 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-iv-resit/6/a/solution

ρ = m / (π a^{2} l)

. A thin cylindrical shell of radius

s

and thickness

d s

has mass

d m = 2 π ρl s d s

. Its contribution to the axial moment of inertia is

s^{2} d m

. Therefore

I_{33} = \int_{0}^{a} s^{2} (2 π ρl s) d s = 2 π ρl \frac{a ^{4}}{4} = \frac{1}{2} m a^{2} .

The attached wheels are massless, so they add no moment of inertia. This is the moment of inertia of a uniform solid cylinder about its central longitudinal axis of rotation.

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