In order to be a master of mechanics, all you need is to be able to effectively use these three equations for a variety of systems. One who knows how to efficiently and accurately evaluate both sides of these equations in terms of knowns and reasonably defined unknowns is an ace mechanic. One who, in addition, knows how to solve these equations is an ace programmer and/or mathematician.

Our emphasis in this book is on systems that can be idealized as a particle or a rigid body or a system of rigid bodies. So, once you know how to get the forces and moments from the free-body diagram and once you know the masses and moments of inertia, then you need to find the accelerations of the centers of mass, the angular velocities, and angular accelerations of the various bodies.

For each body,

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm},$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/C}\times\,% \mathrm{m}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}+\overbrace{{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times\underbrace{{\accentset% {\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}}_{\text{${\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}=[\boldsymbol{I}^{\mathrm{cm}}]% \cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$}}+[\boldsymbol{I}^{% \mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}}^{\dot{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}},\mbox{ and}$
$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}+\frac{1}{2}{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}\cdot\left[[\boldsymbol{I}^{\mathrm{cm}% }]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\right]$

So, as repeatedly mentioned in this chapter, the trick is to evaluate ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ in terms of the natural variables in the problem.

The keys to make appropriate use of these formulae are:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal F}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F% }+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/\cal F}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B/\cal F% }+{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal C/\cal B}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal C/\cal B},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/\cal F}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{0^{\prime}/\cal F}% \par+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{P/\cal B},\mbox{ and}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P/\cal F}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{0^{\prime}/\cal F}% +{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times({% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}})+^{\cal F}\dot{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/O^{\prime}}$
		$\displaystyle+{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P/\cal B}+2{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{P/\cal B}.$

Alternatively, the second, third, and fourth formulae may be effectively derived on an ad-hoc basis by use of the more fundamental formulae

{}^{\cal F}\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}\times{\accentset{% \rightharpoonup}{\boldsymbol{Q}}}\qquad\mbox{ (for any ${\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ fixed in $\cal B$)}

on various base vectors.

Box 21.4 $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ for a rigid body

We can now derive the formula for rate of change of angular momentum of a rigid body $\cal B$ about some point $C$ using the ‘ $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula’

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern% -2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt% \hbox{$\textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{% \lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}})$
	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% cm/C}\times m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}+\overbrace{% \sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/cm}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{i/cm}m_{i}}^{[\boldsymbol{I}^{\mathrm{cm}}]% \cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}}\right]$
	$\displaystyle=$	$\displaystyle\underbrace{\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm% /C}\times m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{cm}}_{{\accentset{% \rightharpoonup}{\boldsymbol{0}}}}+{\accentset{\rightharpoonup}{\boldsymbol{r}% }}_{cm/C}\times m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}+\frac{d}{% dt}\left[[\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\right]$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/C}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}\dots$
		$\displaystyle+\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \cal B}\times\left[[\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\right]\quad+\ ^{\cal B}\frac{d% }{dt}\left[[\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\right]}_{\text{ $\frac{d}{dt}\left[[\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\right]$ using the $\dot{\accentset{\rightharpoonup}{\boldsymbol{Q}}}$ formula}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/C}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{\cal B}\times\left[[\boldsymbol{I}^{\mathrm{cm}}]\cdot% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\right]$

			$\displaystyle\quad+\underbrace{\ {}^{\cal B}\frac{d}{dt}[\boldsymbol{I}^{% \mathrm{cm}}]}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}\text{ $[\boldsymbol{I}^{\mathrm{cm}}]$ is constant in $\cal B$}}+[\boldsymbol{I}^{% \mathrm{cm}}]\cdot\ ^{\cal B}\dot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\cal B}$
		$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/C}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{\cal B}\times\left[[\boldsymbol{I}^{\mathrm{cm}}]\cdot% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\right]+[% \boldsymbol{I}^{\mathrm{cm}}]\cdot\ ^{\cal B}\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}$
		$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/C}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}_{\cal B}\times\left[[\boldsymbol{I}^{\mathrm{cm}}]\cdot% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\right]+[% \boldsymbol{I}^{\mathrm{cm}}]\cdot\dot{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}_{\cal B}$

In the last step of calculation, we have used the important fact about angular acceleration of a body:

	${}^{\cal F}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}$	$\displaystyle=$	${}^{\cal B}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F% }+\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F% }\times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}}_{{% \accentset{\rightharpoonup}{\boldsymbol{0}}}}$
		$\displaystyle=$	${}^{\cal B}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}.$

That is, the derivative with respect to time of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B/\cal F}$ is the same in $\cal F$ as in $\cal B$ .

Now, we look at finding the linear and angular momenta for a rigid body using the 3-D example of the crane in fig. 21.32. The crane is a system of connected bodies moving in well defined ways relative to each other and to the ground.

Example: Mechanics of a rigid body in a mechanism (3-D): A crane, again

Filename:tfigure8-syst-bods

Figure 21.32:

We would like to find the rate of change of linear momentum of the boom, $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}_{\cal A}$ , and the rate of change of angular momentum of the boom about point $O$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\cal A}/O}$ . Assume the mass $m_{\cal A}$ and the moment of inertia matrix $[\boldsymbol{I}^{\mathrm{cm}}]_{\cal A}$ are both known. We have

	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}_{\cal A}$	$\displaystyle=$	$\displaystyle m_{\cal A}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G},% \mbox{ and}$
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\cal A}O}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}\times m_{\cal A% }{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}+[\boldsymbol{I}^{\mathrm{cm% }}]_{\cal A}\cdot{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}\times\left[[% \boldsymbol{I}^{\mathrm{cm}}]_{\cal A}\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal A}\right].$

So, we must find ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A}$ . Here we go. First, we find ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}$

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{\prime}/O}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O^{\prime}}$
		$\displaystyle=$	$\displaystyle\overbrace{d\hat{\boldsymbol{\imath}}+(2s+h)\hat{\boldsymbol{% \jmath}}}^{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{O^{\prime}/O}}+% \overbrace{\frac{\ell}{3}(\cos\phi\hat{\boldsymbol{\imath}}^{\prime}+\sin\phi% \hat{\boldsymbol{\jmath}}^{\prime})}^{{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{G/O^{\prime}}}.$

Next, we find ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}$

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal A/\cal B}=\dot{\theta}\hat{\boldsymbol{\jmath}}^{% \prime}+\dot{\phi}\hat{\boldsymbol{k}}^{\prime}.

Then, we find ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A}$

{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A}={\accentset{% \rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}+{\accentset{\rightharpoonup}{% \boldsymbol{\alpha}}}_{\cal C/\cal B}+{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \cal C/\cal B}=\ddot{\theta}\hat{\boldsymbol{\jmath}}^{\prime}+\ddot{\phi}\hat% {\boldsymbol{k}}^{\prime}+\dot{\theta}\dot{\phi}\hat{\boldsymbol{\imath}}^{% \prime}.

Finally, we find ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{O^{\prime}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}_{\cal A}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O^{\prime}})+{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A/\cal F}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O^{\prime}}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G/\cal B}+2{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{v% }}}_{G/\cal B}.\par

At this point, we see several terms that we need to evaluate to plug into this formula. Here they are:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{O^{\prime}}$	$\displaystyle=$	$\displaystyle\ddot{d}\hat{\boldsymbol{\imath}},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G/\cal B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A/\cal B% }\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O^{\prime}},$
	$\displaystyle=$	$\displaystyle\frac{\ell}{3}\dot{\phi}(\cos\phi\hat{\boldsymbol{\jmath}}^{% \prime}-\sin\phi\hat{\boldsymbol{\imath}}^{\prime}),\mbox{ and}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G/\cal B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A/\cal B% }\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A/\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O^{\prime}})+{% \accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A/\cal B}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O^{\prime}},$
	$\displaystyle=$	$\displaystyle\frac{\ell}{3}(-\ddot{\phi}\sin\phi+\dot{\phi}^{2}\cos\phi)\hat{% \boldsymbol{\imath}}^{\prime}+\frac{\ell}{3}(\ddot{\phi}\cos\phi-\dot{\phi}^{2% }\sin\phi)\hat{\boldsymbol{\jmath}}^{\prime}.$

So, we can plug in and find the absolute acceleration of the center-of-mass of body $\cal A$

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}\par$	$\displaystyle=$	$\displaystyle\ddot{d}\hat{\boldsymbol{\imath}}+\frac{\ell}{3}\left[-(\ddot{% \phi}\sin\phi+(\dot{\phi}^{2}+\dot{\theta}^{2})\cos\phi)\hat{\boldsymbol{% \imath}}^{\prime}\right.$
			$\displaystyle\left.+(\ddot{\phi}\cos\phi-\dot{\phi}^{2}\sin\phi)\hat{% \boldsymbol{\jmath}}^{\prime}+(2\dot{\phi}\dot{\theta}\sin\phi-\ddot{\theta}% \cos\phi)\hat{\boldsymbol{k}}^{\prime}\right].$

Thus, for the crane boom, now we have ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/O}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$ in terms of known dimensions, position vectors, and the angular rates $\dot{\phi}$ , and $\dot{\theta}$ , and their derivatives $\ddot{\phi}$ and $\ddot{\theta}$ . So, now we can evaluate $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}_{\cal A}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\cal A}/O}$ .

