Question

A cockroach of mass m lies on the rim of a uniform disk of mass 4.00 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 0.230 rad/s. Then the cockroach walks half way to the center of the disk.

What then is the angular velocity of the cockroach-disk system? What is the ratio K/K₀ of the new kinetic energy of the system to its initial kinetic energy? What accounts for the change in the kinetic energy?

Answer 1

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A cockroach of mass m lies on the rim of a uniform disk of mass4.0m that can rotate freely about its center like a merry-go-round.Initially the cockroach and disk rotate together with an angularvelocity of .260 rad/s. Then the cockroach walks halfway to thecenter of the disk. (a) What then is the angular velocity of thecockroach-disk system? (b) What is the ratio K/Ko of the newkinetic energy of the system to its initial kinetic energy? (c)What accounts for the change in the kinetic energy?

cockroach mass = m kg

Merry go round mass M = (4)m kg; andit radius is R

moment of inertia
Im=0.5MR^2
=2mR^2

moment of inertial of cockroach

initial moment of inertia of the system is
I1=2mR^2+mR^2
=3mR^2

initial angular speed w1=0.260 rad/s

when the cockroach moves half way to center its moment ofinertia is

moment of inertia of the system then is
I2=2*mR^2+(mR^2)/4
=(9/4)mR^2

angular speed =

conservation of angular momentum gives

3mR^2*0.260
=(9/4)mR^2 *w2

w2=0.34666 rad/s

initial kinetic energy

final kinetic energy is K=

solve to get