Question

A thin uniform rod $AB$ of mass $m$ undergoes pure translation with an acceleration $a$ towards right when two antiparallel forces act on it as shown in the figure. If the distance between $F_1$ and $F_2$ is $b$, then the length of the rod is

• A. $\frac{2F_{2}b}{ma}$
• B. $\frac{2F_{1}b}{ma}$
• C. $\frac{4F_{2}b}{ma}$
• D. $\frac{4F_{1}b}{ma}$

Answer 1

The correct option is A $\frac{2F_{2}b}{ma}$

As the rod is in only translational motion, so, net torque about its centre of mass will be zero.


$\Rightarrow +\frac{F_{1}L}{2}-F_2\left(\frac{L}{2}-b\right)=0$
$[$Taking anticlockwise sense of rotation as $+]$
$\Rightarrow \frac{F_{1}L}{2}-\frac{F_{2}L}{2}+F_{2}b=0$
$\Rightarrow F_1-F_2+\frac{2F_{2}b}{L}=0$ ...........$\left(i\right)$
For translational motion,
$F_2-F_1=ma$ ...........$\left(ii\right)$
On adding $\left(i\right)$ and $\left(ii\right)$,
$\frac{2F_{2}b}{L}=ma$
$\Rightarrow L=\frac{2F_{2}b}{ma}$

$\begin{array}{c}\text{Why this question?}\\ \text{For pure translational motion, always remember}\\ \text{that net torque is zero about centre of mass.}\end{array}$