Yes – this situation is a mess! Rigid body calculations in three dimensions lead to big messy equations. There are a few special cases where things simplify. Some of these situations are shown in the sample problems and the homework.

A note of caution: When back substituting with the formulae given, be sure that the same base vectors are used for $[\boldsymbol{I}^{\mathrm{cm}}]$ as for ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal A}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal A}$ .

SAMPLE 21.7

Filename:sfig8-7-2

Figure 21.33:

An axisymmetric body with precession and spin. A uniform circular disk of mass $m=2\,\mathrm{kg}$ is mounted on gimbals which in turn is mounted on a horizontal shaft. The disk spins about its normal axis of symmetry. The spin axis makes an angle $\phi=15^{{}^{\circ}}$ with the horizontal shaft. The shaft rotates at a constant speed $\omega_{1}=60\,{\rm rpm}.$ The disk spins with respect to the gimbals at a constant speed $\omega_{2}=120\,{\rm rpm}.$ The moment of inertia of the disk about its diameter is $A=10\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$ and the moment of inertia of the disk about its normal is $B=20\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}.$ At the instant shown, find

1.

the angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$ of the disk,
2.

the rate of change of the angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$ of the disk.

Solution Let $x y z$ be the coordinate axes attached to the inertial frame and $x^{\prime}y^{\prime}z^{\prime}$ be the coordinate axes attached to the gimbals $\cal G$ such that $z^{\prime}$ is aligned with the spin axis of the disk. Therefore, at the instant shown,

\hat{\boldsymbol{\imath}}^{\prime}=\cos\phi\hat{\boldsymbol{\imath}}-\sin\phi% \hat{\boldsymbol{k}},\quad\hat{\boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{% \jmath}},\quad\hat{\boldsymbol{k}}^{\prime}=\sin\phi\hat{\boldsymbol{\imath}}+% \cos\phi\hat{\boldsymbol{k}}.

(21.12)

Filename:sfig8-7-2a

Figure 21.34:

1.

Angular momentum: The angular momentum of the disk is given by

${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=[\boldsymbol{I}^{\mathrm{cm% }}]\mathord{\cdot}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal% {D}}$

where $[\boldsymbol{I}^{\mathrm{cm}}]$ is the moment of inertia matrix of the disk about its center-of-mass and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ is the absolute angular velocity of the disk.

Now the absolute angular velocity of the disk

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal G}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{{\cal D}/{\cal G}}$

$\displaystyle\rm{where}\quad{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% _{\cal G}$ $\displaystyle=$ $\displaystyle\omega_{1}\hat{\boldsymbol{k}}=60\,{\rm rpm}\hat{\boldsymbol{k}}=% 2\pi\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}},$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal D}/{% \cal G}}$ $\displaystyle=$ $\displaystyle\omega_{2}\hat{\boldsymbol{k}}^{\prime}=120\,{\rm rpm}\hat{% \boldsymbol{k}}^{\prime}=4\pi\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}^{% \prime}.$

The principal moments of inertia for the disk are given. We have chosen the body axes $x^{\prime}y^{\prime}z^{\prime}$ along the principal axes (i.e. $x^{\prime}y^{\prime}$ along the two diameters of the disk and $z^{\prime}$ along the normal of the disk). Therefore in the $x^{\prime}y^{\prime}z^{\prime}$ coordinate system

$[\boldsymbol{I}^{\mathrm{cm}}]_{x^{\prime}y^{\prime}z^{\prime}}=\left[\begin{% array}[]{ccc}A&0&0\\ 0&A&0\\ 0&0&B\end{array}\right]$

Where $A=10\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$ and $B=20\,\mathrm{kg}\mathord{\cdot}\,\mathrm{m}^{2}$ . Since the inertia matrix is written in the body coordinate system, we need to express ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ also in the body coordinate system to carry out the product $[\boldsymbol{I}^{\mathrm{cm}}]\mathord{\cdot}{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{D}}.$

$\rm{Now}\quad{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}=% \omega_{1}\hat{\boldsymbol{k}}+\omega_{2}\hat{\boldsymbol{k}}^{\prime},$

and from Fig. 21.33 we can write $\hat{\boldsymbol{k}}=-\sin\phi\hat{\boldsymbol{\imath}}^{\prime}+\cos\phi\hat{% \boldsymbol{k}}^{\prime},$ therefore,

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}=(-\omega_{1}% \sin\phi)\hat{\boldsymbol{\imath}}^{\prime}+(\omega_{2}+\omega_{1}\cos\phi)% \hat{\boldsymbol{k}}^{\prime}.$

So,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$ $\displaystyle=$ $\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]\mathord{\cdot}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}=\left[\begin{array}[]{ccc% }A&0&0\\ 0&A&0\\ 0&0&B\end{array}\right]\left\{\begin{array}[]{c}-\omega_{1}\sin\phi\\ 0\\ \omega_{2}+\omega_{1}\cos\phi\end{array}\right\}$ (21.19)

$\displaystyle=$ $\displaystyle(-A\omega_{1}\sin\phi)\hat{\boldsymbol{\imath}}^{\prime}+B(\omega% _{2}+\omega_{1}\cos\phi)\hat{\boldsymbol{k}}^{\prime}.$ (21.20)

Substituting the given values of $\omega_{1},\;\omega_{2},\;A,\;B$ and $\phi$ we get

${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=[-16.26\hat{\boldsymbol{% \imath}}^{\prime}+372.71\hat{\boldsymbol{k}}^{\prime}]\,\mathrm{kg}\mathord{% \cdot}\,\mathrm{m}^{2}/\,\mathrm{s}$

Although we have the answer we need, it is expressed in terms of the body coordinate system’s basis vectors. We can easily express ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$ in the fixed coordinate system using Eq 21.12:

${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=[-16.26(\cos\phi\hat{% \boldsymbol{\imath}}-\sin\phi\hat{\boldsymbol{k}})+372.71(\sin\phi\hat{% \boldsymbol{\imath}}+\cos\phi\hat{\boldsymbol{k}})]\,{\mbox{N}}\mathord{\cdot}% \,\mathrm{m}\mathord{\cdot}\,\mathrm{s}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=(80.76\hat{\boldsymbol{% \imath}}+364.22\hat{\boldsymbol{k}})\,{\mbox{N}}\mathord{\cdot}\,\mathrm{m}% \mathord{\cdot}\,\mathrm{s}$

Rate of change of angular momentum: The rate of change of angular momentum of the disk about point G is given by

\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=[\boldsymbol{I}^{% \mathrm{cm}}]\mathord{\cdot}\dot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\mathcal{D}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{D}}\times[\boldsymbol{I}^{\mathrm{cm}}]\mathord{\cdot}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}

(21.21)

Thus, to compute $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$ , we need the angular acceleration of the disk $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ .

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{% D}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal G}+% \dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal D}/{\cal G}}$
	$\displaystyle=$	$\displaystyle\underbrace{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}% }_{1}}_{0}\hat{\boldsymbol{k}}+\underbrace{\underbrace{\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{2}}_{0}\hat{\boldsymbol{k}}^{\prime}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}\hat{\boldsymbol{k}}% \times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{2}\hat{\boldsymbol{k% }}^{\prime}}_{\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal D}% /{\cal G}}}$
	$\displaystyle=$	$\displaystyle\omega_{1}\omega_{2}(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{k% }}^{\prime})=\omega_{1}\omega_{2}(-\sin\phi\hat{\boldsymbol{\imath}}^{\prime}+% \cos\phi\hat{\boldsymbol{k}}^{\prime})\times\hat{\boldsymbol{k}}^{\prime}=% \omega_{1}\omega_{2}\sin\phi\hat{\boldsymbol{\jmath}}^{\prime}.$

Here, we calculated $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ in the body coordinate system because we need to calculate $[\boldsymbol{I}^{\mathrm{cm}}]\mathord{\cdot}\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{D}}$ and we know $[\boldsymbol{I}^{\mathrm{cm}}]$ in the body coordinate system. Thus the first term in $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$ is

[\boldsymbol{I}^{\mathrm{cm}}]\mathord{\cdot}\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{D}}=\left[\begin{array}[]{ccc}A&0&0\\ 0&A&0\\ 0&0&B\end{array}\right]\left\{\begin{array}[]{c}0\\ \omega_{1}\omega_{2}\sin\phi\\ 0\\ \end{array}\right\}=A\omega_{1}\omega_{2}\sin\phi\hat{\boldsymbol{\jmath}}^{% \prime}.

