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Question

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


A
mvRmR2+I
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B
mvRI
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C
mR2+ImvR
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D
ImvR
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Solution

The correct option is A $$\displaystyle \frac{mvR}{mR^{2}+I}$$
Since there is no external torque acting on the system, angular momentum is conserved.
$$L_i=L_f$$
$$ \therefore mvR + 0 =( I+mR^2)\omega$$ where $$mvR$$ is the angular momentum of the child wrt the center of the merry-go-round and angular momentum of the merry-go-round is zero initially since it is at rest.
$$ \omega = \dfrac{mvR}{( I+mR^2)}$$

Physics

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