Question

A rigid equilateral triangular frame made of three identical thin rods (mass = m & length = l) is free to rotate smoothly in 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 alternative(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: (A), (B), (D)

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

The individual weights each of mg act downward a points x, y and z as shown in the figure.
The moment of each force mg about point H is given as
$M_X=mg\frac{l}{2}$
$M_Y=mg\frac{l}{2}\sin {30}^0$
$M_Z=mg\frac{\sqrt{3}}{2}l\cos {30}^0$
All the three torques exerted by the weights are in clockwise direction about point H.
${\tau }_{total}=mg(\frac{l}{2}+\frac{l}{2}\sin {30}^0+\frac{\sqrt{3}}{2}l\cos {30}^0=mgl(\frac{1}{2}+\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2})=\frac{3}{2}mgl$
MI $I_{HA}=\frac{m{l}^2}{3}$
MI $I_{HB}=\frac{m{l}^2}{3}$
MI $I_{AB}=\frac{m{l}^2}{12}+m(\frac{\sqrt{3}}{2}l{)}^2=\frac{5m{l}^2}{6}$
$I_{total}=I_{HA}+I_{HB}+I_{AB}=\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+\frac{5m{l}^2}{6}=\frac{3}{2}m{l}^2$
Equating the total torque ${\tau }_{total}$ to $I_{total}\omega$
$\frac{3}{2}mgl=\frac{3}{2}m{l}^2\omega$
$\Rightarrow \omega =\frac{g}{l}$
Now for the Hinge Force we have
$N_1=3mg\cos {60}^{\circ }=\frac{3}{2}mg$
$3mg\sin {60}^{\circ }-N_2=3m\times \frac{g}{l}\times \frac{l}{\sqrt{3}}\Rightarrow N_2$$=\frac{\sqrt{3}}{2}mg$
Hinge Force $=\sqrt{N_1^2+N_2^2}=\sqrt{3}mg$