The second term in $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$ is obtained by taking the cross product of ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ expressed in the body coordinate system with the expression for ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$ given by Eq 21.20:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}% \times[\boldsymbol{I}^{\mathrm{cm}}]\mathord{\cdot}{\accentset{\rightharpoonup% }{\boldsymbol{\omega}}}_{\mathcal{D}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}% \times{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$
	$\displaystyle=$	$\displaystyle[(-\omega_{1}\sin\phi)\hat{\boldsymbol{\imath}}^{\prime}+(\omega_% {2}+\omega_{1}\cos\phi)\hat{\boldsymbol{k}}^{\prime}]\times[(-A\omega_{1}\sin% \phi)\hat{\boldsymbol{\imath}}^{\prime}+B(\omega_{2}+\omega_{1}\cos\phi)\hat{% \boldsymbol{k}}^{\prime}]$
	$\displaystyle=$	$\displaystyle(B-A)\omega_{1}\sin\phi[\omega_{1}\cos\phi+\omega_{2}]\hat{% \boldsymbol{\jmath}}^{\prime}.$

Adding the two terms together we get

	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$	$\displaystyle=$	$\displaystyle A\omega_{1}\omega_{2}\sin\phi\hat{\boldsymbol{\jmath}}^{\prime}+% (B-A)\omega_{1}\sin\phi[\omega_{1}\cos\phi+\omega_{2}]\hat{\boldsymbol{\jmath}% }^{\prime}$		(21.22)
		$\displaystyle=$	$\displaystyle[(B-A)\omega_{1}\cos\phi+B\omega_{2}]\omega_{1}\sin\phi\hat{% \boldsymbol{\jmath}}^{\prime}.$		(21.22)

Substituting $B=20\,\mathrm{kg}\cdot\,\mathrm{m}^{2},\;A=10\,\mathrm{kg}\cdot\,\mathrm{m}^{2% },\;\omega_{1}=2\pi\,\mathrm{rad}/\mathrm{s},\;\omega_{2}=4\pi\,\mathrm{rad}/% \mathrm{s}\;\rm{and}\;\phi=15^{{}^{\circ}}$ we get the answer

Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=507.41\,{\rm N}\mathord% {\cdot}{\rm m}\;\hat{\boldsymbol{\jmath}}$ .

SAMPLE 21.8

Filename:sfig8-7-2again

Figure 21.35: The origin G is at the center-of-mass of the disk in this problem.

Torque free motion of an axisymmetric rigid body Consider the gimbal-mounted disk of sample 21.32 again. Assume that the net torque on the disk is zero. That is, for what motions could the gimbals be eliminated and have the motion proceed?

1.

Derive an expression for the ratio of the two angular speeds, $\frac{\omega_{1}}{\omega_{2}}$ , for the torque free motion of the disk.
2.

Assume $\phi$ to be very small. What is the ratio of the angular speeds for the given disks (or any flat axisymmetric object with $B=2A$ )?

Solution

1.

The angular momentum balance for the disk about point G gives:

$\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{G}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{G}$

We are given that the net torque on the disk must be zero, i.e., $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{G}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{G}={\accentset{\rightharpoonup}{\boldsymbol% {0}}}$ . We found an expression for $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{G}$ in the previous sample problem (see Eq 21.22 of Sample 21.32) to be:

$\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{G}=[(B-A)\omega_{1}\cos\phi% +B\omega_{2}]\omega_{1}\sin\phi\hat{\boldsymbol{\jmath}}.$

Now under the torque free condition,

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{G}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad(B-A)\omega_{1}\cos\phi$ $\displaystyle=$ $\displaystyle-B\omega_{2}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\frac{\omega_{1}}{% \omega_{2}}$ $\displaystyle=$ $\displaystyle\frac{B}{(A-B)\cos\phi}$ (21.23)
2.

for small $\phi$ , $\cos\phi\approx 1.$ Also, for the disk, $B=2A$ . Substituting these values into the above expression for $\frac{\omega_{1}}{\omega_{2}}$ , we get

$\frac{\omega_{1}}{\omega_{2}}=\frac{2A}{A-2A}=-2$

That is, the rate of precession is twice the rate of relative spin. Also, if ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{1}=\omega_{1}\hat{% \boldsymbol{k}}$ , then ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{2}=-\frac{\omega_{1}}{2}% \hat{\boldsymbol{k}}^{\prime}$ .

Answer: $\frac{\omega_{1}}{\omega_{2}}=-2$

Comments: The results obtained here are more general than they seem. In particular, it can be shown that the most general motion of a torque free axisymmetric body satisfies Eq 21.23. If you throw a coin in the air it wobbles according to the motion described as constant rate motion about an axis that itself rotates at a constant rate.

In case (b), the absolute angular velocity is $\omega_{1}\hat{\boldsymbol{k}}$ + $\omega_{2}\hat{\boldsymbol{k}}^{\prime}$ which is $\omega_{1}\boldsymbol{k}+(-\frac{\omega_{1}}{2})\boldsymbol{k}^{\prime}$ which is approximately $\frac{\omega_{1}}{2}\boldsymbol{k}^{\prime}$ . So, the wobble rate $\omega_{1}$ is twice the spin rate $\frac{\omega_{1}}{2}$ for flat round objects in torque free motion.

SAMPLE 21.9

Filename:sfig8-7-2disks

Figure 21.36:

Calculation of energy. In the figures shown, the rigid arm AB rotates with constant angular velocity $\Omega=5\,\mathrm{rad}/\mathrm{s}$ and the disk $\mathcal{D}$ rotates with respect to the arm at constant angular speed $\omega=10\,\mathrm{rad}/\mathrm{s}$ . Assume that the rigid arm is massless. The uniform disk has a mass of $2\,\mathrm{kg}$ and measures $50\;\rm{cm}\;$ along the diameter. Find the kinetic energy of the system in each case.

Solution

Case (a): The motion of the disk is in 2-D . Therefore, the kinetic energy of the disk is

{E_{\mathrm{K}}}=\frac{1}{2}mv_{\rm{cm}}^{2}+\frac{1}{2}I_{zz}^{\rm{cm}}\omega% ^{2}

Where $v_{\rm{cm}}$ is the speed of the center-of-mass of the disk and $\omega$ is the magnitude of the absolute angular velocity of the disk. Since the center-of-mass of the disk executes constant-speed circular motion with radius $L=1\,\mathrm{m}$ ,

v_{\rm{cm}}=\Omega L=5\,\mathrm{rad}/\mathrm{s}\cdot 1\,\mathrm{m}=5\,\mathrm{% m}/\mathrm{s}.

The absolute angular velocity of the disk is

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}=\Omega\hat{% \boldsymbol{k}}+\omega\hat{\boldsymbol{k}}=(\Omega+\omega)\hat{\boldsymbol{k}}.

Therefore, the angular speed of the disk is $\omega_{\mathcal{D}}=\Omega+\omega=15\,\mathrm{rad}/\mathrm{s}.$ Substituting these values and $m=2\,\mathrm{kg}$ and $I_{zz}^{\rm{cm}}=\frac{1}{2}mr^{2}=0.0625\,\mathrm{kg}\,\mathrm{m}^{2},\;$ we get

{E_{\mathrm{K}}}=\frac{1}{2}\cdot 2\,\mathrm{kg}\cdot(5\,\mathrm{m}/\mathrm{s}% )^{2}+\frac{1}{2}\cdot(0.0625\,\mathrm{kg}\,\mathrm{m}^{2})\cdot(15\,\mathrm{% rad}/\mathrm{s})^{2}=32\,{\rm N}\mathord{\cdot}{\rm m}

Answer: ${E_{\mathrm{K}}}=32\,{\rm N}\mathord{\cdot}{\rm m}$

Case (b): The motion of the disk is in 3-D. The kinetic energy is now given by

{E_{\mathrm{K}}}=\frac{1}{2}mv_{\rm{cm}}^{2}+\frac{1}{2}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}\cdot([\boldsymbol{I}^{% \mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal% {D}})

The translational part of the energy is the same as in case (a). For the rotational part, note that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}=\Omega\hat{% \boldsymbol{k}}+\omega\hat{\boldsymbol{\jmath}}^{\prime}=\Omega\hat{% \boldsymbol{k}}+\omega\hat{\boldsymbol{\jmath}},\;$ and ^†^†margin: Without calculation, could you have predicted that the kinetic energy in case (a) would be greater than in case (b)?

