Physics › Rotational Motion › Moment of inertia of rigid bodies
A solid sphere of $80 \mathrm{~kg}$ and radius $15 \mathrm{~m}$ moving in a space becomes a circular disc of…
A solid sphere of $80 \mathrm{~kg}$ and radius $15 \mathrm{~m}$ moving in a space becomes a circular disc of radius $20 \mathrm{~m}$ in $1 \mathrm{~h}$. The rate of change of moment of Inertia in this process is
$\frac{30}{9} \mathrm{~kg}^2 \mathrm{~m}^2 \mathrm{~s}^{-1}$
$\frac{25}{9} \mathrm{~kg}^{-\mathrm{m}^2 \mathrm{~s}^{-1}}$
$\frac{10}{9} \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$
$\frac{22}{9} \mathrm{~kg}^{-\mathrm{m}^2 \mathrm{~s}^{-1}}$ Solution
Given, mass of solid sphere $=80 \mathrm{~kg}$ radius of solid sphere, $R_s=15 \mathrm{~m}$
radius of circular disc, $R_c=20 \mathrm{~m}$ and time $=1$ hour $=60$ minute $=60 \times 60 \mathrm{sec}$
$\therefore$ Moment of inertia of solid sphere, $I_s=\frac{2}{5} M R^2$
$
\begin{aligned}
& =\frac{2}{5} \times 80 \times(15)^2 \\
& =7200 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
$
Similarly,
$
\text { moment of inertia of the disc, } \begin{aligned}
I_c & =\frac{1}{2} M R_c^2 \\
& =\frac{1}{2} \times 80 \times(20)^2 \\
& =16000 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
$
$
\begin{aligned}
\text { Rate of change of moment of Inertia } & =\frac{I_c-I_s}{t} \\
& =\frac{16000-7200}{60 \times 60} \\
& =\frac{22}{9} \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}
\end{aligned}
$
Asked in: BITSAT 2022
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