question_answer

A thin rod AB is held horizontally so that it can freely rotate in a vertical plane about the end A as shown in the figure. The potential energy of the rod when it hangs vertically is taken to be zero. The end B of the rod is released from rest from a horizontal position. At the instant the rod makes an angle \[\theta \] with the horizontal

A)  the speed of end B is proportional to \[\sqrt{\sin \theta }\]

B)  the potential energy is proportional to \[(1-cos\,\theta )\]

C)  the angular acceleration is proportional to \[cos\,\theta \]

D)  the torque about A remains the same as its initial value

Correct Answer: A

Solution :

 Loss in potential energy = gain in kinetic energy \[\Rightarrow \] \[mg\frac{L}{2}\sin \theta =\frac{1}{2}I{{\omega }^{2}}\] So, \[\omega \propto \sqrt{\sin \theta }\] and \[v\propto \sqrt{\sin \theta }\] \[\left( \because \,K=\frac{1}{2}I{{\omega }^{2}} \right)\] The speed of end B is proportional to \[\sqrt{\sin \theta }\] and \[U=mgh=mg\frac{L}{2}(1-sin\theta )\]      \[(\because \,\tau =I\alpha )\] \[\Rightarrow \] \[mg\frac{L}{2}\cos \theta =\frac{m{{I}^{2}}}{3}\times \alpha \] \[\alpha \propto \cos \theta \] The angular acceleration is proportional to \[\theta \].