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 the horizontal direction. Find the speed of particle P at a distance $\frac{l}{6}$ from the centre towards A of the rod after time t = $\frac{\pi ml}{12J}$.

Answer 1

Let $v$ and $w$ be the linear and angular velocities of the rod applying an impulse J. Then

Impulse, $J$ = change in the Linear momentum

$J=(m{v}_{cm}-0)⟹v_{cm}=\frac{J}{m}$...........(i)

Angular Impuls$=$Change in angular momentum

$I\omega =J\frac{l}{2}$ (about C) (as shown in fig(i))

$⟹\frac{m{l}^2}{12}\omega =\frac{Jl}{2}$ [as $I=\frac{m{l}^2}{12}$]

$⟹\omega =\frac{6J}{ml}................\left(ii\right)$

This is the angular velocity of the rod.

After the given time t, the rod will rotate through an angle

$\theta =\omega t=\frac{6J}{ml}\times \frac{\pi ml}{12J}=\frac{\pi }{2}$

From equation (2)

$\therefore v_p^{\prime }=\frac{l}{6}\omega =\frac{l}{6}\times \frac{6J}{ml}=\frac{J}{m}$

(this velocity is due to angular rotation of the rod and it will be at ${90}^{\circ }$ to the velocity of $v_{cm}$)

Now total velocity of Point P

$V_P=\sqrt{V_{cm}+v_p^{\prime }}=\sqrt{\frac{J}{m}+\frac{J}{m}}$

$V_p=\sqrt{2}\frac{J}{m}$