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A man stands on a rotating platform, with his arms stretched horizontally holding a $5 \mathrm{~kg}$ weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from $90 \mathrm{~cm}$ to $20 \mathrm{~cm}$. The M.I. of the man together with the platform may be taken to be constant and equal to $7.6 \mathrm{kgm}^2$. What is his new angular speed? (Neglect friction). Is K.E. conserved in the process? If not, from where does the change come about?
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Verified Answer
Let the initial moment of inertia $=I_1$ $=7.6+2 \times 5(0.9)^2$ $=15.7 \mathrm{~kg} \mathrm{~m}^2$
Final moment of inertia $=I_2=7.6+2 \times 5(0.2)^2=8 \mathrm{kgm}^2$ $\omega_1=30 \mathrm{rpm}$ and $\omega_2=$ ?
$\because \quad$ No external torque is acting is acting on the system
$$
\begin{aligned}
&\therefore \quad L=I \omega=\text { constant } \Rightarrow I_1 \omega_1=I_2 \omega_2 \Rightarrow \omega_2=\frac{I_1}{I_2} \omega_1 \\
&=\frac{15.7 \times 30}{8}=58.88 \mathrm{rpm} .
\end{aligned}
$$
Final moment of inertia $=I_2=7.6+2 \times 5(0.2)^2=8 \mathrm{kgm}^2$ $\omega_1=30 \mathrm{rpm}$ and $\omega_2=$ ?
$\because \quad$ No external torque is acting is acting on the system
$$
\begin{aligned}
&\therefore \quad L=I \omega=\text { constant } \Rightarrow I_1 \omega_1=I_2 \omega_2 \Rightarrow \omega_2=\frac{I_1}{I_2} \omega_1 \\
&=\frac{15.7 \times 30}{8}=58.88 \mathrm{rpm} .
\end{aligned}
$$
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