Question

A uniform rod of mass m, length l is placed over a smooth horizontal surface along y-axis and is at rest as shown in the 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 center 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

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

As torque $=$ Change in angular momentum

$\therefore F△E=mv$ ( linear ) $⟶\left(1\right)$

and $\left(\frac{Fl}{2}\right)△t=\frac{m{l}^2}{12}\omega$ $\left(angular\right)⟶\left(2\right)$

Dividing $\left(1\right)$ and $\left(2\right)$

$\Rightarrow 2=\frac{12v}{wl}$

$\Rightarrow w=\frac{6v}{l}$

Using $S=ut$

Displacement of COM $=\frac{\pi }{2}=wt=\left(\frac{6v}{l}\right)t$

and $x=vt$

Dividing $:\frac{2x}{x}=\frac{l}{6}\to x=\frac{\pi l}{12}$

$\therefore x$ co-ordinate of A :

$\Rightarrow \frac{l}{2}+\frac{l\pi }{12}=\frac{l}{2}[1+\frac{\pi }{6}]$

Hence, the answer is $\frac{l}{2}[1+\frac{\pi }{6}].$