Moment of inertia of continuous bodies

Model a uniform 0.80 kg rod 1.50 m long as n equal beads, each holding M/n at the middle of its slice, and turn it about a perpendicular axis through its centre. Four beads give Σ mᵢxᵢ² = 0.1406 kg m², ten give 0.1485 kg m², and a hundred give 0.149985 kg m². The sums follow (ML²/12)(1 − 1/n²) and close on 0.1500 kg m². A bead always misplaces its slice's mass: part lies nearer the axis and part farther, and because x² curves upward the far part counts for more, so every sum falls short. That shortfall is (M/n)(L/n)²/12 per slice, so the total error shrinks as 1/n² and the limit is the integral I = ∫ r⊥² dm. Read that error term again and notice which length appears in it: only L/n, the slice's extent across the direction in which r⊥ varies. No other dimension of the slice enters, so making a slice smaller in any other direction buys no accuracy at all, and that is the licence every element below spends.

An integral over mass cannot be evaluated until dm is written in a coordinate. For a uniform rod the linear density is λ = M/L, so a slice of length dx has dm = λ dx. The 0.80 kg, 1.50 m rod has λ = 0.533 kg m⁻¹. With x measured from the centre, I = λ∫x² dx from −0.75 m to 0.75 m = λL³/12 = ML²/12 = 0.150 kg m², and the units check, since kg m⁻¹ × m³ = kg m². Move the axis to one end and only the limits change: I = λ∫x² dx from 0 to 1.50 m = ML²/3 = 0.600 kg m², four times as much, and the far half of the rod alone supplies seven-eighths of it. Areas and volumes work the same way with σ = M/A and ρ = M/V, provided the element's size is written correctly: a ring of radius r has dA = 2πr dr, a strip of height H has dA = H dx, a disc slice of radius y has dV = πy² dz, and a cylindrical shell of radius r and height H has dV = 2πrH dr.

Turn a uniform disc about the axis through its centre, perpendicular to its face. Every point of a thin ring of radius r and width dr lies at r⊥ = r, so the ring is a legal element even though it is 2πr long. Its mass is dm = σ 2πr dr with σ = M/(πR²), and I = ∫₀ᴿ r² σ 2πr dr = 2πσR⁴/4 = ½MR². For a 2.4 kg disc of radius 0.125 m that is 0.0188 kg m², while a thin hoop of the same mass and radius, with all of it at R, has 0.0375 kg m². The disc's integrand grows as r³, one power from the ring's circumference and two from the distance, so rim material dominates: the inner half of the radius holds a quarter of the mass and a sixteenth of the I. The freedom works along the axis too. Sliding mass parallel to the axis leaves every r⊥ unchanged, so a solid cylinder of any length has I = ½MR² about its axis, and only its mass records how long it is.

About a diameter, the most convenient element for a solid sphere is a shape already solved: a thin disc, whose ½ dm y² can be used as it stands. Slice perpendicular to the axis at height z. The slice is a disc of radius y, with y² = R² − z², thickness dz and mass dm = ρπy² dz, and it contributes dI = ½ dm y² = ½ρπ(R² − z²)² dz. Expanding and integrating from z = −R to R gives ∫(R² − z²)² dz = (16/15)R⁵, so I = (8π/15)ρR⁵, and dividing by M = (4π/3)ρR³ leaves I = (2/5)MR². The formula hands over the thin shell for free. Add a layer dR to the sphere and dI = (8π/3)ρR⁴ dR while dM = 4πρR² dR, so the layer carries dI = (2/3)R² dM. A 7.26 kg solid shot of radius 0.060 m therefore has I = 1.05 × 10⁻² kg m², and a hollow shell of the same mass and radius has 1.74 × 10⁻² kg m². All of the shell's mass is R from the centre, but not R from the axis, which is why its factor is 2/3 and not 1.

A tapered boom is thickest at its pivot and thins to nothing at the tip, so model its linear density as λ = λ₀(1 − x/L), with x measured from the pivot. The mass is M = ∫₀ᴸ λ₀(1 − x/L) dx = λ₀L/2, and the moment of inertia about the pivot is I = ∫₀ᴸ x²λ₀(1 − x/L) dx = λ₀L³(1/3 − 1/4) = λ₀L³/12. Only now can I be written in terms of M: substituting λ₀ = 2M/L gives I = (1/6)ML². For a 12 kg boom 3.0 m long, λ₀ = 8.0 kg m⁻¹ and I = 18 kg m², half the 36 kg m² of a uniform boom of the same mass and length, because its mass crowds the pivot. The common slip is to treat λ₀L as the mass, which turns λ₀L³/12 into (1/12)ML² and reports 9.0 kg m², wrong by a factor of two. The rule behind both numbers is general: the factor in front of ML² or MR² falls when mass moves toward the axis and rises when it moves away.

An integral that gives a wrong answer usually gives a checkably wrong one. First, units: I must come out in kg m², and writing dm = M dx where (M/L) dx belongs leaves kg m³. Second, bounds: if no mass lies farther than d from the axis, then 0 ≤ I ≤ Md². A rod about its centre has d = L/2, so any answer above ¼ML², such as the end-axis (1/3)ML² carried to the wrong axis, is impossible. Third, limits: the annulus result ½M(R₁² + R₂²) must become ½MR² as R₁ → 0 and MR² as R₁ → R₂, and it does. Fourth, ranking: for equal M and R the factors must follow how far the mass sits from the axis, and they do, from the thin hoop at 1 through the thin spherical shell at 2/3, the disc at 1/2 and the solid sphere at 2/5 to the solid cone at 3/10. A result that breaks that order, or a cylinder whose I about its own axis grows with length at fixed mass, has an error in the element or in the limits.