Question

The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of $\sqrt{3gl}$. Find the angle rotated by the string before it becomes slack.

Answer 1

The horizontal velocity imparted to the bob is $u=\sqrt{3gl}$

Using conservation of energy:

The change in the kinetic energy is equal to the change in the potential energy.

$\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=-mgh$

The height of the bob from the ground will be :

$h=l+lcos\theta$

Substitute the values:

$v^2=u^2-2gl(1+\cos \theta )$

$v^2=3gl-2gl(1+\cos \theta )$

$\frac{m{v}^2}{l}=mg\cos \theta$

$v^2=gl\cos \theta$

$3gl-2gl-2gl\cos \theta =gl\cos \theta$

$\cos \theta =\frac{1}{3}$

$\theta ={\cos }^{-1}\left(1/3\right)$

At the time string becomes slack,

$\cos ({180}^{\circ }-\theta )=1/3$

$\theta ={\cos }^{-1}(-1/3)$