Question

A body of mass \( m \) connected to a massless and unstretchable string goes in a vertical circle of radius \( R \) under gravity \( g \). The other end of the string is fixed at the center of the circle. If velocity at the top of the circular path is \( v = \sqrt{ngR} \), where \( n \geq 1 \), then the ratio of kinetic energy of the body at bottom to that at top of the circle is:

• A. \( \frac{n^2 + 4}{n^2} \)
• B. \( \frac{n}{n + 4} \)
• C. \( \frac{n+4}{n} \)
• D. \( \frac{n^2}{n^2 + 4} \)

Hint:

In vertical circular motion, kinetic energy varies with the position along the path due to the potential energy associated with gravity.

Answer 1

The Correct Option is A

The objective is to calculate the velocities at the highest and lowest points of circular motion and determine their ratio. The steps are as follows:

1. Velocity at the Apex:
The velocity at the apex of the circular motion is determined by the formula:

$ V_{\text{Top}} = \sqrt{n^2 g R} $

2. Velocity at the Nadir:
The velocity at the nadir of the circular motion is enhanced by the gravitational potential energy. Its value is given by:

$ V_{\text{Bottom}} = \sqrt{n^2 g R + 4gR} $

3. Velocity Ratio:
To ascertain the ratio of the squared velocities, we compute:

$ \text{Ratio} = \frac{V_{\text{Bottom}}^2}{V_{\text{Top}}^2} $

Substituting the squared velocity expressions:

$ V_{\text{Top}}^2 = n^2 g R $

$ V_{\text{Bottom}}^2 = n^2 g R + 4gR $

$ \text{Ratio} = \frac{n^2 g R + 4gR}{n^2 g R} $

Factoring out $ gR $ from the numerator:

$ \text{Ratio} = \frac{gR (n^2 + 4)}{gR n^2} $

Simplifying the expression yields:

$ \text{Ratio} = \frac{n^2 + 4}{n^2} $

Conclusion:
The ratio of the squares of the velocities is:

$ \boxed{\frac{n^2 + 4}{n^2}} $