A 40 kg child stands on the edge of a rotating merry-go-round with a radius of 2

Question

A 40 kg child stands on the edge of a rotating merry-go-round with a radius of 2m. The moment of inertia of the merry-go-round is 200 kg-m^2.The merry-go-round is rotating counterclockwise with an angular speed of 0.5 rad/sec. The child catches a ball of mass 2.0 kg. Thrown at her. The velocity of the ball just before it is caught is 15 m/sec. The ball makes an angle with 60 degrees with the tangent outer edge of the merry-go-round.

b. What is the energy loss during this process?

c. If the catch takes 0.4 seconds, find the average torque on the merry-go-round during the catch.

Explanation / Answer

To answer this question, use conservation of momentum:

The ball has momentum of mv = (2kg)(15 m/s) = 30 kg m/s

After the boy on the merry-go-round 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(merry-go-round + boy)ωi = I(total)ωf

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

I(merry-go-around) = 200 kg m^2

I(boy) = m(boy)r^2 = (40)(2)^2 = 160 kg m^2

I(ball) = m(ball)(rsinθ)^2 = (2) (2)^2 (sin 60°)^2 = 6 kg m^2

I(total) = (200) + (160) + (6) = 366 kg m^2

Now ω = (mv + I(merry-go-round + boy)ωi) /I(total) = (30 + 360 * 0.5)/(366) = 0.57 rad/sec

Energy gained = I(total)wf^2 - I(merry-go-round + boy)wi^2 = 118.91 - 90

= 18.91 J

(Energy loss = 0)

Now, Torque = d(Iw)/d(t) = 28.62/0.4

= 71.55 Nm

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