JEE Main 2020 · 9 January · evening · Single correct · +4 / −1

A spring mass system (mass m, spring constant k and natural length l) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity ω, (k ≫ mω ²) the relative change in the length of the spring is best given by the option :

  1. (a) (mω ²)/3k
  2. (b) (mω ²)/k
  3. (c) (2mω ²)/k
  4. (d) √(2/3) ( (mω ²)/k )
Answer and solution

Answer (b) (mω ²)/k

Solution

Physics sir · 4 steps

The spring stretches until its pull is exactly the centripetal force the mass needs.

  1. Set up the circle. When the disc spins, the mass moves in a circle about the centre. Let the spring stretch by x, so the radius of the circle becomes l + x.
  2. Force balance. The only horizontal force on the mass is the spring force kx, directed towards the centre. It must supply the centripetal force mω² (l + x).

    kx = mω² (l + x) ⇒ x = (mω² l)/(k - mω²)

  3. Use k much larger. Since k ≫ mω², the mω² in the denominator is negligible, so x ≈ (mω² l)/k.
  4. So. The relative change in length is x/l = mω²/k, which is option (b).

The idea to keep. For a mass held in a circle by a spring, write spring force = mω² r with r the stretched length, then drop small terms only at the end.

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 2020.