Question

A meter scale is suspended freely from one of its ends. Its another end is given a horizontal velocity $v$ such that it completes one revolution in the vertical circle. The value of $v$ is:

• A. $\pi \sqrt{3}m/s$
• B. $\pi \sqrt{6}m/s$
• C. $\pi \sqrt{2}m/s$
• D. $\pi m/s$

Answer 1

The correct option is B $\pi \sqrt{6}m/s$

For one complete revolution of meter scale increase in $PE=MgL$, where $M$ is the mass of the meter scale. therefor
$MgL=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2$
therefor $\omega =\sqrt{\frac{6g}{L}}$. Now $v=L\omega =L\sqrt{\frac{6g}{L}}=\sqrt{6gL}=\sqrt{6g}$ since $L=1m$
and $\sqrt{g}=\sqrt{9.81}=3.132\approx \pi$
hence $v=\pi \sqrt{6}m/s$