Question
A wheel of moment of inertia $50 \mathrm{~kg}-\mathrm{m}^2$ about its own axis is revolving at a rate of 5 revolutions per second. What is its angular momentum?

Answer

Here, $I = 50 \mathrm{~kg}-\mathrm{m}^2$,$\omega=5\text{rps}=5\times2\pi\text{rad/s}$
$\therefore$ Angular momentum $\text{L}=\text{I}\omega=50\times10\pi\ 500\pi\text{J-s}$

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