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Q.
A rod of mass $m$ and length $l$ is made to stand at an angle of $60^{\circ}$ with the vertical. Potential energy of the rod in this position is
Work, Energy and Power
Solution:
For any uniform rod, the mass is supposed to be concentrated at its centre.
Height of the mass from ground is
$h=(l / 2) \sin 30^{\circ}$
Potential energy of the rod
$=m g h \times m \times g \times \frac{l}{2} \sin 30^{\circ} $
$=m \times g \times \frac{l}{2} \times \frac{1}{2}=\frac{m g l}{4}$
