A girl of mass Mstands on the rim of a frictionless merry-go-round of radius Rand rotational inertia Ithat is not moving. She throws a rock of mass mhorizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is v. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

Mass of girl is M

Merry go round of radius is R

Rotational inertia is I

Using the formula$\mathbf{L}\mathbf{=}\mathbf{I\omega }$and$\mathbf{v}\mathbf{=}\mathbf{r\omega }$, find the angular speed of the merry-go-round and the linear speed of the girl.

Formulae are as follow:

$$\begin{gathered} \mathrm{L}=\mathrm{I\omega } \\ \mathrm{v}=\mathrm{R\omega } \end{gathered}$$

Where,$\mathrm{\omega }$is angular frequency, L is angular momentum, Iis moment of inertia, v is velocity and R is radius.

Initially, the merry-go-round is not moving; therefore the initial angular momentum of the system is zero.

The final angular momentum the girl and merry-go-round is,

$\mathrm{L}=\left(\mathrm{I}+{\mathrm{MR}}^2\right)\mathrm{\omega }$

The final angular momentum associated with the thrown rock is$-\mathrm{mRv}$.

Therefore, according to the conservation of the angular momentum,

$$\begin{gathered} 0=\left(\mathrm{I}+{\mathrm{MR}}^2\right)\mathrm{\omega }-\mathrm{mRv} \\ \mathrm{\omega }=\frac{\mathrm{mRv}}{\left(\mathrm{I}+{\mathrm{MR}}^2\right)} \end{gathered}$$

Hence, the angular speed of the merry-go-round is $\mathrm{\omega }=\frac{\mathrm{mRv}}{\left(\mathrm{I}+{\mathrm{MR}}^2\right)}$.

The linear speed of the girl is,

$\mathrm{v}=\mathrm{R\omega }$

Hence,the linear speed of the girl is$\mathrm{v}=\mathrm{R\omega }=\frac{{\mathrm{mR}}^2\mathrm{v}}{\mathrm{I}+{\mathrm{MR}}^2}$.

Therefore, using the formula for the angular momentum and relationship between linear and angular velocity, the expression for linear and angular velocity can be found.