Parallel- & Perpendicular-Axis Theorems

Fix a direction for the axis and pick two axes with that direction: one through the centre of mass, one displaced from it by a perpendicular vector d. A mass element at perpendicular position r′ from the first axis sits at r′ + d from the second, so its contribution to the second moment of inertia is |r′ + d|² dm = r′² dm + 2d⋅r′ dm + d² dm. Integrate over the body. The first term is I_cm. The last is d² times the total mass, Md². The middle term is 2d⋅∫r′ dm, and ∫r′ dm is M times the centre-of-mass position measured from an axis that already passes through the centre of mass, which is zero. So I_P = I_cm + Md², with nothing approximated. Try it on a uniform disc, M = 3.0 kg, R = 0.20 m: I_cm = ½MR² = 0.060 kg m², and about a parallel axis along the rim, d = R, I = 0.060 + 3.0 × 0.040 = 0.18 kg m² = (3/2)MR². The Md² term is never negative, which is why the centre-of-mass axis is the minimum among all axes of its direction: the theorem proves the parabola that plotting I against axis position only suggested.

The cross term vanished because r′ was measured from the centre of mass. Measure from any other axis A and ∫r′ dm is M times the CM offset, no longer zero, so I_B = I_A + M(AB)² is simply false. Test it on a uniform rod, M = 1.0 kg, L = 1.0 m, moving from the end axis to an axis 0.25 m in from that end. The tempting shift gives (1/3) + 1.0 × 0.25² = 0.396 kg m²; the equally tempting subtraction gives 0.333 − 0.0625 = 0.271 kg m². Both are wrong. Go through the centre: the new axis is 0.50 − 0.25 = 0.25 m from the midpoint, so I = ML²/12 + M(0.25)² = 0.0833 + 0.0625 = 0.146 kg m². The general route is I_B = I_A + M(d_B² − d_A²), each d its own distance from the centre-of-mass axis: 0.333 + 1.0 × (0.0625 − 0.25) = 0.146 kg m² again. Moving an axis in toward the centre must lower I, and the correct rule does; the wrong addition raises it.

For a thin sheet lying in the xy-plane, an element at (x, y) is a distance y from the x-axis, x from the y-axis and √(x² + y²) from the z-axis normal to the sheet. So I_z = ∫(x² + y²) dm = ∫y² dm + ∫x² dm = Iₓ + I_y, provided the three axes meet at one point. Symmetry then does most of the work. A thin disc has I_z = ½MR² and, by symmetry, Iₓ = I_y, so each diameter carries ¼MR². A rectangular plate a × b has Iₓ = Mb²/12 and I_y = Ma²/12 about its central axes, hence I_z = M(a² + b²)/12 about its centre without a new integral. The flatness is not decoration. A solid sphere has I = (2/5)MR² about every axis through its centre, so Iₓ + I_y = (4/5)MR² is twice I_z; the theorem fails because a sphere's elements have a z-coordinate that adds to their distance from the x- and y-axes and nothing to their distance from the z-axis. Reserve the perpendicular-axis theorem for bodies thin enough that z² is negligible.

A body built from standard pieces needs no integral of its own. For each part write three numbers: its own moment of inertia about its own centre of mass, from the table; its mass; and the perpendicular distance from its own centre of mass to the axis the problem uses. The part contributes I_cm, i + mᵢdᵢ², and the body's I is the sum. Take a wheel: a thin rim of mass 1.20 kg and radius 0.300 m plus four uniform spokes, each 0.100 kg and 0.300 m long from hub to rim, about the axle. The rim is a hoop about its own centre, 1.20 × 0.300² = 0.108 kg m². Each spoke is a rod turning about one end: (1/12)mL² + m(L/2)² = (1/3)mL² = (1/3) × 0.100 × 0.0900 = 0.00300 kg m², so the four together add 0.0120 kg m². The wheel's I is 0.120 kg m². Notice the ranking the arithmetic produced: the spokes hold a quarter of the mass but a tenth of the inertia, because the theorem weights every part by where its mass sits.

Material removed is a part with negative mass, and it enters the same sum with a minus sign, including both its own I_cm and its own shift term. A uniform disc, M₀ = 4.00 kg, R = 0.300 m, has a hole of radius 0.100 m drilled with its centre 0.150 m from the disc's centre; find I about the disc's central perpendicular axis. The complete disc gives ½M₀R² = 0.180 kg m². The removed material is a smaller disc whose mass is M₀ times the area ratio, 4.00 × (0.100/0.300)² = 0.444 kg. About its own centre it had ½ × 0.444 × 0.100² = 0.00222 kg m², and its shift term is 0.444 × 0.150² = 0.0100 kg m², so drilling removes 0.0122 kg m² and leaves I = 0.168 kg m². The hole took 11.1% of the mass but only 6.8% of the inertia, because it sits close to the axis. One caution: the axis here is the geometric centre of the disc, which after drilling is no longer the centre of mass. That was fine, because the disc and the hole were each shifted from their own centres of mass, and neither shift used the body's overall CM.

The efficient habit is to pick the axis before computing anything. A rod swinging from a pin, a wheel rolling on the road, a door on its hinges: the physics happens about the pin, the contact line, the hinge line, so the shift goes to that axis and the table supplies I_cm. The theorem also runs backward. Hang an irregular plate from a pivot, time its small swings, and the period returns I about the pivot; subtract Md², with d the measured distance from pivot to centre of mass, and you have I_cm, a number no integral could reach for a shape without a formula. Know the boundaries. The parallel-axis theorem keeps the axis direction fixed, while the perpendicular-axis theorem changes direction but only among three mutually perpendicular axes meeting at one point in a sheet; neither tilts an axis to an arbitrary new direction, which is the inertia tensor's job in a later course, and I_P = I_cm + Md² says nothing about it. And every I here assumes a rigid body: if the parts move relative to one another, d changes and so does the answer.