 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-b/6/a/solution

Choose three mutually perpendicular axes through the same point, and label coordinates

x, y, z

I_{x} = \int (y^{2} + z^{2}) d m, I_{y} = \int (z^{2} + x^{2}) d m, I_{z} = \int (x^{2} + y^{2}) d m .

Subtracting gives

I_{x} + I_{y} - I_{z} = 2 \int z^{2} d m \geq 0.

For a thin planar lamina lying in

z = 0

, the right-hand side vanishes, proving the perpendicular axis theorem:

I_{z} = I_{x} + I_{y} .

The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities are

I_{x} + I_{y} \geq I_{z}, I_{y} + I_{z} \geq I_{x}, I_{z} + I_{x} \geq I_{y},

with equality in the first precisely when all mass lies in the plane perpendicular to its

z

axis, and cyclically for the others. None of these identities requires principal axes.

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