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: (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. If multiple situations rank equally, use the same rank for each, then exclude the intermediate ranking (i.e. if objects A, B, and C must be ranked, and A and B must both be ranked first, the ranking would be A:1, B:1, C:3). If all situations rank equally, rank each as '1'.

Answer 1

Answer:

A = B = C > D

Explanation:

Work done to stop the disc is given as

$W = \Delta K$

$W = \frac{1}{2}I(\omega_f^2 - \omega_i^2)$

so we have

(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.

So we have

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

$W_d = \frac{1}{2}I(2^2 - 5^2)$

so we have

A = B = C > D