Question

A playground merry-go-round has a moment of inertia of 600 kg
m^{2}. When the merry-go-round is at rest, a 20 kg boy runs
at 5.9 m/s along a line tangential to the rim and jumps on, landing
on the rim a distance of 3.0 m from the rotation axis of the
merry-go-round. The angular velocity of the merry-go-round is
then:

A.1.2 rad/s

B.0.38 rad/s

C.0.45 rad/s

D.0.56 rad/s

E.0.72 rad/s

Answer #1

Moment of inertia of the merry-go-round = I = 600
kg.m^{2}

Mass of the boy = m = 20 kg

Speed at which the boy is running before jumping on the merry-go-round = V = 5.9 m/s

Distance the boy jumps on the merry-go-round from the axis = R = 3 m

Angular velocity of the merry-go-round after the boy jumps on it =

By conservation of angular momentum,

= 0.45 rad/s

**Angular velocity of the merry-go-round after the boy
jumps on it = 0.45 rad/s**

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