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

We are tasked with determining the velocities at the top and bottom of a circular motion and finding the ratio of these velocities. The solution proceeds as follows:

1. Velocity at the Top:
The velocity at the top of the circular motion is given by:

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

2. Velocity at the Bottom:
The velocity at the bottom of the circular motion includes an additional contribution due to gravitational potential energy. It is given by:

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

3. Ratio of Velocities:
To find the ratio of the squares of the velocities, we compute:

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

Substitute the expressions for $ V_{\text{Top}}^2 $ and $ V_{\text{Bottom}}^2 $:

$ 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} $

Factor out $ gR $ from the numerator:

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

Simplify the expression:

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

Final Answer:
The ratio of the squares of the velocities is:

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