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

Step 1: Apply condition of zero tension.
At the given point where the string makes an angle \(30^\circ\) with the horizontal, the tension in the string is zero. Hence, the centripetal force is provided only by the component of gravitational force towards the centre.
\[ \frac{mv^2}{r} = mg\cos 60^\circ \] \[ \frac{mv^2}{r} = \frac{mg}{2} \Rightarrow v^2 = \frac{gr}{2} \]
Step 2: Apply conservation of mechanical energy.
Let the velocity at the given point be \( v \) and velocity at the lowest point \( A \) be \( v_A \).
The vertical height difference between the two positions is \[ h = r(1 + \sin 30^\circ) = r\left(1 + \frac{1}{2}\right) = \frac{3r}{2} \]
Using energy conservation,
\[ \frac{1}{2}mv_A^2 = \frac{1}{2}mv^2 + mgh \] \[ \frac{1}{2}mv_A^2 = \frac{1}{2}m\left(\frac{gr}{2}\right) + mg\left(\frac{3r}{2}\right) \] \[ \frac{1}{2}mv_A^2 = \frac{gr}{4}m + \frac{3gr}{2}m \]
Step 3: Simplify and calculate velocity at the bottom.
\[ \frac{1}{2}mv_A^2 = \frac{7gr}{4}m \Rightarrow v_A^2 = \frac{7gr}{2} \] \[ v_A = \sqrt{\frac{7}{2}gr} \]