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 $m_{2}, m_{2} \neq = m_{1}$ , is attached to the particle, hanging below L with $θ$ 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 $θ$ are (g denotes the acceleration due to gravity)

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

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.

X is cyclic so its equation is a conservation law

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