Question

A cockroach of mass m lies on the rim of a uniform disk of mass 7.00m that can rotate freely about its center like a merry-go-round. Initially the cockroach and disk rotate together with an angular velocity of 0.230 rad/s. Then the cockroach walks half way to the center of the disk. (a) What then is the angular velocity of the cockroach-disk system? (b) What is the ratio K/K0 of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the kinetic energy? (Select all that apply.)

Answer 1

Mass of cockoach = m
Mass of disk = 7.0 m
w = 0.230 rad/s
Let the radius of disk be r.
Moment of inertia of disk, = 1/2 * (7m)*r^2
Initial Moment of inertia of cockoach, = m*r^2
Final Moment of inertia of cockoach, = m*(r/2)^2 = 1/4 * m*r^2

Using Conservation of angular momentum,
Initial Angular momentum = Final Anagular momentum
(1/2 * (7m)*r^2 + m*r^2) * wi = (1/2 * (7m)*r^2 + 1/4 * m*r^2) * wf
(1/2 * 7 + 1) * 0.230 = (1/2 *7 + 1/4 ) * wf
wf = 0.276 rad/s
Angular velocity of the cockroach-disk system, wf = 0.276 rad/s

(b)
Initial Kinetic Energy of the system, Ko = 1/2* Ii * wi^2
Final Kinetic Energy of the system, K = 1/2* If * wf^2

K/Ko = 1/2* If * wf^2 / 1/2* Ii * wi^2
K/Ko = ((1/2 *7 + 1/4 ) * 0.276^2 )/ ((1/2 * 7 + 1) * 0.230^2 )
K/Ko = 1.2

(c)
As cockroach moves towards the center of the disc, it works against centrifugal force and does positive work against the disc !!