Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity of $u=\sqrt{3gl}$, where $l$ is the length of the pendulum. Select the correct statement.

• A. Bob will complete the circle.
• B. String will become slack at top most point.
• C. String will become slack after rotating an angle ${\cos }^{-1}\frac{1}{3}$.
• D. String will become slack after rotating an angle $\pi -{\cos }^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is D String will become slack after rotating an angle $\pi -{\cos }^{-1}\left(\frac{1}{3}\right)$


Let string slack at B.
Applying conservation of energy at point A and B.
$\Delta U+\Delta K=0$
$mg(l+l\cos \theta )+[\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2]=0$
$-mg[l+l\cos \theta ]=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$
$v^2=u^2-2g[l+l\cos \theta ]....\left(1\right)$
When string slack, tension in the string becomes zero. The component of weight in radial direction balance the centrifugal force at this position.

$\frac{m{v}^2}{l}=mg\cos \theta$
$v^2=gl\cos \theta ......\left(2\right)$
From equation (1) and (2),
$gl\cos \theta =3gl-2gl[1+\cos \theta ]$
$(\because u=\sqrt{3gl})$
$3\cos \theta =1\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{3}\right)$
Hence angle rotated $\alpha$ is
$\alpha =\pi -\theta =\pi -{\cos }^{-1}\left(\frac{1}{3}\right)$