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A thin circular disc of mass $12 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$ rotates with an angular velocity of $100 \mathrm{rad} / \mathrm{s}$. The rotational kinetic energy of the disc is
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$7.5 \mathrm{~kJ}$
Given, mass of a circular disc, $M=12 \mathrm{~kg}$,
radius, $R=0.5 \mathrm{~m}$,
angular velocity, $\omega=100 \mathrm{rad} / \mathrm{s}$
$\therefore$ Rotational kinetic energy of the disc,
$\begin{aligned} & K=\frac{1}{2} I \omega^2 \\ = & \frac{1}{2} \cdot \frac{1}{2} M R^2 \cdot \omega^2 \quad\left[\because I=\frac{1}{2} M R^2\right]\end{aligned}$
Putting the given values in the above relation,
$\begin{aligned} & =\frac{1}{4} \times 12 \times(0.5)^2 \times 100 \times 100 \\ & =7500 \mathrm{~J}=7.5 \times 10^3 5 \mathrm{~J}=7.5 \mathrm{~kJ}\end{aligned}$
Hence, the rotational kinetic energy of the disc is $7.5 \mathrm{~kJ}$.
radius, $R=0.5 \mathrm{~m}$,
angular velocity, $\omega=100 \mathrm{rad} / \mathrm{s}$
$\therefore$ Rotational kinetic energy of the disc,
$\begin{aligned} & K=\frac{1}{2} I \omega^2 \\ = & \frac{1}{2} \cdot \frac{1}{2} M R^2 \cdot \omega^2 \quad\left[\because I=\frac{1}{2} M R^2\right]\end{aligned}$
Putting the given values in the above relation,
$\begin{aligned} & =\frac{1}{4} \times 12 \times(0.5)^2 \times 100 \times 100 \\ & =7500 \mathrm{~J}=7.5 \times 10^3 5 \mathrm{~J}=7.5 \mathrm{~kJ}\end{aligned}$
Hence, the rotational kinetic energy of the disc is $7.5 \mathrm{~kJ}$.
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