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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1/1/b/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2015 / Paper 1 / 1 / b / Solution by

Codex  0 2026-10-20

The derivatives needed in the Euler-Lagrange equation are

\frac{\partial L}{\partial θ ˙} = m l^{2} \dot{θ} + m l (\overset{a}{˙}_{x} cos θ + \overset{a}{˙}_{y} sin θ),

\frac{\partial L}{\partial θ} = m l \dot{θ} (- \overset{a}{˙}_{x} sin θ + \overset{a}{˙}_{y} cos θ) - m g l sin θ .

The terms proportional to

\overset{a}{˙}_{i} \dot{θ}

cancel after differentiation, leaving

l \ddot{θ} + \overset{a}{¨}_{x} cos θ + (g + \overset{a}{¨}_{y}) sin θ = 0.

With

l = l (sin θ, - cos θ, 0)

and

g = - g \hat{y}

,

[l \times (g - \ddot{a})]_{z} = - l [\overset{a}{¨}_{x} cos θ + (g + \overset{a}{¨}_{y}) sin θ] = l^{2} \ddot{θ} .

Hence the vector equation is $l^{2} \ddot{θ} \hat{z} = l \times (g - \ddot{a})$ . In the translating support frame,

- m \ddot{a}

is the inertial force. The rod's tension passes through the hinge and contributes no torque about it. The same torque equation therefore has an immediate mechanical interpretation.

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