Question

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 the following.
(a) -2 rad/s, 5 rad/s;
(b) 2 rad/s, 5 rad/s;
(c) -2 rad/s, -5 rad/s; and
(d) 2 rad/s, -5 rad/s.

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

Answer 1

Answer:

The work done by the torque is the same in each case.

Explanation:

Work-Energy Theorem can be used to calculate the work done by the torque in each case.

$W = \Delta K\\W = K_2 - K_1 = \frac{1}{2}I\omega_2^2 - \frac{1}{2}I\omega_1^2$

Let's apply this theorem to each case:

$W_a = \frac{1}{2}I(5)^2 - \frac{1}{2}I(-2)^2 = \frac{1}{2}I(21)\\W_b = \frac{1}{2}I(5)^2 - \frac{1}{2}I(2)^2 = \frac{1}{2}I(21)\\W_c = \frac{1}{2}I(-5)^2 - \frac{1}{2}I(-2)^2 = \frac{1}{2}I(21)\\W_d = \frac{1}{2}I(-5)^2 - \frac{1}{2}I(2)^2 = \frac{1}{2}I(21)$

As can be seen from the results, in each case the work done by the torque is the same.