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
B
mvRI
C
mR2+ImvR
D
ImvR

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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