91Ó°ÊÓ

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

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:

$\mathbf{L}\mathbf{=}\mathbf{mvr}\mathbf{.}\mathbf{²õ¾\pm ²Ôθ}$

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

According to law of conservation of angular momentum:

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

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

$\sin (90−\mathrm{Ï•})=\mathrm{³¦´Ç²õÏ•}$

${\mathrm{L}}_{\mathrm{i}}=\mathrm{mvr³¦´Ç²õÏ•}$

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

Hence,

$\begin{array}{c}19.17=274×\mathrm{Ó\neg }\\ \mathrm{Ó\neg }=0.07\text{rad}/\text{s}\end{array}$

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

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