MixedJEE Physics · Original learning card10 original chapter questions

Rigid Bodies and Constrained Spring Systems

For a small generalized displacement q, a constrained elastic system satisfies I_eff q double-dot + k_q q = 0, where k_q follows from U = k_q q^2/2.

Why this shows up in the exam

Rods attached to springs · Rolling discs or rings with springs · Amplitude of an internal point in a spring assembly

Learn the idea

For rotational or constrained motion, convert each spring extension into the selected generalized coordinate. A rod or rolling body makes different attachment points move by different amounts, so geometry weights each spring's restoring effect.

🧠 Memory hook: Geometry first, spring algebra second.

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

Formulas & facts to keep ready

  • omega = sqrt(k_q/I_eff) — small angular frequency in a rotational coordinate
  • U = sum_i k_i (Delta l_i)²/2 — total spring energy under the constraint
  • Delta l_i approximately equals c_i q — small-displacement geometric relation for each attachment

How to approach it

  1. 1Choose the independent coordinate
  2. 2Relate all point motions to it
  3. 3Form total kinetic and potential energies and compare coefficients

Common slip-ups that cost marks

  • •Using translational mass instead of moment of inertia
  • •Giving every attachment the same extension
  • •Omitting rolling kinetic energy

🌟 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

A mass of 1 kg is attached to a spring of force constant 100 N/m. Find the angular frequency of small oscillations.

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