Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

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From $\textbf{conservation of angular momentum }$ we know that : $\textbf{ the angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net torque around this point }$ .

\sum \tau = \dfrac{d L }{dt} =0

Or :

\overrightarrow L = \overrightarrow l_{1} + \overrightarrow l_{2} + \overrightarrow l_{3} + .............+ \overrightarrow l_{N} = \text{constant}

\begin{gather*} L_{i} = L_{f} \\ \sum I_{i} \omega_{i} = \sum I_{f} \omega_{f} \\ \end{gather*}

From $\textbf{the moment of inertia of the merry - go round }$ we know that :

I =\frac{1}{2} mr^2

From $\textbf{the moment of inertia of the boy }$ we know that :

m r^2

So,

I_{i} =(\frac{1}{2} m_{m} + m_{1} +m_{2} + m_{3} ) r^2 \quad , \quad I_{f} =( \frac{1}{2} m_{m} + m_{1}+ m_{3} ) r^2

And from $\textbf{the kinematics of the rotational motion }$ we know that :

v = r \omega

From $\textbf{the kinematics of rotational motion }$ we know that :

\left [ \frac{ 20 \ \text{rev}}{ \text{min}} \cdot \frac{ 2 \pi}{ \text{rad}} \cdot \frac{ \text{rad}}{ 60\ \text{s}} = 2.094 \mathrm{rad/s} \right ]

\begin{gather*} L_{f} = L_{i} \\ ( \frac{1}{2} m_{m} + m_{1}+ m_{3} ) r^2 \omega_{f} = (\frac{1}{2} m_{m} + m_{1} +m_{2} + m_{3} ) r^2 \omega_{i} \\ \omega_{f} = \dfrac{(\frac{1}{2} m_{m} + m_{1} +m_{2} + m_{3} ) r^2 \omega_{i} }{ ( \frac{1}{2} m_{m} + m_{1}+ m_{3} ) r^2 }\\ \end{gather*}

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Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

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