Question

A rigid equilateral triangular frame made of three identical thin rods (mass = $m$ & length = $l$) is free to rotate smoothly in a vertical plane. Frame is hinged at one of its vertices $H$. Frame is released from rest from the position shown in the figure. Then select the correct atternative (s).

• A. Net initial torque about point $H$ is $\frac{3}{2}mgl$.
• B. Initial angular acceleration of the frame is $\frac{g}{l}$.
• C. Initial force of hinge on the frame is $3mg$.
• D. Initial force of hinge on the frame is $\sqrt{3}mg$

Answer 1

Answer: (B), (D)

Initial force of hinge on the frame is $\sqrt{3}mg$


Initially,
${\tau }_H=3mg\times \frac{l}{2}$

Moment of inertia of rod $ab$ and $ac$ about $H$= $\frac{m{l}^2}{3}$
Moment of inertia of rod $bc$ about $H$= $\frac{m{l}^2}{12}+m\frac{(\sqrt{3}l}{2}{)}^2$
(using parallel axis theorem)

Therefore, Moment of inertia of system anout $H$

$I_H=\frac{2m{l}^2}{3}+\frac{m{l}^2}{12}+m{\left(\frac{\sqrt{3}l}{2}\right)}^2$
$=\frac{3}{2}m{l}^2$

${\tau }_H=I_H\alpha$
$\Rightarrow \alpha =\frac{g}{l}$

FBD of frame.


$N_1=3mg\cos {60}^{\circ }=\frac{3}{2}mg$
$3mg\sin {60}^{\circ }-N_2=3m\times a$
$=3m\times \alpha \times \frac{l}{\sqrt{3}}$
$=3m\times \frac{g}{l}\times \frac{l}{\sqrt{3}}$
$\Rightarrow N_2=\frac{\sqrt{3}}{2}mg$
Hinge force = $\sqrt{N_1^2+N_2^2}=\sqrt{3}mg$