Question

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

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

Answer 1

Answer: (A)

By law of conservation of angular momentum
$\textit{mvR}=\left(\left(\textit{I}\right)_{\textit{system}}\right)\textit{ω}$
$\Rightarrow \textit{mvR}=\left(\textit{I} + \left(\textit{mR}\right)^{2}\right)\textit{ω}$
$\Rightarrow \textit{ω}=\frac{\textit{mvR}}{\textit{I} + \textit{mR}^{2}}$