Solution

Convert both radii to metres before evaluating the moment of inertia:

I = \frac{1}{2} m (a^{2} + b^{2}) = 1.1770375 \times 1 0^{- 3} kg m^{2} .

(1)

Its linear differential is

d I = \frac{1}{2} (a^{2} + b^{2}) d m + ma d a + mb d b .

(2)

For independent standard uncertainties, add the three contributions in quadrature:

σ_{I} = [\frac{1}{2} (a^{2} + b^{2}) σ_{m}]^{2} + (ma σ_{a})^{2} + (mb σ_{b})^{2} = 2.81 \times 1 0^{- 5} kg m^{2}, I = (1.177 \pm 0.028) \times 1 0^{- 3} kg m^{2} .

(3)

If the quoted errors are instead bounds, their first-order absolute contributions add, giving $I ≃ (1.177 \pm 0.042) \times 1 0^{- 3} kg m^{2}$ . Their statistical interpretation is not specified; the two conventions should remain distinct.

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section A / 3 / Solution

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