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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2016/paper-1/1/b/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2016 / Paper 1 / 1 / b / Solution by

Codex  0 2026-10-20

Δ = θ - θ_{B}

and

u = z - b θ_{B}

. Applying the Euler-Lagrange equations separately to

r, θ, z, θ_{B}

gives

m (\overset{r}{¨} - r \dot{θ}^{2}) \frac{d}{d t} (m r^{2} \dot{θ}) m \overset{z}{¨} M (a^{2} + b^{2}) \ddot{θ}_{B} = - k (r - a cos Δ), = - ka r sin Δ, = - k u, = k (a r sin Δ + b u) .

(1)

For example,

\partial L / \partial θ = - ka r sin Δ

whereas

\partial L / \partial θ_{B} = ka r sin Δ + kb u

; these signs also follow from equal and opposite spring forces projected on the permitted bead motion. The angular particle equation can equivalently be written

m (r^{2} \ddot{θ} + 2 r \overset{r}{˙} \dot{θ}) = - ka r sin Δ

. The factor

M (a^{2} + b^{2})

is the bead's inertia with respect to the helical coordinate, rather than its moment of inertia about the axis alone.

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