Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap14 Straight line motion 16 Circular motion of a rigid object

Chapter 15 Circular motion of a particle

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15.0Introductionstudy PDF
15.1Circular motion kinematicsstudy PDF + solutions
15.2Dynamics of particle circular motionstudy PDF + solutions

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After movement on straight lines the next important special motion is rotation on a circular path. Polar coordinates and base vectors are introduced in this simplest possible context. The key new idea is that not just coordinates, but base vectors, can change with time.

We covered the special case of straight-line motion in the previous chapter. But an unconstrained particle, such as a thrown ball, generally moves on a curved path as pushed by gravity and aerodynamic forces. Also, when a rigid object moves, it translates and rotates while the points on the object move on complicated curved paths. Now we consider the archetypal curved motion, motion along a circular path. Circular motion deserves special attention because

•

the most common connection between moving parts on a machine is with a bearing (or hinge or axle) (fig. 16.1), if the axle on one part is fixed then all points on the part move in circles;
•

circular motion is the simplest case of curved-path motion;
•

circular motion provides a simple way to introduce time-varying base vectors;
•

circular motion includes most of the conceptual ingredients of more general curved motions;
•

at least in 2 dimensions, the only way two particles on one rigid object can move relative to each other is by circular motion (no matter how the object is moving); and
•
circular motion is the simplest case with which to introduce two important rigid-object concepts:
- –
  
  angular velocity, and
- –
  
  moment of inertia.

Many useful calculations can be made by approximating the motion of particles as circular. For example, the motions of points on a jet engine’s turbine blade, a car engine’s crank shaft, a car’s wheel, a windmill’s propeller, the earth spinning about its axis, a clock pendulum or watch balance wheel, all the points on a bicycle when it is going around a corner, a satellite orbiting the earth or a spinning satellite going around its spin axis, might all be approximately described as having circular motion about some appropriate point or axis.

This short chapter concerns only motion in two dimensions, namely the kinematics and mechanics of a single particle going in circles. The next chapter concerns the kinematics and mechanics of rigid objects. A later chapter discusses circular motion, which is always planar, in a three-dimensional context.

15.1 Kinematics of a particle in planar circular motion

Filename:tfigure8-rel-ang-vel

Figure 15.1: Trajectory of particle for circular motion.

This section concerns the position, velocity and accelerations of one point going in circles. The essence of the content here is this:

If $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ is a unit vector in the plane that is rotating counter-clockwise (CCW) at a rate of $\dot{\theta}$ its rate of change is

\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}

where $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ is a unit vector given by rotating $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ $90^{\circ}$ CCW.

If you learn this idea inside and out then either you will have picked up all the other key facts on the way, or you will be able to learn them in a flash. Note, we use ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ interchangeably. Likewise for $\hat{\boldsymbol{e}}_{r}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ .

Filename:tfigure8-ang-vel-ex

Figure 15.2: Plots of

x

versus

t

and

y

versus

t

for a particle going in a circle of radius

R

at constant rate. For simplicity we assumed constant

\dot{\theta}

with

\theta=\dot{\theta}t

. So both

x

and

y

vary as sinusoidal functions of time:

x=R\cos(\dot{\theta}t)

and

y=R\sin(\dot{\theta}t)

Filename:tfigure8-ang-vel

Figure 15.3: Plot of

x

and

y

versus time for a particle going in circles at constant rate.

x

versus

t

is a cosine curve,

y

versus

t

is a sine curve. Together they make up the 3D helix.

Circular motion

The position of a particle going in circles around the origin on the $x y$ plane is

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

with the radius $R$ a constant. Or, in terms of components,

x=R\cos\theta\qquad\mbox{and}\qquad y=R\sin\theta.

A natural graphical representation of this motion is a circle (fig. 15.1). Unfortunately, a picture of the circular trajectory doesn’t give any information about the speed of the particle on the circle. A plot of a particle moving in circles slowly looks just like a plot of a particle moving quickly.

To get a sense of how position changes in time one can plot the functions $x(t)$ and $y(t)$ (fig. 15.2). Unfortunately this figure only indirectly conveys that the particle is going in circles.

If you want to see both the trajectory and the time history of both variables one can make a 3-D plot of $x y$ position versus time (fig. 15.3). The shadows of this helix on the three coordinate planes are the three graphs just discussed.

Finally, rather than representing time as a spatial coordinate, one can represent time with time itself. How? Make an animated movie showing a particle on the $x y$ plane as it moves. Move your finger around in circles on the table. That’s it. Similarly, you can make a dot move in circles on your screen.

How do you make all these plots? Using a calculator or computer you can evaluate $x$ and $y$ for a range of values of $t$ . Then, using pencil and paper, a plotting calculator, or a computer, plot $x$ vs $t$ , $y$ vs $t$ , and $y$ vs $x$ . For animations plot $x$ and $y$ over and over again for a sequence of values of $t$ , and show these on your screen at a sequence of times.

Polar coordinates $R$ and $\theta$ and unit vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

Filename:tfigure8-ang-accel

Figure 15.4: The position vector

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

of the particle relative to the center of the circle is

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

(or

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

) which is both

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}

and

{\accentset{\rightharpoonup}{\boldsymbol{r}}}=R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}

(or

r\hat{\boldsymbol{e}}_{r}

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

makes an angle

\theta

measured counter-clockwise from the positive

x

-axis. The unit vectors

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

and

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

are in the radial and tangential directions, the directions of increasing

R

and increasing

\theta

Filename:tfigure8-ang-accel-ex

Figure 15.5: You can think of the unit vectors

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

and

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

as the set

\hat{\boldsymbol{\imath}}

and

\hat{\boldsymbol{\jmath}}

rotated counter-clockwise by the angle

\theta

Filename:tfigure8-ang-accel1

Figure 15.6: Projections of

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

and

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

in the

x

and

y

directions. From this picture you can immediately extract that

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

and that

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

.
A similar picture showing the projections of

\hat{\boldsymbol{\imath}}

and

\hat{\boldsymbol{\jmath}}

in the

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

and

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

d directions would show that

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

and that

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

Especially for circular motion, it is convenient to represent position, velocity and acceleration with polar, rather than rectangular, coordinates. With polar coordinates we use polar base vectors which, unlike the fixed $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , rotate as the particle goes around.

Let’s redraw fig. 15.1 and show the unit base vectors

$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ (‘e R’) and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ (‘e theta’).

The radial unit vector $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ is directed from the center of the circle towards the point of interest and the transverse vector $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , perpendicular to $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ , is tangent to the circle at that point in the direction of increasing $\theta$ . As the particle goes around, its $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ unit vectors change accordingly. Two different particles both going in circles with the same center at the same rate each have their own $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ vectors.

To be precise we can define $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ as

\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\equiv{\accentset{\rightharpoonup}% {\boldsymbol{R}}}/|{\accentset{\rightharpoonup}{\boldsymbol{R}}}|={\accentset{% \rightharpoonup}{\boldsymbol{R}}}/R\par\qquad\mbox{and}\qquad\hat{\boldsymbol{% e}}_{\scriptscriptstyle\!\theta}\equiv\hat{\boldsymbol{k}}\times\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}.

(15.1)

Note that also $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}=\hat{\boldsymbol{k}}$ .

Filename:sfig8-2-3

Figure 15.7: A close up view of the unit vectors

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

and

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

. They make an angle

\theta

with the positive

x

and

y

-axis, respectively. As the particle advances an amount

\Delta\theta

both

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

and

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

change. In particular, for small

\Delta\theta

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

is approximately in the

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

direction and

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

is approximately in the

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

direction.

The velocity and acceleration of a point going in circles, using polar coordinates

In dynamics we are interested in velocity and acceleration so we need to know how to represent these in polar coordinates.

First, observe that the position of the particle is (see fig. 15.4)

{\accentset{\rightharpoonup}{\boldsymbol{R}}}=R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}.

(15.2)

That is, the position vector is the distance from the origin times a unit vector in the direction of the particle’s position. Given the position, it is just a matter of careful differentiation to find velocity and acceleration.

Here is one of many possible ways to derive the polar-coordinate expressions for velocity and acceleration.

First, velocity is the time derivative of position, so

{\accentset{\rightharpoonup}{\boldsymbol{v}}}={d\over dt}{\accentset{% \rightharpoonup}{\boldsymbol{R}}}={d\over dt}(R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R})=\underbrace{\dot{R}}_{0}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+R\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}.

(15.3)

Because a circle has constant radius $R$ , $\dot{R}$ is zero. But how do we calculate the rate of change of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ with respect to time, $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}$ ,?

Derivatives of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

To find the velocity in polar coordinates we were just confronted with the problem of finding $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}$ .

Method 1: One way to find $d\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}/dt=\dot{\hat{\boldsymbol{e}}}_{% \scriptscriptstyle\!R}$ uses the geometry of fig. 15.7 and the informal calculus of finite differences (represented by $\Delta$ ).

$\Delta\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ is evidently (about) in the direction $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ and has magnitude $\Delta\theta$ so $\Delta\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\approx(\Delta\theta)\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ . Dividing by $\Delta t$ , we have $\Delta\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}/\Delta t\approx(\Delta% \theta/\Delta t)\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ . So, using this sloppy calculus, we get $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ . Similarly, and we will need this shortly, we could get ${\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}}=-\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}$ .

Method 2 This method is a little less geometric and a little more algebraic. We start with the decomposition of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ into cartesian coordinates. These decompositions are found by looking at the projections of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ in the $x$ and $y$ -directions (see fig. 15.6).

	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle\phantom{-}\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}$		(15.4)
	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{% \jmath}}$

We can find $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}$ by differentiating, taking into account that $\theta$ is changing with time but that the unit vectors $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ are fixed (so they don’t change with time).

	$\displaystyle\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle{d\over dt}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}})=-\dot{\theta}\sin\theta\hat{\boldsymbol{\imath}}+\dot{% \theta}\cos\theta\hat{\boldsymbol{\jmath}}=\dot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}$
	$\displaystyle\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle{d\over dt}(-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}})=-\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$

We had to use the chain rule, that is

{d\sin\theta(t)\over dt}={d\sin\theta\over d\theta}{d\theta(t)\over dt}=\dot{% \theta}\cos\theta.

Now, two different ways, we know

\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!R}=\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}\hbox to36.135pt{\hfil and\hfil}% \dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}=-\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}.

(15.5)

Continuing the quest for velocity and acceleration

Now that we know how $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ changes in time we can continue our quest for ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ . Continuing from eqn. (15.3) we now have

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

(15.6)

Similarly we can find the acceleration $\ddot{\accentset{\rightharpoonup}{\boldsymbol{R}}}$ by differentiating once again,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{\accentset{\rightharpoonup% }{\boldsymbol{R}}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{v}}}={d\over dt% }(R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})=\underbrace{% \dot{R}\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}}_{{% \accentset{\rightharpoonup}{\boldsymbol{0}}}}+{R\ddot{\theta}\hat{\boldsymbol{% e}}_{\scriptscriptstyle\!\theta}}+R\dot{\theta}\dot{\hat{\boldsymbol{e}}}_{% \scriptscriptstyle\!\theta}

(15.7)

The first term on the right hand side is zero because $\dot{R}$ is $0$ for circular motion. The third term is evaluated using the formula we just found for the rate of change of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ : $\dot{\hat{\boldsymbol{e}}}_{\scriptscriptstyle\!\theta}=-\dot{\theta}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}$ . So, using that ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=R\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}$ ,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\dot{\theta}^{2}{\accentset{% \rightharpoonup}{\boldsymbol{R}}}+R\ddot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}=-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+R\ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta}

(15.8)

The velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ for a particle going in circles at constant rate are shown in fig. 15.8.

Filename:sfig8-2-3a

Figure 15.8: The directions of velocity

{\accentset{\rightharpoonup}{\boldsymbol{v}}}

and acceleration

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

are shown for a particle going in circles at constant rate. The velocity is tangent to the circle and the acceleration is directed towards the center of the circle.

Example: A person standing on the earth’s equator

Filename:sfig8-2-3b

Figure 15.9:

A person standing on the equator has velocity

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{\theta}R\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle\approx$	$\displaystyle\left(\frac{2\pi\,{\rm rad}}{24\,{\mathrm{hr}}}\right)4000\,{\rm mi% }\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
		$\displaystyle\approx$	$\displaystyle 1050\,\mathrm{mph}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta}\approx 1535\,{\mathrm{ft/s}}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

and acceleration

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}=-\dot{\theta}^{2}R% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$	$\displaystyle\approx$	$\displaystyle-\left(\frac{2\pi\,{\rm rad}}{24\,{\mathrm{hr}}}\right)^{2}4000\,% {\rm mi}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$
		$\displaystyle\approx$	$\displaystyle-274\,{\rm mi}/\,{\mathrm{hr}}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\approx-0.11\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}.$

The velocity of a person standing on the equator, due to the earth’s rotation, is about $1000\,\mathrm{mph}$ tangent to the earth. Her acceleration is about $0.11\,{\mathrm{ft}}/{\mathrm{s}}^{2}\approx 0.03\,\mathrm{m}/\mathrm{s}^{2}$ towards the center of the earth, about $1/300$ of $g$ , about $1/300$ the acceleration of an object in near-earth-surface frictionless free-fall.

Alternate expressions for the velocity and acceleration formulas

Note that we can define a scalar velocity $v=R\dot{\theta}$ . We informally call this scalar the speed even though it can be positive or negative. So

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

Similarly the acceleration is

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=-R\dot{\theta}^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+R\ddot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}=-{v^{2}\over R}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\dot{v}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta% }.\par

where $\dot{v}$ is the rate of change of tangential speed^†^†margin: Caution: Note that the rate of change of speed is not the magnitude of the acceleration: $\dot{v}\neq|{\accentset{\rightharpoonup}{\boldsymbol{a}}}|$ or in other words: ${d\over dt}|{\accentset{\rightharpoonup}{\boldsymbol{v}}}|\neq|{d\over dt}{% \accentset{\rightharpoonup}{\boldsymbol{v}}}|$ . Consider the case of a car driving in circles at constant rate. Its rate of change of speed is zero, yet it has an acceleration. .

Thus the acceleration is made of two terms. One proportional to the speed squared and directed towards the center of the circle, and one proportional to the rate of change of speed and directed tangent to the circle.

Centripetal acceleration

The term

-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=-\dot{\theta}^{% 2}{\accentset{\rightharpoonup}{\boldsymbol{R}}}

is called the centripetal acceleration.

Why, intuitively, is the centripetal acceleration proportional to the speed squared? Well, the acceleration is the change in the velocity vector per unit time. There are two effects

1.

If the speed is twice as big then the velocity is twice as big.
2.

And, for a given radius, the angle it rotates per unit time is twice as big.

These two proportionalities with speed, the size of the velocity vector which rotates and the rate at which it rotates both apply. So if the speed is twice as big the acceleration is 4 times as big. Hence the $v^{2}$ .