[\boldsymbol{I}^{\mathrm{cm}}]=\frac{1}{4}mr^{2}\left[\begin{array}[]{ccc}1&0&% 0\\ 0&1&0\\ 0&0&2\end{array}\right].

Therefore, the rotational kinetic energy is

$\displaystyle\frac{1}{2}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \mathcal{D}}\cdot([\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}})$	$\displaystyle=$	$\displaystyle\frac{1}{2}(\Omega\hat{\boldsymbol{k}}+\omega\hat{\boldsymbol{% \jmath}})\cdot\frac{1}{4}mr^{2}\left[\begin{array}[]{ccc}1&0&0\\ 0&1&0\\ 0&0&2\end{array}\right]\left\{\begin{array}[]{c}0\\ \omega\\ \Omega\end{array}\right\}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(\Omega\hat{\boldsymbol{k}}+\omega\hat{\boldsymbol{% \jmath}})\cdot\frac{1}{4}mr^{2}(\omega\hat{\boldsymbol{\jmath}}+2\Omega\hat{% \boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle\frac{1}{8}mr^{2}(2\Omega^{2}+\omega^{2})$
	$\displaystyle=$	$\displaystyle 2.34\,{\rm N}\mathord{\cdot}{\rm m}$

Therefore,

{E_{\mathrm{K}}}=\frac{1}{2}\cdot 2\,\mathrm{kg}\cdot(5\,\mathrm{m}/\mathrm{s}% )^{2}+2.34\,{\rm N}\mathord{\cdot}{\rm m}\approx 27\,{\rm N}\mathord{\cdot}{% \rm m}.

Answer: ${E_{\mathrm{K}}}=27\,{\rm N}\mathord{\cdot}{\rm m}$

SAMPLE 21.10

Filename:sfig8-4-4

Figure 21.37:

Dynamic reactions. A uniform disk of mass $m=2\,\mathrm{kg}$ and radius $r=0.5\,\mathrm{m}$ is mounted on frictionless bearings on a ‘L’ shaped massless shaft OBC. The disk spins about its centroidal axis at a constant rate $\omega_{2}=15\,\mathrm{rad}/\mathrm{s}$ while the shaft rotates about O in the vertical plane at a constant rate $\omega_{1}=3\,\mathrm{rad}/\mathrm{s}$ . Given that $L=1\,\mathrm{m},\,R=0.5\,\mathrm{m}$ , find the dynamic reactions at the support O at the instant shown in Fig. 21.37.

^†^†margin:

Solution

Filename:sfig8-4-4a

Figure 21.38: Free-body diagram of the shaft and the disk system.

{\accentset{\rightharpoonup}{\boldsymbol{R}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{M}}}

are the unknown dynamic reactions at the support O.

Let us take the disk together with the massless shaft as our system. The free-body diagram of this system is shown in Fig. 21.38. Since we are interested only in dynamic reactions, the force on the system due to gravity, i.e., the weight of the disk is ignored (because this force induces static reactions). Now let us carry out the momentum balance for our system.

From the linear momentum balance for the system,

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm cm}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{R}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}.$		(21.25)

Thus, we need to find the acceleration of point C, the center-of-mass of the disk, to evaluate ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ . Since point C is at the end of the shaft OBC, it rotates about point O with constant speed $\omega_{1}$ , i.e. it executes constant rate circular motion with radius OC (see Figure 21.39) and its motion lies in the $x y$ -plane. Therefore,

Filename:sfig8-4-4b

Figure 21.39: The center-of-mass of the disc, point C, executes constant rate circular motion about point O.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}$	$\displaystyle=$	$\displaystyle\underbrace{\dot{\omega}_{1}}_{0}\,\hat{\boldsymbol{k}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}-\omega_{1}^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{C}$
	$\displaystyle=$	$\displaystyle-\omega_{1}^{2}(L\hat{\boldsymbol{\imath}}+R\hat{\boldsymbol{% \jmath}})$
	$\displaystyle=$	$\displaystyle-9(\,\mathrm{rad}/\mathrm{s})^{2}\cdot(1\,\mathrm{m}\hat{% \boldsymbol{\imath}}+0.5\,\mathrm{m}\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-(9.0\hat{\boldsymbol{\imath}}+4.5\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}^{2}.$

Substituting the result in Eqn.( 21.25) we get

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{R}}}$	$\displaystyle=$	$\displaystyle 2\,\mathrm{kg}\cdot(-9.0\hat{\boldsymbol{\imath}}-4.5\hat{% \boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}^{2}$
		$\displaystyle=$	$\displaystyle-18\,{\mbox{N}}\hat{\boldsymbol{\imath}}-9\,{\mbox{N}}\hat{% \boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=-18\,{\mbox{N}}\hat{\boldsymbol{% \imath}}-9\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$

To find ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ let us do angular momentum balance for the system about point 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}$}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}$	$\displaystyle=$	$\displaystyle[\boldsymbol{I}^{\mathrm{cm}}]\cdot\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}\times([\boldsymbol{I}^{% \mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal% {D}})+\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}}_{={\accentset{% \rightharpoonup}{\boldsymbol{0}}}}$		(21.26)

where the last term is zero because ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}||{\accentset{\rightharpoonup% }{\boldsymbol{a}}}_{C}$ (see Fig .21.39). Thus, to find ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ we need to find the absolute angular acceleration $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ , the absolute angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ and $[\boldsymbol{I}^{\mathrm{cm}}]$ of the disk.

Let us attach a frame $\cal B$ to the shaft OBC. Thus frame $\cal B$ rotates with the shaft with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\omega_{1}% \hat{\boldsymbol{k}}$ . For calculations in the rotating frame, we attach a coordinate system $x^{\prime}y^{\prime}z^{\prime}$ with basis vectors $\hat{\boldsymbol{\imath}}^{\prime},\hat{\boldsymbol{\jmath}}^{\prime},\hat{% \boldsymbol{k}}^{\prime}$ to the shaft (and not to the disk) at point C (see Fig .21.40). Note that the moment of inertia matrix $[\boldsymbol{I}^{\mathrm{cm}}]$ of the disk in the $x^{\prime}y^{\prime}z^{\prime}$ coordinate system remains constant due to the symmetry of the disk about these axes and this matrix in the primed coordinate system is written as

[\boldsymbol{I}^{\mathrm{cm}}]=\frac{mr^{2}}{4}\left[\begin{array}[]{lll}1&0&0% \\ 0&2&0\\ 0&0&1\end{array}\right].

Filename:sfig8-4-4c

Figure 21.40: The primed coordinate system

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

is glued to the rotating frame

\cal B

which rotates with the shaft OBC with angular velocity

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=\omega_{1}% \hat{\boldsymbol{k}}

. At the instant shown, the two coordinate axes

x y z

and

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

are parallel to each other.

Now we find the absolute angular velocity and angular acceleration of the disk:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}+% {\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal{B}}$
	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}+\omega_{2}\hat{\boldsymbol{\jmath}}.$
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{% D}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{% B}}+\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/% \mathcal{B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}/\mathcal% {B}}$
	$\displaystyle=$	$\displaystyle\underbrace{\dot{\omega}_{1}}_{0}\hat{\boldsymbol{k}}+\underbrace% {\dot{\omega}_{2}}_{0}\hat{\boldsymbol{\jmath}}+\omega_{1}\hat{\boldsymbol{k}}% \times\omega_{2}\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}.$

But, for the matrix and vector products in Eqn. (21.26) to be defined both quantities must be expressed in the same coordinate system. Since $[\boldsymbol{I}^{\mathrm{cm}}]$ is in the primed coordinate system, we must express ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}$ in the primed coordinate system. Fortunately, at the instant of interest

\hat{\boldsymbol{\imath}}^{\prime}=\hat{\boldsymbol{\imath}},\quad\hat{% \boldsymbol{\jmath}}^{\prime}=\hat{\boldsymbol{\jmath}},\quad\hat{\boldsymbol{% k}}^{\prime}=\hat{\boldsymbol{k}}.

Therefore, the components of any vector in the two coordinate systems are trivially the same. Hence,

[\boldsymbol{I}^{\mathrm{cm}}]\cdot\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{D}}=\frac{mr^{2}}{4}\left[\begin{array}[]{lll}% 1&0&0\\ 0&2&0\\ 0&0&1\end{array}\right]\left\{\begin{array}[]{c}-\omega_{1}\omega_{2}\\ 0\\ 0\end{array}\right\}=-\frac{1}{4}mr^{2}\omega_{1}\omega_{2}\hat{\boldsymbol{% \imath}},

and similarly,

[\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{\mathcal{D}}=\frac{mr^{2}}{4}(2\omega_{2}\hat{\boldsymbol{\jmath}}+% \omega_{1}\hat{\boldsymbol{k}}).

