Question

A uniform rod of mass $m$ and length $l$ is placed over a smooth horizontal surface along the y-axis and is at rest as shown in figure an impulsive force $F$ is applied for a small time $\Delta t$ along x-direction at point A. The x-coordinate of end A of the rod when the rod becomes parallel to x-axis for the first time is :

[Initially, the coordinate of centre of mass of the rod is (0, 0)]

• 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

Answer: (D)

$\frac{l}{2}(1+\frac{\pi }{6})$

As torque $=$ change in angular momentum
$F\Delta t=mv$ (linear momentum) ......(i)
and $\left(F\frac{l}{2}\right)\Delta t=\frac{m{l}^2}{12}\omega$ (angular momentum).....(ii)
Dividing Eqs. (i) and (ii), we get
$2=\frac{12v}{\omega l}\Rightarrow \omega =\frac{6v}{l}$
Using $S=ut$,
Displacement of CM is $\frac{\pi }{2}=\omega t=\left(\frac{6v}{l}\right)t$
and $x=vt$
Dividing, we get
$\frac{2x}{\pi }=\frac{l}{6}\Rightarrow x=\frac{\pi l}{12}$
Coordinates of A will be $[\frac{\pi l}{12}+\frac{l}{2},0]$.