Question

In Figure, a $\mathbf{30}\mathbf{kg}\mathbf{}$ child stands on the edge of a stationary merry-go-round of radius$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$the rotational inertia of the merry-go-round about its rotation axis is$\mathbf{150}\mathbf{}\mathbf{kg}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}\mathbf{.}$the child catches a ball of mass$\mathbf{1.0}\mathbf{}\mathbf{kg}\mathbf{.}$thrown by a friend. Just before the ball is caught, it has a horizontal velocity$\overrightarrow{\mathbf{v}\mathbf{}}$of magnitude$\mathbf{12}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, at angle$\mathbf{\phi }\mathbf{=}\mathbf{}\mathbf{37}\mathbf{°}$with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught?

Short answer

The angular speed of merry-go-round is $\mathrm{\omega }=0.07\text{rad/s}$.

Step-by-step solution

Step 1: Given

$\begin{array}{c}{\mathrm{I}}_{\mathrm{m}}=150\text{kg}.{\text{m}}^2\\ \mathrm{r}=2\text{m}\\ \mathrm{\varphi }={37}^0\end{array}$

Determining the concept

Firstly, find the unknown speed using the law of conservation of angular momentum. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

$\mathbf{L}\mathbf{=}\mathbf{mvr}\mathbf{.}\mathbf{sin\theta }$

where,m is mass, v is velocity, L is angular momentum and r is radius.

Determining the angular speed of merry-go-round

According to law of conservation of angular momentum:

$\text{Initialangularmomentum}=\text{Finalangularmomentum}$

$\begin{array}{l}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ {\mathrm{L}}_{\mathrm{i}}=\mathrm{mvrsin}(90-\mathrm{\varphi })\end{array}$

$\sin (90-\mathrm{\varphi })=\mathrm{cos\varphi }$

${\mathrm{L}}_{\mathrm{i}}=\mathrm{mvrcos\varphi }$

$\begin{array}{l}{\mathrm{L}}_{\mathrm{i}}=1\times 12\times 2\times \cos \left({37}^0\right)=19.17{\text{kg.m}}^{\text{2}}\text{/s}\\ {\mathrm{L}}_{\mathrm{f}}=\mathrm{I\omega }\\ {\mathrm{L}}_{\mathrm{f}}=(150+30\times 2^2+1\times 2^2)\mathrm{\omega }\\ {\mathrm{L}}_{\mathrm{i}}=274\times \mathrm{\omega }\end{array}$

Hence,

$\begin{array}{c}19.17=274\times \mathrm{\omega }\\ \mathrm{\omega }=0.07\text{rad}/\text{s}\end{array}$

Hence,the angular speed of merry-go-round is $\mathrm{\omega }=0.07\text{rad/s}$.

Therefore, using the law conservation of angular momentum, the unknown angular speed can be calculated.