JEE Main 2023 · 6 April · evening · Single correct · +4 / −1

A child of mass 5 kg is going round a merry-go-round that makes 1 rotation in 3.14 s. The radius of the merry-go-round is 2 m. The centrifugal force on the child will be

  1. (a) 50 N
  2. (b) 80 N
  3. (c) 100 N
  4. (d) 40 N
Answer and solution

Answer (d) 40 N

Solution

Physics sir · 3 steps

Get ω from the time period first. Then centrifugal force is just mω² r, pointing outward.

  1. Angular speed. One rotation is 2π radians, so ω = 2π/T = (2 × 3.14)/3.14 = 2 rad s⁻¹.
  2. Which force. In the child's rotating frame, a pseudo force of size mω² r acts outward, away from the centre. This is the centrifugal force. It is equal in size to the centripetal force.

    F = mω² r = 5 × 2² × 2 = 40 N

  3. So. The centrifugal force on the child is 40 N, option (d).

The idea to keep. Whenever a time period is given, turn it into ω = 2π/T first. Centrifugal force has the same size as centripetal force, mω² r, but it acts outward in the rotating frame.

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

Study Circular Motion

Asked about 70 times in JEE Main since 2020. This question is from 2023.