Box 15.1 Summary: the motion quantities

We can use our results for velocity and acceleration to better evaluate the momenta and energy quantities. These results will allow us to do mechanics problems associated with circular motion. For one particle in circular motion we have:


$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}m$	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}m,$

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}m$	$\displaystyle=$	$\displaystyle(-\dot{\theta}^{2}{\accentset{\rightharpoonup}{\boldsymbol{R}}}+R% \ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})m,$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/0}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}m$	$\displaystyle=$	$\displaystyle R^{2}\dot{\theta}m\hat{\boldsymbol{k}},$

Rate of change of angular momentum:

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

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/0}\times{% \accentset{\rightharpoonup}{\boldsymbol{a}}}m

\displaystyle=

\displaystyle R^{2}\ddot{\theta}m\hat{\boldsymbol{k}},


$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle{1\over 2}v^{2}m$	$\displaystyle=$	$\displaystyle{1\over 2}R^{2}\dot{\theta}^{2}m,$	and

$\displaystyle\dot{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{a}}}\,m$	$\displaystyle=$	$\displaystyle mR^{2}\dot{\theta}\ddot{\theta}$

Box 15.2 d’Alembert’s mechanics: beginners beware

This box is an aside. It will not help you do dynamics problems. As warned on page 2.3,it’s worse than that. The material in this box usually harms more than it helps. The d’Alembert approach to mechanics, described here, cannot be well-absorbed by beginners. Students attempting to use d’Alembert methods make frequent mistakes. We advise against the use of d’Alembert mechanics for beginners. We don’t allow its use in homework and exams.

But you might be curious about this forbidden fruit. To demystify the taboo, we briefly describe the approach. You might as well learn it here instead of somewhere else.

The d’Alembert approach has an intuitive appeal to experts. And the d’Alembert equations are the first step in deriving the more advanced (e.g., Lagrangian, Hamiltonian, ‘method of virtual speed’, and ‘Kane’) approaches to dynamics.

How does it go?

First, label the free-body diagram: ‘free-body diagram including inertial forces.’ Then, in addition to the applied forces draw pseudo-forces equal to $-m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ for every mass particle $m$ . These pseudo-forces are shown in the ‘FBD’ of a falling ball. The pseudo-forces are sometimes called ‘inertial’ forces or ‘d’Alembert forces.

Filename:tfigure1-dalembert-fbd

For the d’Alembert approach, instead of momentum balance equations you write ‘pseudo-statics’ equations of ‘force’ balance and ‘moment’ balance

\overbrace{\qquad\underbrace{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}% }_{\scriptsize\mbox{including inertial forces}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}}^{\mbox{pseudo-statics force balance}}

\overbrace{\underbrace{\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{$\scriptscriptstyle\mathrm{C}$}}}}_{\scriptsize\mbox{including % torques from inertial forces}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}}^{\mbox{pseudo-statics % moment balance}}

These equations include the actual forces as well as the ‘inertial’ forces shown on the ‘d’Alembert free-body diagram’.

By these means, the dynamics equations have been reduced to statics equations. Linear momentum balance is replaced by pseudo-statics force balance. Angular momentum balance is replaced by pseudo-statics moment balance.

The moving of the inertial terms from the right side of the equation to the left leads to both conceptual simplicity and puts the equations of dynamics in a form that is closer to most people’s intuitions. The simplification is not so great as it may seem at first sight. Accelerations still need to be calculated and the sums involved in calculation of rate of change of linear and angular momentum still need to be calculated, only now they are sums of pseudo inertial forces.

Consider the example of sitting in a car as the car rounds a corner to the left. In the momentum balance approach, we write

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\underbrace{m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}}_{\dot{\accentset{\rightharpoonup}{% \boldsymbol{L}}}}

and say the force from the car on you to the left is equal to the rate of change of your linear momentum as you accelerate to the left. In the d‘Alembert approach, we write

{\accentset{\rightharpoonup}{\boldsymbol{F}}}-\underbrace{m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}}_{\mbox{inertia force}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}

and think the inertia force to the right is balanced by the interaction force of the car on your body to the left.

It is a puzzle of human consciousness why such a trivial algebraic manipulation, namely,

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\,\mathrm{m}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}\ \ \Rightarrow\ \ {\accentset{% \rightharpoonup}{\boldsymbol{F}}}-m{\accentset{\rightharpoonup}{\boldsymbol{a}% }}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

should lead to such a great conceptual confusion. But, it is an empirical fact that most of us are susceptible to this confusion.

That is, if you follow your likely first intuition and think of $m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ as a force you will probably join the ranks of many other talented students who consequently make many sign errors.

Every teacher of mechanics has encountered the confusion in their students about whether $-m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ is or is not a force (and most likely in themselves as well.) To avoid such confusion, many teachers or texts take a firm stand and say

•

‘ $m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ is not a force!’; but, as if believing in a different god, others will say with equal conviction
•

‘- $m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ is a force!’.

In this book, we take the former approach. We take the equation

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\,\mathrm{m}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}

to mean:

forces from interactions =

m\cdot

(acceleration of mass).

If you insist on working with the d‘Alembert approach instead, you must do so confidently and clearly. To repeat,

•

instead of labeling your free-body diagram ‘FBD’, label it ‘FBD including inertial forces’,
•

instead of using ‘Linear Momentum Balance’, use ‘Pseudo-Force Balance’, and
•

instead of using ‘Angular Momentum Balance’ use ‘Pseudo-Moment Balance’

We do not recommend d‘Alembert mechanics to beginners, but if you insist, good luck to you and don’t blame us for your (almost inevitable) sign errors!

SAMPLE 15.1 The velocity vector in circular motion. A particle executes circular motion in the $x y$ plane with constant speed $v=5\,\mathrm{m}/\mathrm{s}$ . At $t=0$ the particle is at $\theta=0$ . Given that the radius of the circular orbit is $2.5\,\mathrm{m}$ , find the velocity of the particle at $t=2\sec$ .

Solution

It is given that

$\displaystyle R$	$\displaystyle=$	$\displaystyle 2.5\,\mathrm{m}$
$\displaystyle v$	$\displaystyle=$	$\displaystyle{\rm constant}=5\,\mathrm{m}/\mathrm{s}$
$\displaystyle\theta{(t=0)}$	$\displaystyle=$	$\displaystyle 0.$

The velocity of a particle in constant-rate circular motion is:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle\mbox{where}\quad\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{% \jmath}}.$

Since $R$ is constant and $v=|{\accentset{\rightharpoonup}{\boldsymbol{v}}}|=R\dot{\theta}$ is constant,

\displaystyle\dot{\theta}=\frac{v}{R}=\frac{5\,\mathrm{m}/\mathrm{s}}{2.5\,% \mathrm{m}}=2\,\mathrm{rad}/\mathrm{s}

is also constant. Thus,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}{(t=2\,\mathrm{s})}=\underbrace{R% \dot{\theta}}_{v}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}\Bigr{|}_{t=% 2\,\mathrm{s}}=5\,\mathrm{m}/\mathrm{s}\ \hat{\boldsymbol{e}}_{\theta}(t=2\,% \mathrm{s}).

Clearly, we need to find $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ at $t=2\sec$ .

$\displaystyle\mbox{Now}\quad\quad\dot{\theta}$	$\displaystyle\equiv$	$\displaystyle\frac{d\theta}{dt}=2\,\mathrm{rad}/\mathrm{s}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\int^{\theta}_{0}d\theta$	$\displaystyle=$	$\displaystyle\int^{2\,\mathrm{s}}_{0}2\,\mathrm{rad}/\mathrm{s}\ dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\quad\quad\theta$	$\displaystyle=$	$\displaystyle(2\,\mathrm{rad}/\mathrm{s})\ t\Bigr{\|}^{2\,\mathrm{s}}_{0}$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{rad}/\mathrm{s}\cdot 2\,\mathrm{s}$
	$\displaystyle=$	$\displaystyle 4\,{\rm rad}.$

Therefore,

Filename:sfig8-3-1

Figure 15.10: The velocity vector

{\accentset{\rightharpoonup}{\boldsymbol{v}}}

t=2\,\mathrm{s}

	$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle-\sin{4}\hat{\boldsymbol{\imath}}+\cos{4}\hat{\boldsymbol{\jmath}}$
		$\displaystyle=$	$\displaystyle 0.76\hat{\boldsymbol{\imath}}-0.65\hat{\boldsymbol{\jmath}},$

and

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}{(2\,\mathrm{s})}$	$\displaystyle=$	$\displaystyle 5\,\mathrm{m}/\mathrm{s}(0.76\hat{\boldsymbol{\imath}}-0.65\hat{% \boldsymbol{\jmath}})$
		$\displaystyle=$	$\displaystyle(3.78\hat{\boldsymbol{\imath}}-3.27\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=(3.78\hat{\boldsymbol{\imath}}-3% .27\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}$

SAMPLE 15.2 Basic kinematics: A point mass executes circular motion with angular acceleration $\ddot{\theta}=5\,\mathrm{rad}/\mathrm{s}^{2}.$ The radius of the circular path is 0.25 $\,\mathrm{m}$ . If the mass starts from rest at $\theta=0^{{}^{\circ}}$ , find and draw

1.

the velocity of the mass at $\theta=0^{{}^{\circ}},\;30^{{}^{\circ}},\;90^{{}^{\circ}},$ and $210^{{}^{\circ}}$ ,
2.

the acceleration of the mass at $\theta=0^{{}^{\circ}},\;30^{{}^{\circ}},\;90^{{}^{\circ}},$ and $210^{{}^{\circ}}$ .

Solution

We are given, $\ddot{\theta}=5\,\mathrm{rad}/\mathrm{s}^{2}$ , and $R=0.25\,\mathrm{m}$ .

The velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ in circular (constant or non-constant rate) motion is given by:

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

So, to find the velocity at different positions we need $\dot{\theta}$ at those positions. Here the angular acceleration is constant, i.e., $\ddot{\theta}=5\,\mathrm{rad}/\mathrm{s}^{2}$ . Therefore, we can use the formula ^†^†margin: We use this formula because we need $\dot{\theta}$ at different values of $\theta$ . In elementary physics books, the same formula is usually written as $\dot{\theta}^{2}=\dot{\theta}_{0}^{2}+2\alpha\theta$ where $\alpha$ is the constant angular acceleration and $\dot{\theta}(=\dot{\theta})$ is the angular speed.

\dot{\theta}^{2}=\dot{\theta}_{0}^{2}+2\ddot{\theta}\theta

to find the angular speed $\dot{\theta}$ at various $\theta$ ’s. But $\dot{\theta}_{0}=0$ (mass starts from rest), therefore $\dot{\theta}=\sqrt{2\ddot{\theta}\theta}$ . Now we make a table for computing the velocities at different positions:

Filename:sfig8-3-1a

Figure 15.11: Velocity of the mass at

\theta~{}=~{}0^{{}^{\circ}},\;30^{{}^{\circ}},\;90^{{}^{\circ}},

and

210^{{}^{\circ}}.

Position ( $\theta$ )	$\theta$ in radians	$\dot{\theta}=\sqrt{2\ddot{\theta}\theta}$	${\accentset{\rightharpoonup}{\boldsymbol{v}}}=R\dot{\theta}\hat{\boldsymbol{e}% }_{\scriptscriptstyle\!\theta}$
$0^{{}^{\circ}}$	0	0 rad/s	⇀ 0
$30^{{}^{\circ}}$	$\pi/6$	$\sqrt{10\pi/6}=2.29\,\mathrm{rad}/\mathrm{s}$	$0.57\,\mathrm{m}/\mathrm{s}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$90^{{}^{\circ}}$	$\pi/2$	$\sqrt{10\pi/2}=3.96\,\mathrm{rad}/\mathrm{s}$	$0.99\,\mathrm{m}/\mathrm{s}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$210^{{}^{\circ}}$	$7\pi/6$	$\sqrt{70\pi/6}=6.05\,\mathrm{rad}/\mathrm{s}$	$1.51\,\mathrm{m}/\mathrm{s}\,\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

The computed velocities are shown in Fig. 15.11.

The acceleration of the mass is given by

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\overbrace{a_{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}}^{% \rm radial}+\overbrace{a_{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta}}^{\rm tangential}$
		$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R% \ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}.$

Since $\ddot{\theta}$ is constant, the tangential component of the acceleration is constant at all positions. We have already calculated $\dot{\theta}$ at various positions, so we can easily calculate the radial (also called the normal) component of the acceleration. Thus we can find the acceleration. For example, at $\theta=30^{{}^{\circ}}$ ,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+R% \ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle-0.25\,\mathrm{m}\cdot\frac{10\pi}{6}\frac{1}{{\,\mathrm{s}}^{2}}% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+0.25\,\mathrm{m}\cdot 5\frac{1}{{% \,\mathrm{s}}^{2}}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
	$\displaystyle=$	$\displaystyle-1.31\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+1.25\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}.$

Similarly, we find the acceleration of the mass at other positions by substituting the values of $R,\;\ddot{\theta}$ and $\dot{\theta}$ in the formula and tabulate the results in the table below.

Position ( $\theta$ )	$a_{r}=-R\dot{\theta}^{2}$	$a_{\theta}=R\ddot{\theta}$	${\accentset{\rightharpoonup}{\boldsymbol{a}}}=a_{r}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+a_{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$0^{{}^{\circ}}$	0	$1.25\,\mathrm{m}/\mathrm{s}^{2}$	$1.25\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$30^{{}^{\circ}}$	$-1.31\,\mathrm{m}/\mathrm{s}^{2}$	$1.25\,\mathrm{m}/\mathrm{s}^{2}$	$(-1.31\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+1.25\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})\,\mathrm{m}/\mathrm{s}^{2}$
$90^{{}^{\circ}}$	$-3.93\,\mathrm{m}/\mathrm{s}^{2}$	$1.25\,\mathrm{m}/\mathrm{s}^{2}$	$(-3.93\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+1.25\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})\,\mathrm{m}/\mathrm{s}^{2}$
$210^{{}^{\circ}}$	$-9.16\,\mathrm{m}/\mathrm{s}^{2}$	$1.25\,\mathrm{m}/\mathrm{s}^{2}$	$(-9.16\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+1.25\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})\,\mathrm{m}/\mathrm{s}^{2}$

The accelerations computed are shown in Fig. 15.12. The acceleration vector as well as its tangential and radial components are shown in the figure at each position.

Filename:sfig8-5-wiper

Figure 15.12: Acceleration of the mass at

\theta=0^{{}^{\circ}},\;30^{{}^{\circ}},\;90^{{}^{\circ}},

and

210^{{}^{\circ}}

. The radial and tangential components are shown with grey arrows. As the angular velocity increases, the radial component of the acceleration increases; therefore, the total acceleration vector leans more and more towards the radial direction.

SAMPLE 15.3 In an experiment, the magnitude of angular deceleration of a rotating ball is found to be proportional to its angular speed $\dot{\theta}$ (i.e., $\ddot{\theta}\propto-\dot{\theta}$ ). Assume that the proportionality constant is $k$ .

1.

