Question

Determine the tension in the left string immediately after the right string is cut

Answer 1

Tension of both string is same at this position and $T=\frac{Mg}{2}$

In the second the string connected to rod at $b$ is cut.Let the angular acceleration of rod about point $a$ is $\alpha$

Acceleration of point $b$ is $a_b=\alpha l_b$

Now torque about point $a$ is

$=I_a\alpha$

$=\frac{1}{3}m{l}^2\alpha$

Where

$I_a=$moment of inertia of rod about point $a$$=\frac{1}{3}M{l}^2$

Torque about point $a=$moment of weight about point $a$

$\Rightarrow \frac{1}{3}m{l}^2\alpha =\frac{mgl}{2}$

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

Acceleration of centre of mass of rod,

$a_c=\alpha .\frac{l}{2}=\frac{3g}{4}$

So,considering the linear motion of COM we get,

$mg-T^{\prime }=m{a}_c$

$\Rightarrow mg-T^{\prime }=m.\frac{3g}{4}$

$\Rightarrow T^{\prime }=\frac{mg}{4}$