Find the moment of inertia
Pick an object’s shape and the axis it spins about, then type its mass and size. The calculator gives the moment of inertia in kg·m² and lb·ft², and the radius of gyration.
- Moment of inertia (I)
- 1,000
The moment of inertia is 1,000 kg·m² (I = ½·M·R²).
- In lb·ft²lb·ft²
- 23,730.4
- About the shape’s own axiskg·m²
- 1,000
- Radius of gyration (k)
- 1.41421 m
- Formula used
- I = ½·M·R²
Moment of inertia (I): 1,000. The moment of inertia is 1,000 kg·m² (I = ½·M·R²).
How to calculate
Finds the moment of inertia of a point mass, hoop, disk, hollow cylinder, rod, sphere, shell or plate from its mass and size, about its own axis or an axis moved by the parallel-axis theorem.
Example with the default inputs (Shape and axis Solid cylinder or disk, about its axis, Mass (M) 500 kg, Radius (R) 2 m): The moment of inertia is 1,000 kg·m² (I = ½·M·R²).
Method: I from the shape’s formula (M·R², ½·M·R², ½·M·(R₁² + R₂²), M·L² ÷ 12, M·L² ÷ 3, ⅖·M·R², ⅔·M·R², ¼·M·R² + M·L² ÷ 12, M·(a² + b²) ÷ 12); with an offset axis, I = I_cm + M·d²; k = √(I ÷ M).
- Each object is rigid with its mass spread evenly (uniform density); a hoop, rod or shell is thin.
- The offset axis is parallel to the shape’s own axis, which passes through the center of mass (not for a point mass or a rod about its end).
- The model works in kilograms and meters; 1 lb·ft² = 0.0421401 kg·m².
Worked examples
Each example is checked against the calculator on every build.
- Shape and axis Solid cylinder or disk, about its axis, Mass (M) 500 kg, Radius (R) 2 m gives Moment of inertia (I) 1,000.Source: OpenStax, University Physics Volume 1, §10.5 Calculating Moments of Inertia (rod ML²/12 and ML²/3, disk mR²/2, parallel-axis theorem I = I_cm + md²; Example 10.11: 1,025 kg·m²). https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia, retrieved 2026-10-02: ½ × 500 × 2² = 1,000 kg·m²
- Shape and axis Thin rod, about one end, Mass (M) 2 kg, Length (L) 0.5 m gives Moment of inertia (I) 0.166667.Source: OpenStax, University Physics Volume 1, §10.5 Calculating Moments of Inertia (rod ML²/12 and ML²/3, disk mR²/2, parallel-axis theorem I = I_cm + md²; Example 10.11: 1,025 kg·m²). https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia, retrieved 2026-10-02, Example 10.12: (1/3)(2.0 kg)(0.50 m)² = 0.167 kg·m²
- Shape and axis Solid sphere, about a diameter, Mass (M) 1 kg, Radius (R) 0.2 m, Axis offset (d) 0.7 m gives About the shape’s own axis 0.016, Moment of inertia (I) 0.506.Source: OpenStax, University Physics Volume 1, §10.5 Calculating Moments of Inertia (rod ML²/12 and ML²/3, disk mR²/2, parallel-axis theorem I = I_cm + md²; Example 10.11: 1,025 kg·m²). https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia, retrieved 2026-10-02, Example 10.12: (2/5)(1.0)(0.2)² + (1.0)(0.7)² = 0.016 + 0.490; Wikipedia, List of moments of inertia (hoop mr², disk mr²/2, rod mL²/12 and mL²/3, sphere 2mr²/5, shell 2mr²/3, tube m(r₁² + r₂²)/2, plate m(h² + w²)/12, cylinder about a diameter m(3r² + h²)/12). https://en.wikipedia.org/wiki/List_of_moments_of_inertia, retrieved 2026-10-02
- Shape and axis Hollow (annular) cylinder, about its axis, Mass (M) 3 kg, Radius (R) 0.1 m, Inner radius (R₁) 0.08 m gives Moment of inertia (I) 0.0246.Source: Wikipedia, List of moments of inertia (hoop mr², disk mr²/2, rod mL²/12 and mL²/3, sphere 2mr²/5, shell 2mr²/3, tube m(r₁² + r₂²)/2, plate m(h² + w²)/12, cylinder about a diameter m(3r² + h²)/12). https://en.wikipedia.org/wiki/List_of_moments_of_inertia, retrieved 2026-10-02; OpenStax, University Physics Volume 1, §10.4 Moment of Inertia and Rotational Kinetic Energy, Figure 10.20 (moments of inertia of common shapes). https://openstax.org/books/university-physics-volume-1/pages/10-4-moment-of-inertia-and-rotational-kinetic-energy, retrieved 2026-10-02
