Question

A uniform rod of mass $m$ and length $l$ is hinged at its upper end. It is released from a horizontal position. When it becomes vertical, what force does it exert on the hinge ?

• A. $\frac{3}{2}mg$
• B. $2mg$
• C. $\frac{5}{2}mg$
• D. $mg$

Answer 1

Answer: (C)

$\frac{5}{2}mg$

By force balance we have
$N-mg=\frac{m{v}^2}{r}$
Also balancing the torque about the hinge we get

$\frac{1}{2}I{\omega }^2=mg\frac{l}{2}$
or

$\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2=mg\frac{l}{2}$
or

$\frac{m{v}^2}{l}=3mg$
or
$\frac{m{v}^2}{r}=\frac{3}{2}mg$

Substituting this in force balance equation we get

$N=\frac{3}{2}mg+mg=\frac{5}{2}mg$