WBJEE · Physics · Motion In Two Dimensions
A simple pendulum of length \(L\) swings in a vertical plane. The tension of the string when it makes an angle \(\theta\) with the vertical and the bob of mass \(m\) moves with a speed \(v\) is \(\mathrm{fg}\) is the gravitational acceleration)
- A \(m v^{2} / L\)
- B \(m g \cos \theta+m v^{2} / L\)
- C \(m g \cos \theta-m v^{2} / L\)
- D \(m g \cos \theta\)
Answer & Solution
Correct Answer
(B) \(m g \cos \theta+m v^{2} / L\)
Step-by-step Solution
Detailed explanation
The situation is given below For motion along a vertical circular track, the required centripetal force is along the radius and towards the centre of the circle is given by \(T-m g \cos \theta=\frac{m v^{2}}{L}\) \(\Rightarrow \quad T=\frac{m v^{2}}{L}+m g \cos \theta\)
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