Question

A bob of mass $M$ is suspended through a massless string of length $L$. The horizontal velocity at position $A$ is sufficient enough to make it reach at point $B$. The angle at which the speed of bob is half of that at $A$, satisfies

• A. $\theta =\frac{\pi }{4}$
• B. $\frac{\pi }{4}<\theta <\frac{\pi }{2}$
• C. $\frac{\pi }{2}<\theta <\frac{3\pi }{4}$
• D. $\frac{3\pi }{4}<\theta <\pi$

Answer 1

The correct option is D $\frac{3\pi }{4}<\theta <\pi$

The bob attached to string just reaches at $B$ that means $v=0$ there, hence for just completing the circular loop, velocity at bottom most point:
$v=\sqrt{5gL}....\left(i\right)$
At some position $C$ string makes angle $\theta$ with vertical and velocity of bob is $\frac{v}{2}$.


Height above bottom point $A$ is $h$ at position $C$
$h=L-L\cos \theta =L(1-\cos \theta )...\left(ii\right)$
If an object is rising along the vertical circular path with initial velocity $u$, then at height $h$ by applying mechanical energy conservation final velocity $v$ is given by:
$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mgh$
$v^2=u^2-2gh...\left(iii\right)$
for position $A$ to $C$
$u=v,v=\frac{v}{2}$
From Eq. $\left(ii\right)$ and $\left(iii\right)$
$\Rightarrow {\left(\frac{v}{2}\right)}^2=v^2-2gL(1-\cos \theta )$

$\Rightarrow \frac{3v^2}{4}=2gL(1-\cos \theta )$
Putting value of $v$ from Eq. $\left(i\right)$,
$\frac{15gL}{4}=2gL(1-\cos \theta )$
$\frac{15}{8}=1-\cos \theta$
$\therefore \cos \theta =\frac{-7}{8}=-0.875$
It indicates that value of $\cos \theta$ is $-ve$ and close to $-1$, so $\theta$ must be an angle of $2^{\text{nd}}$ quadrant which is near to ${180}^{\circ }\text{or}\pi$
$\therefore$ Hence range $(\frac{3\pi }{4}<\theta <\pi )$ suits most.