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6.System of Particles and Rotational Motion
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A flywheel is making $\frac{3000}{\pi}$ revolutions per minute about its axis. If the moment of inertia of the flywheel about that axis is $400\, kgm^2$, its rotational kinetic energy is
A
$2 \times 10^6\, J$
B
$3 \times 10^3\, J$
C
$500\,\pi^2 \,J$
D
$12 \times 10^3\, J$
Solution
Revolutions per minute $(\mathrm{N})=\frac{3000}{\pi} \mathrm{rpm}$
so, frequency, $n=\frac{N}{60}=\frac{3000}{60 \times \pi}=\frac{50}{\pi} \mathrm{Hz}$
and $\quad \omega=2 \pi n=2 \pi \times \frac{50}{\pi}=100 \mathrm{rad} / \mathrm{sec}$
$\mathrm{I}=400 \mathrm{kg} \mathrm{m}^{2}$
so rotational kinetic energy $(K)$ is
$\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \times 400 \times(100)^{2}=2 \times 10^{6} \mathrm{Joule}$
Standard 11
Physics
