JEE Main 2021 · 20 July · evening · Single correct · +4 / −1

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :

  1. (a) Solid sphere
  2. (b) Solid cylinder
  3. (c) Hollow cylinder
  4. (d) Ring
Answer and solution

Answer (b) Solid cylinder

Solution

Physics sir · 3 steps

In rolling, rotational over translational KE is k²/R², so find the body where that is one half.

  1. Rolling condition. Without slipping, v = ω R. Write the moment of inertia as I = mk², where k is the radius of gyration.

    Kᵣₒₜ/Kₜᵣₐₙₛ = (1/2Iω²)/(1/2mv²) = (mk² (v/R)²)/mv² = k²/R²

  2. Use the given ratio. Rotational KE is 50% of translational KE, so k²/R² = 1/2. That means I = 1/2mR².
    BodyI about its axisk²/R²
    Solid sphere2/5mR²2/5, that is 40%
    Solid cylinder1/2mR²1/2, that is 50%
    Hollow cylindermR²1, that is 100%
    RingmR²1, that is 100%
  3. So. Only the solid cylinder has I = 1/2mR², so the body is (b).

The idea to keep. For any body rolling without slipping, Kᵣₒₜ/Kₜᵣₐₙₛ = k²/R². Know the I values of the standard bodies and every ratio question becomes a lookup.

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

Study Rotational Motion

Asked about 180 times in JEE Main since 2020. This question is from 2021.