Question

Consider a uniform rod $PQ$ of mass $M$ and length $L$ lying on a smooth horizontal surface. An impulse $J$ is applied to the end $Q$ in the horizontal direction as shown in the figure. The speed of a particle $R$ at a distance $\frac{L}{6}$ from the centre towards end $P$ of the rod after time $t=\frac{\pi ML}{12J}$ is

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

Answer 1

The correct option is A $\frac{\sqrt{2}J}{M}$


Let $V$ and $\omega$ be the linear and angular velocities of rod after impact.
Using Impulse - momentum theorem for translatory motion:
$MV=J$ or $V=\frac{J}{M}$ ....(1)

Using Impulse - momentum theorem for rotational motion:
$I\omega =\frac{JL}{2}\Rightarrow \frac{M{L}^2}{12}\omega =\frac{JL}{2}$
or $\omega =\frac{6J}{ML}$ ......(2)

After the given time t = $\frac{\pi ML}{12J}$, the rod will rotate an angle $\theta =\omega t$
$\theta =\frac{6J}{ML}\times \frac{\pi ML}{12J}=\frac{\pi }{2}$


Velocity of $R$ will be calculated considering linear and angular velocity.

Velocity due to translational motion will be equal to $V$ or $\frac{J}{M}$.

Velocity due to rotation, $V_{R^{{}^{\prime }}}=\frac{L}{6}\omega$
$\Rightarrow V_{R^{{}^{\prime }}}=\frac{L}{6}\times \frac{6J}{ML}=\frac{J}{M}=V$ (From equation (2) and (1))
So, the net speed of point $R$ is given as
$|\overrightarrow{V_R}|=\sqrt{V^2+V^2}=\sqrt{2}V=\sqrt{2}\frac{J}{M}$