First slide

Vertical circular motion

Question

A simple pendulum bob is given a horizontal velocity $\large \sqrt{\frac72gl}$ at the bottom. The string slackens after swinging through an angle θ with the vertical. The θ is

Easy

A

30⁰
B

$150^{0}$
C

90⁰
D

120⁰

Solution

At B, tension is zero. $\large \therefore \frac {mv^2}{l}=mgsin\alpha...(1)$
Conserving energy $\large \frac 12mv^2+mg(l+lsin\alpha)=\frac 12mu^2$
$\large \Rightarrow \frac {mv^2}{l}+2mg(1+sin\alpha)=\frac 72mg$ $\large \Rightarrow sin\alpha=\frac 12\Rightarrow \alpha=30^0$
$\large \theta =90^0+30^0=120^0$

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