Find $\dot{\theta}$ as a function of $t$ , given that $\dot{\theta}(t=0)=\dot{\theta}_{0}$ .
2.

Given that $k=0.1/\,\mathrm{s}$ , how much time does it take for $\dot{\theta}$ to reduce to half the initial value?

Solution

The equation given is:

\ddot{\theta}=\frac{d\dot{\theta}}{dt}=-k\dot{\theta}.

(15.9)

We can solve this equation in a couple of ways.

Method-1: Let us guess a solution of the exponential form with arbitrary constants and plug it into eqn. (15.9) to check if our solution works. Let $\dot{\theta}(t)=C_{1}e^{C_{2}t}$ . Substituting in eqn. (15.9), we get

$\displaystyle C_{1}C_{2}e^{C_{2}t}$	$\displaystyle=$	$\displaystyle-kC_{1}e^{C_{2}t}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}C_{2}$	$\displaystyle=$	$\displaystyle-k,$
$\displaystyle\mbox{also,}\qquad\dot{\theta}(0)$	$\displaystyle=$	$\displaystyle\dot{\theta}_{0}=C_{1}e^{C_{2}\cdot 0}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}C_{1}$	$\displaystyle=$	$\displaystyle\dot{\theta}_{0}.$
Therefore,
$\displaystyle\dot{\theta}(t)$	$\displaystyle=$	$\displaystyle\dot{\theta}_{0}e^{-kt}.$	(15.10)

Answer: $\dot{\theta}(t)=\dot{\theta}_{0}\,e^{-kt}$

Method-2: Equation (15.9) can also be solved by direct integration as follows.

$\displaystyle\frac{d\dot{\theta}}{\dot{\theta}}$	$\displaystyle=$	$\displaystyle-k\,dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{\dot{\theta}_{0}}^% {\dot{\theta}(t)}\frac{d\dot{\theta}}{\dot{\theta}}$	$\displaystyle=$	$\displaystyle-\int_{0}^{t}k\,dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ln\dot{\theta}\Bigr{\|}_% {\dot{\theta}_{0}}^{\dot{\theta}(t)}$	$\displaystyle=$	$\displaystyle-kt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ln\left(\frac{\dot{% \theta}(t)}{\dot{\theta}_{0}}\right)$	$\displaystyle=$	$\displaystyle-kt$
Therefore,
$\displaystyle\dot{\theta}(t)$	$\displaystyle=$	$\displaystyle\dot{\theta}_{0}\,e^{-kt},$

which is the same solution as equation (15.10).

Filename:sfig8-5-wiper-a

Figure 15.13: Plot of

\dot{\theta}(t)/\dot{\theta}_{0}=e^{-kt}

for

k=0.1/\,\mathrm{s}

. The angular speed

\dot{\theta}

reduces to half of its initial value in 6.93

\,\mathrm{s}

. Note that this time is the same for

\dot{\theta}

to reduce to half its value at any given time

t

(not just at

t=0

2.

We need to find $t$ for $\dot{\theta}=\dot{\theta}_{0}/2$ , given that $k=0.1$ . From eqn. (15.10), we get

$\displaystyle\frac{\dot{\theta}}{\dot{\theta}_{0}}$ $\displaystyle=$ $\displaystyle e^{-kt}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t$ $\displaystyle=$ $\displaystyle\frac{1}{-k}\ln\left(\frac{\dot{\theta}}{\dot{\theta}_{0}}\right)$

$\displaystyle=$ $\displaystyle\frac{1}{-0.1}\ln\left(\frac{1}{2}\right)=\frac{-0.693}{-0.1/\,% \mathrm{s}}=6.93\,\mathrm{s}.$

Answer: $t=6.93\,\mathrm{s}\mbox{ for }\dot{\theta}(t)=\dot{\theta}_{0}/2$

SAMPLE 15.4 Using kinematic formulae: The spinning wheel of a stationary exercise bike is brought to rest from $100\,{\rm rpm}$ by applying brakes over a period of 5 seconds.

1.

Find the average angular deceleration of the wheel.
2.

Find the number of revolutions it makes during the braking.

Solution

We are given,

\dot{\theta}_{0}=100\,{\rm rpm},\quad\dot{\theta}_{\rm final}=0,\hbox to36.135% pt{\hfil and\hfil}t=5\,\mathrm{s}.

1.

Let $\alpha$ be the average (constant) deceleration. Then

$\dot{\theta}_{\rm final}=\dot{\theta}_{0}-\alpha t.$

Therefore,

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle\frac{\dot{\theta}_{0}-\dot{\theta}_{\rm final}}{t}$

$\displaystyle=$ $\displaystyle\frac{100\,{\rm rpm}-0\,{\rm rpm}}{5\,\mathrm{s}}$

$\displaystyle=$ $\displaystyle\frac{100\,\mbox{rev}}{60\,\mathrm{s}}\cdot\frac{1}{5\,\mathrm{s}}$

$\displaystyle=$ $\displaystyle 0.33\frac{\,\mbox{rev}}{{\,\mathrm{s}}^{2}}.$

Answer: $\alpha=0.33\frac{\,\mbox{rev}}{{\,\mathrm{s}}^{2}}$
2.

To find the number of revolutions made during the braking period, we use the formula

$\theta(t)=\underbrace{\theta_{0}}_{0}+\dot{\theta}_{0}t+\frac{1}{2}(-\alpha)t^% {2}=\dot{\theta}_{0}t-\frac{1}{2}\alpha t^{2}.$

Substituting the known values, we get

$\displaystyle\theta$ $\displaystyle=$ $\displaystyle\frac{100\,\mbox{rev}}{60\,\mathrm{s}}\cdot 5\,\mathrm{s}-\frac{1% }{2}\,0.33\frac{\,\mbox{rev}}{{\,\mathrm{s}}^{2}}\cdot 25{\,\mathrm{s}}^{2}$

$\displaystyle=$ $\displaystyle 8.33\,\mbox{rev}-4.12\,\mbox{rev}$

$\displaystyle=$ $\displaystyle 4.21\,\mbox{rev}.$

Answer: $\theta=4.21\,\mbox{rev}$

Comments:

•

Note the negative sign used in both the formulae above. Since $\alpha$ is deceleration, that is, a negative acceleration, we have used negative sign with $\alpha$ in the formulae.
•

Note that it is not always necessary to convert $\,{\rm rpm}$ in rad/s. Here we changed $\,{\rm rpm}$ to $\,\mbox{rev}/\,\mathrm{s}$ because time was given in seconds.

SAMPLE 15.5

Filename:sfig8-5-2disks

Figure 15.14: Time varying angular acceleration,

\ddot{\theta}(t)=c\sin\beta t

Non-constant acceleration: A particle of mass $500\,{\mathrm{grams}}$ executes circular motion with radius $R=100\,\mathrm{cm}$ and angular acceleration $\ddot{\theta}(t)=c\sin\beta t$ , where $c=2\,\mathrm{rad}/\mathrm{s}^{2}$ and $\beta=2\,\mathrm{rad}/\mathrm{s}$ .

1.

Find the position of the particle after 10 seconds if the particle starts from rest, that is, $\theta(0)=0$ .
2.

How much kinetic energy does the particle have at the position found above?

Solution

We are given $\ddot{\theta}(t)=c\sin\beta t$ , $\dot{\theta}(0)=0$ and $\theta(0)=0$ . We have to find $\theta(10\,\mathrm{s})$ . Basically, we have to solve a second order differential equation with given initial conditions.

$\displaystyle\ddot{\theta}\equiv\frac{d}{dt}(\dot{\theta})$	$\displaystyle=$	$\displaystyle c\sin\beta t$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{\dot{\theta}_{0}=0% }^{\dot{\theta}(t)}d\dot{\theta}$	$\displaystyle=$	$\displaystyle\int_{0}^{t}c\sin\beta\tau\,d\tau$
$\displaystyle\dot{\theta}(t)$	$\displaystyle=$	$\displaystyle\left.{-\frac{c}{\beta}}\cos\beta\tau\right\|_{0}^{t}=\frac{c}{% \beta}(1-\cos\beta t).$

Filename:sfig8-4-1

Figure 15.15: Angular speed,

\dot{\theta}(t)=\frac{c}{\beta}(1-\cos\beta t)

, plotted against time for

c=2\,\mathrm{rad}/\mathrm{s}^{2}

and

\beta=2\,\mathrm{rad}/\mathrm{s}

Thus, we get the expression for the angular speed $\dot{\theta}(t)$ . We can solve for the position $\theta(t)$ by integrating once more:

$\displaystyle\dot{\theta}\equiv\frac{d}{dt}(\theta)$	$\displaystyle=$	$\displaystyle\frac{c}{\beta}(1-\cos\beta t)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\int_{\theta_{0}=0}^{% \theta(t)}d\theta$	$\displaystyle=$	$\displaystyle\int_{0}^{t}\frac{c}{\beta}(1-\cos\beta\tau)$
$\displaystyle\theta(t)$	$\displaystyle=$	$\displaystyle\frac{c}{\beta}\left[\tau-\frac{\sin\beta\tau}{\beta}\right]_{0}^% {t}$
	$\displaystyle=$	$\displaystyle\frac{c}{\beta^{2}}(\beta t-\sin\beta t).$

Filename:sfig8-4-1a

Figure 15.16: Angular position,

\theta(t)=\frac{c}{\beta^{2}}(\beta t-\sin\beta t)

, plotted against time for

c=2\,\mathrm{rad}/\mathrm{s}^{2}

and

\beta=2\,\mathrm{rad}/\mathrm{s}

Now substituting $t=10\,\mathrm{s}$ in the last expression along with the values of other constants, we get

	$\displaystyle\theta(10\,\mathrm{s})$	$\displaystyle=$	$\displaystyle\frac{2\,\mathrm{rad}/\mathrm{s}^{2}}{(2\,\mathrm{rad}/\mathrm{s}% )^{2}}[2\,\mathrm{rad}/\mathrm{s}\cdot 10\,\mathrm{s}-\sin(2\,\mathrm{rad}/% \mathrm{s}\cdot 10\,\mathrm{s})]$
		$\displaystyle=$	$\displaystyle 9.54\,{\rm rad}.$

Answer: $\theta=9.54\,{\rm rad}$

The kinetic energy of the particle is given by

$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}mv^{2}=\frac{1}{2}m(R\dot{\theta})^{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}mR^{2}[\underbrace{\frac{c}{\beta}(1-\cos\beta t)}_{% \dot{\theta}(t)}]^{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\,0.5\,\mathrm{kg}\cdot 1\,\mathrm{m}^{2}\cdot\left[% \frac{2\,\mathrm{rad}/\mathrm{s}^{2}}{2\,\mathrm{rad}/\mathrm{s}}\cdot(1-\cos(% 20))\right]^{2}$
	$\displaystyle=$	$\displaystyle 0.086\,\mathrm{kg}\cdot\,\mathrm{m}^{2}\cdot\,\mathrm{s}^{2}=0.0% 86{\rm Joule}.$

Answer: ${E_{\mathrm{K}}}=0.086{\rm J}$

Problems for 15.1 Kinematics of a particle in circular motion

Preparatory Problems

15.1.1 A particle goes on a circular path with radius $R$ making the angle $\theta=ct$ measured counter clockwise from the positive $x$ axis. Assume $R=5\,\mathrm{cm}$ and $c=2\pi\,\mathrm{s}^{-1}$ .

(a)

Plot the path.
(b)

What is the angular rate in revolutions per second?
(c)

Put a dot on the path for the location of the particle at $t=t^{*}=1/6\,\,\mathrm{s}$ .
(d)

What are the $x$ and $y$ coordinates of the particle position at $t=t^{*}$ ? Mark them on your plot.
(e)

Draw the vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ at $t=t^{*}$ .
(f)

What are the $x$ and $y$ components of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ at $t=t^{*}$ ?
(g)

What are the $R$ and $\theta$ components of $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ at $t=t^{*}$ ?
(h)

Draw an arrow representing both the velocity and the acceleration at $t=t^{*}$ .
(i)

Find the $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ components of position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ at $t=t^{*}$ .
(j)
Find the $x$ and $y$ components of position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ at $t=t^{*}$ . Find the velocity and acceleration two ways:
1. (a)
  
  Differentiate the position given as ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=x\hat{\boldsymbol{\imath}}+y\hat% {\boldsymbol{\jmath}}$ .
2. (b)
  
  Differentiate the position give as ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=r\hat{\boldsymbol{e}}_{r}$ and then convert the results to Cartesian coordinates.

15.1.2 A bead goes around a circular track of radius $1\,\mathrm{ft}$ at a constant speed. It makes it around the track in exactly $1\,{\rm s}$ .

(a)

Find the speed of the bead. Does this vary in time?
(b)

Find the magnitude of acceleration of the bead. Does this vary in time?
(c)

Is the magnitude of the acceleration the derivative of the speed (i.e., $|{\accentset{\rightharpoonup}{\boldsymbol{a}}}|\mathrel{\mathop{=}\limits^{?}}% {d\over dt}|{\accentset{\rightharpoonup}{\boldsymbol{v}}}|)$

15.1.3 If a particle moves along a circle at constant rate (constant $\dot{\theta}$ ) following the equation

{\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)=R\cos(\dot{\theta}t)\hat{% \boldsymbol{\imath}}+R\sin(\dot{\theta}t)\hat{\boldsymbol{\jmath}}

which of these things are true and why? If not true, explain why.

(a)

${\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$
(b)

${\accentset{\rightharpoonup}{\boldsymbol{v}}}=$ constant
(c)

$|{\accentset{\rightharpoonup}{\boldsymbol{v}}}|=$ constant
(d)

${\accentset{\rightharpoonup}{\boldsymbol{a}}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$
(e)

${\accentset{\rightharpoonup}{\boldsymbol{a}}}=$ constant
(f)

$|{\accentset{\rightharpoonup}{\boldsymbol{a}}}|=$ constant
(g)

${\accentset{\rightharpoonup}{\boldsymbol{v}}}\perp{\accentset{\rightharpoonup}% {\boldsymbol{a}}}$

15.1.4 A particle moves according to:

	$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle R\cos(ct),$
	$\displaystyle y(t)$	$\displaystyle=$	$\displaystyle R\sin(ct).$

where $R=1\,\mathrm{m}$ and $c=5\,\mathrm{rad}/\mathrm{s}$ .

(a)

Show that the speed of the particle is constant.
(b)

How much time does the particle take to go from P at $(0,1\,\mathrm{m})$ to Q at $(1\,\mathrm{m},0)$ ?
(c)

What is the acceleration of the particle at point Q?

More-Involved Problems

15.1.5 A $200\,{\rm mm}$ diameter gear rotates at a constant speed of $100\,{\rm rpm}$ .

(a)

What is the speed of a peripheral point on the gear?
(b)

If no point on the gear is to exceed the centripetal acceleration of $25\,\mathrm{m}/\mathrm{s}^{2}$ , find the maximum allowable angular speed (in $\,{\rm rpm}$ ) of the gear.

