Question

A boy of mass $M$ stands on a platform of radius $R$ capable to rotate about its axis. The moment of inertia of the platform is $I$. The system is at rest. The friend of the boy throws a ball of mass $m$ with a velocity $v$ horizontally. The boy on the platform catches it. Find the angular velocity of the system in the process.

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

Answer 1

The correct option is D $\frac{mvR}{I+(M+m)R^2}$

The system according to the question can be represented as:


Here the net external torque is zero$\left(0\right)$ as the forces acting upon the boy are internal. Hence we can conserve the angular momentum $\left(\text{i.e}\right)$

Initial angular momentum $=$ Final angular momentum

Initial angular momentum $=mvR$,
Final angular momentum $=[I+(M+m\left)R^2\right]\omega$, As we know that $L=I\omega$, where $I$ is the moment of inertia about the centre.

$mvR=[I+(M+m\left)R^2\right]\omega$
or $\omega =\frac{mvR}{I+(M+m)R^2}$

Hence option D is the correct answer.