A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies (K . E)_A/(K . E)_B is :

Solution

Physics sir · 3 steps

At B the string just goes slack, so gravity alone supplies the centripetal force there.

1. Condition at B. With the minimum speed, the tension at the top B just becomes zero. Gravity alone provides the centripetal force: mg = mv_B²/L, so v_B² = gL.
2. Energy conservation. The bob rises by 2L from A to B. So 1/2mv_A² = 1/2mv_B² + mg(2L), which gives v_A² = gL + 4gL.

   v_A² = 5gL, v_B² = gL

3. So. Kinetic energy goes as v², so (K.E)_A/(K.E)_B = 5gL/gL = 5. The ratio is 5 : 1, option (a).

The idea to keep. For a bob on a string, just completing the top means T = 0 there, so vₜₒₚ = √gL and v_(bottom) = √5gL. Memorise this pair. It turns up in many problems.

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