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 Incoming links: Past exam of the physics 1 paper of the University of Oxford / 2017 / 8 / a

Past exam of the physics 1 paper of the University of Oxford / 2017 / 8 / b / Solution 2026-10-25

Take the rotation axis as

z

. A slice of the uniform sphere at coordinate

z

is a disc of radius

R (z) = a^{2} - z^{2}

and mass

d m = ρ π (a^{2} - z^{2}) d z

, where

ρ = 3 M / (4 π a^{3})

is its mass density. Applying a to the slice gives

d I = \frac{1}{2} R (z)^{2} d m = \frac{ρ π}{2} (a^{2} - z^{2})^{2} d z .

The slices' axes coincide, so no displaced-axis correction is needed. Integrating,

I = \frac{ρ π}{2} \int_{- a}^{a} (a^{4} - 2 a^{2} z^{2} + z^{4}) d z = \frac{ρ π}{2} (2 a^{5} - \frac{4}{3} a^{5} + \frac{2}{5} a^{5}) = \frac{8 π ρ a ^{5}}{15} .

Thus the moment of inertia of a uniform solid sphere is

I = \frac{2}{5} M a^{2} .

The sphere's rotational symmetry makes this valid about every central axis. A thin spherical shell would have a different mass distribution and a different moment of inertia.

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