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. Find the normal reaction on the rod as soon as the rod is released

• A. $\frac{3mg}{29}$
• B. $\frac{3mg}{31}$
• C. $\frac{4mg}{27}$
• D. $\frac{27mg}{31}$

Answer 1

The correct option is C

$\frac{4mg}{27}$

The torque applied by gravity about the edge $O=T=mg\left(\frac{31}{4}\right)$

$\Rightarrow I_o\propto =\frac{3}{4}mgl$ ...(i) where $I_g+m{\left(\frac{3l}{4}\right)}^2=\frac{m{l}^2}{12}+\frac{9}{16}m{l}^2=\frac{m{l}^2}{4}(\frac{1}{3}+\frac{9}{4})I_o=\frac{31m{l}^2}{48}$ ...(ii)

(i) And (ii) yield, $\propto =\frac{\frac{3mgl}{4}}{\frac{31m{l}^2}{48}}$ $\Rightarrow \propto =\frac{144g}{124l}=\frac{36g}{31l}$ ....(iii)

Newton's second law of motion : mg - N = ma

Kinematics: $\alpha =\left(\frac{3l}{4}\right)\propto =\frac{36g}{31}$
Using (iii) & (iv) we obtain, $N=mg-m\left(\frac{27g}{31}\right)\Rightarrow N=\frac{4mg}{27}$.