Work done by Torque Due to Force

Question

This is a homework problem I had that I couldn't quite figure out the reasoning behind.

A force is applied to the rim of a disk that can rotate like a merry-go-round, so as to change its angular velocity. Its initial and final angular velocities, respectively, for four situations are:

(a) -2 rad/s, 5 rad/s ($\Delta\omega = 7$)

(b) 2 rad/s, 5 rad/s ($\Delta\omega = 3$)

(c) -2 rad/s, -5 rad/s ($\Delta\omega = -3$)

(d) 2 rad/s, -5 rad/s ($\Delta\omega = -7$)

Rank the situations according to the work done by the torque due to the force, greatest first.

The equation for Work done by torque is $W=\tau\Delta\theta.$ Because of this, I would assume that the greater $\Delta\omega$ would result in a greater torque, because of the greater $\Delta\theta $ that accompanies it. Yet the answer lists work done by the torque as equal in all four cases. I can only assume this is because the acceleration is zero for all cases. But how can you tell?

Answer 1

The work done is the same in all 4 cases. The easy way of seeing this is to use the Work-Energy Theorem : work done equals change in energy. Here there is only kinetic energy, and since this depends on $\omega^2$ the minus signs make no difference, hence the change in KE is the same in each case.

However, in some cases there is a bigger change $\Delta\omega$. There is a bigger change in momentum, and the force is applied for a longer time while the object rotates through a larger angle. So isn't more work being done?

Exactly the same puzzle occurs in linear motion. It takes the same amount of work to change velocity from -2m/s to +5m/s as from +2m/s to +5m/s. An object with an initial velocity of -2m/s continues moving left retarded by the force until its velocity is brought to 0m/s. Then it moves the same distance back to the right while being accelerated, arriving back where it started but now with a velocity of +2m/s. The distance moved by the point of application of the force is greater than if the starting velocity was +2m/s. Surely that means more work is done?

No. The explanation is that work is force times displacement, which is a vector, not force times distance. The displacement is the same in both cases.

While the velocity was brought to 0m/s, the object did work on the agent which provided the force. The force did -ve work on the object. This work could have been stored, eg in a spring. The stored work was returned to the object to give it a velocity of +2m/s. It takes zero net work to change the direction of the velocity (think of work done during circular motion), only to change the magnitude of velocity.

Answer 2

There is an accelerations since the velocity changes...

But, remember that energy goes like the square of the velocity!

So $\Delta \omega$ in this case is not pretty useful, rather it is the square of the velocities. The energy is $\propto I\omega^2$. Since the work is the difference of energy you get it equal for all four cases.