Case 1
- Inertia about the centre of mass (kg·m²)
- 2kg·m²
- Mass of the body (kg)
- 5kg
- Distance between the two axes (m)
- 3m
47.00kg·m²
Open with these valuesParallel Axis Theorem Calculator
Change any value to update the result.
47.00kg·m²
Inertia about the new axis: 47.00 kg·m²Examples
How the result moves
Distance between the two axes (m) (m) → Inertia about the new axis (kg·m²)Your current valid inputs, with one value varied across the range.
How the result moves
Held fixed: Inertia about the centre of mass (kg·m²) 2.00000 kg·m², Mass of the body (kg) 5.000 kg.
| Distance between the two axes (m) (m) | Inertia about the new axis (kg·m²) |
|---|---|
| 0.000 | 2.00 |
| 2.000 | 22.00 |
| 3.000Your value | 47.00 |
| 4.000 | 82.00 |
| 6.000 | 182.00 |
Shift the axis away from the centre of mass and the inertia grows by m × d²: 2 kg·m² plus 5 kg at 3 m gives 47 kg·m². The distance is measured between the two parallel axes, not to the edge of the body. Because d is squared, the centre-of-mass axis always has the smallest inertia of any parallel axis.
Worked examples
Case 2
- Inertia about the centre of mass (kg·m²)
- 0kg·m²
- Mass of the body (kg)
- 10kg
- Distance between the two axes (m)
- 2m
40.00kg·m²
Open with these valuesCase 3
- Inertia about the centre of mass (kg·m²)
- 100kg·m²
- Mass of the body (kg)
- 50kg
- Distance between the two axes (m)
- 10m
5,100.00kg·m²
Open with these valuesHow it's calculated
I = I_cm + m × d²
- StepEnter the moment of inertia about the centre-of-mass axis.Inertia about the centre of mass (kg·m²)=2kg·m²
- StepEnter the total mass of the body.Mass of the body (kg)=5kg
- StepEnter how far the new axis sits from the centre-of-mass axis.Inertia about the centre of mass (kg·m²)=2kg·m²
- ResultRead the inertia about the new axis; it is never smaller than I_cm.Inertia about the centre of mass (kg·m²)=2kg·m²→47.00kg·m²
Reference table
| I_cm, mass, distance | What it shows | Inertia |
|---|---|---|
| 5, 3, 0 | The axes coincide, nothing is added | 5 |
| 1.5, 2, 4 | A light body pushed well out | 33.5 |
| 0, 10, 2 | A pure point mass, 2 m out | 40 |
| 2, 5, 3 | The worked example | 47 |
| 2, 5, 6 | Twice the distance, four times the m·d² term | 182 |
| 100, 50, 10 | A heavy body, far off axis | 5100 |
Questions
What is the parallel axis theorem?
It gives the moment of inertia about any axis parallel to one through the centre of mass: I = I_cm + m × d². Here m is the mass and d the perpendicular distance between the two parallel axes.
Which distance do I enter?
The perpendicular distance between the new axis and the parallel axis through the centre of mass — not the distance to an edge or a corner. The two axes must genuinely be parallel; otherwise the simple theorem does not apply.
Where do I get the centre-of-mass inertia?
You enter it, because it depends on the shape. A point mass or thin ring is mr², a solid disc ½mr², a solid sphere ⅖mr², a rod about its centre 1/12 mL².
Why does moving the axis always increase the inertia?
The added term m × d² is never negative, since mass is positive and the distance is squared. The centre-of-mass axis is therefore the axis of least inertia in any given direction.
What happens when the distance is zero?
The two axes coincide, the m × d² term vanishes and the result equals I_cm unchanged. That is the fifth row of the table above.