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A body is rolling without slipping on a horizontal plane. If the rotational kinetic energy of the body $50 \%$ of its total kinetic energy, then the body is
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The correct answer is:
Hollow sphere
For without slipping
$\mathbf{v}=\mathbf{r} \omega$
Total energy, $\mathrm{E}=\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{~m} \mathrm{v}^2$
$$
=\frac{1}{2} \mathrm{I}^2+\frac{1}{2} \mathrm{mr}^2 \omega^2
$$
Rotational kinetic energy, $\mathrm{E}_{\mathrm{A}}=\frac{1}{2} \mathrm{I} \omega^2$
Given, $\mathrm{E}_{\mathrm{R}}=\frac{50 \mathrm{E}}{100}$
$$
\begin{aligned}
& \frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left[\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{r}^2\right] \\
& \frac{1}{2} \mathrm{I} \omega^2=\frac{1}{4} \mathrm{I} \omega^2+\frac{1}{4} \mathrm{~m} \omega^2 \mathrm{r}^2 \\
& \mathrm{I}=\frac{1}{2} \mathrm{I}+\frac{1}{2} \mathrm{mr^{2 }} \\
& \frac{\mathrm{I}}{2}=\frac{1}{2} \mathrm{mr} \\
& \Rightarrow \mathrm{I}=\mathrm{m} \mathrm{r}^2
\end{aligned}
$$
It's moment of inertia of thin circular ring.
$\mathbf{v}=\mathbf{r} \omega$
Total energy, $\mathrm{E}=\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{~m} \mathrm{v}^2$
$$
=\frac{1}{2} \mathrm{I}^2+\frac{1}{2} \mathrm{mr}^2 \omega^2
$$
Rotational kinetic energy, $\mathrm{E}_{\mathrm{A}}=\frac{1}{2} \mathrm{I} \omega^2$
Given, $\mathrm{E}_{\mathrm{R}}=\frac{50 \mathrm{E}}{100}$
$$
\begin{aligned}
& \frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left[\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{r}^2\right] \\
& \frac{1}{2} \mathrm{I} \omega^2=\frac{1}{4} \mathrm{I} \omega^2+\frac{1}{4} \mathrm{~m} \omega^2 \mathrm{r}^2 \\
& \mathrm{I}=\frac{1}{2} \mathrm{I}+\frac{1}{2} \mathrm{mr^{2 }} \\
& \frac{\mathrm{I}}{2}=\frac{1}{2} \mathrm{mr} \\
& \Rightarrow \mathrm{I}=\mathrm{m} \mathrm{r}^2
\end{aligned}
$$
It's moment of inertia of thin circular ring.
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