A solid sphere of radius $R$, rotating with an angular velocity $\omega$ about its diameter, suddenly stops rotating and $75 \%$ of its $\mathrm{KE}$ is converted into heat. If $\mathrm{c}$ is the specific heat capacity of the material in SI units, show that the temperature of $3 \mathrm{R}^2 \mathrm{CO}^2$ the sphere rises by $\frac{3 R^2 \omega^2}{20 c}$.
Q 96.11
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The $\mathrm{Ml}$ of a solid sphere about its diameters, $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2$ where $M$ is its mass.
The rotational KE of the sphere,
$
\begin{aligned}
E=\frac{1}{2} I \omega^2 & =\frac{1}{2}\left(\frac{2}{5} M R^2\right) \omega^2 \\
& =\frac{1}{5} M R^2 \omega^2
\end{aligned}
$
If $\Delta \theta$ is the rise in temperature,
$
\begin{aligned}
& \quad M c \Delta \theta=\frac{3}{4} E=\frac{3}{4}\left(\frac{1}{5} M R^2 \omega^2\right) \\
& \therefore \Delta \theta=\frac{3 R^2 \omega^2}{20 c}
\end{aligned}
$
The rotational KE of the sphere,
$
\begin{aligned}
E=\frac{1}{2} I \omega^2 & =\frac{1}{2}\left(\frac{2}{5} M R^2\right) \omega^2 \\
& =\frac{1}{5} M R^2 \omega^2
\end{aligned}
$
If $\Delta \theta$ is the rise in temperature,
$
\begin{aligned}
& \quad M c \Delta \theta=\frac{3}{4} E=\frac{3}{4}\left(\frac{1}{5} M R^2 \omega^2\right) \\
& \therefore \Delta \theta=\frac{3 R^2 \omega^2}{20 c}
\end{aligned}
$
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