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An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm. The moment of inertia of the flywheel is
Detailed Solution
$\omega_i=60\times\frac{2\pi}{60}=2\pi$ rad/s and $\omega_f=360\times\frac{2\pi}{60}=12\pi$ rad/s. Energy spent $=\Delta KE=\frac{1}{2}I(\omega_f^2-\omega_i^2)$, so $484=\frac{1}{2}I[(12\pi)^2-(2\pi)^2]=70\pi^2 I$, giving $I=\frac{484}{70\pi^2}\approx0.7$ kg m$^2$.