Question

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 :

• A. $\tan^{-1}(\sqrt{2})$
• B. $2\tan^{-1}\left(\frac{1}{2}\right)$
• C. $\tan^{-1}\left(\frac{1}{2}\right)$
• D. $2\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Answer 1

The Correct Option is B

Loss in kinetic energy equals gain in potential energy:

\(\frac{1}{2}mv^2 = mg\ell(1 - \cos \theta).\)

From the equation:

\(v^2 = 2g\ell(1 - \cos \theta).\)

Acceleration at the lowest point is given by:

\(\text{Acceleration} = \frac{v^2}{\ell} = 2g(1 - \cos \theta).\)

Acceleration at the extreme point:

\(a = g \sin \theta.\)

Equating the magnitudes of acceleration:

\(2g(1 - \cos \theta) = g \sin \theta \implies \sin \theta = 2(1 - \cos \theta).\)

Simplifying gives:

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

The Correct answer is: $2\tan^{-1}\left(\frac{1}{2}\right)$