A cockroach of mass m lies on the rim of a uniform disk of mass 6.50 m that can

Question

A cockroach of mass m lies on the rim of a uniform disk of mass 6.50 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 0.200 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.)

friction
cockroach does negative work on the disc
gravity cockroach does positive work on the disc
centrifugal force
centripetal force



OK so I cramstered the solution for this from the textbook and I couldnt get the right answer. they also used 1/2mr^2 for the inertia of the cockroach instead mr^2 for initial and final which I dont understand. So please explain that as well. Thanks :)

Explanation / Answer

correcting my earlier mistake

(a)

assume the radius of the disk is R, the moment of inertia of the disk is

I = 1/2 (6.5m) R2

the angular mometum of the disk plus cockroach is (remember v0 = 0R)

I 0 + R * mv0 = (13/4)mR20 + mR2 0 = (17/4)mR2 0

assume the angular velocity is after cockroach walks half way

the angular momentum becomes

I + (R/2)mv = (13/4)mR2 + (1/4)mR2 = (14/4)mR2

from conservation of energy, we have

(17/4)mR2 0 = (14/4)mR2

= (17/14)0 = (17/14) *0.2 = 0.243 rad/s    

(b)

K0 = (1/2)I02 + (1/2)mv02 = (13/8)mR2 02 + (1/2)mR2 02 = (17/8)mR2 02

K = (1/2)I2 + (1/2)mv2 = (13/8)mR2 2 + (1/2)m(R/2)2 = (7/4)mR2 2

K/K0 = (14/17) (/0)2 = (14/17)(17/14)2 = 17/14 = 1.21

(c)cockroach does positive work on the disc

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