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{D}}% \times([\boldsymbol{I}^{\mathrm{cm}}]\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\mathcal{D}})$	$\displaystyle=$	$\displaystyle(\omega_{1}\hat{\boldsymbol{k}}+\omega_{2}\hat{\boldsymbol{\jmath% }})\times\frac{mr^{2}}{4}(2\omega_{2}\hat{\boldsymbol{\jmath}}+\omega_{1}\hat{% \boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle\frac{mr^{2}}{4}(\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}-2% \omega_{1}\omega_{2}\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle-\frac{1}{4}mr^{2}\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}.$

Now substituting these expressions in Eqn. (21.26) we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}$	$\displaystyle=$	$\displaystyle-\frac{1}{4}mr^{2}\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}-% \frac{1}{4}mr^{2}\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-\frac{1}{2}mr^{2}\omega_{1}\omega_{2}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-\frac{1}{2}(2\,\mathrm{kg})\cdot 0.25\,\mathrm{m}^{2}\cdot 15\,% \mathrm{rad}/\mathrm{s}^{2}\cdot 3\,\mathrm{rad}/\mathrm{s}$
	$\displaystyle=$	$\displaystyle-11.25\,{\rm N}\mathord{\cdot}{\rm m}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}=-11.25\,{\rm N}\mathord{\cdot}{% \rm m}$

Problems for 21.2 Instantaneous dynamics and “inverse dynamics” in 3-D

21.2.1 Under what circumstances is the linear momentum of a system conserved (that is, does not change with time)?

21.2.2 For a continuous system, where at a given instant in time velocity depends on position, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}({\accentset{\rightharpoonup}{\boldsymbol{x}}})$ , (a) how is linear momentum defined? (b) How is rate of change of linear momentum defined?

21.2.3 For what points $C$ , with positions ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ , are the balance of angular momentum equations $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{C}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{% H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$ , correct:

(a)

the origin,
(b)

the center of mass, and
(c)

any point anywhere?

21.2.4 Bead on a stationary 3-D wire. A bead with mass $m$ slides on a frictionless wire that is fixed in space. The wire is twisted and curved in complicated ways. There is no gravity. The only force on the bead comes from the wire. The initial speed of the bead is $v_{0}$ . Justify your answers to the following questions with text and/or equations. Add any extra assumptions if you feel they are required.

(a)

Is it true that ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ (where ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ is the total force on the bead and $\boldsymbol{a}$ its acceleration?)?
(b)

Is the momentum of the bead, $m{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ , constant?
(c)

Is the kinetic energy of the bead, ${1\over 2}mv^{2}$ , constant?
(d)

What is the speed of the bead at $t=3\,{\rm s}$ ?
(e)

Is the acceleration always such that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{\boldsymbol{e}}_{t}=0$ (where $\hat{\boldsymbol{e}}_{t}$ is the unit tangent to the wire)?
(f)

Is the acceleration always such that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}\cdot\hat{\boldsymbol{e}}_{n}=0$ (where $\hat{\boldsymbol{e}}_{n}$ is the unit normal to the wire)?
(g)

Using any notation you like to describe the wire shape, write an expression for the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ on the bead from the wire.
(h)

For a circular wire, what is the direction of the force from the bead on the wire?
(i)

For a helical wire, (say, ${{\accentset{\rightharpoonup}{\boldsymbol{r}}}}=r_{0}\cos\theta{\hat{% \boldsymbol{\imath}}}+r_{0}\sin\theta{\hat{\boldsymbol{\jmath}}}+C\theta{\hat{% \boldsymbol{k}}})$ , what is the direction of the force from the bead on the wire? Answer: $-(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})$ .

21.2.5 A $500\,\mathrm{kg}$ roller coaster car travels along a track defined by the conical spiral $r=z$ , $\theta=-2z$ . If $\dot{\theta}=2\,\mathrm{rad}/\mathrm{s}$ is constant, determine the force exerted on the car by the track when $z=5\,\mbox{meters}$ . Assume that the gravitational force acts in the negative $z$ direction. Remember to express the force in vector form.

21.2.6 When do you use which term in ${\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\rm O}}}$ formula for a rigid body? To do the angular momentum balance about an arbitrary point $O$ we write $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\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}$}}}$ where the most general formula for ${\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\rm O}}}$ for one rigid body is

{\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\rm O}}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm cm/o}\times m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\rm cm}+[{\bf I}^{\rm cm}]\cdot\dot{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}+{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}\times([{\bf I}^{\rm cm}]\cdot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}).

Give examples of motions of a rigid body (you may also use a point mass) in which

(a)

only one term at a time in the above formula for ${\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{{\rm O}}}$ survives (three examples) and
(b)

only two terms at a time in the formula survive (three examples).

21.2.7 (See also problems 21.27, 21.29, and 21.2.) Find the reaction supplied by the environment to the system of bodies $\cal A\cal B\cal D$ so that each body spins at a constant rate in problem 21.21. Answer: Net force: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m(-omega_{0}^{2}L)\hat{% \boldsymbol{\jmath}}$ , Net moment: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/G}=\frac{1}{2}mr^{2}(-\omega_{% 0}\omega_{1}\hat{\boldsymbol{\imath}}+\omega_{0}\omega_{2}\hat{\boldsymbol{% \jmath}}-\omega_{1}\omega_{2}\hat{\boldsymbol{k}})$ .

21.2.8 Disk suspended at corner by a ball and socket joint. See also problem 21.2. A disk with negligible mass has two point masses glued to it. It is stationary and horizontal when it is allowed to fall, but it is held back by a ball-and-socket joint at one point on its perimeter. Just after the disk is released, what is the acceleration of the point P? Is it up or down?

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=0.732g\hat{\boldsymbol{k}}.$

Filename:pfigure-s94h13p2

Figure 21.41

21.2.9 A car driving in circles counter-clockwise at constant speed so that its left rear tire has speed $v_{o}$ (the middle of the tire, that is, or the location of the ground contact point.) The radius of the circle that the tire travels on is $R_{o}$ . The radius of the tire is $r_{o}$ . Right next to the tire is the car fender (i.e., the fender overlaps the outer face of the tire). In fact the fender is rubbing on the tire just a little, but not quite enough to disturb any insects that might be in the neighborhood).

(a)

What is the angular velocity of the car? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{car}=\frac{v_{o}}{R}\hat{% \boldsymbol{k}}$ .
(b)

What is the angular velocity of the tire relative to the car? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{tire/car}=-\frac{v_{o}}{r}% \hat{\boldsymbol{e}}_{r}$ , where $\hat{\boldsymbol{e}}_{r}$ is parallel to the axle of the wheel and moves with the car and $k$ is perpendicular to the ground.
(c)

What is the angular velocity of the tire? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{tire}=\frac{v_{o}}{R}\hat{% \boldsymbol{k}}-\frac{v_{o}}{r}\hat{\boldsymbol{e}}_{r}$ .
(d)

What is the total torque applied to the car (not including the tires as part of the car, the rotating engine is also massless) relative to the car center of mass? Answer: total torque, ${\accentset{\rightharpoonup}{\boldsymbol{M}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ .
(e)

What is the total torque on the car relative to the center of the circle around which it is traveling? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ .
(f)

What is the angular momentum of the tire relative to its center of mass (use any sensible model of the tire, say a rigid disk.)? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}=\frac{mv_{o}r^{2}}% {4}\left[-\frac{2}{r}\hat{\boldsymbol{e}}_{r}+\frac{1}{R}\hat{\boldsymbol{k}}\right]$ .
(g)

What is the rate of change of angular momentum of the tire (relative to its center of mass)? Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}=-\frac{mv_{o}^{2}r}% {2R}\hat{\boldsymbol{e}}_{t}$ .
(h)

What is the total torque of all the forces applied to the tire from the road and the car as calculated at the center of the tire? Answer: $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}=-\frac{mv_{o}^% {2}r}{2R}\hat{\boldsymbol{e}}_{t}$ .
(i)

A bug is climbing straight up on the car fender named at speed $v_{g}$ . At the instant of interest she is right next to the center of the wheel. What is her velocity and acceleration? What is the total force acting on her? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{bug}=-\frac{v_{o}^{2}}{R}\hat{% \boldsymbol{e}}_{r}+\frac{v_{o}v_{g}}{r}\hat{\boldsymbol{e}}_{t}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{bug}=m_{bug}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{bug}$ .
(j)

Another bug is climbing on a straight line marked on the tire at constant speed $v_{b}$ . He is in the middle of the tire at the instant of interest. What is his absolute velocity and acceleration? What is the total force acting on him? (Both bugs are at the same place at the instant of interest and walking on lines that are, at the moment, parallel.)
(k)

Redo the problem, questions (a) -(j), but this time the car speed is increasing with acceleration $a_{o}$ . Questions (a) - (j) of this problem are a special case of question (k). You should check that your answers from (k) reduce to the answers in questions (a) - (j). Or, if you are really cocky, you should skip the first questions and only do (k). Then, display your answers for the special case $a_{o}=0$ .

