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  4. A playground merry-go-round of radius R=2.00 m has a moment of inertia I=250 kg ⋅ m 2 and is rotating at 10.0 rev / min about a frictionless, vertical axle. Facing the axle, a 25.0- kg child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is

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Q. A playground merry-go-round of radius $R=2.00 \,m$ has a moment of inertia $I=250 \,kg \cdot m ^{2}$ and is rotating at $10.0 \,rev / min$ about a frictionless, vertical axle. Facing the axle, a $25.0- kg$ child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is

System of Particles and Rotational Motion

Solution:

From conservation of angular momentum,
$I_{i} \omega_{i} =I_{f} \omega_{f}\left(250 \,kg \cdot m ^{2}\right)(10.0 rev / min ) $
$=\left[250 \,kg \cdot m ^{2}+(25.0 \,kg )(2.0 \,m )^{2}\right] \omega_{2}$
$\omega_{2} =7.14 \,rev/min$