 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-4/10c/c/solution

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

Neglecting terms of order

Ω^{2}

and linearizing in

θ

gives

R \ddot{θ} = - g θ - 2Ω \overset{y}{˙}, \overset{y}{¨} = 2Ω R \dot{θ} .

The second equation and the initial data imply

\overset{y}{˙} = 2Ω R (θ - θ_{0}) .

Its contribution to the first equation is of order

Ω^{2}

, so the retained equation is

\ddot{θ} + \frac{g}{R} θ = 0.

Thus

θ (t) = θ_{0} cos (ω_{0} t), ω_{0} = \frac{g}{R} .

Integrating the associated

y

velocity and using

y (0) = 0

gives

y (t) = 2Ω R θ_{0} (\frac{sin ( ω _{0} t )}{ω _{0}} - t) .

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