21.2.10 The moment of inertia matrix. Reconsider the system of problem 21.2 — a disk with negligible mass that has two point masses glued to it.

(a)

Find the moment of inertia matrix $[{\bf I}^{O}$ ] of the system.
(b)

Find the angular acceleration of the system when released from rest in the configuration shown. You may use your answer from problem 21.2.

21.2.11 A square plate in space. A uniform square plate with mass $m$ and sides with length $2L$ is floating stationary in space. A force of magnitude $F$ is suddenly applied to it at point A. Using the coordinates shown the moment of inertia matrix for this plate is:

[I_{/cm}]={mL^{2}\over 3}\left[\begin{matrix}1&0&0\cr 0&1&0\cr 0&0&2\cr\end{% matrix}\right]

(a)

What is the acceleration of the center of the plate ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=\frac{F}{m}\hat{\boldsymbol{% k}}$ .
(b)

What is the angular acceleration of the plate $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\frac{3F}{mL}(\hat{% \boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}})$ .
(c)

What is the acceleration of point B ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\frac{4F}{m}\hat{\boldsymbol% {k}}$ .

Filename:pfigure-f93f5

Figure 21.42

21.2.12 Force on a rectangular plate in space. A uniform rectangular plate of mass $m=10\,{\mathrm{lbm}}$ is floating in space. A force $F=5\,{\rm lbf}$ is suddenly applied at a corner that is perpendicular to the plate.

(a)

Find the acceleration of the center of mass of the plate. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\frac{4F}{m}\hat{\boldsymbol% {k}}$ . ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-16\,{\mathrm{ft}}/{\mathrm{s}}^% {2}\hat{\boldsymbol{k}}$
(b)

Find the angular acceleration of the plate. Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=(96\hat{\boldsymbol{% \imath}}+48\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/\mathrm{s}^{2}$
(c)

Some of the plate accelerates in the direction of the force and some accelerates back towards the tail of the force vector. What part of the plate accelerates in the opposite direction of the applied force?

Filename:pfigure-s94h13p3

Figure 21.43

21.2.13 Force on a square plate. A constant force with magnitude $F$ is applied, in the direction shown in the $y z$ plane, to the corner $B$ of a uniform square plate with side length $d$ and mass $m$ . Before the force is applied the plate is at rest in space with no other forces applied to it. Consider the time immediately after the time of the force application.

(a)

What, in terms of $m$ , $F$ , $d$ , $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , $\hat{\boldsymbol{k}}$ , and $\phi$ is the acceleration of point $B$ ?
(b)

For what angle $\phi$ is the acceleration of point B in the direction of the force?
(c)

Find another direction for $F$ so that the force at $B$ and the acceleration at $B$ are parallel (not necessarily in the $\hat{\boldsymbol{\jmath}}$ - $\hat{\boldsymbol{k}}$ plane). [Physical reasoning will probably work more easily than mathematical reasoning.]
(d)

Find yet another direction for $F$ so that the force at $B$ and the acceleration at $B$ are parallel (not necessarily in the $\hat{\boldsymbol{\jmath}}$ - $\hat{\boldsymbol{k}}$ plane).

Filename:p-s96-p3-3

Figure 21.44

21.2.14 (See also problems 21.26, 21.28, and 21.44.) Find the reaction supplied by the environment to the system of bodies $\cal A\cal B\cal D$ so that each body spins at a constant rate in problem 21.25.

21.2.15 (See also problems 21.26, 21.28, and 21.44.) Reconsider the system in problem 21.25. Bodies $\cal A$ and $\cal B$ are of negligible mass and body $\cal D$ is a uniform disc of mass $m$ . Find the angular momentum and rate of change of angular momentum of the system of bodies $\cal A\cal B\cal D$ about:

(a)

point $G$
(b)

point $E$ .

Make reasonable assumptions about moments of inertia.

Filename:bikefork1-ang-mom

Figure 21.45

21.2.16 (See also problems 21.27, 21.29, and 21.2.) Reconsider the system in problem 21.21. Bodies $\cal A$ and $\cal B$ are of negligible mass and body $\cal D$ is a uniform disc of mass $m$ . Find the angular momentum and rate of change of angular momentum of the system of bodies $\cal A\cal B\cal D$ about:

(a)

point $G$ Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=\frac{1}{2}mr^{2}\omega_{2}% \hat{\boldsymbol{\imath}}+\frac{1}{4}mr^{2}\omega_{1}\hat{\boldsymbol{\jmath}}% +m\omega_{0}(L^{2}+\frac{1}{4}r^{2})\hat{\boldsymbol{k}}$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}=\frac{1}{2}mr^{2}(-% \omega_{0}\omega_{1}\hat{\boldsymbol{\imath}}+\omega_{0}\omega_{2}\hat{% \boldsymbol{\jmath}}-\omega_{1}\omega_{2}\hat{\boldsymbol{k}})$ .
(b)

point $E$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/E}=\frac{1}{2}mr^{2}\omega_{2}% \hat{\boldsymbol{\imath}}+\frac{1}{4}mr^{2}\omega_{1}\hat{\boldsymbol{\jmath}}% +\frac{1}{4}mr^{2}\omega_{0}\hat{\boldsymbol{k}}$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/E}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/G}=\frac{1}{2}mr^{2}(-\omega_{0}\omega_{1}% \hat{\boldsymbol{\imath}}+\omega_{0}\omega_{2}\hat{\boldsymbol{\jmath}}-\omega% _{1}\omega_{2}\hat{\boldsymbol{k}})$ .

Make reasonable assumptions about moments of inertia.

Filename:bikefork-ang-mom

Figure 21.46

21.2.17 Reconsider problem 21.53. A thin uniform disk $\cal D$ of mass $m=2\,\mathrm{kg}$ and radius $r=0.2\,\mathrm{m}$ is mounted on a massless shaft AB as shown in the figure. Rod CO is welded to the shaft and has negligible mass. The shaft rotates at a constant angular speed $\Omega=120\,{\rm rpm}$ about the positive $x$ -axis. The disk spins about its own axis at a constant rate $\omega=60\,{\rm rpm}$ with respect to the shaft (or bar CO). Find the dynamic reactions on the bearings at A and B at the instant shown. Answer: $A_{y}=-3.16\,{\mbox{N}}$ , $B_{y}=3.16\,{\mbox{N}}$ , $A_{z}=B_{z}=-63.16\,{\mbox{N}}$

Filename:summer95f-5-a

Figure 21.47

21.2.18 A bead $P$ rides on a frictionless thin circular hoop of radius $R$ . The hoop rotates about its diameter at a constant angular speed $\omega$ in the vertical plane. Find, in terms of $\theta,\dot{\theta}$ and $\omega$ and base vectors you define (a) the velocity of the bead (a vector) and (b) the acceleration of the bead (a vector).

Filename:pfigure4-2-rp10

Figure 21.48

21.2.19 A thin rod of mass $m$ and length $L$ moves while making contact with the frictionless surface (the $\hat{\boldsymbol{e}}_{1}-\hat{\boldsymbol{e}}_{2}$ plane) at its tip A, as shown in the figure. $\theta$ is assumed to be constant and the rod always remains in the $\hat{\boldsymbol{e}}_{3},\hat{\boldsymbol{e}}_{r}$ plane.

(a)

What is the kinematical relation between the velocity of the center of mass, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}$ , and the angular velocity of the rod, ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ ?
(b)

Obtain the equations of motion of the rod in terms of the speed of point A and the angle $\phi$ .

