Question

A kid of mass $50\text{kg}$ stands at the edge of a platform of radius $1\text{m}$ which can be freely rotated about its axis. The moment of inertia of the platform is $0.02{\text{kg-m}}^2$. The system is at rest. The kid throws a ball of mass $2\text{kg}$ horizontally to his friend (friend standing on the ground) in a direction tangential to the rim with a speed $10\text{m/s}$ as seen by his friend. The angular velocity with which the platform will start rotating in $\text{rad/s}$ is

Answer 1

Given,
Moment of inertia of platform (disc) $=I=0.02{\text{kg-m}}^2$
Radius of platform $R=1\text{m}$
Velocity of the ball $v=10\text{m/s}$
Mass of ball $m=2\text{kg}$
Mass of boy $M=50\text{kg}$


As we see, there is no external torque on the system. So the angular momentum of the system about the any axis will remain constant or conserved.
Hence, $\overrightarrow{L_i}=\overrightarrow{L_f}$
As we know, $L=I\omega$
MOI of the system (platform + boy) after throwing the ball is $=I+M{R}^2$
Hence, $L_i=L_f$
$\Rightarrow 0=mvR-(I+M{R}^2)\omega$
Because boy and platform will start moving in the opposite direction of the ball as shown in figure,
$\omega =\frac{mvR}{I+M{R}^2}=\frac{2\times 10\times 1}{0.02+50\times 1^2}$
$\omega =0.4\text{rad/s}$