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Part BCSIR NET December 2023

Lagrangian formalism and generalised coordinates

Let g denote the acceleration due to gravity and a > 0. A particle of mass m glides without friction on the cycloid x=a(θ−sin⁡θ),y=a(1+cos⁡θ),0≤θ≤2πx = a(\theta - \sin \theta), y = a(1 + \cos \theta), 0 \le \theta \le 2\pi. The equation of motion of the particle is

  1. A.(1−cos⁡θ)θ(1 - \cos \theta)\thetä + ½ sin⁡θ(θ\sin \theta (\thetȧ)2−(g/2a)sin⁡θ=0)^{2} - (g/2a) \sin \theta = 0✓ Correct
  2. B.(1−2cosθ)θ(1 - 2cos \theta)\thetä +sin⁡θ(θ+ \sin \theta (\thetȧ)2−(g/a)sin⁡θ=0)^{2} - (g/a) \sin \theta = 0
  3. C.m(1−2cosθ)θm(1 - 2cos \theta)\thetä +sin⁡θ(θ+ \sin \theta (\thetȧ)2+(g/a)sin⁡θ=0)^{2} + (g/a) \sin \theta = 0
  4. D.m(1−2cosθ)θm(1 - 2cos \theta)\thetä +(m/2)sin⁡θ(θ+ (m/2) \sin \theta (\thetȧ)2−(g/a)sin⁡θ=0)^{2} - (g/a) \sin \theta = 0

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: The generalised momentum ∂L/∂q̇ always equals mass × velocity

More on this topic

The chapter behind this: Lagrangian mechanics — free to read

From Classical Mechanics › Lagrangian formalism and generalised coordinates

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