Question

A uniform rod of mass M and length l is held in position shown in the figure. Than find the potential energy of the rod

• A. $U=\frac{Mgcos\theta l}{2}$
• B. $U=\frac{Mgsin\theta l}{2}$
• C. $U=\frac{Mgtan\theta l}{2}$
• D. None of these

Answer 1

The correct option is B $U=\frac{Mgsin\theta l}{2}$

$\int dU={\int }_0^l\frac{M}{l}dx.g.x.sin\theta$
$\therefore U=\frac{Mgsin\theta l}{2}$
Note that centre of mass of the rod is at height $\left(\frac{l}{2}sin\theta \right)$ from the ground.