- Shape and axis Rectangular plate, about a perpendicular axis through its center, Mass (M) 6 kg, Side a 1 m, Side b 0.5 m gives Moment of inertia (I) 0.625, Radius of gyration (k) 0.322749 m.Source: Wikipedia, List of moments of inertia (hoop mr², disk mr²/2, rod mL²/12 and mL²/3, sphere 2mr²/5, shell 2mr²/3, tube m(r₁² + r₂²)/2, plate m(h² + w²)/12, cylinder about a diameter m(3r² + h²)/12). https://en.wikipedia.org/wiki/List_of_moments_of_inertia, retrieved 2026-10-02
How it works
For mass M:
| Shape and axis | Moment of inertia |
|---|---|
| Point mass at distance r | M·r² |
| Hoop or thin ring, about its axis | M·R² |
| Solid cylinder or disk, about its axis | ½·M·R² |
| Hollow (annular) cylinder, inner radius R₁ and outer R₂, about its axis | ½·M·(R₁² + R₂²) |
| Thin rod of length L, about its center | M·L² ÷ 12 |
| Thin rod of length L, about one end | M·L² ÷ 3 |
| Solid sphere, about a diameter | ⅖·M·R² |
| Thin spherical shell, about a diameter | ⅔·M·R² |
| Hoop, about a diameter | ½·M·R² |
| Solid cylinder of length L, about a diameter through its center | ¼·M·R² + M·L² ÷ 12 |
| Rectangular plate a × b, about the perpendicular axis through its center | M·(a² + b²) ÷ 12 |
Parallel-axis theorem. For an axis parallel to the shape’s own axis through its center of mass, at a distance d from it: I = I_cm + M·d². The offset is not offered for a point mass or for a rod about its end, whose axes do not pass through the center of mass.
Radius of gyration: k = √(I ÷ M). In lb·ft²: I ÷ (0.45359237 × 0.3048²) = I ÷ 0.0421401.
Rules
- The mass is more than 0 and at most 10¹² kg; each size is more than 0 and at most 1,000 km; the offset is from 0 to 1,000 km.
- For a hollow cylinder the inner radius may not be larger than the outer radius. Radii typed in different units that differ by less than one part in 10¹² count as equal (an inner radius equal to the outer one gives a thin tube, M·R²).
- Uniform, rigid objects; a hoop, rod or shell is thin.
Output format. I in kg·m² and lb·ft², and k in meters, to 6 significant figures.
Worked examples by hand
Disk, 500 kg, R = 2 m. I = ½ × 500 × 2² = 1,000 kg·m² (OpenStax Example 10.11, the merry-go-round).
Rod about one end, 2 kg, L = 0.5 m. I = 2 × 0.5² ÷ 3 = 0.5 ÷ 3 = 0.1667 kg·m² (OpenStax Example 10.12).
Sphere, 1 kg, R = 0.2 m, axis 0.7 m from the center. I_cm = ⅖ × 1 × 0.04 = 0.016. I = 0.016 + 1 × 0.7² = 0.016 + 0.49 = 0.506 kg·m² (OpenStax Example 10.12).
Hollow cylinder, 3 kg, R₂ = 0.1 m, R₁ = 0.08 m. I = ½ × 3 × (0.01 + 0.0064) = 0.0246 kg·m².
Plate 1 m × 0.5 m, 6 kg. I = 6 × (1 + 0.25) ÷ 12 = 0.625 kg·m². k = √(0.625 ÷ 6) = 0.3227 m.
Other questions people ask
What is the moment of inertia?
A measure of how hard it is to change an object’s rotation, the rotational version of mass. It depends on the mass and on how far that mass is from the axis: I = Σ m·r² over every bit of the object.
What is the moment of inertia of a disk?
A solid disk or cylinder about its axis has I = ½·M·R². A 500 kg merry-go-round of radius 2.0 m has I = ½ × 500 × 2² = 1,000 kg·m².
What is the moment of inertia of a rod?
A thin rod of length L has I = M·L² ÷ 12 about its center and M·L² ÷ 3 about one end. A 2.0 kg rod 0.50 m long, turned about one end, has I = 2 × 0.25 ÷ 3 = 0.167 kg·m².
What is the parallel-axis theorem?
To find the moment of inertia about an axis parallel to one through the center of mass, add M·d², where d is the distance between the axes: I = I_cm + M·d². A 1 kg sphere of radius 0.2 m on an axis 0.7 m from its center has I = 0.016 + 0.49 = 0.506 kg·m².
Why does a hoop have more inertia than a disk of the same mass?
All of a hoop’s mass is at the rim, at distance R, so I = M·R². A disk has much of its mass close to the axis, so I = ½·M·R², half as much.
What is the radius of gyration?
The distance k from the axis at which the whole mass, put in one place, would give the same moment of inertia: k = √(I ÷ M). For a solid disk, k = R ÷ √2.