Question

A cockroach of mass m lies on the rim of a uniform disk of mass 5.00 m that can rotate freely about its center like a merry-go-round. Initially the cockroach and disk rotate together with an angular velocity of 0.460 rad/s. Then the cockroach walks halfway 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?

Answer 1

(a). Let the cockroach mass be m. Mass of the merry go round is, M = 5m Moment of inertia of merry go round is, I_m = 1/2MR^2 = 5m R^2/2 Moment of inertia of cockroach is, I_c = m R^2 Initial moment of inertia of cockroach is, I_1 = 5m R^2/2 + m R^2 = m R^2(5/2 + 1) = 7/2 m R^2 Initial angular velocity is, omega_1 = 0.460 rad/s When the cockroach moves half way to the center its moment of inertia is, I_c = m(R/2)^2 = m R^2/4 Moment of inertia of the system is, I_2 = 5m R^2/2 + m R^2/4 = m R^2(5/2 + 1/4) = 11/4 m R^2 \r\n According to the conservation of angular momentum is,I_1 omega_1 = I_2 omega_2 (7/2 mR^2)(0.460) = (11/4 m R^2)omega_2 omega_2 = (7/2 mR^2)(0.460)/(11/4 m R^2) = 0.585 rad/s Angular velocity of the cockroach-disk system is 0.585 rad/s Initial kinetic energy is, K_0 = 1/2 I_1 omega_1^2 Final kinetic energy is,