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 Incoming links: Eccentric circular-hole disk inertia

Past exam of the physics 1 paper of the University of Oxford / 2014 / 9 / a / Solution 2026-10-25

For the eccentric circular-hole disk inertia, let

D = π ρ R^{2}

be the mass of the complete disc before removing material,

h = π ρ r_{o}^{2}

the missing mass, and

W = D - h

the wheel's remaining mass. Denote the complete disc's centre by C and the axle/hole centre by H;

C H = r_{o}

. Assume the hole fits inside the disc,

0 < r_{o} < R /2

. The parallel axis theorem gives the full disc's moment of inertia about H as

\frac{1}{2} D R^{2} + D r_{o}^{2}

. Subtract the absent little disc about its own centre H:

I_{H} = \frac{1}{2} D R^{2} + D r_{o}^{2} - \frac{1}{2} h r_{o}^{2} = \frac{π ρ}{2} (R^{4} + 2 R^{2} r_{o}^{2} - r_{o}^{4}) .

(1)

The axle is a support, not part of the rotating wheel's stated mass. Its frictionless bearing does not alter this moment of inertia. For fixed-axis angular velocity

ω

T_{H} = \frac{1}{2} I_{H} ω^{2} = \frac{π ρ ω ^{2}}{4} (R^{4} + 2 R^{2} r_{o}^{2} - r_{o}^{4}) .

(2)

The missing material must be subtracted with its actual displacement relative to the axis; treating the wheel as a concentric annulus would give the wrong result.

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