A 30-kg child runs around the edge of playground merry-go-round. The merry-go-ro

Question

A 30-kg child runs around the edge of playground merry-go-round. The merry-go-round has mass of 200 kg and radius of 2.0 m. Consider it to be uniform solid disk for which = 1/2 mr^2. The child is observed to be running, relative to the Earth, at speed of 3.0 m/s while the merry-go-round rotates in the opposite direction making one revolution every 12 sec. If the child stops running, what is the angular speed of the child plus merry-go-round combination? How much work is done as the child stops running?

Explanation / Answer

initial angular momentum Li = Idisk*wi - mchild*v*R

wi = 1 rev/ 12 s = 2pi/12 rad/s = pi/6 rad/s

final angular momentum Lf = (Idisk + mchild*R^2)*wf

from momentum conservation

Lf = Li

(Idisk + mchild*R^2)*wf = Idisk*wi - mchild*v*R

(((1/2)*200*2^2) + (30*2^2))*wf = (1/2)*200*2^2*(pi/6) - (30*3*2)

wf = 0.057 rad/s

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part(b)

work = change in KE

W = Ki - Kf

Kf = (1/2)*(Idisk + mchild*R^2)*wf^2 = (1/2)*( (1/2)*200*2^2 + (30*2^2) )*0.057^2 = 0.845 J

Ki = (1/2)*Idisk*wi^2 + (1/2)*mchild*v^2

Ki = (1/2)*(1/2)*200*2^2*(pi/6)^2 + (1/2)*30*3^2 = 190 J

W = 190-0.845 = 189 J

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