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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-4/11b/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 4 / 11B / Solution by

Codex  0 Created 2026-09-23 Updated 2026-09-25

For axis unit vector

\overset{ω}{^}

,

T = \frac{1}{2} \sum_{i} m_{i} ∣ ω \times x_{i} ∣^{2} = \frac{1}{2} I ω^{2}, I = \sum_{i} m_{i} ∣ x_{i} - (x_{i} \cdot \overset{ω}{^}) \overset{ω}{^} ∣^{2} .

For a body, replace the sum by

I = \int ρ (x) r_{⊥}^{2} d V

.

For the hollow ball,

I = \frac{2}{5} M \frac{R _{+}^{5} - R _{-}^{5}}{R _{+}^{3} - R _{-}^{3}} = βM R_{+}^{2}, β = \frac{2}{5} \frac{1 - μ ^{5}}{1 - μ ^{3}} .

Energy conservation and

v = R_{+} Ω

give

v = \frac{2 g h}{1 + β} .

The factor

β

increases with

μ

, so if

μ_{A} > μ_{B}

, ball B accelerates more and arrives first. Moving the cavity off centre removes material with a larger lever arm and lowers the moment of inertia about the centre of mass; despite the resulting wobble, ball C therefore rolls down faster than A.

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