Question

A uniform rod AB of mass 'm' and length 'l' at rest on a smooth horizontal surface. An impulse P is applied to the end B. The time taken by the rod to turn through a right angle is:

• A. $2\pi \frac{p}{2m}$
• B. $2\pi \frac{ml}{p}$
• C. $\frac{\pi ml}{12p}$
• D. $\frac{\pi p}{ml}$

Answer 1

The correct option is C $\frac{\pi ml}{12p}$

$I_{cm}=\frac{1}{12}m{l}^{2}L=P.\frac{l}{2}\Rightarrow \frac{1}{12}m{l}^2\omega =P\frac{l}{2}\Rightarrow \omega =\frac{6P}{ml}\theta =\frac{\pi }{2}=\omega t\Rightarrow t=\frac{\pi }{2\omega }=\frac{\pi ml}{2.6P}=\frac{\pi ml}{12P}$