 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/8b/a/solution

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

z

from the vertex, the cone has cross-sectional area

π R^{2} z^{2} / l^{2}

. Hence

z_{CM} = \frac{\int _{0}^{l} z ^{3} d z}{\int _{0}^{l} z ^{2} d z} = \frac{3 l}{4}, r_{CM} = a e_{3}, a = \frac{3 l}{4} .

I_{O} = I_{CM} + M {∣ ma t hb f a ∣^{2} 1 - a a^{T}} .

The displacement is along

e_{3}

, so it changes the two transverse moments but not the axial one. Thus, writing the vertex moments as

I_{1} = I_{2} = I

and

I_{3} = J

I = I_{1}^{CM} + M a^{2} = \frac{π R ^{2} lρ}{20} (R^{2} + 4 l^{2}) = \frac{3 M}{20} (R^{2} + 4 l^{2}),

and

J = I_{3}^{CM} = \frac{π R ^{4} lρ}{10} = \frac{3 M R ^{2}}{10} .

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