drawn along the axis of rotation and having a length proportional to the amount of the angular velocity, and pointing in the direction that a person must look along the axis to see the body rotating in a clockwise direction. For example, if the rotating disc shown in figure 65 has an angular velocity 10, it will be represented by a vector 10 units long drawn in the direction of the arrow.

133. Change in Direction of Angular Velocity.-If the axis of rotation is changing in direction the angular velocity at one instant might be represented by the vector B and a short time later by C (Fig. 66). The change in angular velocity would then be represented by the vector D, for this combined with B

FIG. 65.

B

C

FIG. 66.

D-At

gives C according to the composition of vectors. If t is the time during which the change has taken place, then D= At where A is the angular acceleration. An angular acceleration of this character is found in the motion of a top. ($146.)

134. Rotation with Constant Acceleration. The equations for rotation with constant acceleration are exactly analogous to those for simple translation (§96), as may be seen thus:

Problems.

I. If a wheel revolves 1800 times per minute, what is its angular velocity; and if it is 6 in. in diameter what is the linear velocity of a point on its periphery?

2. A locomotive rounds a curve having a radius of 800 ft. at 15 miles per hour; what is its angular velocity?

3. A wheel is given a speed of 100 revolutions per min. in 2 min.; what is its angular acceleration in radian-seconds?

4. How many revolutions will a fly wheel make while its angular velocity is changing from 3 to 10, in 20 sec., if the acceleration is constant?

5. A body rotates about an axis with constant angular acceleration 8; how many turns will it have made in 10 sec. from the start?

6. How many revolutions will a body make starting from rest with angular acceleration 4 before it will be revolving at the rate of 20 turns per sec.?

7. What is the linear velocity of a point 1 ft. from the axis of a wheel making 2.5 turns per sec.? Also the velocity of a point 1.4 ft. from axis? What is the angular velocity of each?

8. Find angular velocity of a wheel in which a point 6 in. from the axis has a velocity of 4 ft. per sec.

KINETICS OF ROTATION ABOUT A FIXED Axis.

135. Angular Acceleration Caused by Torque.-Suppose the bar shown in figure 67 is acted

on by a force F at a distance

Axis

FIG. 67.

2

d from the axis; it is required to find how rapidly the speed f, of rotation of the bar about the axis will increase in consequence of the moment of force, or torque F d.

Imagine the bar divided into little masses m1, m2, M3,

etc., and suppose the effect of the force F is to cause an angular acceleration A in the rotation of the bar; that is, its angular velocity is increased at the rate of A radians per second. The linear acceleration of the mass m, at distance r, from the axis will then be r1A, and consequently the force acting on m1 must be m,r,A and may be represented by f1. This force f, is due to

F and is transmitted to m, by the rigidity of the bar. So also m, must be acted on by a force f2=m,r,A since it has the acceleration r2A. And similarly every one of the masses m1, m2, m3, etc., into which the bar is divided is acted on by the force needed to give it its acceleration, as indicated by the small arrows in the figure.

Now, if a force equal and opposite to f, is applied to m1, and a force equal and opposite to f2 is applied to m2, and so on, applying to each of the little masses a force just such as to counteract its acceleration, it is clear that there will be no acceleration and the bar will be in equilibrium. That is, a system of forces equal and opposite to f1, f2, etc., will just balance the turning moment of the force F about the axis O. Consequently the sum of the moments of f1, f2, etc., about O must be equal to the moment of F about that axis. Thus,

The quantity in the parenthesis, which depends only on the mass of the body and its distribution with reference to the given axis is called the moment of inertia of the body about that axis, and may be represented by the symbol I.

The torque or sum of the moments of whatever forces may be acting to rotate the body around the given axis may be represented by L, and we then have,

That is, the angular acceleration caused by a given torque is equal to the torque divided by the moment of inertia of the body about the given axis.

Notice the analogy to the formula F=ma, moment of force or torque corresponds to force, moment of inertia corresponds to mass, and angular acceleration corresponds to linear acceleration.

M'

The effect of torque in causing angular acceleration may be illustrated by the apparatus shown in the figure. A light bar carrying two masses M and M' is mounted on a horizontal axis perpendicular to the bar and is set in motion by a weight W hung from a cord wrapped around a drum on the axis. When the masses M and M' are in the position shown, the bar gains angular velocity slowly, for the farther the masses are from the axis, the greater the moment of inertia of the rotating system. When the masses are close to the axis the moment of inertia is smaller and the bar gains angular velocity very much more rapidly than before. The calculation of moments of inertia will be discussed in paragraphs 138 to 140. 136. Angular Momentum.-The formula of the last paragraph

M

FIG. 68.

The product of the moment of inertia by the angular velocity about an axis is known as the angular momentum of the rotating body about that axis, and equation (1) above, states that the change in the angular momentum of a body about any axis is equal to the moment of force or torque about that axis multiplied by the time during which it acts.

When the axis of torque is perpendicular to the axis of rotation of the body its only effect is to change the direction of the axis of rotation, but the amount of the angular momentum remains unchanged. This is illustrated by the top (§146).

137. Kinetic Energy of a Rotating Body.-When all parts of a body have the same velocity the kinetic energy of the body as we have already seen is Mv2 where M is the mass of the body and v its velocity. But in case of a rotating rigid body the

velocity of any part depends on its distance from the axis. In this case we may imagine the whole mass to be divided into small portions, and calculate the kinetic energy of each of these portions separately and then add them together to find the total energy of rotation.

The body represented in figure 69 is supposed to rotate about an axis perpendicular to the paper. Imagine the whole body cut up into little rods parallel to the axis. whose ends are seen as the reticulation in the diagram. Let the mass of one of these rods be m, its distance from the axis r, and its velocity due to the rotation of the body be v. Then its kinetic energy is mv2.

ω

AXIS

FIG. 69.

But if is the angular velocity of the body, or will be the linear velocity of a mass at a distance r from the axis.

Thus, wr=v and mv2=mw2r2.

Now, let m, represent the mass of another of the rods into which the body has been imagined divided and r1 its distance from the axis, then its kinetic energy is m,w2r12, and so the total kinetic energy of the body is

there being one term for each part into which the body is conceived to be divided. Or we may write

K. E. w2 (m r2+m, r12 + etc.),

=

since the angular velocity of every part of the body is the same. But the quantity in parenthesis is the moment of inertia I of the bar about the axis, therefore

K. E. = I w2.

Notice again the analogy between this expression and the formula for kinetic energy of translation M v2.

Moment of inertia corresponds to mass.

Angular velocity corresponds to linear velocity.

138. Moment of Inertia of a Rod.-The method of computing moments of inertia may be illustrated by the case of a straight uniform rod with the axis at one end. Let I be the length and M the mass of the rod. Conceive it to be divided into n equal parts, each part having a