Question

Immediately after the right string is cut, what is the linear acceleration of the free end of the rod ?

Answer 1

Torque about A

${\tau }_A=I\alpha$ , where I is moment of inertia about A , is $I=\frac{m{l}^2}{3}$

Center of mass of the rod acts its geometrical center i.e. $AC=l/2$

So $mg\times \frac{l}{2}=\frac{m{l}^2}{3}\alpha$

$\alpha =\frac{3g}{2l}$

Acceleration of free end B is $a_B=\alpha \times BA=\frac{3g}{2l}\times l$

$a_B=\frac{3g}{2}$