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A hoop of radius $2 \mathrm{~m}$ weighs $100 \mathrm{~kg}$. It rolls along a horizontal floor so that its centre of mass has a speed of $20 \mathrm{~cm} / \mathrm{s}$. How much work has to be done to stop it?
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Verified Answer
$M=100 \mathrm{~kg}, R=2 \mathrm{~m}, v=20 \mathrm{~cm} / \mathrm{s}=0.2 \mathrm{~m} / \mathrm{s}$
Work required to stop the hoop $=$ Total energy
$$
\begin{aligned}
&W=\frac{1}{2} M v^2+\frac{2}{5} I \omega^2 \quad\left[\because I=M R^2\right] \\
&=\frac{1}{2} M v^2+\frac{2}{5} M R^2 \omega^2=\frac{1}{2} M v^2+\frac{1}{2} M v^2 \quad[\because v=r \omega] \\
&=M v^2=100(0.2)^2=4 \mathrm{~J} .
\end{aligned}
$$
Work required to stop the hoop $=$ Total energy
$$
\begin{aligned}
&W=\frac{1}{2} M v^2+\frac{2}{5} I \omega^2 \quad\left[\because I=M R^2\right] \\
&=\frac{1}{2} M v^2+\frac{2}{5} M R^2 \omega^2=\frac{1}{2} M v^2+\frac{1}{2} M v^2 \quad[\because v=r \omega] \\
&=M v^2=100(0.2)^2=4 \mathrm{~J} .
\end{aligned}
$$
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