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 in magnitude. Angle $\theta $ of thread deflection in the extreme position will be:

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

Answer 1

Answer: (A)

$a_{A}=g \, sin\theta \, $ (only tangential)
$a_{B}=\frac{v^{2}}{l}$ (only radial)
$K.E. \, + \, P.E. \, =$ $\frac{1}{2}m0^{2}+mgl \, \left(1 - cos \theta \right)=\frac{1}{2}mv^{2}$
$v^{2}=2gl \, \left(1 - cos \theta \right)$ ...........(i)
$\therefore a_{B}=\frac{v^{2}}{l}=2g \, \left(\right.1-cos\theta \left.\right)$
Since, $a_{A}=a_{B}$
$\therefore g \, sin\theta =2g \, \left(1 - c o s \theta \right)\Longrightarrow 2sin \frac{\theta }{2} \, cos ⁡ \frac{\theta }{2}=2 \, \times 2\left(sin\right)^{2} ⁡ \frac{\theta }{2}$
$\Longrightarrow tan \frac{\theta }{2} = \frac{1}{2} \, \, \Longrightarrow \, \theta = 2 \left(tan\right)^{- 1} ⁡ \left(\frac{1}{2}\right)$