Question

A uniform rod of mass m and length $\ell$ is placed horizontally on a smooth horizontal surface. An impulse $P$ is applied at one end perpendicular to the length of rod. Find the velocity of centre of mass and angular velocity of rod just after the impulse is applied.

Answer 1

(Refer image.1)

Angular impulse =Impulse $\times$ perpendicular distance

change is momentum = P (here)

$P_i=0$ $P_f=mv$

$\alpha =0$ $v=\frac{P}{m}$ $\theta =\frac{\pi }{2}$,

$w=\frac{6P}{ml}$

$s=ut+\frac{1}{2}a{t}^2$

$\frac{\pi }{2}=\frac{6Pt}{ml}$

$t=\frac{\pi ml}{12P}$

$w=\frac{6P}{ml}$ by $Li=Lf$

( Refer image.2)

$L_i=L_f\Rightarrow P.\frac{l}{2}=Iw+mv$

$\frac{Pl}{2}=\frac{m{l}^2}{12}w$

Hence, $\Rightarrow w=\frac{6P}{ml}$