Part BCSIR NET December 2024x-is-cyclic-so-its-equation-is-a-conservation-law

X is cyclic so its equation is a conservation law

Consider a particle of mass $m_{1}$ moving along a horizontal line L that is perpendicular to a vertical wall. Let x denote the distance of the particle from the wall. Suppose a simple pendulum of length l having mass $m_{2}, m_{2} \ne m_{1}$ , is attached to the particle, hanging below L with $\theta$ measured from the downward vertical. If the pendulum oscillates in a plane containing L, then the equations of motion in terms of the generalized coordinates x and $\theta$ are (g denotes the acceleration due to gravity)

A. $(m_{1} + m_{2})$ ẍ + lm $_{2}(d/$ dt $)(\theta$ ̇ $\cos \theta) = 0$ and $l\theta$ ̈ + (d/dt)(ẋ $\cos \theta) +$ ẋ $\theta$ ̇ $\sin \theta + g \sin \theta = 0$ ✓ Correct
B. $(m_{1} + m_{2})$ ẍ + lm $_{2}\theta$ ̈ $\cos \theta = 0$ and $l\theta$ ̈ + ẍ $\cos \theta +$ ẋ $\theta$ ̇ $\sin \theta + g \sin \theta = 0$
C. $(m_{1} + m_{2})$ ẍ + lm $_{1}(d/$ dt $)(\theta$ ̇ $\cos \theta) = 0$ and $l\theta$ ̈ $+ m_{2}(d/$ dt)(ẋ $\cos \theta) +$ ẋ $\theta$ ̇ $\sin \theta + g \sin \theta = 0$
D. $(m_{1} + m_{2})$ ẍ + l(d/dt $)(\theta$ ̇ $\cos \theta) +$ ẋ $\theta$ ̇ $\sin \theta + g \sin \theta = 0$ and $l\theta$ ̈ + (d/dt)(ẋ $\cos \theta) + \theta$ ̇ $\cos \theta = 0$

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, 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

X is cyclic so its equation is a conservation law

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