A cockroach of mass m lies on the rim of a uniform disk of mass 6.00 m that can

Question

A cockroach of mass m lies on the rim of a uniform disk of mass 6.00 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 0.210 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?rad/s

(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?

Help is greatly appreciated!

Explanation / Answer

(a) Angular velocity v=r =v/r When the cockroach goes half way then r1=r/2 =2*v/r ==> v/r =/2 = 0.210/2 = 105 rad/s Hence its angular velocity reduces to half of it. (b) In this case the mass of the whole system is m = 6m The moment of inertia I=mR2 Initial KE=1/2*I*2 When the cockroach goes halfway through it then its radiusbecomes R=R/2.Then I=I/4 Final KE=1/2*I/4*2/4 Its ratio is k/k0=16/1 When the cockroach goes halfway through it then its radiusbecomes R=R/2.Then I=I/4 Final KE=1/2*I/4*2/4 Its ratio is k/k0=16/1 (c) cockroach does negative work on the disc centripetal force these are the factors of it.
these are the factors of it.

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