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 \[\theta \].You need to login to perform this action.
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