WBJEE · Physics · Rotational Motion
A thin rod \(A B\) 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
Answer & Solution
Correct Answer
(C) the angular acceleration is proportional to \(\cos \theta\)
Step-by-step Solution
Detailed explanation
Loss in potential energy = gain in kinetic energy \(\Rightarrow \quad m g \frac{L}{2} \sin \theta=\frac{1}{2} I \omega^{2}\) So \(\omega \propto \sqrt{\sin \theta}\) and \(\quad v \propto \sqrt{\sin \theta} \quad\left(\because K=\frac{1}{2} \mathrm{H} \omega^{2}\right)\) The…
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