Question

A uniform steel rod of mass $m$ and length $\ell$ is pivoted at one end. If it is inclined with the horizontal at an angle $\theta$, find its potential energy.

• A. $\frac{1}{2}mg\ell \cos \theta$
• B. $\frac{1}{2}mg\ell (1-\sin \theta )$
• C. $mg\ell \cos \theta$
• D. $mg\ell \sin \theta$

Answer 1

The correct option is B $\frac{1}{2}mg\ell (1-\sin \theta )$

Given that,

Mass = $M$

Length =$l$

Angle = $\theta$

We know that,

The height of center of mass from the ground is $\frac{l}{2}$

So, the potential energy of rod

$P.E=\frac{-mgl}{2}$

Now, when the rod is displaced is through angle θ

The height of center of mass from the ground is $\frac{l}{2}\sin \theta$

So, the final potential energy is

$P.E_f=\frac{-mgl}{2}\sin \theta$

Now, the change in potential energy is

$=P.E_i-P.E_f$

$=\frac{mgl}{2}(1-\sin \theta )$

Hence, the potential energy is $\frac{mgl}{2}(1-\sin \theta )$