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. $2ta{n}^{-1}\frac{1}{2}$
• B. $ta{n}^{-1}\frac{1}{2}$
• C. $ta{n}^{-1}\sqrt{2}$
• D. $ta{n}^{-1}2$

Answer 1

Answer: (A)

$a_A=gsin\theta$ (only tangential)
$a_B=\frac{v^2}{l}$ (only radial)
$K.E.+P.E.=$ $\frac{1}{2}m{0}^2+mgl\left(1-cos\theta \right)=\frac{1}{2}m{v}^2$
$v^2=2gl\left(1-cos\theta \right)$ ...........(i)
$\therefore a_B=\frac{v^2}{l}=2g(1-cos\theta )$
Since, $a_A=a_B$
$\therefore gsin\theta =2g\left(1-cos\theta \right)⟹2sin\frac{\theta }{2}cos\frac{\theta }{2}=2\times 2{\left(sin\right)}^2\frac{\theta }{2}$
$⟹tan\frac{\theta }{2}=\frac{1}{2}⟹\theta =2{\left(tan\right)}^{-1}\left(\frac{1}{2}\right)$