Question

A small ball of mass $m$ is attached at one end of a thread. The thread is held taut and horizontal, and the ball is released from this position. Choose the correct option(s).

• A. Angle between the thread and vertical when the tension in thread is equal to the weight of the ball in magnitude is ${\cos }^{-1}\left(1/3\right)$.
  
• B. Angle between the thread and vertical when the net acceleration becomes horizontal for mass $m$ is ${\tan }^{-1}\left(\sqrt{2}\right)$.
• C. Speed of the ball when tension in the thread is equal to weight of the ball is $\sqrt{\frac{2gR}{3}}$.
• D. Speed of the ball when net acceleration of the ball is horizontal is $\sqrt{\frac{2gR}{3}}$.

Answer 1

Answer: (A), (B), (C)

Speed of the ball when tension in the thread is equal to weight of the ball is $\sqrt{\frac{2gR}{3}}$.

By conservation of mechanical energy:
$mgR\cos \theta =\frac{1}{2}m{v}^2$



$\Rightarrow v=\sqrt{2gR\cos \theta }$ ...(1)

or $v^2/R=2g\cos \theta$

Radial force equation: $T=mg\cos \theta +\frac{m{v}^2}{R}$ .... (2)
Substituting the value of $v$ in equation (2), we get
$T=mg\cos \theta +\frac{m}{R}\left(2gR\cos \theta \right)$

If $T=mg$
$\Rightarrow mg=3mg\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{3}$

and $v=\sqrt{2gR\cos \theta }=\sqrt{\frac{2gR}{3}}$


When net acceleration becomes horizontal:
$\tan \theta =\frac{2g\cos \theta }{g\sin \theta }=2\cot \theta$
or $\theta ={\tan }^{-1}\sqrt{2}$

i.e $\cos \theta =\frac{1}{\sqrt{3}}$
Then, velocity $v=\sqrt{2gR\cos \theta }=\sqrt{\frac{2gR}{\sqrt{3}}}$

Thus, options (a), (b) & (c) are correct.