Mass Moment of Inertia Calculator

Model a toilet paper roll as a hollow cylinder to find its mass moment of inertia (I = ½m(r₁² + r₂²)), radius of gyration, and how fast it rolls downhill.

Quick Facts

Hollow cylinder formula
I = ½m(r₁² + r₂²)
A toilet paper roll is a hollow cylinder: r₁ is the cardboard tube, r₂ is the outer edge of the paper.
Special cases
Disk: r₁=0 → I=½mr₂². Hoop: r₁=r₂ → I=mr₂².
A full roll's inertia sits between a solid disk and a thin hoop.
Rolling acceleration
a = g·sinθ / (1 + I/(mR²))
Lower I/(mR²) means faster acceleration down a ramp — this decides the race.

Your Results

Calculated
Moment of Inertia (I)
-
I = ½m(r₁² + r₂²), about the central axis
Radius of Gyration (k)
-
k = √(I / m)
Rolling Inertia Ratio
-
I / (mR²); 0.5 = solid disk, 1.0 = thin hoop
Downhill Rolling Acceleration
-
a = g sinθ / (1 + I/(mR²)), rolling without slipping

Ready

Enter the roll's mass and radii, then press Calculate.

Formula and Method for a Toilet Paper Roll's Moment of Inertia

A toilet paper roll is, geometrically, a hollow cylinder (a "thick-walled tube" or annular cylinder): a solid ring of paper wound between an inner radius r₁ (the cardboard tube) and an outer radius r₂ (the outside of the roll), rotating about its own central axis. For any hollow cylinder of mass m, the mass moment of inertia about that central axis is I = ½m(r₁² + r₂²). This calculator also reports the radius of gyration, the dimensionless rolling-inertia ratio I/(mR²), and the linear acceleration the roll would have rolling down an incline — the physics behind a "toilet paper race."

How the calculation works

Enter the roll's mass and its inner and outer radii (with units), and the calculator converts everything to SI units (kilograms and meters) before applying I = ½m(r₁² + r₂²). It then derives the radius of gyration k = √(I/m) — the distance from the axis where a thin ring of the same mass would have identical inertia — and the ratio I/(mR²) using the outer radius R = r₂ as the rolling (contact) radius. Finally, for a roll released on an incline of angle θ and rolling without slipping, Newton's second law along the incline combined with the rotational equation of motion gives the linear acceleration a = g·sinθ / (1 + I/(mR²)), where g = 9.80665 m/s². A lower I/(mR²) ratio means less of the roll's gravitational potential energy is diverted into spin, so it accelerates faster — which is exactly why a full roll tends to out-roll a nearly empty one of the same outer size.

Common mistakes and edge cases

  • Mixing units: enter both radii in the same unit selector; do not enter the outer radius in centimeters and the inner radius in inches.
  • Inner radius must be smaller than outer radius: r₁ = r₂ describes a thin hoop (I = mr₂²), not a roll with paper on it — if they are equal there is no wound paper left.
  • Setting r₁ = 0: this reduces the formula to a solid disk or cylinder, I = ½mr₂² — useful for modeling a bare spindle or a fully solid roll with no core hole.
  • Rolling without slipping assumption: the downhill acceleration formula assumes the roll rolls (not slides) and ignores air resistance and rolling friction losses, so a real race may be slightly slower than the ideal result.

Frequently Asked Questions

What is the moment of inertia of a toilet paper roll?
A toilet paper roll is shaped like a hollow cylinder (an annulus extruded along its axis), so its mass moment of inertia about the central axis is I = ½m(r₁² + r₂²), where m is the mass, r₁ is the inner radius (the cardboard tube), and r₂ is the outer radius of the wound paper.
Why does a fuller roll win a race down a ramp?
For rolling without slipping, linear acceleration down an incline is a = g sinθ / (1 + I/(mR²)). A fresh, full roll has a lower I/(mR²) ratio than a nearly empty roll with the same outer radius, so more of its gravitational energy goes into moving forward rather than spinning, and it accelerates faster.
What is the radius of gyration and why does it matter?
The radius of gyration k = √(I/m) is the distance from the axis at which the roll's entire mass could be concentrated as a thin ring and still have the same moment of inertia. It lets you compare rolls of different mass and size on equal footing without repeating the full formula.
What happens to the moment of inertia as the roll unwinds?
As paper unwinds, both the mass m and the outer radius r₂ shrink while the inner radius r₁ (set by the cardboard tube) stays fixed. Because I depends on m and on r₂², the moment of inertia drops sharply as the roll empties, which is why an almost-empty roll spins up and decelerates the linear motion much faster than a full one.