 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/15e/b/solution

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

Put

z = x_{3}

R (z) = z 1 - \frac{z}{h}, 0 \leq z \leq h .

Its volume and uniform mass density are

V = π \int_{0}^{h} R (z)^{2} d z = π \int_{0}^{h} (z^{2} - \frac{z ^{3}}{h}) d z = \frac{π h ^{3}}{12}, ρ = \frac{12 M}{π h ^{3}} .

By the continuous center of mass formula, the distance of the center of mass from the pointed origin is

l = \frac{ρ π}{M} \int_{0}^{h} z R (z)^{2} d z = \frac{12}{h ^{3}} (\frac{h ^{4}}{4} - \frac{h ^{4}}{5}) = \frac{3 h}{5} .

For a disk of radius

R

\int r^{2} d A = π R^{4} /2

. Therefore the axial moment is

I_{3} = \frac{ρ π}{2} \int_{0}^{h} R (z)^{4} d z = \frac{ρ π}{2} h^{5} \int_{0}^{1} u^{4} (1 - u)^{2} d u = \frac{2}{35} M h^{2} .

For the transverse

x_{1}

-axis,

\int x_{2}^{2} d A = π R^{4} /4

, while every slice also contributes

z^{2}

times its mass. Hence

I_{1} = ρ π \int_{0}^{h} (\frac{R ( z ) ^{4}}{4} + z^{2} R (z)^{2}) d z = ρ π h^{5} (\frac{1}{4 \cdot 105} + \frac{1}{30}) = \frac{3}{7} M h^{2} .

I_{2} = I_{1}

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