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. $\frac{mvR}{m{R}^2+I}$
• B. $\frac{mvR}{I}$
• C. $\frac{m{R}^2+I}{mvR}$
• D. $\frac{I}{mvR}$

Answer 1

The correct option is A $\frac{mvR}{m{R}^2+I}$

Since there is no external torque acting on the system, angular momentum is conserved.
$L_i=L_f$
$\therefore mvR+0=(I+m{R}^2)\omega$ 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.
$\omega =\frac{mvR}{(I+m{R}^2)}$