A 30.3 kg child stands on the edge of a stationary merry-go-round of mass 195. k

Question

A 30.3 kg child stands on the edge of a stationary merry-go-round of mass 195. kg and radius 2.83 m. The rotational inertia of the merry-go-round about its axis of rotation is 315. kg*m2. The child catches a ball of mass 1.47 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of 12.0 m/s that makes an angle of 37.0° with a line tangent to the outer edge of the merry-go-round, as shown in the overhead view of figure 12-44. What is the angular speed of the merry-go-round just after that ball is caught? rad/s

Explanation / Answer

To answer this question, use conservation of momentum:

The ball has momentum of mv = (1.47kg)(12 m/s) = 17.64 kg m/s

After the boy on the merry-go-around catches the ball, that momentum is transfered to the moments of inertia of the merry-go-around, plus the boy and the ball:

mv = I(total)?

where I(total) = I(merry-go-around) + I(boy) + I(ball)

I(merry-go-around) = 315 kg m^2

I(boy) = m(boy)r^2 = (30.3)(2.83)^2 = 242.66 kg m^2

I(ball) = m(ball)(rsin?)^2 = (1.47)(2.83)^2(sin 37

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