Question

A uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $J$ is applied to the end $B$ perpendicular to the rod in horizontal direction. Speed of particle $P$ at a distance $\frac{l}{6}$ from the centre towards $A$ of the rod after time $\text{t}=\frac{\pi \text{ml}}{12\text{J}}$ is

• A. $2\frac{\text{J}}{\text{m}}$
• B. $\frac{\text{J}}{\sqrt{2}\text{m}}$
• C. $\frac{\text{J}}{\text{m}}$
• D. $\sqrt{2}\frac{\text{J}}{\text{m}}$

Answer 1

Answer: (D)

Let v and ω be the linear and angular speeds of the rod
after applying an impulse J at B. Then from :
impulse = change in momentum
we have
$\text{mv}=\text{J or}\text{v}=\frac{\text{J}}{\text{m}}$ ...(1)
$\text{I}\omega =\text{J}\cdot \frac{\text{l}}{2}$
or $\frac{\text{m}{\text{l}}^2}{12}\omega =\text{J}\cdot \frac{\text{l}}{2}\text{or}\omega =\frac{6\text{J}}{\text{m}\text{l}}$ ...(2)
After the given time $\text{t}=\frac{\pi \text{m}\text{l}}{12\text{J}}\text{,}$ the rod will rotate an angle
$\theta =\omega \text{t}=\left(\frac{6\text{J}}{\text{m}\text{l}}\right)\left(\frac{\pi \text{m}\text{l}}{12\text{J}}\right)=\frac{\pi }{2}$
$\frac{\text{l}}{6}\text{.}\omega =\left(\frac{\text{l}}{6}\right)\left(\frac{6\text{J}}{\text{m}\text{l}}\right)=\frac{\text{J}}{\text{m}}=\text{v}$
∴ $\left|{\text{v}}_{\text{P}}^{\overrightarrow{}}\right|=\sqrt{2}\text{v}=\sqrt{2}\frac{\text{J}}{\text{m}}$