Parallel & Perpendicular Axis Theorems

Two shortcuts that turn one known moment of inertia into many.

11 minUpdated 21 October 2026parallel axis theoremperpendicular axis theoremradius of gyration

Lesson at a glanceShow the key points

Parallel axis theorem: I = I(cm) + Md², where d is the distance between the given axis and a parallel axis through the centre of mass.
I is smallest about an axis through the centre of mass.
Perpendicular axis theorem (plane bodies only): Iz = Ix + Iy for x and y axes in the plane and z perpendicular, all through one point.
Ring about a diameter: MR²/2; disc about a diameter: MR²/4; solid cylinder about a line on its surface: (3/2)MR²; solid sphere about a tangent: (7/5)MR².
Radius of gyration k: I = Mk²; for a disc about its axis k = R/√2.

Integrating for every new axis would be exhausting. Two theorems let one known moment of inertia generate many others.

The parallel axis theorem

Parallel axes

I = I_{cm} + Md^2

I about any axis = I about the parallel axis through the centre of mass + M × (distance between them)²

Why it works. Put the centre of mass C at the origin with the z-axis through it, and let the given axis be the parallel line through (d, 0, 0). A particle at (x, y, z) has squared distance $x^2 + y^2$ from the z-axis and $(x - d)^2 + y^2$ from the given axis. So

I = \sum m_i\big[(x_i - d)^2 + y_i^2\big] = \sum m_i(x_i^2 + y_i^2) + Md^2 - 2d\sum m_i x_i

The first sum is $I_{cm}$ . The last sum is M times the x-coordinate of the centre of mass, which is zero because the centre of mass is at the origin. Done.

Interactive

Moving the axis away from the centre of mass

A 1 kg body turns about an axis perpendicular to the screen (the ⊙ marker). Drag the axis. I = I(cm) + Md²: the curve is a parabola with its minimum at the centre of mass. Faint copies show the body swung about the chosen axis.

\tfrac1{12}ML^2

0.083kg·m²

Md^2

0.250kg·m²

I = I_{cm} + Md^2

0.333kg·m²

k = \sqrt{I/M}

0.577m

Body

Axis distance from the centre of mass

d

0.50m

Worked example

A rod swung from one end

Find the moment of inertia of a uniform 2.0 kg rod, 1.2 m long, about a perpendicular axis through one end.

The perpendicular axis theorem

This one works only for flat (plane) bodies. Put the body in the x–y plane with the z-axis perpendicular to it, all three axes through one point:

Perpendicular axes (plane bodies)

I_z = I_x + I_y

Why. For a particle in the plane, its squared distance from the z-axis is $x^2 + y^2$ , the sum of its squared distances from the y-axis (x²) and the x-axis (y²). Multiply by mass and add over all particles.

Interactive

Perpendicular axes: Iz = Ix + Iy

A flat 1 kg plate lies in the x–y plane; z points out of the screen. Ix depends only on the plate's spread in y (its width b), Iy only on its spread in x (length a). Their sum is exactly Iz, because every particle's r² = x² + y².

Ix + Iy

I_x = Mb^2/12

0.0300kg·m²

I_y = Ma^2/12

0.1200kg·m²

I_z

0.1500kg·m²

Length along x

a

1.20m

Width along y

b

0.60m

Worked example

A disc about a diameter, then about a tangent

A uniform disc has mass M and radius R. Find its moment of inertia about (a) a diameter, (b) a tangent lying in its plane.

Two more results that follow from the parallel axis theorem:

Solid cylinder about a line on its curved surface, parallel to its axis: $\tfrac12MR^2 + MR^2 = \tfrac32MR^2$ .
Solid sphere about a tangent: $\tfrac25MR^2 + MR^2 = \tfrac75MR^2$ .

Radius of gyration

Definition · Radius of gyration

The distance k defined by

I = Mk^2

: if all the body’s mass were packed into a thin ring of radius k about the same axis, it would have the same moment of inertia.

For a disc about its axis, $Mk^2 = MR^2/2$ , so $k = R/\sqrt2 \approx 0.71R$ . For a solid sphere about a diameter, $k = \sqrt{2/5}\,R \approx 0.63R$ .