15.1.6 A particle is in circular motion in the $x y$ -plane at the constant angular speed of $\dot{\theta}=2\,\mathrm{rad}/\mathrm{s}$ at radius $0.5\,\mathrm{m}$ . At $t=0$ the particle is at $\theta=0$ .

(a)

Draw the path and mark the position of the particle at $t=0.5\,\mathrm{s}$ and $t=15\,\mathrm{s}$ .
(b)

Find the velocity and acceleration of the particle at $t=0.5\,{\rm s}$ and $t=15\,{\rm s}$ .

15.1.7 A particle undergoes constant rate circular motion in the $x y$ -plane. At some instant $t_{0}$ , its velocity is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}(t_{0})=-3\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}+4\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$ and after $5\,{\rm s}$ the velocity is $v(t_{0}+5\,\mathrm{s})=(5/\sqrt{2})\,\mathrm{m}/\mathrm{s}(\hat{\boldsymbol{% \imath}}+\hat{\boldsymbol{\jmath}})$ . If the particle has not yet completed one revolution between the two instants, find

(a)

the angular speed of the particle,
(b)

the distance traveled by the particle in $5\,{\rm s}$ , and
(c)

the acceleration of the particle at the two instants.

15.1.8 A bead on a circular path of radius $R$ in the $x y$ -plane has rate of change of angular speed $\alpha=bt^{2}$ . The bead starts from rest at $\theta=0$ .

(a)

What is the bead’s angular position $\theta$ (measured from the positive $x$ -axis) and angular speed $\dot{\theta}$ as a function of time ?
(b)

What is the angular speed as function of angular position?

15.1.9 A bead on a circular wire has an angular speed given by $\dot{\theta}=c\theta^{1/2}$ . The bead starts from rest at $\theta=0$ . What is the angular position and speed of the bead as a function of time? [This problem is subtle because it has multiple solutions. One answer you can find with a quick guess. Another you can find by separation of variables. The full general solution is an appropriate mixture of these two.)] Answer: One solution is $\theta=0$ for all $t$ . Another set of solutions is $\theta=\left({c(t-t_{0})\over 2}\right)^{2}$ where $t_{0}$ is an arbitrary constant. To make sense of this second solution set one needs to have $\theta=0$ until $t=t_{0}$ .

15.1.10 Solve $\ddot{\theta}=C$ , given $\dot{\theta}(0)=\dot{\theta}_{0}$ , $\theta(0)=\theta_{0}$ and that $C$ is a constant. That is, find $\theta$ in terms of some or all of $C$ , $\dot{\theta}_{0}$ , $\theta_{0}$ and $t$ .

15.1.11 Given that $\ddot{\theta}-\lambda^{2}\theta=0$ , $\theta(0)=\pi/2$ , $\dot{\theta}(0)=0$ , and $\lambda=3/\,\mathrm{s}$ find the value of $\theta$ at $t=1\,\mathrm{s}$ .

15.1.12 Two runners run on a circular track side-by-side at the same constant angular rate $\dot{\theta}=0.25\,\mathrm{rad}/\mathrm{s}$ about the center of the track. The inside runner is in a lane of radius $r_{i}=35\,\mathrm{m}$ and the outside runner is in a lane of radius $r_{o}=37\,\mathrm{m}$ . What is the velocity of the outside runner relative to the inside runner?

15.1.13 A particle oscillates on the arc of a circle with radius $R$ according to the equation $\theta=\theta_{0}\cos(\lambda t)$ . What are the conditions on $R,\theta_{0}$ , and $\lambda$ so that the maximum acceleration in this motion occurs at $\theta=0$ . “Acceleration” here means the magnitude of the acceleration vector.

15.1.14 A particle moves on a circular arc starting from rest at $\theta_{0}=0$ . As $\theta$ increases, the magnitude of the acceleration is constant. Assume, all in consistent units, that $R=1$ and $|{\accentset{\rightharpoonup}{\boldsymbol{a}}}|=1$ .

(a)

Write the statement ‘the magnitude of acceleration is constant’ as an equation in terms of $\dot{\theta}$ and $\ddot{\theta}$ .
(b)

Find a solution to the equation with the given initial conditions (analytically or numerically).
(c)

Find and plot $\dot{\theta}$ vs $t$ and $\theta$ vs $t$ .
(d)

In circular motion does $|{\accentset{\rightharpoonup}{\boldsymbol{a}}}|$ =constant necessarily mean that the motion is at or is gradually approaching constant rate circular motion? Is so, why? If not show a counter-example.

15.1.15 A particle moves in a circle so that its acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ always makes a fixed angle $\phi$ with the position vector $-{\accentset{\rightharpoonup}{\boldsymbol{r}}}$ , with $0\leq\phi\leq\pi/2$ . For example, $\phi=0$ would be constant rate circular motion. Assume $\phi=\pi/4$ , $R=1\,\mathrm{m}$ and $\dot{\theta}_{0}=1\,{\rm rad}/\,\mathrm{s}$ . How long does it take the particle to reach

(a)

the speed of sound ( $\approx 300\,\mathrm{m}/\mathrm{s}$ )?
(b)

the speed of light ( $\approx 3\cdot 10^{8}\,\mathrm{m}/\mathrm{s}$ )?
(c)

$\infty$ ?

15.2 Dynamics of a particle in circular motion

Filename:sfig8-6-2

Figure 15.17: Point mass spinning in circles. Sketch of system and a free-body diagram.

The simplest examples of circular motion concern the motion of a particle constrained by a massless connection to be a fixed distance from a support point.

Example: Rock spinning on a string

Neglecting gravity, we can now deal with the familiar problem of a point mass being held in constant circular-rate motion by a massless string or rod.

Linear momentum balance for the mass gives:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$
$\displaystyle\Rightarrow-T\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle\left\{-T\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\right.$	$\displaystyle=$	$\displaystyle\left.m(-\dot{\theta}^{2}\ell\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R})\right\}$
$\displaystyle\left\{\right\}\cdot\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\Rightarrow$		$\displaystyle T=\dot{\theta}^{2}\ell m=(v^{2}/\ell)m$

The force required to keep a mass in constant rate circular motion is $mv^{2}/\ell$ (sometimes remembered as $mv^{2}/R$ ).

The simplest example of ‘celestial mechanics’ is also circular motion.

Example: Geosynchronous orbit

Assuming a spherical earth, the centrally acting force of earth’s gravity on a satellite is $m g$ at the earth’s surface and decays with radius squared so is

F=mg{R_{e}^{2}\over r^{2}}

where $R_{e}$ is the radius of the earth and $r$ is the distance of the satellite from the center of the earth.

Linear momentum balance for the mass gives:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$
$\displaystyle\Rightarrow-mg{R_{e}^{2}\over r^{2}}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle\left\{-mg{R_{e}^{2}\over r^{2}}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\right.$	$\displaystyle=$	$\displaystyle\left.m(-\dot{\theta}^{2}r\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R})\right\}$
$\displaystyle\left\{\right\}\cdot\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\Rightarrow$		$\displaystyle r=\left({gR_{e}^{2}\over\dot{\theta}^{2}}\right)^{1/3}.\par$

Communication satellites in ‘geosynchronous’ orbits go around once a day (staying in the sights of millions of satellite dishes). So, using $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ , $R_{e}\approx 6400\,{\mbox{km}}$ and $\dot{\theta}\approx$ 1 rev/day, we get $r=42600\,{\mbox{km}}$

^†^†margin: There are various errors (approximations) in this satellite calculation, of course. The earth doesn’t rotate once per day, but a little more because it goes around once per day relative to a line connecting the earth and sun and that line is itself rotating relative to the ‘fixed stars’ (the period of a geosynchronous satellite is one part in 365.25 shorter than a day). And the force of gravity on a near-earth mass is a bit more than

m g

because ‘

g

’ actually measures the force it takes to hold up a mass on the earth’s surface, which is the gravity force less the acceleration from going in circles on the surface of the earth (the actual acceleration of gravity at the earth’s surface is about 0.5% more than

g

). And the earth isn’t exactly spherical, and so on. The actual geosynchronous radius is close to

42164\,{\mbox{km}}

. So the example calculation is off by 436 kmor about 1%.

Similar calculations can find the motion of low altitude satellites, the motion of the moon around the earth and of the earth around the sun.

Centripetal and centrifugal forces

Because the centrally directed part of a particle’s acceleration is called the ‘centripetal’ acceleration, the centrally directed force needed to keep a particle in circular motion is sometimes called the ‘centripetal’ force. Thus, in the first example above the tension in the string is a centripetal force, and in the satellite problem the gravity force is a centripetal force. On the other hand, the ‘centrifugal’ force outwards is not really a force at all and is best dropped as a concept, at least for beginners.

Non-constant rate circular motion

Situations in which the circular rate is not constant are just slightly more complex. In these cases the part of the acceleration tangent to the circular motion is non-zero,

a_{\theta}=\ddot{\theta}r=\dot{v}_{\theta},

so the net force on the particle has a component tangent to the circle.

Linear momentum balance in polar coordinates

The equation of linear momentum balance for a particle ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ in polar coordinates can be written as follows:

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
	$\displaystyle\sum F_{r}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+\sum F_{% \theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle m\left(a_{r}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+a_{% \theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}\right).$		(15.11)

For circular motion we have, from Section 15.1, that

a_{r}=-\dot{\theta}^{2}r=-{v^{2}\over r}\qquad\mbox{and}\qquad a_{\theta}=% \ddot{\theta}r=\dot{v}.

Angular momentum

In general one can use angular momentum balance with respect to any point you like. But for circular motion with the circle center at 0 one is almost always concerned with angular momentum balance about 0. In this case the various torque and angular momentum expressions are particularly simple, for example

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/0}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{/0}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}\times\left(\sum{\accentset{\rightharpoonup}{% \boldsymbol{F}}}\right)$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times(m{\accentset{% \rightharpoonup}{\boldsymbol{a}}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}r\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\times\left(\sum F_{r}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\sum F_{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta}\right)$	$\displaystyle=$	$\displaystyle r\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times m(a_{r}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+a_{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta})$
$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}={\accentset{\rightharpoonup}{% \boldsymbol{0}}},\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}=\hat{\boldsymbol{k}}\hbox to36.13% 5pt{\hfil$\Rightarrow$\hfil}\left\{r\sum F_{\theta}\hat{\boldsymbol{k}}\right.$	$\displaystyle=$	$\displaystyle\left.rma_{\theta}\hat{\boldsymbol{k}}\right\}$
$\displaystyle\left\{\right\}\cdot\hat{\boldsymbol{k}}\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}r\sum F_{\theta}$	$\displaystyle=$	$\displaystyle rma_{\theta}.$	(15.13)

Energy

Kinetic energy is also particularly simple in polar coordinates for circular motion because there is only one degree of freedom:

	$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle{mv^{2}\over 2}={mv_{\theta}^{2}\over 2}$		(15.14)
		$\displaystyle=$	$\displaystyle m\dot{\theta}^{2}r^{2}/2.\par$		(15.14)

The simple pendulum

Filename:sfig8-6-2a

Figure 15.18: The ideal simple pendulum.

Filename:efig1-2-28

Figure 15.19: Free-body diagram of the mass in a simple pendulum.

Perhaps the most famous mechanics example of circular motion at non-constant rate is a simple pendulum.

As a child’s swing, the inside of a grandfather clock, a hypnotist’s device, or a gallows, the motion of a simple pendulum is a clear image to all of us. Galileo studied the simple pendulum before Newton created Newton’s laws, and the pendulum is a core topic in high-school and freshman physics.

For starters, we consider a 2-D pendulum of fixed length with no forcing other than gravity. All mass is concentrated at a point. Of primary interest is the motion of the pendulum and the tension in the string. First we find governing differential equations (the equations of motion).

First, the tension in the pendulum rod (or string) acts along the length because the rod is a massless two-force body. At least that is the idealization. For any real pendulum, where the rod is not precisely massless and where the mass is not precisely concentrated at a point, there is a small force transmitted that is not along the rod. We neglect this ‘shear’ force in the treatment of the ideal pendulum.

One way to get the equation of motion is to use linear momentum balance in polar coordinates,

eqn. (LABEL:eq:fismapolar),

and dot both sides with $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ to get

	$\displaystyle-T\underbrace{\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\cdot% \hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}}_{0}+mg\underbrace{\hat{% \boldsymbol{\imath}}\cdot\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}}_{-% \sin\theta}$	$\displaystyle=m\left(\ell\ddot{\theta}\underbrace{\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}\cdot\hat{\boldsymbol{e}}_{\scriptscriptstyle\!% \theta}}_{1}-\ell\dot{\theta}^{2}\underbrace{\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\cdot\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}}_% {0}\right)$
	$\displaystyle\Rightarrow-mg\sin\theta$	$\displaystyle=m\ell\ddot{\theta}$
	$\displaystyle\mbox{so}\qquad\ddot{\theta}$	$\displaystyle=-\frac{g}{\ell}\sin\theta.$

Small angle approximation (linearization)

For small angles, $\sin\theta\approx\theta$ , so we have

\ddot{\theta}=-\frac{g}{\ell}\theta

for small oscillations. This equation describes a harmonic oscillator with $\sqrt{\frac{g}{\ell}}$ replacing the $\sqrt{\frac{k}{m}}$ coefficient in a spring-mass system. Thus the general solution is

\par\theta=A\cos\sqrt{g/l}t+B\sin\sqrt{g/l}t\par

(15.15)

where $A=\theta_{0}$ and $B\sqrt{g/l}=\dot{\theta}_{0}$ . This solution has the famous property, Galileo loved this, that the frequency is the same for big as for small oscillations. Thus, a pendulum of a given length that swings back and forth 1 degree makes about the same number of swings per minute as one that swings with an amplitude of 10 degrees. How big is the error in this constant frequency result? Well, something less than the error in the approximation that $\sin\theta=\theta$ .

\par\%\mbox{ error}=100\cdot{\theta-\sin\theta\over\sin\theta}\approx 100\cdot% {\theta^{3}/3\over\theta}\approx{\theta^{2}\over 3}\approx 1\%

for $\theta=10^{\circ}\approx 1/6\,{\rm rad}$ . The actual error in the period is less than this, as you can find by numerically solving the non-linear pendulum equation.

The inverted pendulum

A pendulum with the mass-end up is called an inverted pendulum.

By methods just like we used for the regular pendulum, we find the equation of motion to be

\ddot{\theta}=\frac{g}{\ell}\sin\theta

which, for small $\theta$ , is well approximated by

Filename:pfigure-blue-118-2

Figure 15.20: The inverted pendulum

\ddot{\theta}=\frac{g}{\ell}\theta.

As opposed to the simple pendulum, which has oscillatory solutions, this differential equation has exponential solutions

\theta=C_{1}e^{\sqrt{g/\ell}\ t}+C_{2}e^{-\sqrt{g/\ell}\ t},

one term of which has exponential growth (the implicit “ $+$ ” in front of the argument of the $\sqrt{g/\ell}\ t$ ), indicating the inherent instability of the inverted pendulum. That is, as is intuitively obvious, an inverted pendulum has tendency to fall over when slightly disturbed from the vertical position^†^†margin: After the pendulum falls a ways, say past 30 degrees from vertical, the exponential solution is not an accurate description, but the actual motion (as viewed by an experiment, a computer simulation, or the exact elliptic integral solution of the equations) shows that the pendulum keeps falling. .

