Question

A uniform rod of mass $m$, length $l$ is placed at rest on a smooth horizontal surface along $y-$ axis as shown in figure. An impulsive force $F$ is applied for a very small time interval $\Delta t$ along $x$ direction at point $A$. What will be the $x-$ coordinate of the end $A$ of the rod when the rod becomes parallel to $x-$ axis for the first time? Take origin at the mid point of the rod.

• A. $\frac{\pi l}{12}$
• B. $\frac{l}{2}(1+\frac{\pi }{12})$
• C. $\frac{l}{2}(1-\frac{\pi }{6})$
• D. $\frac{l}{2}(1+\frac{\pi }{6})$

Answer 1

The correct option is D $\frac{l}{2}(1+\frac{\pi }{6})$

Let the velocity of centre of mass of the rod after receiving the impulse be $v$,
Then, $J=P_f-P_i=mv-0$
or $J=mv$
Initial angular momentum, $L_i=0$
Final angular momentum
$L_f=L_{\text{CM}}+m{v}_{\text{CM}}r$
Moment of linear momentum about $\text{COM}$ is zero i.e $m{v}_{\text{CM}}r=0$
$\Rightarrow L_f=L_{\text{CM}}=I_{\text{CM}}\omega$
$\therefore \text{Angular impulse}=\text{change in angular momentum}..\left(ii\right)$
$J\times \left(\frac{l}{2}\right)=I_{\text{CM}}\omega ...\left(ii\right)$
$\Rightarrow \frac{mvl}{2}=\frac{m{l}^2}{12}\times \omega$
Or, $\omega =\frac{6v}{l}$

When rod becomes parallel to $x-$ axis, it rotates through an angle $\frac{\pi }{2}$.
$\alpha =0$, since impulsive force acts for a very short interval and there is no other external torque on system.
$\Rightarrow \theta =\omega t$
$\frac{\pi }{2}=\frac{6v}{l}t$
$\therefore t=\frac{\pi l}{12v}$
Distance travelled by $\text{COM}$ of rod,
$S=v.t$
$S=v.\frac{\pi l}{12v}$
$\Rightarrow S=\frac{\pi l}{12}$


Initially $x_A=0$
$x-$ coordinate of end $A$ when rod becomes horizontal,
$x_A=0+(\frac{\pi l}{12}+\frac{l}{2})$
$\therefore x_A=\frac{l}{2}(1+\frac{\pi }{6})$