First slide

Vertical circular motion

Question

The length of a simple pendulum is 1m. The bob is given a velocity 7m/s in horizontal direction from mean position. During upward motion of bob, if the string is burnt when it is horizontal, then the maximum vertical height of ascent of bob from rest position is

Moderate

A

2.5 m
B

2 m
C

3 m
D

3.5 m

Solution

From conservation of energy
$\frac{1}{2} m v^{2} = \frac{1}{2} m v_{1}^{2} + m g r (v_{1} i s t h e v e l o c i t y a t t h e h o r i z o n t a l p o s i t o n) v_{1}^{2} = v^{2} - 2 g r F r o m t h e r e m a x i m u m h e i g h t i s \frac{v_{1}^{2}}{2 g} = \frac{v^{2} - 2 g r}{2 g} = \frac{v^{2}}{2 g} - r T h e r e f o r e t o t a l h e i g h t = r + \frac{v^{2}}{2 g} - r = \frac{7^{2}}{2 \times 9.8} m = 2.5 m$

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