Question

A rod of mass $m$ and length $l$ (is made to stand at an angle of ${{60}^{o}}$ 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)

$\frac{mgl}{4}$

Answer 2

Centre of the mass of the rod lies at the midpoint and when the rod is displaced through an angle ${{60}^{o}}$ it rises to point B. From the figure $\sin {{30}^{o}}=\frac{BC}{AB}$ Or $\sin {{30}^{o}}=\frac{L}{l/2}$ Or $\frac{1}{2}=\frac{L}{l/2}$ Or $L=\frac{l}{4}$
Then potential energy of the rod in this position is $U=mgL$ $U=mg\frac{l}{4}$