A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely wit

Question

A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.744 rev/s. A 59.4-kg person running tangential to the rim of the merry-go-round at 3.99 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. (a) Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round? decrease increase stay the same (b) Calculate the initial and final kinetic energies for this system.

Explanation / Answer

Initial angular momemtum :

I*w + I'*w'

I = moment of inertia of the merry go roung

I' = moment of inertia of the man

It's a disk : I = M*R^2 / 2

155*(2.63)^2 / 2*0.641*(2pi/3600) + 59.4*(2.63)^2*(3.41/2.63) = Initial angular momentum :

Final angular momentum :

I*w' + I'*w' >>> angular velocity is the same :

155*(2.63)^2 / 2*w' + 59.4*(2.63)^2*w'

So :

(155*(2.63)^2 / 2)*0.744*(2pi) + 59.4*(2.63)^2*(3.99/2.63) = 155*(2.63)^2 / 2*w' + 59.4*(2.63)^2*w'

253.9+ 623.32578 = 946.92361*w'

w' = 0.9264 rad/s

initial kinetic energy

0.5*59.4*(3.99)^2 +(0.25*155*2.63^2*(4.6747)^2 = 6330 J ==>initiak KE

Final kinetic energy

((0.25*155*2.63^2 + 59.4*(3.99)^2 ) *(0.9264)^2 = 1041.6 J

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