91Ó°ÊÓ

Mass of girl is M

Merry go round of radius is R

Rotational inertia is I

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

Formulae are as follow:

$$\begin{gathered} \mathrm{L}=\mathrm{IÓ\neg } \\ \mathrm{v}=\mathrm{RÓ\neg } \end{gathered}$$

Where,$\mathrm{Ó\neg }$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,

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

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

Therefore, according to the conservation of the angular momentum,

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

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

The linear speed of the girl is,

$\mathrm{v}=\mathrm{RÓ\neg }$

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

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.