 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-3/9e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Put the cylinder's centre at the origin, with

- h /2 \leq z \leq h /2

0 \leq r \leq a

, and uniform density

ρ = M / (π a^{2} h)

I_{1} = I_{2}

. Direct integration yields

I_{3} = ρ \int_{- h /2}^{h /2} d z \int_{0}^{a} r^{3} d r \int_{0}^{2 π} d θ = \frac{M a ^{2}}{2} .

For a transverse principal moment of inertia, the two contributions are

\int y^{2} d m = ρ h \frac{a ^{4}}{4} \int_{0}^{2 π} sin^{2} θ d θ = \frac{M a ^{2}}{4}, \int z^{2} d m = ρ π a^{2} \frac{h ^{3}}{12} = \frac{M h ^{2}}{12} .

Therefore

I_{1} = I_{2} = M (a^{2} /4 + h^{2} /12), I_{3} = M a^{2} /2

 New to topics? Read the docs here!