 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-4/3c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The parallel axis theorem says that for an axis through

x

, parallel to the centre-of-mass axis in direction

\hat{d}

I_{T} (x) = I_{T} (x_{T}) + M_{T} ∣ (x - x_{T}) \times \hat{d} ∣^{2} .

Let

ρ_{A}^{2} = ∣ (x_{A} - x_{C}) \times \hat{d} ∣^{2}, ρ_{B}^{2} = ∣ (x_{B} - x_{C}) \times \hat{d} ∣^{2} .

Moments about the axis through

x_{C}

add, so

I_{C} (x_{C}) = I_{A} (x_{A}) + M_{A} ρ_{A}^{2} + I_{B} (x_{B}) + M_{B} ρ_{B}^{2} .

Using

M_{A} = M_{C} - M_{B}

and solving gives

M_{B} = \frac{I _{C} ( x _{C} ) - I _{A} ( x _{A} ) - I _{B} ( x _{B} ) - M _{C} ρ _{A}^{2}}{ρ _{B}^{2} - ρ _{A}^{2}}

when the denominator is nonzero. In the degenerate equal-distance case, this inertia equation alone does not determine

M_{B}

; the centre-of-mass equation

M_{C} x_{C} = M_{A} x_{A} + M_{B} x_{B}

supplies the remaining information.

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