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 circle. If velocity at top of circular path is n√(g R) , where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is :

Solution

Physics sir · 3 steps

Going down from top to bottom, gravity adds mg(2R) of kinetic energy. Add it to the top KE and divide.

1. KE at the top. With vₜₒₚ = n√gR, we get vₜₒₚ² = n² gR. So Kₜₒₚ = 1/2m n² gR.
2. Energy conservation. The string tension is always perpendicular to the velocity, so it does no work. Only gravity does work, and the body falls through a height 2R from top to bottom.

   K_(bottom) = Kₜₒₚ + mg(2R) = 1/2mgR(n² + 4)

3. So. K_(bottom)/Kₜₒₚ = (1/2mgR(n²+4))/(1/2mgR n²) = (n²+4)/n², which is option (a).

The idea to keep. In a vertical circle, tension does no work, so energy conservation links any two points. Bottom and top always differ by mg(2R) of kinetic energy.

Answer from the NTA final answer key. Solution by PrepWiser.