Question

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

• A. $mgl$
• B. $\frac{mgl}{2}$
• C. $\frac{mgl}{3}$
• D. $\frac{mgl}{4}$

Answer 1

Answer: (D)

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

$=mgh\times m\times g\times \frac{l}{2}\sin 30^{\circ }$
$=m\times g\times \frac{l}{2}\times \frac{1}{2}=\frac{mgl}{4}$