Question

One fourth length of a uniform rod of length $2l$ and mass $m$ is placed on a horizontal table and the rod is held horizontal. The rod is released from rest. If the normal reaction on the rod as soon as the rod is released is $\frac{4mg}{x}$, find the value of $x$.

Answer 1

The torque applied by gravity about the edge $O=\tau =mg\left(\frac{3l}{4}\right)$
$\Rightarrow I_0\alpha =\frac{3}{4}mg$ ....(i)
where $I_0$ = $I_g+m{\left(\frac{3l}{4}\right)}^2$
$=\frac{m{l}^2}{12}+\frac{9}{16}m{l}^2$
$=\frac{m{l}^2}{4}(\frac{1}{3}+\frac{9}{4})$
$I_0=\frac{31m{l}^2}{48}$ .....(ii)

(i) and (ii) yield,

$\alpha =\frac{3gml/4}{31m{l}^2/48}$
Newton's second law of motion,
$mg-N=ma$ - (iv)
Kinematics: $a=\left(\frac{3l}{4}\right)\alpha$ - (v)
Using (iv) and (v) we obtain
$N=mg-m\left(\frac{27g}{31}\right)$ $\Rightarrow N=\frac{4mg}{31}$