Question

A simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle $\theta$ with the vertical, then the angle $\alpha$, which the acceleration vector of the bob makes with the string is equal to

• A. $0$
• B. $ta{n}^{-1}\left(2tan\theta \right)$
• C. $ta{n}^{-1}\left(\frac{tan\theta }{2}\right)$
• D. $\pi /2$

Answer 1

Answer: (C)

$ta{n}^{-1}\left(\frac{tan\theta }{2}\right)$

On applying energy conservation taking the bottom line as reference,

Potential Energy at top = (Potential Energy + Kinetic Energy) at given instant

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

$mglcos\theta =\frac{m{v}^2}{2}$

So $v=\sqrt{2glcos\theta }$

tangential acceleration of bob is $a_T=gsin\theta$

Radial acceleration $a_R=\frac{v^2}{l}=2gcos\theta$

acceleration vector angle with string is $tan\alpha =\frac{a_T}{a_R}=\frac{gsin\theta }{2gcos\theta }$

$\alpha ={\tan }^{-1}\left(\frac{tan\theta }{2}\right)$