Set 8 Problem number 9

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

Problem

A ball rolling down a hill undergoes a uniform angular acceleration of 1.3 radians/second ^ 2.

• How long will it take for its angular velocity to increase from 2.9 radians/second to 3 radians/second?
• Through what angular displacement will the ball rotate during this time?

Solution

This situation is reasoned out in the same way as if we were measuring positions in meters and velocities in meters per second. In this case, though, we talk about angular positions in radians, angular velocities in radians / second and angular accelerations in rad/sec^2..

• The change in angular velocity is .09999 radians/second.
• At the rate of 1.3 radians/second/second, the change will require ( .09999 radians/second)/ ( 1.3 radians/second/second) = .07691 seconds.

• During this time, since the angular acceleration is uniform the average velocity of the object will be ( 2.9 radians/second  + 3 radians/second) / 2 = 2.95 radians/second.
• At this average rate the angular displacement during .07691 seconds will be .2268 radians.

Generalized Solution

If the initial and final velocities were v0 and vf, and the acceleration a, we would easily see that the time `dt is related to these quantities by the definition of acceleration:

• aAve = (vf - v0) / `dt.
• This relationship is easily rearranged to give us `dt = (vf - v0) / a.

In the present case, instead of v0, vf, `dt and a,   we have angular velocities `omega0 and `omegaf, and angular acceleration `alpha.

• The only difference is that whereas v0, vf and a were expressed in terms of meters and seconds, `omega and `alpha quantities are expressed in radians and seconds. However the concepts are practically identical.
• The angular acceleration of an object is change in angular velocity divided by the time required for the change.
• Thus we have `alphaAve = (`omegaf - `omega0) / `dt, which we rearrange to obtain `dt = (`omegaf - `omega0) / `alphaAve.