Question

A playground merry-go-round has a radius R and a rotational inertia I. When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then:

Answer 1

Answer:

The angular velocity of the merry-go-round is $\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}$.

Explanation:

After a careful reading of the statement, we notice that merry-go-round-child system is a system with no external force exerting on it, such that the Principle of Angular Momentum Conservation can applied to analyze the system:

$m_{C}\cdot \left(\frac{v}{R} \right) = (I+m_{C}\cdot R^{2})\cdot \omega$ (Eq. 1)

Where:

$m_{C}$ - Mass of the child, measured in kilograms.

$v$ - Initial speed of the child, measured in meters per second.

$R$ - Radius of the playground merry-go-round, measured in meters.

$I$ - Moment of inertia, measured in kilograms per square meter.

$\omega$ - Angular velocity of the merry-go-round-child system, measured in radians per second.

Then, we clear the final angular speed:

$\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}$

The angular velocity of the merry-go-round is $\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}$.