The unit vectors $\hat{\boldsymbol{e}}_{1}^{*}$ and $\hat{\boldsymbol{e}}_{3}^{*}$ lie in the $(\hat{\boldsymbol{e}}_{3}-\boldsymbol{e}_{r})$ plane and $\hat{\boldsymbol{e}}_{2}^{*}$ is normal to the $(\hat{\boldsymbol{e}}_{3}-\hat{\boldsymbol{e}}_{r})$ plane. The starred basis vectors are written in terms of the standard basis as follows:

$\displaystyle\hat{\boldsymbol{e}}_{r}$	$\displaystyle=$	$\displaystyle\cos\theta\hat{\boldsymbol{e}}_{1}+\sin\theta\hat{\boldsymbol{e}}% _{2},$
$\displaystyle\hat{\boldsymbol{e}}_{1}^{*}$	$\displaystyle=$	$\displaystyle\sin\phi\hat{\boldsymbol{e}}_{3}+\cos\phi\hat{\boldsymbol{e}}_{r},$
$\displaystyle\hat{\boldsymbol{e}}_{3}^{*}$	$\displaystyle=$	$\displaystyle\cos\phi\hat{\boldsymbol{e}}_{3}+\sin\phi\hat{\boldsymbol{e}}_{r}% ,\qquad\mbox{and}\qquad$
$\displaystyle\hat{\boldsymbol{e}}_{2}^{*}$	$\displaystyle=$	$\displaystyle-\sin\theta\hat{\boldsymbol{e}}_{1}+\cos\theta\hat{\boldsymbol{e}% }_{1}.$

Filename:pfigure-blue-125-2

Figure 21.49

21.2.20 A cosmonaut who is otherwise free to move in 3-dimensional space finds himself stuck to the surface of a giant stationary cylinder with slippery magnets. There is no friction and (net) gravity is negligible. $\theta$ and $z$ are cylindrical coordinates that are aligned with the cylinder and mark the position of the cosmonaut relative to the cylinder. The cylinder has radius $a$ . A nearby astronaut notes that at time $t=0$ : $\theta=\theta_{0}$ , $\dot{\theta}=\dot{\theta}_{0}$ , $z=z_{0}$ , and ${\dot{z}}={\dot{z}}_{0}$ . What does she predict for the location of the cosmonaut (the values of $\theta$ and $z$ ) at an arbitrary later time $t$ ? You can assume that both the cylinder and the astronaut were stationary in the same Newtonian (inertial) frame.

Filename:pfigure-blue-68-1

Figure 21.50

21.2.21 You are an astronaut in free space and your jet pack is no longer operating. You wish to reorient your body in order to face the central government buildings so as to salute your national leaders before reentering the atmosphere. Discuss whether this maneuver can be done in terms of momentum, energy, or other dynamical concepts. (You are free to move your arms and legs, but you are not allowed to throw anything.) [For aliens and those less patriotic, the same problem arises in the case of a trampoline jumper who decides to change direction in mid-air. Is it possible?]

21.2.22 A collar slides on the rigid rod $A B C$ that has a $30^{\circ}$ bend at $B$ as shown in the figure. The collar is currently at point $D$ on the rod. The rod $A B C$ is held by fixed hinges that are aligned with the $z$ -axis, and rotates at the constant rate $\omega_{1}=2\,\mathrm{rad}/\mathrm{s}$ . The collar has mass $m_{D}=1\,\mathrm{kg}$ .

(a)

Assume a mechanism not shown constrains the collar to slide out along the rod at the constant rate $u=1\,\mathrm{m}/\mathrm{s}$ ( $u$ is the distance along the rod traveled per unit time). What is the net force that acts on the collar when it is $1\,\mathrm{m}$ from $B$ ?
(b)

Assume that the contact between the collar and the rod is frictionless and that no forces act on the collar besides those due to the rod (e.g. neglect gravity). At the instant shown in the picture, assume that $u=1\,\mathrm{m}/\mathrm{s}$ (because of some initial conditions not discussed here) but is not necessarily constant in time. What is the acceleration of the collar?

Filename:pfigure-blue-58-1

Figure 21.51

21.2.23 A bifilar pendulum is formed of a rod of length $a$ and mass $m$ , supported by two filaments of length $\ell$ and separation $2b$ . What is the frequency of small ‘torsional’ oscillations about a vertical axis though the rod’s mid-point 0?

Filename:pfigure-blue-157-1

Figure 21.52

21.2.24 A trifilar pendulum is formed of a thin horizontal disk of radius $a$ and mass $m$ , hung by three wires a distance $\ell$ below the ceiling. The wire attachments are equally spaced on the perimeter of the disk. The attachment points on the ceiling form an equilateral triangle.

(a)

If the support wires are vertical what is the period of oscillation?
(b)

How big should the triangle on the ceiling be to minimize the period of oscillation?
(c)

To maximize the period of oscillation?

21.2.25 Sometimes thrown footballs (and frisbees too!) wobble and sometimes they spin smoothly. Describe as best you can why these motions differ.

21.2.26 See also problem 21.47. A thin uniform disk $\cal D$ of mass $m=2\,\mathrm{kg}$ and radius $r=0.2\,\mathrm{m}$ is mounted on a massless shaft AB as shown in the figure. Rod CO is welded to the shaft and has negligible mass. The shaft rotates at a constant angular speed $\Omega=120\,{\rm rpm}$ about the positive $x$ -axis. The disk spins about its own axis at a constant rate $\omega=60\,{\rm rpm}$ with respect to the shaft (or bar CO). Find the absolute angular velocity and acceleration of the disk. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D}=2\pi(2\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/\mathrm{s}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal D}=8\pi^{2}\hat{% \boldsymbol{k}}\,\mathrm{rad}/\mathrm{s}^{2}$

Filename:summer95f-5

Figure 21.53

21.2.27 A thin circular disk of mass $m$ and radius $R$ is mounted on a thin light axle, coincident with its axis of symmetry. The axle is supported by bearings A and B attached to a turntable, as shown in the figure. The turntable spins at a constant speed $\Omega$ about $\hat{\boldsymbol{e}}_{3}$ , while the disk spins about its axle at a constant speed $\mu$ relative to the turntable (via motors not shown). The radial unit vector $\boldsymbol{e}_{r}=\cos(\Omega t)\hat{\boldsymbol{e}}_{1}+\sin(\Omega t)\hat{% \boldsymbol{e}}_{2}$ is fixed to the turntable and aligned with $\hat{\boldsymbol{e}}_{1}$ at time $t=0$ . Compute the vertical components of the reaction forces at A and B. Will the reaction forces change with time?

Filename:pfigure-blue-127-2

Figure 21.54

21.2.28 Steady precession of a spinning disk. A uniform disk of mass $m=2\,\mathrm{kg}$ and radius $R=0.25\,\mathrm{m}$ is rigidly mounted on a massless axle AB which is supported by frictionless bearings at A and B. The axle, in turn, is mounted on a massless semi-circular gimbal as shown in the figure. The gimbal rotates about the vertical $x$ -axis, which passes through the center of the disk D, at a constant rate $\Omega=2\,\mathrm{rad}/\mathrm{s}$ . The axle AB is spinning at a constant rate $\omega=120\,{\rm rpm}$ with respect to the gimbal. At the instant shown, the axle is aligned with the $z$ -axis and the disk D is in the $x y$ -plane. Ignore gravity. At the instant of interest:

(a)

What is the angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D}$ of the disk? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal D}=(2\hat{\boldsymbol% {\imath}}+4\pi\hat{\boldsymbol{k}})\,\mathrm{rad}/\mathrm{s}$
(b)

What is the angular acceleration of the disk? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal D}=-8\pi\,\mathrm{rad% }/\mathrm{s}^{2}\hat{\boldsymbol{\jmath}}$
(c)

What is the net torque required to keep the motion going? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{M}}}=-{\pi\over 2}\,{\rm N}\mathord{% \cdot}{\rm m}\hat{\boldsymbol{\jmath}}$
(d)

What are the reaction forces at the bearings A and B? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}={\pi\over 2}\,{\mbox{N}}\hat% {\boldsymbol{\imath}},\;{\accentset{\rightharpoonup}{\boldsymbol{R}}}_{B}=-{% \accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}$

Filename:pfigure-s94h13p4

Figure 21.55

21.2.29 A cyclist rides a flat, circular track of radius $r=150\,\mathrm{ft}$ at a constant tangential speed of $v=22\,{\mathrm{ft/s}}$ on $27\in$ diameter wheels. Assuming that the bicycle is vertical, determine the total angular velocity and angular acceleration of the wheel of the bike in the sketch (the rest of the bicycle is not shown). Use the coordinates given in the sketch.

Filename:pfigure-blue-90-2

Figure 21.56

21.2.30 Rolling disk presses down. A rigid disk (idealized as flat) with radius $R$ and mass $m$ is rigidly attached (welded) to a rigid massless stick with length $L$ which is connected by a ball-and-socket joint to the support at point $O$ . The wheel rolls on a horizontal surface. Point $O$ is at the height of the center of the wheel. The stick revolves around O at a rate of $\omega$ . Say gravity is negligible and, even though the wheel is rolling, you may assume there are no frictional forces acting between the wheel and the horizontal surface. What is the vertical reaction on the bottom of the wheel?

