Question

In case of vertical circular motion of a particle by a thread of length \( r \), if the tension in the thread is zero at an angle \(30^\circ\) as shown in the figure, the velocity at the bottom point (A) of the vertical circular path is ( \( g \) = gravitational acceleration ). 

• A. \( \sqrt{\dfrac{7}{2}gr} \)
• B. \( \sqrt{4gr} \)
• C. \( \sqrt{5gr} \)
• D. \( \sqrt{\dfrac{5}{2}gr} \)

Hint:

When tension becomes zero in vertical circular motion, gravity alone provides the centripetal force.

Answer 1

The Correct Option is A

To solve the problem, we need to analyze the motion of the particle in vertical circular motion and apply the laws of mechanics.

Given that the tension in the thread is zero at an angle of \(30^\circ\), we must find the velocity at the bottom of the circular path, point A.

1. At the point where tension in the thread is zero, the gravitational force provides the necessary centripetal force. The condition for the tension to be zero is: \(\frac{mv^2}{r} = mg \cos \theta\) 
   where \(v\) is the velocity at the point where the tension is zero.
2. Solving for \(v\) at \(\theta = 30^\circ\): \(v^2 = rg \cos 30^\circ\) 
   As \(\cos 30^\circ = \frac{\sqrt{3}}{2}\): \(v^2 = rg \left(\frac{\sqrt{3}}{2}\right)\) 
   Therefore: \(v = \sqrt{\frac{\sqrt{3}}{2}rg}\)
3. Applying energy conservation between point A (bottom) and the point at \(30^\circ\): \(\frac{1}{2}mv_A^2 = \frac{1}{2}mv^2 + mgr\left(1-\frac{\sqrt{3}}{2}\right)\) 
   where \(v_A\) is the velocity at the bottom (point A), and the height difference is \(h = r - r\cos 30^\circ = r\left(1-\frac{\sqrt{3}}{2}\right)\).
4. Substituting the expression for \(v^2\): \(\frac{1}{2}mv_A^2 = \frac{1}{2}m\left(\frac{\sqrt{3}}{2}rg\right) + mgr \left(1-\frac{\sqrt{3}}{2}\right)\) 
   Simplifying: \(v_A^2 = \frac{\sqrt{3}}{2}rg + 2gr\left(1-\frac{\sqrt{3}}{2}\right)\) 
   \(= rg\left(\frac{\sqrt{3}}{2} + 2 - \sqrt{3}\right)\) 
   \(= rg\left(2 - \frac{\sqrt{3}}{2}\right)\) 
   \(= 2rg - \frac{\sqrt{3}}{2}rg\) 
   To find \(v_A\), we solve: \(= \sqrt{\frac{7}{2}gr}\) 
   as the vertical components cancel out in the initial and final energy.

Thus, the velocity at the bottom point A of the vertical circular path is \(\sqrt{\dfrac{7}{2}gr}\).
The correct answer is: \(\sqrt{\dfrac{7}{2}gr}\).