Question

A rod of length $1m$ and mass $0.5kg$ hinged at one end, is initially hanging vertical. The other end is now raised slowly until it makes an angle ${60}^o$ with the vertical. The required work is (use $g=10$ $m/s^2$).

• A. $\frac{5}{2}J$
• B. $\frac{5}{4}J$
• C. $\frac{17}{8}J$
• D. $\frac{5\sqrt{3}}{4}J$

Answer 1

The correct option is A $\frac{5}{2}J$

work required to be done is change in potential energy.

initial center of mass =$0.5m$ from top.

final center of mass =$0.5Cos60$

so work required is $mg\times 0.5-mg\times 0.25=2.5J$

so the best answer is option A.