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Part BCSIR NET June 2025p-dot-is-the-partial-of-L-in-q-so-a-plus-q-squared-term-pushes-outward

P dot is the partial of L in q so a plus q squared term pushes outward

Suppose a dynamical system has the Lagrangian L = (q̇1)2+(q_{1})^{2} + (q̇2)2+(q1)2+q_{2})^{2} + (q_{1})^{2} + q̇1q_{1}q̇2_{2}. If p1p_{1} and p2p_{2} are momenta conjugate to q1q_{1} and q2q_{2} respectively, then which of the following statements is true?

  1. A.ṗ1=2q1_{1} = 2q_{1}, ṗ2=0_{2} = 0✓ Correct
  2. B.ṗ1=−q1_{1} = -q_{1}, ṗ2=0_{2} = 0
  3. C.ṗ1=−q1/2,p2=q2_{1} = -q_{1}/2, p_{2} = q_{2}
  4. D.ṗ1=q1,p2=−q2_{1} = q_{1}, p_{2} = -q_{2}

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, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

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

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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