Q.A uniform circular disc of mass $2\ \text{kg}$ and radius $0.5\ \text{m}$ has a moment of inertia $I = \tfrac12 MR^2$ about an axis through its centre, perpendicular to its plane. Using the theorem of parallel axes, find its moment of inertia about a parallel axis passing through a point on its rim, also perpendicular to the plane of the disc.

By the theorem of perpendicular axes, for a plane lamina, Iz = Ix + Iy, where Ox and Oy are two mutually perpendicular axes in the plane of the lamina and Oz is the axis perpendicular to the plane, all three meeting at one point.

The theorem of perpendicular axes states: for a planar (flat, two-dimensional) body, the moment of inertia about an axis perpendicular to the plane of the body (Iz) is equal to the sum of its moments of inertia about two mutually perpendicular axes (Ix and Iy) lying in the plane of the body and intersecting at the point where the perpendicular axis passes through.

Derivation sketch: take a mass element dm at position (x, y) in the plane. Its distance from the z-axis (perpendicular to the plane) is r = √(x²+y²), so its contribution to Iz is dm(x²+y²) = dm·x² + dm·y². Integrating, Iz = ∫dm·y² + ∫dm·x² = Ix + Iy (since Ix, about the x-axis, sums the squared perpendicular distance y², and Iy sums x²).

So Iz = Ix + Iy — option (c).

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