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

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 kinetic energy at the top and bottom of the circle is related to the potential energy at these points. By applying the principles of energy conservation and the given velocity at the top, the ratio of kinetic energies at the bottom and the top is given by: \[ \frac{K_{\text{bottom}}}{K_{\text{top}}} = \frac{n^2}{n^2 + 4} \]