Question

A uniform rod of mass m and length L is held horizontally by two vertical strings of negligible mass as shown in the figure, after the string is cut, then?

Answer 1

$2T=mg⟹T=\frac{mg}{2}$

Torque on the rod is

$T=mg\left(\frac{1}{2}\right)⟹I\alpha =\frac{mgl}{2}=\frac{1}{3}m{l}^2\alpha =\frac{mgl}{2}=\alpha =\frac{3g}{2l}$

The acceleration on center of mass falls down

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

$ma=F_{down}-F_{up}⟹m\frac{3g}{4}=mg-T^{\prime }⟹T^{\prime }=mg-\frac{3mg}{4}$

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