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Physics - Rotation Question with Solution | TestHub

PhysicsRotationAngular momentumEasy2 minQB

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A merry-go-round is at rest, pivoted on a frictionless axis. A child of mass $m$ runs along the path tangential to the rim with speed $v$ and jumps on to merry-go-round. If $R$ is the radius of the merry-go-round and $I$ is the moment of inertia, then the angular velocity of the merry-go-round and the child is

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Answer:

Solution:

By law of conservation of angular momentum

$mvR = (I_{system}) ω$

$\Rightarrow mvR = (I + {mR}^{2}) ω$

$\Rightarrow ω = \frac{mvR}{I + {mR}^{2}}$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Angular momentum

⏱ 2mℹ️ Source: QB

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