Part CCSIR NET December 2024the-radial-inertia-term-vanishes-with-r-dot

The radial inertia term vanishes with r dot

Consider a particle of mass m which is moving on a surface due to gravity. Suppose the Lagrangian of the particle in the cylindrical coordinates is given by $L = (m/2)[(1 + 4\alpha^{2}r^{2})$ ṙ $^{2} + r^{2}\theta$ ̇ $^{2}] - m\alpha$ gr $^{2}$ , where $\alpha$ is a positive constant, and g is the acceleration due to gravity. If the particle is in circular motion, then

A.| $\theta$ ̇| $= \sqrt{2/(g\alpha)}$
B.| $\theta$ ̇| $= \sqrt{1/(g\alpha)}$
C.| $\theta$ ̇| $= \sqrt{2g\alpha}$ ✓ Correct
D.| $\theta$ ̇| $= \sqrt{g\alpha}$

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 hypothesis dropped.

See pricing

50 are analysed free — try those first.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

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The radial inertia term vanishes with r dot

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