Question

A uniform rod of mass $m$, length $l$ rests on a smooth horizontal surface. Rod is given a sharp horizontal impulse $p$ perpendicular to the rod at a distance $l/4$ from the centre. The angular velocity of the rod will be

• A. $\frac{3p}{ml}$
• B. $\frac{p}{ml}$
• C. $\frac{p}{2ml}$
• D. $\frac{2p}{ml}$

Answer 1

The correct option is A $\frac{3p}{ml}$

Let $v$ and $w$ be the velocity and angular velocity of the rod finally.

Impulse= change in linear momentum

$P=mv$

$L_O=constant$

$0=mv\frac{l}{4}-I_{CM}w$

$mv\frac{l}{4}=\frac{1}{12}m{l}^{2}w$

$⟹w=\frac{3v}{l}=\frac{3P}{ml}$