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

Question

A 30-kg child runs around the edge of a playground merry-go-round. The merry-go-round has a mass of 200 kg and a radius of 2.0 m. Consider it to be a uniform solid disk for which I = 1/2 mr^2. The child is observed to be running, relative to the Earth, at a 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

(A) momentum of child = m v r

= 30 x 3 x 2 = 180 kg m^2 /s

w = 1 x 2pi / 12 s = 0.524 rad/s

momentum of merry go round = I w

= (200 x 2^2 / 2) (0.524)

= 400 x 0.524 = 209.44 kg m^2 / s in the opposite direction of child

Applying momentum conservation,

180 - 209.44 = (400 + (30 x 2^2 / 2)) wf

wf = 0.064 rad/s Or 1 rev per 98 sec

(B) initial KE = Ki = 30 x 3^2 /2

= 135 J

Kf = (30) (2 x 0.064)^2 /2 = 0.25 J

work done = chANGE IN ke = - 134.75 J

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