More about pendula

Pendula are useful as models of many phenomena from the swing of a leg in walking to the tipping of a chimney in an earthquake. Pendula also serve as a simple example for many more general concepts in mechanics. For example, the pendulum is popular as an example of “chaos”; if you push a pendulum periodically its motions can be wild.

Box 15.3 Other derivations of the pendulum equation

The simplest derivation of the pendulum differential equation is to use linear momentum balance in polar coordinates. Here are two other derivations.

Method one: linear momentum balance in cartesian coordinates

The equation of linear momentum balance is

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\overbrace{\dot{\accentset{% \rightharpoonup}{\boldsymbol{L}}}}^{m{\accentset{\rightharpoonup}{\boldsymbol{% a}}}}

Evaluating the left side (using the free-body diagram) and right side (using the kinematics of circular motion), we get

-T\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}+mg\hat{\boldsymbol{\imath}}=m[% \ell\ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\ell\dot{% \theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}]

(15.16)

From the picture (or recalling) we see that $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}=\cos\theta\hat{\boldsymbol{\imath% }}+\sin\theta\hat{\boldsymbol{\jmath}}$ and

$\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=\cos\theta\hat{\boldsymbol{% \jmath}}-\sin\theta\hat{\boldsymbol{\imath}}$ . So, upon substitution into the equation above, we get

	$\displaystyle-T(\cos\theta\hat{\boldsymbol{\imath}}$	$\displaystyle+\sin\theta\hat{\boldsymbol{\jmath}})+mg\hat{\boldsymbol{\imath}}$
		$\displaystyle=m\left[\ell\ddot{\theta}\left(\cos\theta\hat{\boldsymbol{\jmath}% }-\sin\theta\hat{\boldsymbol{\imath}}\right)-\ell\dot{\theta}^{2}\left(\cos% \theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}}\right)\right]$

Breaking this equation into its $x$ and $y$ components (by dotting both sides with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , respectively) gives

	$\displaystyle-T\cos\theta+mg$	$\displaystyle=$	$\displaystyle-m\ell\left(\ddot{\theta}\sin\theta+\dot{\theta}^{2}\cos\theta% \right)\mbox{ and}$
	$\displaystyle-T\sin\theta$	$\displaystyle=$	$\displaystyle m\ell\left(\ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta% \right).$		(15.17)

Note, when deriving equations of motion, we think of both positions and the rates and velocities as knowns. For example, we take $\theta$ and $\dot{\theta}$ as known. But how do we know them? We don’t. But thinking of them as known helps us write a set of differential equations from which we can eventually find them. Thus the equations above are two simultaneous equations that we can solve for the two unknowns $T$ and $\ddot{\theta}$ to get

	$\displaystyle\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{g}{\ell}\sin\theta$		(15.18)
	$\displaystyle T$	$\displaystyle=$	$\displaystyle m[\ell\dot{\theta}^{2}+g\cos\theta].$		(15.19)

The first equation is the familiar pendulum differential equation, the second allows us to find the tension in the pendulum string.

Method two: angular momentum balance

Using angular momentum balance, we can ‘kill’ (eliminate) the tension term at the start. Taking angular momentum balance about the point $O$ , we get

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\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}$}}}$
$\displaystyle-mg\ell\sin\theta\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\text{$\ell\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$}{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{/O}\times{\accentset{% \rightharpoonup}{\boldsymbol{a}}}\text{ $\ell\ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\ell\dot{% \theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$}\ m$
$\displaystyle-mg\ell\sin\theta\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle m\ell^{2}\ddot{\theta}\hat{\boldsymbol{k}}$
$\displaystyle\Rightarrow\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{g}{\ell}\sin\theta$

since $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}=0$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta}=\hat{\boldsymbol{k}}$ . So, the governing equation for a simple pendulum is

\ddot{\theta}=-\frac{g}{\ell}\sin\theta

Method three: Conservation of energy

The string tension is always orthogonal to the velocity so does no work. The gravity force is conservative. So energy is conserved.

$\displaystyle constant$	$\displaystyle=$	$\displaystyle E_{\mathrm{T}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ constant$	$\displaystyle=$	$\displaystyle{E_{\mathrm{K}}}+{E_{\mathrm{P}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ 0$	$\displaystyle=$	$\displaystyle\dot{E_{\mathrm{K}}}+\dot{E_{\mathrm{P}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ 0$	$\displaystyle=$	$\displaystyle{d\over dt}\left({1\over 2}mv^{2}\right)+{d\over dt}(mgh)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ 0$	$\displaystyle=$	$\displaystyle{d\over dt}\left({1\over 2}m(\ell\dot{\theta})^{2}\right)+{d\over dt% }(-mg\ell\cos\theta)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ 0$	$\displaystyle=$	$\displaystyle m\ell^{2}\ddot{\theta}\dot{\theta}+mg\ell(\sin\theta)\dot{\theta}$

Now $m$ cancels from both sides and we can divide through by $\ell^{2}$ . We can also divide through by $\dot{\theta}$ , but for exceptional instants in time when $\dot{\theta}=0$ . Thus

\ddot{\theta}+{g\over\ell}\sin\theta=0

which is the familiar differential equation for a pendulum. This method lacks some rigor in that the cancelation of $\dot{\theta}$ is not valid at exactly every instant in time. However, it is valid for all but those instants, and happens to give the right answer at the exceptional instants as well.

SAMPLE 15.6

Filename:pfigure-s95f3a

Figure 15.21: The motor rotates the structure at a constant angular speed in the counterclockwise direction.

Circular motion in 2-D. Two bars, each of negligible mass and length $\ell=3\,\mathrm{ft}$ , are welded together at right angles to form an ‘L’ shaped structure. The structure supports a $3.2\,{\rm lbf}\;(=mg)$ ball at one end and is connected to a motor on the other end (see Fig. 15.21). The motor rotates the structure in the vertical plane at a constant rate $\dot{\theta}=10\,\mathrm{rad}/\mathrm{s}$ in the counter-clockwise direction. Take $g=32\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ . At the instant shown in Fig. 15.21, find

1.

the velocity of the ball,
2.

the acceleration of the ball, and
3.

the net force and moment applied by the motor and the support at O on the structure.

Solution The motor rotates the structure at a constant rate. Therefore, the ball is going in circles with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\theta}\hat{% \boldsymbol{k}}=10\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}$ . The radius of the circle is $R=\sqrt{\ell^{2}+\ell^{2}}=\ell\sqrt{2}$ . Since the motion is in the $x y$ plane, we use the following formulae to find the velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ .

Filename:Danef94s1q2

Figure 15.22: The ball follows a circular path of radius

R

. The position, velocity, and acceleration of the ball can be expressed in terms of the polar basis vectors

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

and

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

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

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=-R\dot{\theta}^{2}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}+R\ddot{\theta}\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!\theta},

where $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ are the polar basis vectors shown in Fig. 15.22. In Fig. 15.22, we note that $\theta=45^{{}^{\circ}}$ . Therefore,

$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$	$\displaystyle=$	$\displaystyle\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{% \jmath}}$
	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{2}}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}}),$
$\displaystyle\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{% \jmath}}$
	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{2}}(-\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}}).$

Here, $R=L\sqrt{2}=3\sqrt{2}\,\mathrm{ft}$ is constant, and $\ddot{\theta}=0$ because $\dot{\theta}=10\,\mathrm{rad}/\mathrm{s}=$ constant. Thus,

1.

the velocity of the ball is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ $\displaystyle=$ $\displaystyle R\dot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

$\displaystyle=$ $\displaystyle 3\sqrt{2}\,\mathrm{ft}\cdot 10\,\mathrm{rad}/\mathrm{s}\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

$\displaystyle=$ $\displaystyle 30\sqrt{2}\,{\mathrm{ft/s}}\cdot\frac{1}{\sqrt{2}}(-\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$

$\displaystyle=$ $\displaystyle 30\,{\mathrm{ft/s}}(-\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}}).$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=30\,{\mathrm{ft/s}}(-\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$
2.

The acceleration of the ball is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ $\displaystyle=$ $\displaystyle-R\dot{\theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$

$\displaystyle=$ $\displaystyle-3\sqrt{2}\,\mathrm{ft}\cdot(10\,\mathrm{rad}/\mathrm{s})^{2}\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!R}$

$\displaystyle=$ $\displaystyle-300\sqrt{2}\,{\mathrm{ft}}/{\mathrm{s}}^{2}\cdot\frac{1}{\sqrt{2% }}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$

$\displaystyle=$ $\displaystyle-300\,{\mathrm{ft}}/{\mathrm{s}}^{2}(\hat{\boldsymbol{\imath}}+% \hat{\boldsymbol{\jmath}}).$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=-300\,{\mathrm{ft}}/{\mathrm{s}}% ^{2}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$

Let the net force and the moment applied by the motor-support system be ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ as shown in Fig. 15.23. From the linear momentum balance for the structure,

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}-mg\hat{\boldsymbol{% \jmath}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}+mg\hat{% \boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle\overbrace{\frac{3.2\,{\rm lbf}}{32\,{\mathrm{ft}}/{\mathrm{s}}^{% 2}}}^{m}(-300\sqrt{2}\,{\mathrm{ft}}/{\mathrm{s}}^{2})\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}+\overbrace{3.2\,{\rm lbf}}^{mg}\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-30\sqrt{2}\,{\rm lbf}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R% }+3.2\,{\rm lbf}\hat{\boldsymbol{\jmath}}.$
	$\displaystyle=$	$\displaystyle-30\sqrt{2}\,{\rm lbf}\frac{1}{\sqrt{2}}(\hat{\boldsymbol{\imath}% }+\hat{\boldsymbol{\jmath}})+3.2\,{\rm lbf}\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-30\,{\rm lbf}\hat{\boldsymbol{\imath}}-26.8\,{\rm lbf}\hat{% \boldsymbol{\jmath}}.$

Filename:pfigure-blue-123-1

Figure 15.23: Free-body diagram of the structure.

Similarly, from the angular momentum balance for the structure,

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\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}$}}},$
$\displaystyle\hbox to36.135pt{where\hfil}\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}\!_{\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{$% \scriptscriptstyle\mathrm{/O}$}}}\times mg(-\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}+\underbrace{R\hat{% \boldsymbol{e}}_{\scriptscriptstyle\!R}}_{\ell(\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}})}\times mg(-\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}-mg\ell\hat{% \boldsymbol{k}},$
$\displaystyle\hbox to36.135pt{and\hfil}\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}\!_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}\times m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}$
	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times m(-R\dot{% \theta}^{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})$
	$\displaystyle=$	$\displaystyle-mR^{2}\dot{\theta}^{2}\underbrace{(\hat{\boldsymbol{e}}_{% \scriptscriptstyle\!R}\times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R})}_{{% \accentset{\rightharpoonup}{\boldsymbol{0}}}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}$	$\displaystyle=$	$\displaystyle mg\ell\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle\underbrace{3.2\,{\rm lbf}}_{mg}\cdot\underbrace{3\,\mathrm{ft}}_% {\ell}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle 9.6\,{\rm lbf}\mathord{\cdot}\,\mathrm{ft}\hat{\boldsymbol{k}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=-30\,{\rm lbf}\hat{\boldsymbol{% \imath}}-26.8\,{\rm lbf}\hat{\boldsymbol{\jmath}},\;\;{\accentset{% \rightharpoonup}{\boldsymbol{M}}}=9.6\,{\rm lbf}\mathord{\cdot}\,\mathrm{ft}% \hat{\boldsymbol{k}}$

Note: If there was no gravity, the moment applied by the motor would be zero.

SAMPLE 15.7

Filename:pfigure-blue-119-2

Figure 15.24:

A 50 gm point mass executes circular motion with angular acceleration $\ddot{\theta}=2\,\mathrm{rad}/\mathrm{s}^{2}$ . The radius of the circular path is 200 mm. If the mass starts from rest at $t=0$ , find

1.

Its angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}$ about the center at $t=5\,\mathrm{s}$ .
2.

Its rate of change of angular momentum $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}$ about the center.

Solution

1.

From the definition of angular momentum,

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

$\displaystyle=$ $\displaystyle R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times m\dot{\theta% }R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

$\displaystyle=$ $\displaystyle mR^{2}\dot{\theta}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})$

$\displaystyle=$ $\displaystyle mR^{2}\dot{\theta}\hat{\boldsymbol{k}}$

On the right hand side of this equation, the only unknown is $\dot{\theta}$ . Thus to find ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ at $t=5\,\mathrm{s}$ , we need to find $\dot{\theta}$ at $t=5\,\mathrm{s}$ . Now,

$\displaystyle\ddot{\theta}$ $\displaystyle=$ $\displaystyle\frac{d\dot{\theta}}{dt}$

$\displaystyle d\dot{\theta}$ $\displaystyle=$ $\displaystyle\ddot{\theta}\,dt$

$\displaystyle\int_{\dot{\theta}_{0}}^{\dot{\theta}(t)}d\dot{\theta}$ $\displaystyle=$ $\displaystyle\int_{0}^{t}\ddot{\theta}\,dt$

$\displaystyle\dot{\theta}(t)-\dot{\theta}_{0}$ $\displaystyle=$ $\displaystyle\ddot{\theta}(t_{t}-t_{0})$

$\displaystyle\dot{\theta}$ $\displaystyle=$ $\displaystyle\dot{\theta}_{0}+\ddot{\theta}(t-t_{0})$

Writing $\alpha$ for $\ddot{\theta}$ and substituting $t_{0}=0$ in the above expression, we get $\dot{\theta}(t)=\dot{\theta}_{0}+\alpha t$ , which is the angular speed version of the linear speed formula $v(t)=v_{0}+at$ . ^†^†margin: Be warned that these formulae are valid only for constant rate of change of speed. Substituting $t=5\,\mathrm{s}$ , $\dot{\theta}_{0}=0$ , and $\alpha=2\,\mathrm{rad}/\mathrm{s}^{2}$ we get $\dot{\theta}=2\,\mathrm{rad}/\mathrm{s}^{2}\cdot 5\,\mathrm{s}=10\,\mathrm{rad% }/\mathrm{s}$ . Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$ $\displaystyle=$ $\displaystyle 0.05\,\mathrm{kg}\cdot(0.2\,\mathrm{m})^{2}\cdot 10\,\mathrm{rad% }/\mathrm{s}\hat{\boldsymbol{k}}$

$\displaystyle=$ $\displaystyle 0.02\,\mathrm{kg}.{\,\mathrm{m}}^{2}/\,\mathrm{s}=0.02\,{\rm N}% \mathord{\cdot}{\rm m}\cdot\,\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=0.02\,{\rm N}\mathord{\cdot}{\rm m}\cdot\,% \mathrm{s}.$

Similarly, we can calculate the rate of change of angular momentum:

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{/0}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}$
	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}\times m(R\ddot{% \theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-\dot{\theta}^{2}R\hat% {\boldsymbol{e}}_{\scriptscriptstyle\!R})$
	$\displaystyle=$	$\displaystyle mR^{2}\ddot{\theta}(\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}% \times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta})$
	$\displaystyle=$	$\displaystyle mR^{2}\ddot{\theta}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle 0.05\,\mathrm{kg}\cdot(0.2m)^{2}\cdot 2\,\mathrm{rad}/\mathrm{s}% ^{2}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle 0.004\,\mathrm{kg}\cdot{\,\mathrm{m}}^{2}/{\,\mathrm{s}}^{2}=0.0% 04\,{\rm N}\mathord{\cdot}{\rm m}$

Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}=0.004\,{\rm N}\mathord{\cdot}{\rm m}$

SAMPLE 15.8 The simple pendulum.

