How this instrument works
Moment of inertia measures how stubbornly a spinning body resists a change in its spin rate, and for a hollow cylinder — a tube, pipe, or wound roll with a genuine hole down its centre — that resistance depends on two radii, not one. I = ½m(R₁²+R₂²) treats the object as an annulus: mass sitting between an inner boundary R₂ and an outer boundary R₁, none of it filling the empty core. Push R₂ toward zero and the formula collapses to the familiar solid-disc case, ½mR₁² — the solid disc is just the special case of a hollow cylinder with no hole at all.
The ½(R₁²+R₂²) term is not an approximation; it falls straight out of the calculus. Slice the annulus into thin concentric rings, each carrying mass dm = 2πρr dr, and sum r²dm from the inner radius out to the outer one. Divide that sum by the total mass, m = πρ(R₁²−R₂²), and the density and π cancel cleanly, leaving exactly ½(R₁²+R₂²) — the mass-weighted average of the squared radius across the whole ring of material, not merely the average of the two boundary radii themselves.
The formula is the same one behind a bit of kitchen physics that goes viral every so often: a nearly-finished toilet-paper roll seems to spin out of control compared with a fresh one, and it is not your imagination. As paper leaves the outside, mass and outer radius shrink together, and because R₁ is squared, the resistance to spinning collapses far faster than the pull on the sheet does. Model that roll — or a low wire spool, or a packing-tape core — as a solid disc instead of a hollow cylinder and you overstate I, because a solid disc assumes mass fills the empty tube too.
- Weigh the roll and enter it as Roll mass — grams suit a household roll, kilograms suit a larger core like a carpet or fabric bolt.
- Measure from the centre to the paper's outer edge and enter that as Outer radius (full roll) — half of a tape-measured diameter, never the diameter itself.
- Measure the empty cardboard tube's radius and enter it as Inner radius (cardboard tube); it must stay smaller than the outer radius or the instrument flags the input.
- Read Moment of inertia (hollow cylinder), kg·m² — the figure that sets how much torque it takes to spin the roll up or bring it to a stop.
Worked example — a 200 g toilet-paper roll
A fresh two-ply roll weighs 200 g, measures 6 cm from the centre to its outer edge, and rides on a cardboard tube with a 2 cm radius. Converting to SI gives mass = 0.2 kg, outer radius = 0.06 m, inner radius = 0.02 m, and the formula returns I = ½ × 0.2 × (0.06² + 0.02²) = ½ × 0.2 × (0.0036 + 0.0004) = ½ × 0.2 × 0.004 = 0.0004 kg·m². That single figure stands between a gentle tug and a roll that spins freely off its holder.
Pull the same roll down to 50 g with an outer radius shrunk to 3 cm over the unchanged 2 cm tube, and moment of inertia drops to about 3.25 × 10⁻⁵ kg·m² — roughly a twelfth of the full roll's value, even though the force needed to pull the last sheets off barely changes. That mismatch, resistance to spinning falling away far quicker than the pull force does, is the actual mechanism behind a nearly-empty roll suddenly unspooling three sheets at a time.
Questions
Why use a hollow-cylinder formula instead of a solid disc?
Because a wound roll's mass sits in the paper itself, not in the empty tube at its centre. A solid-disc formula, I = ½mR², implicitly assumes mass fills that whole radius, hollow core included, and so overstates the true moment of inertia. I = ½m(R₁²+R₂²) accounts for the missing material by treating the inner radius as a real boundary of the mass, not just a hole drawn on a diagram.
Why does an almost-empty roll seem to spin out of control?
Because moment of inertia collapses faster than the pulling force does as the roll unwinds. Mass and outer radius shrink together, and since outer radius is squared in the formula, a roll worn down to 50 g and 3 cm has its resistance to spinning fall to roughly a twelfth of a fresh roll's value — so the same tug now spins it up far more violently.
What happens as the inner radius approaches zero?
The formula approaches the solid-disc case, since (R₁²+R₂²) collapses toward R₁² once R₂ is negligible. The Inner radius field enforces a tiny positive minimum rather than an exact zero, because a hollow cylinder with a literal zero-radius hole is really just a solid cylinder — but below a millimetre or so the numeric difference from the solid-disc answer is negligible anyway.
Does this formula apply to spools, pipes, and gears besides paper rolls?
Yes — anything shaped like a uniform tube fits it: wire and cable spools, pipe and conduit stock, gift-wrap and packing-tape cores, even flywheels or pulleys bored out for a shaft. The physics only needs mass distributed uniformly between an inner and outer radius around a fixed spin axis; what the material is or what it's used for makes no difference to the arithmetic.
Where does I = ½m(R₁²+R₂²) actually come from?
From integrating r² over the mass in the annular cross-section. Splitting the roll into thin concentric rings of mass dm = 2πρr dr and summing r²dm from the inner radius to the outer one, then dividing by the total mass m = πρ(R₁²−R₂²), leaves exactly ½(R₁²+R₂²) once the density and π cancel — the same integral that produces ½mR² for a solid disc, just carried out between two radii instead of from zero.
Is this the same moment of inertia engineers use for beam bending?
No. Beam bending uses the second moment of area, measured in units like mm⁴ or in⁴, describing how a cross-section resists bending rather than how a mass resists a change in spin. This instrument returns mass moment of inertia in kg·m². The two share a name and a similar-looking integral — ∫r² dA against ∫r² dm — but they are never interchangeable in an actual calculation.