Q.
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) $\frac{3\pi}{4} < \theta < \pi$
Explanation
(d) Velocity of the bob at the point A
$v = \sqrt{5gL}$ …..(i)
$\left( \frac{v}{2} \right)^{2} = v^{2} - 2gh\ldots.(ii)$
$h = L(1 - \cos{\theta)}\ldots(iii)$
$Solving\ Eqs.(i),\ (ii)and\ (iii),\ we\ get$
$\cos{\theta = - \frac{7}{8}}$
$or\ \ \ \theta = \cos^{- 1}\left( - \frac{7}{8} \right) = 151{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.