Question

Immediately after the right string is cut, what is the linear acceleration of the middle 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 rod acts at its geometric center i.e. AC=$l/2$

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

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

Acceleration at the mid point C is $a_C=\alpha \times CA=\frac{3g}{2l}\times \frac{l}{2}$

$a_C=\frac{3g}{4}$