Question

A body of mass \( m \) connected to a massless and unstretchable string goes in 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 top of circular path is \( v = \sqrt{n g R} \), 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:

For problems involving kinetic energy in circular motion, use conservation of mechanical energy. The total mechanical energy remains constant if only conservative forces like gravity are doing work.

Answer 1

The Correct Option is A

The total mechanical energy at the top of the circular path consists of kinetic energy and potential energy. The kinetic energy at the top of the circle is given by: \[ KE_{{top}} = \frac{1}{2} m v_{{top}}^2 = \frac{1}{2} m (n g R) \] At the bottom of the circle, the kinetic energy is given by: \[ KE_{{bottom}} = \frac{1}{2} m v_{{bottom}}^2 \] where \( v_{{bottom}} \) is the velocity at the bottom, which is higher due to the work done by gravity. Using conservation of energy, we calculate \( v_{{bottom}} \) and then find the ratio of the kinetic energy at the bottom to that at the top. Thus, the ratio is \( \frac{KE_{{bottom}}}{KE_{{top}}} = \frac{n^2}{n^2 + 4} \).