Filename:pfigure-s95q14

Figure 15.25:

A simple pendulum swings about its vertical equilibrium position (2-D motion) with amplitude $\theta_{\max}=10^{{}^{\circ}}$ . Find

1.

the magnitude of the maximum angular acceleration,
2.

the maximum tension in the string.

Solution

1.

The equation of motion of the pendulum is given by (see eqn. (15.18) in the text):

$\ddot{\theta}=-\frac{g}{\ell}\sin\theta.$

We are given that $|\theta|\leq\theta_{\max}$ . For $\theta_{\max}=10^{{}^{\circ}}=0.1745\,{\rm rad},\,\sin\theta_{\max}=0.1736$ . Thus we see that $\sin\theta\approx\theta$ even when $\theta$ is maximum. Therefore, we can safely use linear approximation (although we could solve this problem without it); i.e.,

$\ddot{\theta}=-\frac{g}{\ell}\theta.$

Clearly, $|\ddot{\theta}|$ is maximum when $\theta$ is maximum. Thus,

$|\ddot{\theta}|_{\max}=\frac{g}{\ell}\theta_{\max}=\frac{9.81\,\mathrm{m}/% \mathrm{s}^{2}}{1\,\mathrm{m}}\mathord{\cdot}(0.1745\,{\rm rad})=1.71\,\mathrm% {rad}/\mathrm{s}^{2}.$

Answer: $|\ddot{\theta}|_{\max}=1.71\,\mathrm{rad}/\mathrm{s}^{2}$
2.

The tension in the string is given by (see equation 15.19 of text):

$T=m(\ell\dot{\theta}^{2}+g\cos\theta).$

This time, we will not make the small angle assumption. We can find $T_{\max}$ and the corresponding $\theta$ using conservation of energy. Let the position of maximum amplitude be position 1 and the position at any $\theta$ be position 2. When $\theta=\theta_{\max}$ , the mass comes to rest and switches its direction of motion. Thus, its angular velocity and, hence, its kinetic energy is zero at $\theta_{\max}$ .

Using conservation of energy, we have

$\displaystyle{E_{\mathrm{K}}}_{1}+{E_{\mathrm{P}}}_{1}$ $\displaystyle=$ $\displaystyle{E_{\mathrm{K}}}_{2}+{E_{\mathrm{P}}}_{2}$

$\displaystyle 0+mg\ell(1-\cos\theta_{\max})$ $\displaystyle=$ $\displaystyle\frac{1}{2}m(\ell\dot{\theta})^{2}+mg\ell(1-\cos\theta).$ (15.20)

and solving for $\dot{\theta}$ , we get,

$\dot{\theta}=\sqrt{\frac{2g}{\ell}\left(\cos\theta-\cos\theta_{\max}\right)}.$

Therefore, the tension at any $\theta$ is

$T(\theta)=m(\ell\dot{\theta}^{2}+g\cos\theta)=mg(3\cos\theta-2\cos\theta_{\max% }).$

To find the maximum tension, we set $\frac{dT}{d\theta}=0$ , and find that, for $0\leq\theta\leq\theta_{\max}$ , $T$ is maximum when $\theta=0$ . Now, substituting $\theta=0$ in $T(\theta)$ , we get,

$T_{\max}=mg\bigl{(}3\cos(0)-2\cos(\theta_{\max})\bigr{)}=0.2\,\mathrm{kg}\cdot 9% .81\,\mathrm{m}/\mathrm{s}^{2}\bigl{(}3-1.97\bigr{)}=2.02\,{\mbox{N}}.$

The maximum tension corresponds to maximum speed which occurs at the bottom of the swing where all of the potential energy is converted to kinetic energy.

Answer: $T_{\max}=2.02\,{\mbox{N}}$

SAMPLE 15.9

Filename:Danef94s3q2

Figure 15.26:

The nonlinear pendulum: Consider the simple pendulum of Sample 15.24 again. Let the mass be $m$ and the length of the pendulum $\ell$ . The equation of motion of the pendulum is $\ddot{\theta}=-\frac{g}{\ell}\sin\theta$ as derived in the text (see eqn. (15.18)). This is a nonlinear ordinary differential equation but it can be solved easily numerically. Write a computer code using some ODE solver to solve the equation. Take $g$ and $\ell$ such that $\lambda=\sqrt{g/\ell}=2\pi$ (this makes the time period of the pendulum $T=2\pi/\lambda=1\,\mathrm{s}$ ). Using the code, do the following calculations.

1.

Solve the equation over a time interval of $t=0$ to 4 seconds using the initial conditions $\theta(0)=6^{\circ}$ and $\dot{\theta}(0)=0$ , and plot $\theta$ vs $t$ , $\dot{\theta}$ vs $t$ , and $\dot{\theta}$ vs $\theta$ . How do these plots compare with the solution of the linear equation $\ddot{\theta}=-\frac{g}{\ell}\theta$ ?
2.

Solve the equation again over the same time interval using the initial conditions $[\theta(0),\dot{\theta}(0)]=[18^{\circ},0]$ , and $[30^{\circ},0]$ . Plot $\theta(t)$ starting with all the three initial conditions used so far on the same graph and comment on the time period of oscillations.
3.

Solve the equation again over $t\in[0,4\,\mathrm{s}]$ using $\theta(0)=\pi/2$ and $\pi/1.02$ while keeping $\dot{\theta}(0)=0$ . Again plot $\theta$ vs $t$ , $\dot{\theta}$ vs $t$ , and $\dot{\theta}$ vs $\theta$ , for the three solutions obtained with $\theta(0)=\pi/6,\pi/2$ , and $\pi/1.02$ . Comment on the plots.
4.

For the last three initial conditions, compute ${E_{\mathrm{P}}}$ , ${E_{\mathrm{K}}}$ , and $E_{\mathrm{T}}={E_{\mathrm{P}}}+{E_{\mathrm{K}}}$ from the solutions obtained. For each initial condition, plot ${E_{\mathrm{P}}}$ , ${E_{\mathrm{K}}}$ , and $E_{\mathrm{T}}$ on the same graph and show that the total energy in each case remains constant irrespective of the nature of oscillations.

Solution

Filename:bikefork1-ang-accel

Figure 15.27: Numerical solution of the nonlinear pendulum equation,

\ddot{\theta}=-g/ell\sin\theta

, with the initial conditions,

\theta(0)=\pi/30

and

\dot{\theta}(0)=0

; (a)

\theta(t)

, (b)

\dot{\theta}(t)

, and (c)

\dot{\theta}

\theta

(phase plane).

The equation of motion of the pendulum is (as given)

\ddot{\theta}=-\frac{g}{\ell}\sin\theta.

To solve this second order differential equation numerically, we need to first convert it into a set of two first order equations. Let $\omega=\dot{\theta}$ . Then, we can write

Filename:bikefork-ang-accel

Figure 15.28: Comparison of

\theta(t)

obtained from three different initial launch angles,

\theta(0)=6^{\circ},\ 18^{\circ}

and

30^{\circ}

	$\displaystyle\dot{\theta}$	$\displaystyle=$	$\displaystyle\omega$
	$\displaystyle\dot{\omega}$	$\displaystyle=$	$\displaystyle-\frac{g}{\ell}\sin\theta.$

We are now ready to write a computer program to solve these equations numerically. We use the following pseudocode to accomplish the task.

    ODEs = {thetadot = omega,
            omegadot = -g/l*sin(theta) }
    ICs   = {theta(0) = pi/30, omega(0) = 0 }
    Set  g = 1, l = g/(4*pi^2)
    Solve ODEs with ICs for t=0 to t=4
    plot theta vs t, and omega vs t; plot omega vs theta

1.

Small amplitude oscillations: The solution obtained with $\theta(0)=6^{\circ}=\pi/30$ , and $\dot{\theta}(0)=0$ is shown in fig. 15.27. The plots of $\theta(t)$ and $\dot{\theta}(t)$ clearly show the initial conditions at $t=0$ . From the figure, we see that the motion is sinusoidal and the time period of oscillation is 1 second, as expected.
2.

Deviation from linear equation solution: The new initial conditions involve larger initial angles ( $\theta(0)=18^{\circ}$ and $30^{\circ}$ ). That is the only difference. We use the same program as used before and get the solutions with the new initial conditions. We plot $\theta(t)$ against $t$ for all the three solutions on the same graph. The resulting plot is shown in fig. 15.28.

Now what we observe from this plot is that the three solutions, starting with the three different initial conditions, do not have the same time period of oscillations. The difference is not clearly visible between $\theta(0)=6^{\circ}$ and $\theta(0)=18^{\circ}$ solutions but it is much clearer for $\theta=30^{\circ}$ (see the third peak, marked with $3T$ ). As the initial angle, $\theta(0)$ , increases, the period of oscillation seems to increase.

The dependence of time period (or frequency) of oscillations on the amplitude is the hallmark of nonlinear oscillators. In contrast, linear oscillators have a constant period of oscillation, irrespective of the amplitude of motion. For our pendulum, as long as the initial $\theta$ is so small that $\sin\theta\approx\theta$ , the equation of motion can be replaced by the linear equation, $\ddot{\theta}=-g/\ell\theta$ , and all solutions will have the same time period of oscillation. As $\theta(0)$ becomes larger, the approximation $\sin\theta\approx\theta$ breaks down, and the linear equation of motion is no longer valid.
3.

Large amplitude oscillations: We now run the program with large initial angles, $\theta(0)=\pi/2\,(90^{\circ})$ and $\theta(0)=\pi/1.02\ \ (\approx 176^{\circ}$ , i.e., close to the vertically upright position), and obtain the corresponding solutions. Plots of $\theta(t)$ and $\dot{\theta}(t)$ for three initial conditions, small $\theta$ ( $\pi/30$ ), moderately large $\theta$ ( $\pi/2$ ), and very large $\theta$ ( $\pi/1.02$ ) are shown in fig. 15.29. From the plots it is clear that not only the period of oscillation increases drastically with larger amplitudes, but also the qualitative nature of oscillations changes. For small amplitude (small initial $\theta$ ), oscillations are simple harmonic but for larger amplitudes (large initial $\theta$ ) oscillations are no more simple harmonic. This fact is more evident from the velocity plot, fig. 15.29(b). The phase plot, fig. 15.29(c), shows how the three solution trajectories (also called orbits) look in the phase space. All simple harmonic motions lead to circular orbits (you can show that by writing the solution for $\theta(t)$ and $\dot{\theta}(t)$ and then showing that $\theta^{2}+\dot{\theta}^{2}=$ constant) in this phase space. However, for large amplitude motion, the orbits become oblong and approach a rather strange looking trajectory, called the separatrix, as the amplitude of motion grows. This separatrix marks the boundary of all possible periodic motions of the pendulum. Outside this separatrix, solutions do exist but they correspond to whirling motion of the pendulum which is not periodic (because $\theta(t)$ keeps growing without bounds).

Filename:bikefork1-alt

Figure 15.29: Comparison of pendulum motions when it is released from rest at small angle ( $\theta(0)=\pi/30$ ), at horizontal position (( $\theta(0)=\pi/2$ ), and at almost vertically upright position ( $\theta(0)=\pi/1.02$ ); (a) $\theta$ vs $t$ , (b) $\dot{\theta}$ vs $t$ , and $\dot{\theta}$ vs $\theta$ (phase plane plot).
4.

Energy conservation: Let $\theta^{*}$ and $\dot{\theta}^{*}$ be the values of angular displacement and angular speed of the pendulum at some instant $t^{*}$ . Then, assuming $\theta=0$ to be the datum for potential energy, we can write the expressions for potential energy and kinetic energy as

$\displaystyle{E_{\mathrm{P}}}$ $\displaystyle=$ $\displaystyle mg\ell(1-\cos\theta^{*})$

$\displaystyle{E_{\mathrm{K}}}$ $\displaystyle=$ $\displaystyle\frac{1}{2}m\ell^{2}{\dot{\theta}^{*2}}.$

Therefore, the total energy at $t=t^{*}$ is,

$E_{\mathrm{T}}={E_{\mathrm{P}}}+{E_{\mathrm{K}}}=mg\ell(1-\cos\theta^{*})+% \frac{1}{2}m\ell^{2}{\dot{\theta}^{*2}}.$

From the numerical solutions obtained for the three initial conditions, we have values of $\theta$ and $\dot{\theta}$ at different time instants. Now, using the formulas for ${E_{\mathrm{P}}}$ , ${E_{\mathrm{K}}}$ and $E_{\mathrm{T}}$ , we compute the values of these quantities and plot them as shown in fig. 15.30. We see that for each initial condition, the potential and kinetic energies vary differently with time. However, the total energy remains constant at all times. This is expected as there is no dissipation in the system (not present in our mathematical model). A given initial condition determines the initial energy of the pendulum which must be preserved throughout the motion.

Filename:bikefork-alt

Figure 15.30: Plots of potential energy ${E_{\mathrm{P}}}$ , kinetic energy ${E_{\mathrm{K}}}$ , and total energy $E_{\mathrm{T}}$ during motion under three different initial conditions: (a) $\theta(0)=\pi/30,\,\dot{\theta}(0)=0$ ; (b) $\theta(0)=\pi/2,\,\dot{\theta}(0)=0$ ; and (c) $\theta(0)=\pi/1.02,\,\dot{\theta}(0)=0$ .

Answer:

Problems for 15.2 Dynamics of a particle in circular motion

Preparatory Problems

15.2.1 Force on a person standing on the equator. Find the magnitude of the total force acting on a $150\,{\mathrm{lbm}}$ person standing on the equator. The total force is the gravity force plus the force of the ground on the person (note that these two do not exactly cancel). Neglect the motion of the earth around the sun and of the sun around the solar system, etc. The radius of the earth is $3963\,\mathrm{mi}$ . Give your solution in both pounds ( $\,{\rm lbf}$ ) and Newtons ( N). Answer: $F=0.52\,{\rm lbf}=2.3\,{\mbox{N}}$

15.2.2 Consider a mass $m$ in circular motion. Let $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\sum F_{r}\hat{\boldsymbol{e% }}_{r}+\sum F_{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ . Using $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ , express $\sum F_{r}$ and $\sum F_{\theta}$ in terms of some or all of $\theta,\dot{\theta},\ddot{\theta},\,$ , $r$ , and $m$ .

