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A merry-go-round, made of a ring-like platform of radius $R$ and mass $M$, is revolving with angular speed $\omega$. A person of mass $M$ is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round of afterwards is
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As there is external torque acting on the system, angular momentum should be conserved.
Hence $I \omega=$ constant
where, $I$ is moment of inertia of the system and $\omega$ is angular velocity of the system.
From Eq. (i),
$$
I_1 \omega_1=I_2 \omega_2
$$
where $\omega_1$ and $\omega_2$ are angular velocities before and after jumping.
$\left(\because I=m r^2\right)$ So, given that,
$$
\begin{aligned}
&m_1=2 M, m_2=M, \omega_1=\omega, \omega_2=? \\
&r_1=r_2=R m_1 r_1^2 \omega_1=m_2 r_2^2 \omega_2 \\
&2 M R^2 \omega=M R^2 \omega_2
\end{aligned}
$$
as mass reduced to half, hence, moment of inertia also reduced to half.
$$
\omega_2=2 \omega
$$
Hence $I \omega=$ constant
where, $I$ is moment of inertia of the system and $\omega$ is angular velocity of the system.
From Eq. (i),
$$
I_1 \omega_1=I_2 \omega_2
$$
where $\omega_1$ and $\omega_2$ are angular velocities before and after jumping.
$\left(\because I=m r^2\right)$ So, given that,
$$
\begin{aligned}
&m_1=2 M, m_2=M, \omega_1=\omega, \omega_2=? \\
&r_1=r_2=R m_1 r_1^2 \omega_1=m_2 r_2^2 \omega_2 \\
&2 M R^2 \omega=M R^2 \omega_2
\end{aligned}
$$
as mass reduced to half, hence, moment of inertia also reduced to half.
$$
\omega_2=2 \omega
$$
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