Question

A bob of mass m is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle $\theta$ at which the speed of the 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

$\frac{3\pi }{4}<\theta <\pi$

Refer to the Fig. Here OA=OB=OC = L and $OD=OCcos\theta =Lcos\theta$.

Therefore $h=OA-OD-Lcos\theta$

Or $h=L(1-cos\theta )$.

From conservation of energy, total energy at A = total energy at C, i.e.

$\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{v}{2}\right)}^2+mgL(1-cos\theta )$

$v^2=\frac{8gL}{3}(1-cos\theta )$ (1)

The minimum velocity the bob must have at A so as to reach B is

$v=\sqrt{5gL}.$ Putting this is Eq.(I),

We get $cos\theta =-\frac{7}{8}.$

Therefore $\theta$ lies between $\frac{3\pi }{4}and\pi$