In the figure here, a 25 kg child stands on the edge of a stationary merry-go-ro

Question

In the figure here, a 25 kg child stands on the edge of a stationary merry-go-round of radius 2.1 m. The rotational inertia of the merry-go-round about its rotation axis is 200 kg·m2. The child catches a ball of mass 0.5 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of magnitude 9 m/s, at angle ? = 14 ? with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught?

Explanation / Answer

Applying conservation of angular momentum

=> mball * vball * distance from centre = total moment of inertia * angular speed

where,   total moment of inertia = moment of inertia of merry go round + moment of inertia of ball and child

=>   0.5 * 9 * 2.1 * sin(90-14) = (200 + 25 *2.12 + 0.5 * 2.12) * w

=> 9.17 = (312.455) * w

=> w = 0.02934 rad/sec    --------------------> angular speed of the merry-go-round just after the ball is caught

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