Question

A simple pendulum consisting of a mass $M$ attached to a string of length $L$ is released from rest at an angle $\alpha$ . A pin is located at a distance $l$ below the pivot point. When the pendulum swings down, the string hits the pin as shown in the figure. If the particle doesn't undergo a complete vertical circle, then the maximum angle $\theta \left(<\left(\pi \right)/2\right)$ through which the string swings after hitting the pin is

• A. $co{s}^{-1}\left[\frac{Lcos\alpha +l}{L+l}\right]$
• B. $co{s}^{-1}\left[\frac{Lcos\alpha -l}{L-l}\right]$
• C. $co{s}^{-1}\left[\frac{Lcos\alpha +l}{L-l}\right]$
• D. $co{s}^{-1}\left[\frac{Lcos\alpha -l}{L+l}\right]$

Answer 1

Answer: (B)

At the bottom most point, square of speed of bob,
$v^2=2gL\left(1-cos\alpha \right)$
It will rise further to a height,
$h=\frac{v^2}{2g}=L\left(1-cos\alpha \right)$
or $\left(L-l\right)\left(1-cos\theta \right)=L\left(1-cos\alpha \right)$
$\therefore \theta =co{s}^{-1}\left[\frac{Lcos\alpha -l}{L-l}\right]$