Question

A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is ν horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.

Answer 1

Explanation:

Taking the system {ball+boy+platform}

L=I × omega

MVR = {(M+m)R square + I}× omega

therefore angular speed= MVR/ (M+m)R square + I

hope its helpful !!!!❤️

Answer 2

Given that,

Mass of kid = M

Radius of platform = R

Moment of inertia = I

Mass of ball = m

Velocity of ball= v

If we take the total bodies as a system,

Net external torque is zero.

So, angular momentum is conserved.

We need to calculate the angular speed of the platform

Using formula of angular momentum

$L=I\omega$

Where, I = moment of inertia

$\omega$ = angular velocity

Put the value into the formula

$mvR=(I+(m+M)R^2)\omega$

$\omega=\dfrac{mvr}{I+(m+M)R^2}$

Hence, The angular speed of the platform is $\dfrac{mvr}{I+(m+M)R^2}$