First slide

Vertical circular motion

Question

A simple pendulum crosses the highest point of a vertical circle with critical speed. What will be the centripetal acceleration of the bob, when the string is horizontal?

Moderate

A

g
B

2 g
C

3 g
D

2.5 g

Solution

if it is critical velocity then at the lowest point it is $\sqrt{5 g l}$ at the horizontal position it will be $\sqrt{3 g l}$ from conservation of energy
$\frac{1}{2} m {v^{2}}_{0} = \frac{1}{2} m v^{2} + m g l \Rightarrow \frac{1}{2} m 5 g l - m g l = \frac{3}{2} m g l = \frac{1}{2} m v^{2} c e n t r e p e t a l a c c e l e r a t i o n \frac{v^{2}}{l} = 3 g$

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