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A uniform circular disc of mass $400 \mathrm{~g}$ and radius $4.0 \mathrm{~cm}$ is rotated about one of its diameter at an angular speed of $10 \mathrm{rot} / \mathrm{s}$. The kinetic energy of the disc is
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Verified Answer
The correct answer is:
$3.2 \times 10^{-1} \mathrm{~J}$
Given, mass of disc, $M=400 \mathrm{~g}=0.4 \mathrm{~kg}$
Radius, $R=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$
Angular speed, $\omega=2 \pi \times 10$
$$
=20 \pi \mathrm{rad} / \mathrm{s}
$$
$\therefore$ Moment of inertia of disc about its diameter
$$
I_{d}=\frac{1}{4} M R^{2}
$$
$\therefore$ Kinetic energy of disc
$$
\begin{aligned}
&=\frac{1}{2} I_{d} \omega^{2} \\
&=\frac{1}{2} \times \frac{1}{4} M R^{2} \times \omega^{2}=\frac{1}{8} M R^{2} \omega^{2} \\
&=\frac{1}{8} \times 0.4 \times\left(4 \times 10^{-2}\right)^{2} \times(20 \pi)^{2} \\
&=3.2 \times 10^{-1} \mathrm{~J}
\end{aligned}
$$
Hence, no option is correct.
Radius, $R=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$
Angular speed, $\omega=2 \pi \times 10$
$$
=20 \pi \mathrm{rad} / \mathrm{s}
$$
$\therefore$ Moment of inertia of disc about its diameter
$$
I_{d}=\frac{1}{4} M R^{2}
$$
$\therefore$ Kinetic energy of disc
$$
\begin{aligned}
&=\frac{1}{2} I_{d} \omega^{2} \\
&=\frac{1}{2} \times \frac{1}{4} M R^{2} \times \omega^{2}=\frac{1}{8} M R^{2} \omega^{2} \\
&=\frac{1}{8} \times 0.4 \times\left(4 \times 10^{-2}\right)^{2} \times(20 \pi)^{2} \\
&=3.2 \times 10^{-1} \mathrm{~J}
\end{aligned}
$$
Hence, no option is correct.
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