Question
An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If $$AB=BC,$$ and the angle made by $$AB$$ with downward vertical is $$\theta ,$$ then:
An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If $$AB=BC,$$ and the angle made by $$AB$$ with downward vertical is $$\theta ,$$ then:
A.
$$\tan \theta = \frac{1}{{2\sqrt 3 }}$$
B.
$$\tan \theta = \frac{1}{2}$$
C.
$$\tan \theta = \frac{2}{{\sqrt 3 }}$$
D.
$$\tan \theta = \frac{1}{3}$$
Answer :
$$\tan \theta = \frac{1}{3}$$
Solution :
Given that, the rod is of uniform mass density and $$AB=BC$$
Let mass of one rod is $$m.$$
Balancing torque about hinge point.
$$\eqalign{ & mg\left( {{C_1}P} \right) = mg\left( {{C_2}N} \right) \cr & mg\left( {\frac{L}{2}\sin \,\theta } \right) = mg\left( {\frac{L}{2}cos\,\theta - L\,\sin \,\theta } \right) \cr & \Rightarrow \frac{3}{2}mgL\,\sin \,\theta = \frac{{mgL}}{2}\cos \,\theta \cr & \Rightarrow \frac{{\sin \,\theta }}{{\cos \,\theta }} = \frac{1}{3}\,\,\,\,or,\tan \,\theta = \frac{1}{3} \cr} $$
Given that, the rod is of uniform mass density and $$AB=BC$$
Let mass of one rod is $$m.$$
Balancing torque about hinge point.
$$\eqalign{ & mg\left( {{C_1}P} \right) = mg\left( {{C_2}N} \right) \cr & mg\left( {\frac{L}{2}\sin \,\theta } \right) = mg\left( {\frac{L}{2}cos\,\theta - L\,\sin \,\theta } \right) \cr & \Rightarrow \frac{3}{2}mgL\,\sin \,\theta = \frac{{mgL}}{2}\cos \,\theta \cr & \Rightarrow \frac{{\sin \,\theta }}{{\cos \,\theta }} = \frac{1}{3}\,\,\,\,or,\tan \,\theta = \frac{1}{3} \cr} $$