15.2.3 Using $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ , find the expressions for $\sum F_{x}$ and $\sum F_{y}$ in terms of $\ddot{\theta},\,\dot{\theta},\,r$ , and $\theta$ . [Hint $\hat{\boldsymbol{e}}_{r}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}=-\sin\theta\hat{\boldsymbol{% \imath}}+\cos\theta\hat{\boldsymbol{\jmath}}$ ].

15.2.4 A bead of mass $m$ goes around a circular path of radius $R$ in the $x y$ -plane with angular acceleration $\ddot{\theta}=ct^{3}$ . The bead starts from rest at $\theta=0$ .

(a)

What is the angular momentum of the bead about the origin at $t=t_{1}$ ?
(b)

What is the rate of change of angular momentum about the origin at $t=t_{1}$ ?
(c)

What is the kinetic energy of the bead at $t=t_{1}$ ?
(d)

Does the kinetic energy increase, decrease, or remain constant with time? Why?

15.2.5 A $200$ gm particle goes in circles about a fixed center at a constant speed $v=1.5\,\mathrm{m}/\mathrm{s}$ . It takes $7.5\,{\rm s}$ to go around the circle once.

(a)

Find the angular speed of the particle.
(b)

Find the magnitude of acceleration of the particle.
(c)

Take center of the circle to be the origin of a $x y$ -coordinate system. Find the net force on the particle when it is at $\theta=30^{{}^{\circ}}$ from the $x$ -axis.

15.2.6 A race car cruises on a circular track at a constant speed of $120\,\mathrm{mph}$ . It goes around the track once in three minutes. Find the magnitude of the centripetal force on the car. What applies this force on the car? Does the driver have any control over this force?

15.2.7 A particle moves on a counter-clockwise, origin-centered circular path in the $x y$ -plane at a constant rate. The radius of the circle is $r$ , the mass of the particle is $m$ , and the particle completes one revolution in time $\tau$ .

(a)
Neatly draw the following things:
1. 1.
  
  The path of the particle.
2. 2.
  
  A dot on the path when the particle is at $\theta=0^{\circ}$ , $90^{\circ}$ , and $210^{\circ}$ , where $\theta$ is measured from the $x$ -axis (positive counter-clockwise).
3. 3.
  
  Arrows representing $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ , $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ at each of these points.
(b)

Calculate all of the quantities in part (3) above at the points defined in part (2), (represent vector quantities in terms of the cartesian base vectors $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ ). Answer: For $\theta=0^{\circ}$ ,

$\displaystyle\hat{\boldsymbol{e}}_{r}$ $\displaystyle=$ $\displaystyle\hat{\boldsymbol{\imath}}$

$\displaystyle\hat{\boldsymbol{e}}_{t}$ $\displaystyle=$ $\displaystyle\hat{\boldsymbol{\jmath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ $\displaystyle=$ $\displaystyle\frac{2\pi r}{\tau}\hat{\boldsymbol{\jmath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ $\displaystyle=$ $\displaystyle-\frac{4\pi^{2}r}{\tau^{2}}\hat{\boldsymbol{\imath}},$

for $\theta=90^{\circ}$ ,

$\displaystyle\hat{\boldsymbol{e}}_{r}$ $\displaystyle=$ $\displaystyle\hat{\boldsymbol{\jmath}}$

$\displaystyle\hat{\boldsymbol{e}}_{t}$ $\displaystyle=$ $\displaystyle-\hat{\boldsymbol{\imath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ $\displaystyle=$ $\displaystyle-\frac{2\pi r}{\tau}\hat{\boldsymbol{\imath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ $\displaystyle=$ $\displaystyle-\frac{4\pi^{2}r}{\tau^{2}}\hat{\boldsymbol{\jmath}},$

and for $\theta=210^{\circ}$ ,

$\displaystyle\hat{\boldsymbol{e}}_{r}$ $\displaystyle=$ $\displaystyle-\frac{\sqrt{3}}{2}\hat{\boldsymbol{\imath}}-\frac{1}{2}\hat{% \boldsymbol{\jmath}}$

$\displaystyle\hat{\boldsymbol{e}}_{t}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\hat{\boldsymbol{\imath}}-\frac{\sqrt{3}}{2}\hat{% \boldsymbol{\jmath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$ $\displaystyle=$ $\displaystyle-\frac{\sqrt{3}\pi r}{\tau}\hat{\boldsymbol{\jmath}}+\frac{\pi r}% {\tau}\hat{\boldsymbol{\imath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ $\displaystyle=$ $\displaystyle\frac{2\sqrt{3}\pi^{2}r}{\tau^{2}}\hat{\boldsymbol{\imath}}+\frac% {2\pi^{2}r}{\tau^{2}}\hat{\boldsymbol{\jmath}}.$
(c)

If this motion was imposed by the tension in a string, what would that tension be? Answer: $T=\frac{4m\pi^{2}r}{\tau^{2}}$ .
(d)

Is radial tension enough to maintain this motion or is another force needed to keep the motion going (assuming no friction) ? Answer: Tension is enough.
(e)

Again, if this motion was imposed by the tension in a string, what is $F_{x}$ , the $x$ component of the force in the string, when $\theta=210^{\circ}$ ? Ignore gravity.

15.2.8 The velocity and acceleration of a $1\,\mathrm{kg}$ particle, undergoing constant rate circular motion, are known at some instant $t$ :

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=-10\,\mathrm{m}/\mathrm{s}(\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}),\quad{\accentset{% \rightharpoonup}{\boldsymbol{a}}}=2\,\mathrm{m}/\mathrm{s}^{2}(\hat{% \boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}).

(a)

Write the position of the particle at time $t$ using $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$ base vectors.
(b)

Find the net force on the particle at time $t$ .
(c)

At some later time $t^{\ast}$ , the net force on the particle is in the $-\hat{\boldsymbol{\jmath}}$ direction. Find the elapsed time $t-t^{\ast}$ .
(d)

After how much time does the force on the particle reverse its direction.

15.2.9 A particle of mass $3\,\mathrm{kg}$ moves in the $x y$ -plane so that its position is given by

{\accentset{\rightharpoonup}{\boldsymbol{r}}}(t)=4\,\mathrm{m}\left[\cos\left(% \frac{2\pi t}{s}\right)\hat{\boldsymbol{\imath}}+\sin\left(\frac{2\pi t}{s}% \right)\hat{\boldsymbol{\jmath}}\right]

with respect to point O, the origin of a fixed cartesian coordinate system.

(a)

What is the path of the particle? Show how you know what the path is.
(b)

What is the angular velocity of the particle? Is it constant? Show how you know if it is constant or not.
(c)

What is the velocity of the particle in polar coordinates?
(d)

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

What net force does it exert on its surroundings at $t=0\,\mathrm{s}$ ? Assume the $x$ and $y$ axes are fixed.
(f)

What is the angular momentum of the particle at $t=3\,\mathrm{s}$ about point $O$ ?

15.2.10 A comparison of constant and nonconstant rate circular motion. A 100 gm mass is going in circles of radius $R=20\,\mathrm{cm}$ at a constant rate $\dot{\theta}=3\,\mathrm{rad}/\mathrm{s}$ . Another identical mass is going in circles of the same radius but at a non-constant rate. The second mass is accelerating at $\ddot{\theta}=2\,\mathrm{rad}/\mathrm{s}^{2}$ and at position A, it happens to have the same angular speed as the first mass.

(a)

Find and draw the accelerations of the two masses (call them I and II) at position A.
(b)

Find $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}}$ for both masses at position A. Answer: $(\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/O}$}}})_{I}={\accentset{\rightharpoonup}{% \boldsymbol{0}}},\,(\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2% .0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt% \hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}})_{II}=0.0080\,{\rm N}% \mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}$ .
(c)

Find ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ for both masses at positions A and B. Do the changes in ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}$ between the two positions reflect (qualitatively) the results obtained in (b)? Answer: Position-A: $({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}})_{I}=0.012\,{\rm N}\mathord{\cdot}{\rm m}% \cdot\,\mathrm{s}\hat{\boldsymbol{k}}$ , $({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}})_{II}=0.012\,{\rm N}\mathord{\cdot}{\rm m}% \cdot\,\mathrm{s}\hat{\boldsymbol{k}}$ , Position-B: $({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}})_{I}0.012\,{\rm N}\mathord{\cdot}{\rm m}% \cdot\,\mathrm{s}\hat{\boldsymbol{k}}$ , $({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}})_{II}=0.014\,{\rm N}\mathord{\cdot}{\rm m}% \cdot\,\mathrm{s}\hat{\boldsymbol{k}}$ .]
(d)

If the masses are pinned to the center O by massless rigid rods, is tension in the rods enough to keep the two motions going? Explain.

Filename:pfigure-blue-128-1

Figure 15.31:

More-Involved Problems

15.2.11 A small mass $m$ is connected to one end of a spring. The other end of the spring is fixed to the center of a circular track. The radius of the track is $R$ , the unstretched length of the spring is $\ell_{0}$ (with $\ell_{0}<R$ ), and the spring constant is $k$ .

(a)

With what speed should the mass be launched in the track so that it keeps going at a constant speed?
(b)

If the spring is replaced by another spring of same relaxed length but twice the stiffness, what will be the new required launch speed of the particle?

15.2.12 A bead of mass $m$ is attached to a spring of stiffness $k$ . The bead slides without friction in the tube shown. The tube is driven at a constant angular rate $\dot{\theta}_{0}$ about axis $AA^{\prime}$ by a motor (not pictured). There is no gravity. The unstretched spring length is $r_{0}$ . Find the radial position $r$ of the bead if it is stationary with respect to the rotating tube. Answer: $r=\frac{kr_{o}}{k-m\omega_{o}^{2}}$ .

Filename:tfigure8-syst-bods

Figure 15.32:

15.2.13 A particle of mass $m$ is restrained by a string to move with a constant angular speed $\omega$ around a circle of radius $R$ on a horizontal frictionless table. If the radius of the circle is reduced slowly to $r$ , by pulling the string with a slowly varying force $F$ through a hole in the table, what will the particle’s angular velocity be in the final circular motion? Is kinetic energy changed in moving from circular motion at $R$ to circular motion at $r$ ? Why or why not?

Filename:sfig8-7-2

Figure 15.33:

15.2.14 An ‘L’ shaped rigid, massless, and frictionless bar is made up of two uniform segments of length $\ell=0.4\,\mathrm{m}$ each. A collar of mass $m=0.5\,\mathrm{kg}$ , attached to a spring at one end, slides frictionlessly on one of the arms of the ‘L’. The spring is fixed to the elbow of the ‘L’ and has a spring constant $k=6\,{\rm N}/{\rm m}$ . The structure rotates clockwise at a constant rate $\omega=2\,\mathrm{rad}/\mathrm{s}$ . If the collar is steady at a distance $\frac{3}{4}\ell=0.3\,\mathrm{m}$ away from the elbow of the ‘L’, find the relaxed length of the spring, $\ell_{0}$ . Neglect gravity. Answer: $\ell_{0}=0.2\,\mathrm{m}$ .

Filename:sfig8-7-2a

Figure 15.34:

Forces in constant rate circular motion.

15.2.15 A massless rigid rod with length $\ell$ attached to a ball of mass $M$ spins at a constant angular rate $\omega$ which is maintained by a motor (not shown) at the hinge point. The rod can only withstand a tension of $T_{cr}$ before breaking. Find the maximum angular speed of the ball so that the rod does not break assuming

(a)

there is no gravity, and
(b)

there is gravity (neglect bending stresses).

Filename:sfig8-7-2again

Figure 15.35:

15.2.16 A $1\,\mathrm{m}$ long massless string has a particle of $10{\,{\rm grams}}$ mass at one end and is tied to a stationary point $O$ at the other end. The particle rotates counter-clockwise in circles on a frictionless horizontal plane. The rotation rate is $2\pi\,{{\rm rev}/{\rm sec}}$ . Assume an $x y$ -coordinate system in the plane with its origin at $O$ .

(a)

Make a clear sketch of the system.
(b)

What is the tension in the string (in Newtons)? Answer: $T=0.16\pi^{4}N$ .
(c)

What is the angular momentum of the mass about $O$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}=0.04\pi^{2}\,{\mathrm{kg}}\mathord{\cdot}{% \mathrm{m}}/{\mathrm{s}}\hat{\boldsymbol{k}}$
(d)

When the string makes a $45^{\circ}$ angle with the positive $x$ and $y$ axis on the plane, the string is quickly and cleanly cut. What is the position of the mass 1 sec later? Make a sketch of the particle’s trajectory. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=[\frac{\sqrt{2}}{2}-v\cos(\frac{% \pi t}{4})]\hat{\boldsymbol{\imath}}+[\frac{\sqrt{2}}{2}+v\sin(\frac{\pi t}{4}% )]\hat{\boldsymbol{\jmath}}$ .

15.2.17 A ball of mass $M$ fixed to an inextensible rod of length $\ell$ and negligible mass rotates about a frictionless hinge as shown in the figure. A motor (not shown) at the hinge point accelerates the mass-rod system from rest by applying a constant torque $M_{O}$ . The rod is initially lined up with the positive $x$ -axis. The rod can only withstand a tension of $T_{cr}$ before breaking. At what time will the rod break and after how many revolutions? Neglect bending stresses.

(a)

Neglect gravity.
(b)

Include gravity.

Filename:sfig8-7-2disks

Figure 15.36:

15.2.18 A particle of mass $m$ , tied to one end of a rod whose other end is fixed at point $O$ to a motor, moves in a circular path in the vertical plane at a constant rate. Gravity acts in the $-\hat{\boldsymbol{\jmath}}$ direction.

(a)

Find the difference between the maximum and minimum tension in the rod. Answer: $2mg$ .
(b)

Find the ratio $\frac{\Delta T}{T_{max}}$ where $\Delta T=T_{max}-T_{min}$ . A criterion for ignoring gravity might be if the variation in tension is less than $2\%$ of the maximum tension; i.e., when $\frac{\Delta T}{T_{max}}<0.02$ . For a given length $r$ of the rod, find the rotation rate $\omega$ for which this condition is met. Answer: $\omega=\sqrt{99g/r}$ .
(c)

For $\omega=300\,{\rm rpm}$ , what would be the length of the rod for the condition in part (b) to be satisfied? Answer: $r\approx 1\,\mathrm{m}$ $(r>0.98\,\mathrm{m})$ .

Filename:sfig8-4-4

Figure 15.37:

15.2.19 A massless rigid bar of length $L$ is hinged at the bottom. A force $F$ is applied at point A at the end of the bar. A mass $m$ is glued to the bar at point B, a distance $d$ from the hinge. There is no gravity. What is the acceleration of point A at the instant shown? Assume the angular velocity is initially zero.

