 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/8e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

θ

be the angle from the downward vertical to the hoop's centre, measured at the suspension point, and let

ϕ

be the absolute angle from the downward vertical to the bead's radius from the hoop centre. The bead position is the sum of two vectors of length

a

. Including the hoop's moment of inertia about the pivot gives

T = \frac{m a ^{2}}{2} (3 \dot{θ}^{2} + \dot{ϕ}^{2} + 2 \dot{θ} \dot{ϕ} cos (θ - ϕ)),

while

V = - m g a (2 cos θ + cos ϕ) .

Thus one suitable Lagrangian is

L = \frac{m a ^{2}}{2} (3 \dot{θ}^{2} + \dot{ϕ}^{2} + 2 \dot{θ} \dot{ϕ} cos (θ - ϕ)) + m g a (2 cos θ + cos ϕ) .

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