Solve the following :
Question:

A uniform rod of mass $300 \mathrm{~g}$ and length $50 \mathrm{~cm}$ rotates at a uniform angular speed of $2 \mathrm{rad} / \mathrm{s}$ about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of the rotation, (b) the speed of the centre of the rod and (c) its kinetic energy.

Solution:

(a) $L=l \omega$

$=\frac{m L^{2}}{3}(\omega)$

$=\frac{(0.3)(0.5)^{2}}{3}(2)$

$=\quad \frac{m^{2}}{s}$

$=0.05 \mathrm{~kg}-\frac{1}{3}$

(b) $\mathrm{v}=\mathrm{R \omega}$

$=(0.25)(2)$

$\mathrm{v}=0.5 \mathrm{~m} / \mathrm{s}$

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