Question

A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme position and lowest position are equal. The angle $\theta$ of thread deflection in the extreme position will be:

• A. ${\tan }^{-1}\left(2\right)$
• B. ${\tan }^{-1}\left(\sqrt{2}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• D. $2{\tan }^{-1}\left(\frac{1}{2}\right)$

Answer 1

Answer: (D)

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

At highest point velocity is zero $a_N=0$

$a_T=gsin\theta$

At lowest point $a_N=\frac{v^2}{l}$

$a_T=0$

Applying law of conventional energy

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

$\frac{v^2}{l}=2g(1-cos\theta )$

$gsin\theta =2g(1-cos\theta )$

$2sin\frac{\theta }{2}cos\frac{\theta }{2}=2\times 2{sin}^2\frac{\theta }{2}$

$\theta =2{tan}^{-1}\frac{1}{2}$