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, horizontally. The rod is released from rest. The normal reaction on the rod as soon as the rod is released will be,

Diagram

• A. $\frac{5mg}{27}$
• B. $\frac{4mg}{19}$
• C. $\frac{4mg}{27}$
• D. $\frac{mg}{9}$

Answer 1

The correct option is C $\frac{4mg}{27}$

Step 1 : Draw a labelled diagram



Step 2 : Find the normal reaction on the rod.

The torque applied by gravity about the edge $0=\tau =mg\left(3l/4\right)$

$I_{0}a=\frac{3}{4}mgl$ ...(i)

MOI about point O,

$I_o=I_g+m{\left(\frac{3l}{4}\right)}^2$

$=\frac{m{l}^2}{12}+\frac{9}{16}m{l}^2$

$=\frac{m{l}^2}{4}(\frac{1}{3}+\frac{9}{4})$

$I_o=\frac{31m{l}^2}{48}$ ....(ii)

(i) and (ii) yield ,

$a=\frac{3mgl/4}{31m{l}^2/48}$

$a=\frac{144g}{124l}=\frac{36g}{31l}$ ....(iii)

Newton's second law of motion:

$mg-N=m{a}_t$ .....(iv)

Translational acceleration,

$a_t=\left(\frac{3l}{4}\right)a=\frac{27g}{31}$

Using (iv) and (iii) we obtain,

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

$N=\frac{4mg}{27}$

Final Answer : (b)