A cockroach of mass m lies on the rim of a uniform disk of mass of 7 m that can

Question

A cockroach of mass m lies on the rim of a uniform disk of mass of 7m 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. Then the cockroach walks halfway to the center of the disk. (Use any variable or symbol stated above as necessary.)

(a) What then is the angular velocity of the cockroach–disk system?

32/290

(b) What is the ratio K/K0 of the new kinetic energy of the system to its initial kinetic energy?

1/4

32/290

Explanation / Answer

given that

mass of cockroch   Mc = m

mass of disk     Md = 7*m

angular velocity = w0

by using conservation of angular momentum

p = w*I,

where I is the moment of inertia of the system and w the angular velocity.

When the cockroach moves toward the center of the disc, the moment of inertia changes. to maintain constant angular momentum, the angular velocity must change.

Initial moment of inertia = moment of disc + moment of cockroach.

Ii = Md*r^2/2 + Mc*r^2

Ii = (7*m*r^2) / 2 + m*r^2

then the cockroach reduces its r value to r/2, so final moment is

If = (7*m*r^2)/2 + m*r^2/4

Then

pi = pf

Ii*w0 = If*w

w0*((7*m*r^2)/2 + m*r^2) = w * ((7*m*r^2)/2 + m*r^2/4)

w = w0 * ( (7*m*r^2) / 2 + m*r^2 ) / ( 7*m*r^2)/2 + m*r^2/4)

w = w0 * ( 4.5 ) / (3.75)

w/w0 = 4.5/3.75 ........(eq1)

The initial kinetic energy is

KEi = 0.5*Ii*w0^2

the final KE is

KEf = 0.5 * If * w^2.

The KE ratio is

KEf/KEi = ( If / Ii) * ( w / w0 )^2 .......(eq2)

If/Ii = 3.75/4.5.

put the value of w/w0 from equation 1 in to equation2 .

KEf /KEi= (3.75/4.5) * ((4.5/3.75)^2)

KEf / KEi = 4.5 /3.75

answers

part(a)   w = (4.5/3.75)*w0 = 1.2*w0

part(b)    KEf / KEi = 4.5 /3.75

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