Two uniform rods of equal length but differen | vedclass

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(D) Let the length of each rod be $l$. The center of mass of each rod is at its midpoint,i.e.,at a distance $l/2$ from the pivot $O$.For the rod with

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$\sqrt{3}$

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(D) Let the length of each rod be $l$. The center of mass of each rod is at its midpoint,i.e.,at a distance $l/2$ from the pivot $O$.For the rod with mass $m$,the angle it makes with the vertical is $60^{\circ}$. The torque due to its weight is $	au_m = mg \cdot (l/2) \sin(60^{\circ})$.For the rod with mass $M$,the angle it makes with the vertical is $30^{\circ}$. The torque due to its weight is $	au_M = Mg \cdot (l/2) \sin(30^{\circ})$.In equilibrium,the clockwise torque must equal the anticlockwise torque:$Mg \cdot \frac{l}{2} \sin(30^{\circ}) = mg \cdot \frac{l}{2} \sin(60^{\circ})$$M \cdot \frac{1}{2} = m \cdot \frac{\sqrt{3}}{2}$$\frac{M}{m} = \sqrt{3}$

Two uniform rods of equal length but different masses are rigidly joined to form an $L$-shaped body,which is then pivoted as shown. If in equilibrium the body is in the shown configuration,the ratio $M/m$ will be:

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