MixedJEE Physics · Original learning card10 original chapter questions

Kinetic Energy of Pure Rolling

A rigid body in pure rolling has total kinetic energy one-half M v_cm squared plus one-half I_cm omega squared, with omega = v_cm/R only when the no-slip condition holds.

Why this shows up in the exam

Comparing rolling rings and spheres · Stopping-work calculations · Inferring speed from rolling energy

Learn the idea

Rolling kinetic energy is the sum of centre translation and rotation about the centre. Rolling kinetic energy is the sum of centre translation and rotation about the centre. Start from a clear axis, origin, body, and reference frame; the geometry and constraints then decide which rotational law is safe to use.

🧠 Memory hook: Rolling energy has a glide part and a spin part.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • K = (1/2) M v_cm² + (1/2) I_cm omega² — Translation-plus-spin kinetic energy.
  • K = (1/2)(M + I_cm/R²)v_cm² — Pure-rolling form after using omega=v_cm/R.

How to approach it

  1. 1Separate centre translation and spin
  2. 2Use I about the centre
  3. 3Apply the rolling constraint only if valid

Common slip-ups that cost marks

  • •Counting translational energy only
  • •Using inertia about the contact point while also adding translation
  • •Imposing omega=v/R during slip

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

Masses 1 kg and 3 kg lie at x = 0 and x = 4 m. Find the x-coordinate of their centre of mass.

Take a timed JEE Physics sectional mock