Chapter 8
The Giancoli onLine Tutor

Rotational Motion

I. Angular Variables and Kinematics

Definitions

In this chapter we deal with rotational motion. The true rotational variables are:


Study the movie at the left, which shows a rod rotating with a constant angular velocity. After viewing the movie use the buttons on the right to step through the movie one frame at a time until you understand the displayed calculations. Calculate the angular speed in radians/sec using the displayed data. Show that it is constant. What is its value? Check your answer
If you do not have the quicktime plugin you can view a GIF animation version of the movie here.



Rotational Relationships

A point on the end of the rod is moving along a circle of circumference equal to 2*p*R and its doing so at a constant speed v. The speed equals
2*p*R
T
where T is the time for one orbit. The angle (in radians) subtended by the point in one revolution is 2*p. The angular velocity,w, is consequently
2*p/T.
Comparing the two equations we find that

v=Rw.

Similar arguments would show that,

x=Rq
a=Ra
where "x" is the linear distance travelled by the point as it moves along the circumference and "a" is its linear acceleration, if any. Because of these relationships the rotational motion of a fixed body is exactly equivalent to the motion of a point along a line. Thus the problems here are solved exactly like the problems in Chapter 2 and the problem solving strategy is the same!

The only difference is in the symbols.
Instead of
  1. v=vo+at
  2. x=xo+vo*t+.5*a*t2
  3. v2=vo2+2*a*(x-xo)
we now have,
  1. w=wo+at
  2. q=qo+wo*t+.5*a*t2
  3. w2=wo2+2*a*(q-qo).
Instead of finding 4 elements
in the following table

xo
x
vo
v
a
t
we have to find 4 items in

qo
q
wo
w
a
t


There really is no difference; you work the problems just as you worked the problems in Chapter 2.

Interactive Example


A spinning wheel rotating at 300 rpm gradually slows to a stop in two minutes. What angle does it turn through in that time? Fill in the following table clicking outside each time to obtain your result.

qo
q
wo
w
a
t

When you have found the four needed values, it is time to use the equations. Follow the following strategy:
  1. If the variable you are looking for, is on the left side of an equation check to see if you know everything on the right side of the equation. If so use this equation, if not try the next step.
  2. Identify which equations have the sought after variable, and check to see if all the other variables in either equation are known. If so, use this equation (and algebra) to obtain your answer, if not, try the next step.
  3. Find an equation in which only one variable is unknown. Solve for this variable and enter it in the above table. Now repeat steps 1, 2 and 3.

Interactive Example

Listed below are the three equations available to you. Choose the one you think you should use!

  1. w=wo+at
  2. q=qo+wo*t+.5*a*t2
  3. w2=wo2+2*a*(q-qo)


There are 2p radians in a revolution, how many revolutions does the wheel turn through in coming to a stop. After calculating your result you can check your answer .

II. Angular Dynamics

Force (Torque). Mass (Rotational Inertia) and Acceleration (Angular)

The material covered in Chapter 4 (F=ma, etc.) is also here in a disguised form.

Torque, t, plays the roll of force. Force produces an acceleration along a line and torque is produced by a force acting through a lever arm and equals the product of the length of the lever ( the distance from the point of application of the force to the axis of rotation) and the component of the force perpendicular to the lever arm. See the illustrations below.



Rotational Inertia,I, (aka Moment of Inertia) plays the roll of mass. Mass is resistance to acceleration along a line I is resistance to angular acceleration about an axis.

I=Sm*R2,

where m is each element of mass of the body and R is its distance from the axis. Thus if all the mass in a body is concentrated at a fixed distance from the axis, such as, with a hoop, we have

I= SmR2=R2Sm= R2M= MR2

The Rotational Inertia for more complicated structures, such as, balls and disks, are much harder to work out and generally are simply looked up in the book. See Figure: 8-20 in Giancoli.

Work,Energy and Momentum

All of the formulae from Chapters 2 , 4, 6 and 7, directly translate to this chapter by replacing The "linear" quantity with the analogous rotational quantity. For example, in Chapter 4 we learned that

W=F*(x-xo),

so replace F by t and x by q. The result

W=t *(q-qo).

We arrive then at the following results.

KE=mv2/2
KE=Iw2/2
F= ma
t=Ia
p= mv
L=Iw
L is known as the angular momentum.


The relationships between the various quantities also hold. For Example,

WT= mv22/2- mv12/2
becomes
WT= Iw22/2-Iw12/2.

III. Rotational Problems

There are three basic type of problems encountered in this chapter:

Angular Momentum Problems

The first of these, as mentioned above, are worked just like those in Chapter 2. Conservation of Angular Momentum problems are easier than the analogous problems in Chapter 7. Just as what happened in that chapter when there was no external force, if there are no external torques momentum is conserved, in this case, angular momentum.

