Question

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

• A. \pi ml\12p
• B. \pi ml\6p
• C. ml\6p
• D. None of these

Answer 1

The correct option is A \pi ml\12p

According to Impulse momentum theorem: Impulse provided$=$change in momentum.

$p=m{v}_c$

Also, $p\cdot \frac{l}{2}=I\omega =\frac{m{l}^2}{12}\cdot \omega$

$\Rightarrow p=\frac{ml\omega }{6}$

$\Rightarrow \omega =\frac{6p}{ml}$

Time period of rotation $T=\frac{2\pi }{\omega }$

Rod turns 1 right angle in $\frac{T}{4}$ time $\Rightarrow \frac{T}{4}=\frac{2\pi }{4\omega }=\frac{\pi }{2\omega }=\frac{\pi }{2\times 6p}ml=\frac{\pi ml}{12p}$