Question

Suppose the platform of the previous problem is brought to rest with the ball in the hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed ν as seen by his friend. Find the angular velocity with which the platform will start rotating.

Answer 1

The angular velocity with which the platform will start rotating is $\omega=\frac{m v R}{\left(I+M R^{2}\right)}$

Explanation:

• System’s initial angular momentum is zero. This can be denoted by the formula, $L_1 = 0$
• With an angular velocity ω, let the platform starts rotating.
• Let’s consider mass of the ball as ‘m’ and the velocity of the ball as ‘v’.
• $\text {Total external torque} = 0$
• So, the system’s final angular momentum

                          $L_{2}= I \omega+M R^{2} \omega-m v R$

• And, initial angular momentum is equal to final angular momentum , where initial angular momentum is equal to zero.

                        $1 \omega+M R^{2} \omega-m v R=0$

                         then, $\omega=\frac{m v R}{\left(I+M R^{2}\right)}$

Therefore the angular velocity with which the platform will start rotating is $\omega=\frac{m v R}{\left(I+M R^{2}\right)}$ , where the initial angular momentum is equal to zero.