Filename:sfig8-4-4a

Figure 15.38:

15.2.20 The mass $m$ is attached rigidly to the rotating disk by the light rod AB of length $\ell$ . Neglect gravity. Find $M_{A}$ (the moment on the rod $A B$ from its support point at $A$ ) in terms of $\dot{\theta}$ and $\ddot{\theta}$ . What is the sign of $M_{A}$ if $\dot{\theta}=0$ and $\ddot{\theta}>0$ ? What is the sign if $\ddot{\theta}=0$ and $\dot{\theta}>0$ ?

Filename:sfig8-4-4b

Figure 15.39:

Pendulum problems

15.2.21 Simple pendulum, comprehensive version. This problem covers many aspects of a simple pendulum. A point mass $M$ hangs on a massless string or rod of length $L$ . The gravitational force is $M g$ . The pendulum is in a vertical plane. At any time $t$ , the angle between the straight down line and the pendulum, measured counter clockwise, is $\theta(t)$ . Neglect air friction. When numbers are called for use $M=1\,\mathrm{kg}$ , $L=1\,\mathrm{m}$ and $g=10\,\mathrm{m}/\mathrm{s}^{2}$ .

(a)
Find the equations of motion. That is, assume that you know both $\theta$ and $\dot{\theta}$ , find $\ddot{\theta}$ . There are several ways to do this problem.Answer: $\ddot{\theta}=-(g/L)\sin\theta$ Find the equations using
1. (a)
  
  Linear momentum balance
2. (b)
  
  Angular momentum balance
3. (c)
  
  Conservation of energy
(b)

Tension. Assuming that you know $\theta$ and $\dot{\theta}$ , find the tension $T$ in the string.
(c)

Reaction components. Assuming you know $\theta$ and $\dot{\theta}$ , find the $x$ and $y$ components of the force that the hinge support causes on the pendulum. Clearly define the directions of positive $x$ and $y$ with a sketch.
(d)

Reduction to first order equations. The equation that you found in (a) is a nonlinear second order ordinary differential equation. It can be changed to a pair of first order equations by defining a new variable $\omega\equiv\dot{\theta}$ . Write the equation from (a) as a pair of first order equations. Answer: $\dot{\omega}=-(g/L)\sin\theta,\quad\dot{\theta}=\omega$
(e)

Numerical solution. Given the initial conditions $\theta(t=0)=\pi/2$ and $\omega(t=0)=\dot{\theta}(t=0)=0$ . Using numerical integration, find: $\theta(t),\ \dot{\theta}(t)\ \&\ T(t)$ . Make a single plot, or three vertically aligned plots, of these variables for one full oscillation of the pendulum.
(f)

Maximum tension. Using your numerical solutions, find the maximum value of the tension in the rod as the mass swings. Answer: $T_{max}=30N$
(g)

Plot the $x$ and $y$ reaction components as a function of time.
(h)

Period of oscillation. How long does it take to make one oscillation?
(i)

Other observations. Some questions:
1. (a)
  
  Does the solution to (f) depend on the length of the string?
2. (b)
  
  Is the solution to (f) exactly $30$ or just a number near $30$ ? If it is exact can you find the result analytically?
3. (c)
  
  Is the period found in (h) longer or shorter than the period found by solving the linear equation $\ddot{\theta}+(g/l)\theta=0$ , based on the (inappropriate-to-use in this case) small angle approximation $\sin\theta=\theta$ ? Explain intuitively why you expect the period to be longer or shorter?

15.2.22 Tension in a simple pendulum string. A simple pendulum of length $2\,\mathrm{m}$ with mass $3\,\mathrm{kg}$ is released from rest at an initial angle of $60^{\circ}$ from the vertically down position.

(a)

What is the tension in the string just after the pendulum is released?
(b)

What is the tension in the string when the pendulum has reached $30^{\circ}$ from the vertical?

15.2.23 Cartesian coordinates Find the nonlinear governing differential equation for a simple pendulum

\ddot{\theta}=-{g\over\ell}\sin\theta

using linear momentum in Cartesian coordinates and without using the polar coordinate formulas for velocity and acceleration. Of course you can use that $x=\ell\cos\theta$ and $y=\ell\sin\theta$ for $x$ pointing down.

Filename:sfig8-4-4c

Figure 15.40:

15.2.24 Tension in a rope-swing rope. Model a swinging person as a point mass. The swing starts from rest at an angle $\theta=90^{\circ}$ . When the rope passes through vertical the tension in the rope is higher (it is hard to hang on). A person wants to know ahead of time if she is strong enough to hold on. How hard does she have to hang on compared, say, to her own weight? You are to find the solution two ways. Use the same m, g, and L for both solutions.

(a)

Find $\ddot{\theta}$ as a function of $g,\,L,\theta$ , and $m$ . This equation is the governing differential equation. Write it as a system of first order equations. Solve them numerically. Once you know $\dot{\theta}$ at the time the rope is vertical you can use other mechanics relations to find the tension. If you like, you can plot the tension as a function of time as the mass falls.
(b)

Use conservation of energy to find $\dot{\theta}$ at $\theta=0$ . Then use other mechanics relations to find the tension. Answer: The maximum tension is 3 times the person’s weight.

Filename:pfigure-s94h13p2

Figure 15.41:

15.2.25 Pendulum. A pendulum with a negligible-mass rod and point mass $m$ is released from rest at the horizontal position $\theta=\pi/2$ .

(a)

Find the acceleration (a vector) of the mass just after it is released at $\theta=\pi/2$ in terms of $\ell,m,g$ and any base vectors you define clearly.
(b)

Find the acceleration (a vector) of the mass when the pendulum passes through the vertical at $\theta=0$ in terms of $\ell,m,g$ and any base vectors you define clearly
(c)

Find the string tension when the pendulum passes through the vertical at $\theta=0$ (in terms of $\ell,m\ {\rm and}\ g$ ).

Filename:pfigure-f93f5

Figure 15.42:

15.2.26 Write a computer program to solve the nonlinear pendulum equation, $\ddot{\theta}=-\frac{g}{\ell}\sin\theta$ , over a given time interval (0, $t$ ), and initial conditions $\theta(0)$ , and $\dot{\theta}(0)$ . The output should be a vector of time instants, $t_{i}$ , in the given time interval and the corresponding $\theta_{i}$ and $\dot{\theta}_{i}$ .

Now use your computer program to find the solution of

\ddot{\theta}=-\sin\theta,\qquad\theta(0)=\pi/4,\ \dot{\theta}(0)=0.

Compare the solution obtained with the analytical solution of the corresponding simple pendulum equation, $\ddot{\theta}=-\theta$ with the same initial conditions. In particular,

(a)

Find the difference in the time period of oscillations of the two systems.
(b)

Plot $\dot{\theta}$ obtained from the two solutions against time and comment on the differences.
(c)

Plot $\dot{\theta}$ against $\theta$ from the two solutions on the same plot and compare the two phase portraits. Comment on the differences.

15.2.27 Solve the nonlinear pendulum equation numerically taking 20 different initial angular positions between $\theta=0$ and $\theta=\pi$ , each time releasing the pendulum gently from rest. Find the time period of oscillation, $T$ , from each solution and plot it against the amplitude of motion, i.e., $\theta(0)$ .

(a)

How does the period of oscillation depend on the amplitude for small amplitudes?
(b)

What is the limiting value of the time period for large amplitudes, i.e., $\theta(0)\rightarrow\pi$ ?
(c)

How does $T$ depend on the amplitude over the entire range?

15.2.28 A pendulum of mass $m$ and length $\ell$ is released from rest at $\theta(0)=60^{\circ}$ . It executes oscillatory motion. If the pendulum were to be released from two different positions, $\theta(0)=45^{\circ}$ and $\theta(0)=0^{\circ}$ , with some corresponding initial angular speed such that the ensuing motion were exactly the same as that with $\theta(0)=60^{\circ}$ and $\dot{\theta}(0)=0$ , find the required initial angular speeds.

(a)

First, find the corresponding $\dot{\theta}(0)$ without any computer simulation.
(b)

Verify your answer by plotting computer generated solutions for the three different initial conditions.
(c)

What is the general relationship between $\theta(0)$ and $\dot{\theta}(0)$ that produces a predefined motion generated by, say, a given set of $\theta(0)=\theta_{0}$ and $\dot{\theta}(0)=\dot{\theta}_{0}$ .

15.2.29 Use a computer program to solve the nonlinear pendulum equation, $\ddot{\theta}+\lambda^{2}\sin\theta=0$ , where $\lambda^{2}=1.56/{\,\mathrm{s}}^{2}$ , with the following 11 initial conditions: $[\theta(0),\dot{\theta}(0)]=[1,\ 0]$ , [2, 0], [3, 0], [4, -1], [-4, 1], [4, -1.02], [-4, 1.02], [4, -1.1], [-4, 1.1], [4, 1.4], and [-4, 1.4], where $\theta$ is in radians and $\dot{\theta}$ in $\,\mathrm{rad}/\mathrm{s}$ . Obtain each solution over the time interval $t=0$ to $t=20\,\mathrm{s}$ .

(a)

Plot all solutions in the phase space (i.e., $\theta$ vs $\dot{\theta}$ ) in a single graph.
(b)

What does the extension of the plot beyond $\theta=\pm\pi$ mean?
(c)

Which initial conditions give solutions outside the separatrix? What do these solutions mean? Are these solutions periodic?
(d)

If you added a little bit of viscous damping to the pendulum motion, can you guess what will happen to the solutions inside the separatrix? [Hint: think about energy associated with these solutions.]

$\displaystyle\frac{\dot{\theta}}{\dot{\theta}_{0}}$	$\displaystyle=$	$\displaystyle e^{-kt}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t$	$\displaystyle=$	$\displaystyle\frac{1}{-k}\ln\left(\frac{\dot{\theta}}{\dot{\theta}_{0}}\right)$
	$\displaystyle=$	$\displaystyle\frac{1}{-0.1}\ln\left(\frac{1}{2}\right)=\frac{-0.693}{-0.1/\,% \mathrm{s}}=6.93\,\mathrm{s}.$

$\displaystyle\alpha$	$\displaystyle=$	$\displaystyle\frac{\dot{\theta}_{0}-\dot{\theta}_{\rm final}}{t}$
	$\displaystyle=$	$\displaystyle\frac{100\,{\rm rpm}-0\,{\rm rpm}}{5\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle\frac{100\,\mbox{rev}}{60\,\mathrm{s}}\cdot\frac{1}{5\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 0.33\frac{\,\mbox{rev}}{{\,\mathrm{s}}^{2}}.$

$\displaystyle\theta$	$\displaystyle=$	$\displaystyle\frac{100\,\mbox{rev}}{60\,\mathrm{s}}\cdot 5\,\mathrm{s}-\frac{1% }{2}\,0.33\frac{\,\mbox{rev}}{{\,\mathrm{s}}^{2}}\cdot 25{\,\mathrm{s}}^{2}$
	$\displaystyle=$	$\displaystyle 8.33\,\mbox{rev}-4.12\,\mbox{rev}$
	$\displaystyle=$	$\displaystyle 4.21\,\mbox{rev}.$

$\displaystyle\ddot{\theta}$	$\displaystyle=$	$\displaystyle\frac{d\dot{\theta}}{dt}$
$\displaystyle d\dot{\theta}$	$\displaystyle=$	$\displaystyle\ddot{\theta}\,dt$
$\displaystyle\int_{\dot{\theta}_{0}}^{\dot{\theta}(t)}d\dot{\theta}$	$\displaystyle=$	$\displaystyle\int_{0}^{t}\ddot{\theta}\,dt$
$\displaystyle\dot{\theta}(t)-\dot{\theta}_{0}$	$\displaystyle=$	$\displaystyle\ddot{\theta}(t_{t}-t_{0})$
$\displaystyle\dot{\theta}$	$\displaystyle=$	$\displaystyle\dot{\theta}_{0}+\ddot{\theta}(t-t_{0})$

	$\displaystyle{E_{\mathrm{K}}}_{1}+{E_{\mathrm{P}}}_{1}$	$\displaystyle=$	$\displaystyle{E_{\mathrm{K}}}_{2}+{E_{\mathrm{P}}}_{2}$
	$\displaystyle 0+mg\ell(1-\cos\theta_{\max})$	$\displaystyle=$	$\displaystyle\frac{1}{2}m(\ell\dot{\theta})^{2}+mg\ell(1-\cos\theta).$		(15.20)

	$\displaystyle{E_{\mathrm{P}}}$	$\displaystyle=$	$\displaystyle mg\ell(1-\cos\theta^{*})$
	$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}m\ell^{2}{\dot{\theta}^{*2}}.$

$\displaystyle\hat{\boldsymbol{e}}_{r}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\imath}}$
$\displaystyle\hat{\boldsymbol{e}}_{t}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle\frac{2\pi r}{\tau}\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle-\frac{4\pi^{2}r}{\tau^{2}}\hat{\boldsymbol{\imath}},$

$\displaystyle\hat{\boldsymbol{e}}_{r}$	$\displaystyle=$	$\displaystyle-\frac{\sqrt{3}}{2}\hat{\boldsymbol{\imath}}-\frac{1}{2}\hat{% \boldsymbol{\jmath}}$
$\displaystyle\hat{\boldsymbol{e}}_{t}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\hat{\boldsymbol{\imath}}-\frac{\sqrt{3}}{2}\hat{% \boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}$	$\displaystyle=$	$\displaystyle-\frac{\sqrt{3}\pi r}{\tau}\hat{\boldsymbol{\jmath}}+\frac{\pi r}% {\tau}\hat{\boldsymbol{\imath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}$	$\displaystyle=$	$\displaystyle\frac{2\sqrt{3}\pi^{2}r}{\tau^{2}}\hat{\boldsymbol{\imath}}+\frac% {2\pi^{2}r}{\tau^{2}}\hat{\boldsymbol{\jmath}}.$

Chapter 15 Circular motion of a particle

Download this chapter by section

15.1 Kinematics of a particle in planar circular motion

Circular motion

Polar coordinates R and θ and unit vectors 𝒆ˆR and 𝒆ˆθ

The velocity and acceleration of a point going in circles, using polar coordinates

Derivatives of 𝒆ˆR and of 𝒆ˆθ

Continuing the quest for velocity and acceleration

Alternate expressions for the velocity and acceleration formulas

Centripetal acceleration

Problems for 15.1 Kinematics of a particle in circular motion

Preparatory Problems

More-Involved Problems

15.2 Dynamics of a particle in circular motion

Centripetal and centrifugal forces

Non-constant rate circular motion

Linear momentum balance in polar coordinates

Angular momentum

Energy

The simple pendulum

Small angle approximation (linearization)

The inverted pendulum

More about pendula

Method one: linear momentum balance in cartesian coordinates

Method two: angular momentum balance

Method three: Conservation of energy

Problems for 15.2 Dynamics of a particle in circular motion

Preparatory Problems

More-Involved Problems

Pendulum problems

More circular motion problems

Polar coordinates $R$ and $\theta$ and unit vectors $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$

Derivatives of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!R}$ and of $\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$