Question 32. A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ( \theta ) of thread deflection in the extreme position will be :
(1) \tan ^{-1}(\sqrt{2})
(2) 2 \tan ^{-1}\left(\frac{1}{2}\right)
(3) \tan ^{-1}\left(\frac{1}{2}\right)
(4) 2 \tan ^{-1}\left(\frac{1}{\sqrt{5}}\right)
Answer (2)
Explanation:
Loss in kinetic energy = Gain in potential energy
\Rightarrow \frac{1}{2} mv^{2}=mg \ell(1-\cos \theta) \Rightarrow \frac{v^{2}}{\ell}=2 ~g(1-\cos \theta)Acceleration at lowest point =\frac{v^{2}}{\ell}
Acceleration at extreme point =g \sin \theta
Hence, \frac{v^{2}}{\ell}=g \sin \theta
\therefore \sin \theta=2(1-\cos \theta)