One fourth length of a uniform rod of length 2l and mass m is placed on a horizontal table and the rod is held horizontally. The rod is released from rest. The normal reaction on the rod as soon as the rod is released will be

The torque applied by gravity about the edge 

$\begin{array}{l}O=\tau =mg(3\ell /4)\\ \Rightarrow I_0\alpha =\frac{3}{4}mg ......\left(a\right)\\ \text{ where }{I}_0=I_g+m{\left(\frac{3\ell }{4}\right)}^2\\ =\frac{m{\ell }^2}{12}+\frac{9}{16}m{\ell }^2=\frac{m{\ell }^2}{4}\left(\frac{1}{3}+\frac{9}{4}\right)\end{array}$

$I_0=\frac{31m{\ell }^2}{48} ......\left(b\right)$

$\text{ (a) and (b) yield, }\alpha =\frac{3mg\ell /4}{31m{\ell }^2/48}$

$\Rightarrow \alpha =\frac{144g}{124\ell }=\frac{36g}{31\ell } ......\left(c\right)$

Newton's second law of motion:

$mg-N=ma .....\left(d\right)$

$Kinematics: a=\left(\frac{3\ell }{4}\right)\alpha =\frac{36g}{31}$

Using (d) and (e) we obtain

$N=mg-m\left(\frac{27g}{31}\right)\Rightarrow N=\frac{4mg}{31}$