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{N}}}={\omega^{2}mR\over 2}\hat{% \boldsymbol{k}}$

Filename:s92f1p7

Figure 21.57

21.2.31 In problem 21.23, find the angular momentum and rate of change of angular momentum of the two disks about point $O$ . Assume that the mass of each disk is $m$ .

Filename:twodisks-ang-mom

Figure 21.58

21.2.32 A vehicle of mass $m$ , width $b$ , and height $h$ rolls without slip along a road which runs into the page(along the $z$ -axis.) The inertial properties about the vehicle center of mass are $I_{xx}^{cm}$ , $I_{yy}^{cm}$ , $I_{zz}^{cm}$ , and $I_{xy}^{cm}=I_{xz}^{cm}=I_{yz}^{cm}=0$ . Make additional assumptions as needed to get an answer to the problem. Compute the contact forces acting on each wheel normal to the road in the following cases:

(a)

The vehicle moves at constant speed $v$ along a straight road.
(b)

The vehicle moves at constant speed $v$ along a circular road (radius $\rho$ ) curving to the left. Does the right or left wheel experience the larger force?

Filename:pfigure-blue-90-1

Figure 21.59

21.2.33 A car drives counter-clockwise in circles at constant speed $V$ . The radius of the circle that the left rear wheel travels on is $R$ . The radius of the wheel is $r$ . Just when the wheel is rolling due north a pebble of mass $m$ , which is stuck in the tread, is passing directly above the wheel axle. (You should assume that the wheel axle is horizontal, that the wheel rolls without slip and that the wheel deformation and gravity are negligible.) At this instant, what is the force that the tread causes on the pebble? Resolve this force into east, north and upwards components. [Hint: several intermediate calculations are probably required.]

21.2.34 A child constructs a spinning top from a solid sphere with radius $a$ and mass $m$ attached to a massless rod of length $2d$ . The sphere pivots freely about $O$ . It is set spinning with a constant rate $\dot{\phi}$ and precessing with a constant rate $\dot{\psi}$ . The top of the rod is held by a horizontal cord.

Note: The moments of inertia of a solid sphere of mass $m$ and radius $a$ about any axis through the center is $(2/5)ma^{2}$ . (All products of inertia are zero, due to symmetry.)

(a)

Using an axis system attached as shown ( $z$ lies along the shaft, $x$ is out of the page and $y$ is chosen to produce a right-handed system), write the body’s angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ as well as the angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\Omega}}}$ of the moving coordinate system relative to inertial space.
(b)

What is the sphere’s inertia tensor about $O$ ?
(c)

What is the top’s angular momentum about $O$ ?
(d)

Compute the tension in the cord.

Filename:pfigure-blue-110-1

Figure 21.60

21.2.35 A thin disk of mass $m$ and radius $R$ is connected by a massless shaft of length $L$ to a free swivel joint at $O$ . The disk spins about the shaft at rate $\omega$ and proceeds in a steady slow precession (rate $\Omega$ ) about the vertical.

(a)

Write the disk’s inertia tensor about $O$ . Specify the axes that you have chosen.
(b)

Write the disk’s angular momentum about $O$ , ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ , in the same coordinate system.
(c)

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

Compute the precession rate $\Omega$ for the slow precession case ( $\omega\gg\Omega$ ).

Filename:pfigure-blue-107-1

Figure 21.61

21.2.36 Spinning and precessing top. (hard question) The picture schematically shows what is going on when a spinning top is in steady precession. The top is, approximately, spinning at constant rate about its symmetry axis. This axis is in turn (so to speak) rotating about the $y$ axis at a constant rate. This rotation about the $y$ axis is called precession. The angular momentum vector thus changes in time as the top precesses and the angular momentum vector also precesses. The picture is a reasonable approximation if the top is rotating much faster than it is precessing. The picture is not accurate if the precession rate is not much smaller than the rotation rate. Here are some questions to help you see why.

(a)

If the top is precessing as well as spinning what is the direction of the angular velocity vector?
(b)

What is the direction of the angular momentum vector of the top? Is it parallel to the top’s axis? Is it parallel to the angular velocity vector?
(c)

How does the angular momentum vector differ from that shown in the figure?
(d)

Does this difference affect any predictions about the precession rate of a spinning top?
(e)

Assume the top is on a pedestal and is horizontal as it precesses. In this case the correct precession rate is predicted by the naive/approximate theory that the angular momentum is just that due to spin and that this vector precesses with the top axis. Why does this simplification happen to work?

Filename:pfigure-s94h14p4

Figure 21.62

21.2.37 An amusement park ride consists of two cylinder-shaped cars of length $\ell=8\,\mathrm{ft}$ and radius $r=3\,\mathrm{ft}$ mounted at opposite ends of a long arm of length $2h=20\,\mathrm{ft}$ which is pivoted at its center as shown in the figure. The arm is constrained to rotate about the horizontal axle $O-O$ . Each car can also rotate about its own axis. Car A, presently at the top of its flight, is spinning (much to the discomfort of its occupants) on its own axis as shown at a constant rate of $\omega_{2}=5\,\mathrm{rad}/\mathrm{s}$ relative to the vertical arm. The vertical arm is rotating about axis $O-O$ at the rate $\omega_{1}=0.5\,\mathrm{rad}/\mathrm{s}$ in the direction shown, as seen by a ground observer. The point $P$ , located at $x=a=2\,\mathrm{ft}$ and $z=b=1\,\mathrm{ft}$ and representing a point in the inner ear of a passenger, is undergoing absolute velocity and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ , respectively. Find the vectors ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ . (If you are using a moving coordinate system to determine ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ , state clearly how it moves).

Filename:pfigure-blue-112-1

Figure 21.63

21.2.38 buzzing coin If you drop a round coin it will often go round and round progressively faster before buzzing to a stop. In terms of parameters you define, what is the frequency of this buzzing as a function of the tip angle? This problem is more subtle than a straight-forward calculation as there are many solutions for a coin rolling at a given tip angle in circles. What feature picks out the solutions that are observed with real coins?

21.2.39 Estimate the sideways force (Pounds or Newtons) required for a person to walk a straight marked path over the North pole. Radius of earth $\approx 4000$ miles $\approx 2\times 10^{7}$ feet $\approx 6\times 10^{6}$ meters. Make other simplifying approximations so that you can do the arithmetic without a calculator.

	$\displaystyle\hat{\boldsymbol{\lambda}}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}{\omega}% =\frac{(2\hat{\boldsymbol{k}}-6\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/% \mathrm{s}}{6.32\,\mathrm{rad}/\mathrm{s}}$
		$\displaystyle=$	$\displaystyle 0.32\hat{\boldsymbol{k}}-0.95\hat{\boldsymbol{\jmath}}.$

	$\displaystyle\rm{where}\quad{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% _{\cal G}$	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}=60\,{\rm rpm}\hat{\boldsymbol{k}}=% 2\pi\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}},$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{{\cal D}/{% \cal G}}$	$\displaystyle=$	$\displaystyle\omega_{2}\hat{\boldsymbol{k}}^{\prime}=120\,{\rm rpm}\hat{% \boldsymbol{k}}^{\prime}=4\pi\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}^{% \prime}.$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm arm}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\rm wheel/arm}$
	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}+\omega_{2}(-\hat{\boldsymbol{e}}_{% 2})$
	$\displaystyle=$	$\displaystyle\omega_{1}\hat{\boldsymbol{k}}-\omega_{2}\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle(2\hat{\boldsymbol{k}}-6\hat{\boldsymbol{\jmath}})\,\mathrm{rad}/% \mathrm{s}.$

Chapter 21 Elementary introduction to 3D rigid-object dynamics

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21.1 Velocity and acceleration of points on a rigid object in 3-D: 𝝎⇀ and 𝜶⇀

Angular velocity

Angular velocity in two dimensions

Angular velocity in three dimensions

Relative velocity of two points on a rigid body

Absolute velocity of a point on a rigid body

Acceleration of points on a rigid body

Angular acceleration

Angular acceleration in two dimensions

Relative acceleration of two points on a rigid body

Absolute acceleration of a point on a rigid body

Angular velocity

Angular velocity relative to a body or frame

Absolute angular velocity using data from a moving frame

Is rotation a vector?

An irony

Summary: Rotations and angular velocity

Absolute angular acceleration using data from a moving frame

Summary of angular acceleration

Problems for 21.1 Velocity and acceleration of points on a rigid body: 𝝎⇀ and 𝜶⇀

21.2 3D dynamics of a rigid body

Momenta and energy of rigid bodies

Problems for 21.2 Instantaneous dynamics and “inverse dynamics” in 3-D

21.1 Velocity and acceleration of points on a rigid object in 3-D: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$

Problems for 21.1 Velocity and acceleration of points on a rigid body: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$