Theorems of Parallel and Perpendicular Axes

Two general theorems -- stated here without proof, as directed by the syllabus -- make it possible to

find the moment of inertia of a rigid body about many different axes, starting from just ONE known value,

without ever repeating the underlying integration.

Theorem of parallel axes. For ANY rigid body, of any shape, and about ANY axis whatsoever,

where $I_{cm}$ is the body's moment of inertia about a parallel axis passing through its own centre of

mass, $M$ is the total mass of the body, and $d$ is the perpendicular distance between the two parallel

axes (see the accompanying figure, left panel). This theorem places no restriction at all on the shape of

the body -- it holds for a rod, a ring, a disc, a sphere, or any irregular rigid body, for ANY choice of the

second, offset axis, so long as it remains exactly parallel to the one passing through the centre of mass.

It is exactly this theorem that connects the two rod results quoted in Section 5.11: taking $I_{cm} = \tfrac{1}{12}ML^2$ (the rod's centre) and $d = L/2$ (the distance out to either end) gives $I_{end} = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{12}ML^2 + \tfrac14 ML^2 = \tfrac13 ML^2$, exactly the standard

value quoted for an axis through one end.

Theorem of perpendicular axes. This second theorem is more restrictive: it applies ONLY to a flat,

PLANAR (laminar) rigid body -- one whose thickness can be treated as negligible, so that it lies entirely

within a single plane. For such a body, choosing any two mutually perpendicular axes $Ox$ and $Oy$ lying IN

the plane of the lamina and intersecting at a point $O$, and a third axis $Oz$ perpendicular to the plane

through that same point $O$ (see the accompanying figure, right panel),

the moment of inertia about the axis perpendicular to the plane equals the sum of the moments of inertia

about the two perpendicular in-plane axes through the same point. This theorem is exactly what connects a

uniform disc's diameter value to its central-perpendicular value quoted in Section 5.11: by the disc's own

circular symmetry, its moment of inertia is identical about EVERY diameter (so $I_x = I_y = \tfrac14 MR^2$ …