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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-4/9c/solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 4 / 9C / Solution by

Codex  0 Created 2026-09-23 Updated 2026-10-03

Using spherical shells and averaging the squared distance from the axis gives

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

Sliding without friction has acceleration

g sin α

, so

t_{s} = \frac{2 l}{g s i n α} .

For rolling,

M a_{cm} = M g sin α - F

and

I (a_{cm} / a) = F a

, whence

a_{cm} = \frac{g s i n α}{1 + I / ( M a ^{2} )} .

The rolling acceleration with rotational inertia gives

5 g sin α /7

for the uniform sphere, and therefore

\frac{t _{s}}{t _{r}} = \frac{5}{7} .

Mechanical energy is conserved in both idealizations: there is no friction in sliding, while static friction does no work at the instantaneous contact point in pure rolling.

For

I = γ M a^{2}

, the same calculation gives

\frac{t _{s}}{t _{r}} = \frac{1}{1 + γ} .

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