 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4/9b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let the full cube mass be

M = 8 ρ a^{3}

and the removed cylinder mass

m_{h} = 2 π ρ a r^{2}

. About the rod, the remaining moment is

I = M a^{2} (2/3 + γ^{2}) - \frac{1}{2} m_{h} r^{2} .

The remaining centre-of-mass distance

ℓ

obeys

(M - m_{h}) ℓ = M γa

, so the small-angle equation is

I \ddot{θ} + M g γa θ = 0

. Hence

T = 2 π \frac{a}{g γ} (\frac{2}{3} + γ^{2} - \frac{π}{8} \frac{r ^{4}}{a ^{4}}) .

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