Usually problems involve a single rotating body whose, for one reason or another, Rotational Inertia changes. In that case the problem is simply worked by using

I2w2=I1w1,

where "1" refers to the values before the change and "after". The problem usually involves finding w2. Occasionally you will encounter an inelastic collision between rotating objects. For example when a record changer (remember those?) drops a record on a freely rotating record on a turntable. This type of problem is worked just like the inelastic collision problems in chapter 7, using

I1w1+I2w2=(I1+I2)w'.

We have not discussed the vector nature of these rotational quantities. At this level they are rarely important, but they exist none-the-less. The direction is always parallel to the axis of rotation and the actual direction is determined by the right-hand-rule. With the fingers of your right hand curled in the direction of rotation your thumb is in the direction of the corresponding vector. Perhaps a better way to understand this is to remember that it is in the direction a normal screw would advance if it is turned in the direction of rotation.

This becomes important when using the above equation w is a vector, which is why it is in bold. Therefore you must use "+/-" when you enter a number. Viewed from above an object rotating clockwise is positive; counter-clockwise is negative. If not otherwise stated assume the rotation is clockwise.


Interactive Activities


A 16kg, .5 meter diameter disk is freely rotating at 300 rpm and an initially nonrotating identical disk is dropped on it and sticks to it. To the nearest tenth of a radian/sec what is the angular velocity of the combination?

We will work this in steps:

  1. What is the rotational Inertia of the rotating disk before the other is dropped?

  2. What is the rotational Inertia of the combined rotating disk after the second one is dropped?

  3. What is the angular velocity of the rotating disk before the other is dropped?

  4. What is the angular momentum of the rotating disk before the other is dropped?

  5. What is the angular momentum of the combined rotating disk after the second one is dropped?

  6. What is the angular velocity in radians/sec of the combined rotating disk after the second one is dropped?

  7. What is the angular velocity in rpm of the combined rotating disk after the second one is dropped?


Rolling Problems

The other type of problem is one in which the object is rolling., such as, a bicycle wheel rolling along. In this case you use:
  1. v=vo+at
  2. x=xo+vo*t+.5*a*t2
  3. v2=vo2+2*a*(x-xo)


to describe the motion of the center of mass ( v is the velocity of the center of mass,etc) and
  1. w=wo+at
  2. q=qo+wo*t+.5*a*t2
  3. w2=wo2+2*a*(q-qo).
    to describe how the object is rotating about the center of mass as it rolls along.. The word 'rolling' is the key to solving this type of problem. When an regular object rolls (without slipping) through one complete turn, its axis, and hence its center of mass travels a distance equal to its circumference. This means that the velocity of its center of mass, v, is related to its rotational velocity, w, by the relationship:

    v=R*w.
    The simplest way to work this type of problem is to use Conservation of Mechanical Energy (ME). For a rolling body the Mechanical Energy is given by

    ME=
    (Mechanical Energy)    		 mv2/2 +Iw2/2 +mgh
    Kinetic Energy of Translation +Kinetic Energy of Rotation +Potential Energy
    You proceed as you did in Chapter 6, usually this means setting the ME at the beginning equal to the ME at the end. You have only "one" equation and usually you will only be given one piece of data.

    You might be asked how fast a cylinder is rolling at the bottom of a ramp, if it was released from a height of 10 meters . You are missing information on the mass of the cylinder and its radius. Also you don't know w or v (which you are to find). You also don't know I but that can be found from the appropriate table (Figure 8-20 in Giancoli).

    • Since the cylinder is rolling, you do know that v and w are related as above! So if you know one of them, you can find the other.
    • But the problem of the missing Mass and Radius remains. From Chapter 6 you know that when mass is not given, it will generally cancel out when the problem is actually solved. The same is true here.
    • The Radius will also not be a problem.
      • The rotational kinetic energy term contains both I and w.
      • The rotational inertia I will always be proportional to m*R2;
      • w2 is equal to v2/R2,
      • R will cancel out leaving you with three terms each containing the mass "m" which then, in turn, cancels out.
      • You are left with only v and the height.
    • The problem is solved.

    Interactive Activities


    A Bowling ball rolling at 5 m/s rolls up a ramp and comes to a stop at what height?
    As outlined above the mass and radius will cancel, but for this purpose we will assume a mass of 5 kg and a radius of 10 cm. Use 10 for g

    1. What is the rotational inertia of the bowling ball?

    2. What is the translational kinetic energy of the ball before it reaches the ramp?

    3. What is the angular velocity of the ball?

    4. What is the rotational kinetic energy of the ball before it reaches the ramp?

    5. What is the total kinetic energy of the ball before it reaches the ramp?

    6. What is the final energy of the ball when it stops at the top of the ramp?

    7. What is the height of the ramp?


    © 1997-1999 Carl Adler mailto:Carl@Image-ination.com