Question

A straight rod AB of mass $M$ and length $L$ is placed on a frictionless horizontal surface. A horizontal force having constant magnitude $F$ and $a$ fixed direction starts acting at the end A. The rod is initially perpendicular to the force. The initial acceleration of end B is $xF/M$. Find $x$.

Answer 1

Net force is equal to mass times acceleration of centre of mass .

$\therefore a=\frac{F}{m}$

Writing torque equation about center ,

$F\left(\frac{l}{2}\right)=I\left(\alpha \right)$

$F\left(\frac{l}{2}\right)=\frac{m{l}^2}{12}\left(\alpha \right)$

$\alpha =\frac{6F}{ml}$

$\therefore$ acceleration of b is : $a_b=\frac{l}{2}\alpha -a$

$a_b=\frac{l}{2}\frac{6F}{lm}-\frac{F}{m}=\frac{2F}{m}$

$\therefore x=2$