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A uniform disc of mass $100 \mathrm{~kg}$ and radius $2 \mathrm{~m}$ is rotating at $1 \mathrm{rad} / \mathrm{s}$ about a perpendicular axis passing through its centre. A boy of mass $60 \mathrm{~kg}$ standing at the centre of the disk suddenly jumps to a point which is $1 \mathrm{~m}$ from the centre of the disc. The final angular velocity of the boy (in $\mathrm{rad} / \mathrm{s}$ ) is
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The correct answer is:
0.77
Angular momentum is conserved as there is no external torque.
So, $I_1 \omega_1=I_2 2 \omega_2$
$$
\begin{aligned}
& \Rightarrow\left(\frac{1}{2} M_{\text {disc }} \times R_{\text {disc }}^2\right) \omega_1=\left(\frac{1}{2} M_{\text {disc }} R_{\text {disc }}^2+M_{\text {boy }} R_{\text {boy }}^2\right) \omega_2 \\
& \Rightarrow\left(\frac{1}{2} \times 100 \times 2^2\right) \times 1=\left(\frac{1}{2} \times 100 \times 2^2+60 \times 1^2\right) \times \omega_2 \\
& \Rightarrow \omega_2=\frac{200}{200+60}=0.77 \mathrm{rad}-\mathrm{s}^{-1}
\end{aligned}
$$
So, $I_1 \omega_1=I_2 2 \omega_2$
$$
\begin{aligned}
& \Rightarrow\left(\frac{1}{2} M_{\text {disc }} \times R_{\text {disc }}^2\right) \omega_1=\left(\frac{1}{2} M_{\text {disc }} R_{\text {disc }}^2+M_{\text {boy }} R_{\text {boy }}^2\right) \omega_2 \\
& \Rightarrow\left(\frac{1}{2} \times 100 \times 2^2\right) \times 1=\left(\frac{1}{2} \times 100 \times 2^2+60 \times 1^2\right) \times \omega_2 \\
& \Rightarrow \omega_2=\frac{200}{200+60}=0.77 \mathrm{rad}-\mathrm{s}^{-1}
\end{aligned}
$$
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