What to remember

1Parallel axes: $I = I_{cm} + Md^2$ ; I is least through the centre of mass.
2Perpendicular axes (flat bodies only): $I_z = I_x + I_y$ .
3Ring about a diameter MR²/2; disc about a diameter MR²/4; rod about an end ML²/3.
4Radius of gyration: $I = Mk^2$ .

Check yourself

Can the perpendicular axis theorem give a solid sphere's I about a diameter from its I about another diameter?

A body has I = 0.5 kg·m² about an axis through its centre of mass. Its mass is 2 kg. About a parallel axis 0.5 m away, I is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchParallel axis theoremMIT OpenCourseWare 8.01SC ReadParallel axis theoremHyperPhysics ReadPerpendicular axis theoremHyperPhysics ReadParallel axis theoremWikipedia ReadPerpendicular axis theoremWikipedia ReadRadius of gyrationWikipedia

PreviousCalculating Moment of Inertia Next Rolling: Rotation Plus Translation

Parallel & Perpendicular Axis Theorems

Two shortcuts that turn one known moment of inertia into many.

11 minUpdated 21 October 2026parallel axis theoremperpendicular axis theoremradius of gyration

Lesson at a glanceShow the key points

Parallel axis theorem: I = I(cm) + Md², where d is the distance between the given axis and a parallel axis through the centre of mass.
I is smallest about an axis through the centre of mass.
Perpendicular axis theorem (plane bodies only): Iz = Ix + Iy for x and y axes in the plane and z perpendicular, all through one point.
Ring about a diameter: MR²/2; disc about a diameter: MR²/4; solid cylinder about a line on its surface: (3/2)MR²; solid sphere about a tangent: (7/5)MR².
Radius of gyration k: I = Mk²; for a disc about its axis k = R/√2.

Integrating for every new axis would be exhausting. Two theorems let one known moment of inertia generate many others.

The parallel axis theorem

Parallel axes

I = I_{cm} + Md^2

I about any axis = I about the parallel axis through the centre of mass + M × (distance between them)²

I = \sum m_i\big[(x_i - d)^2 + y_i^2\big] = \sum m_i(x_i^2 + y_i^2) + Md^2 - 2d\sum m_i x_i

The first sum is $I_{cm}$ . The last sum is M times the x-coordinate of the centre of mass, which is zero because the centre of mass is at the origin. Done.

Interactive

Moving the axis away from the centre of mass

\tfrac1{12}ML^2

0.083kg·m²

Md^2

0.250kg·m²

I = I_{cm} + Md^2

0.333kg·m²

k = \sqrt{I/M}

0.577m

Body

Axis distance from the centre of mass

d

0.50m

Worked example

A rod swung from one end

Find the moment of inertia of a uniform 2.0 kg rod, 1.2 m long, about a perpendicular axis through one end.

The perpendicular axis theorem

This one works only for flat (plane) bodies. Put the body in the x–y plane with the z-axis perpendicular to it, all three axes through one point:

Perpendicular axes (plane bodies)

I_z = I_x + I_y

Interactive

Perpendicular axes: Iz = Ix + Iy

Ix + Iy

I_x = Mb^2/12

0.0300kg·m²

I_y = Ma^2/12

0.1200kg·m²

I_z

0.1500kg·m²

Length along x

a

1.20m

Width along y

b

0.60m

Worked example

A disc about a diameter, then about a tangent

A uniform disc has mass M and radius R. Find its moment of inertia about (a) a diameter, (b) a tangent lying in its plane.

Two more results that follow from the parallel axis theorem:

Solid cylinder about a line on its curved surface, parallel to its axis: $\tfrac12MR^2 + MR^2 = \tfrac32MR^2$ .
Solid sphere about a tangent: $\tfrac25MR^2 + MR^2 = \tfrac75MR^2$ .

Radius of gyration

Definition · Radius of gyration

The distance k defined by

I = Mk^2

: if all the body’s mass were packed into a thin ring of radius k about the same axis, it would have the same moment of inertia.

For a disc about its axis, $Mk^2 = MR^2/2$ , so $k = R/\sqrt2 \approx 0.71R$ . For a solid sphere about a diameter, $k = \sqrt{2/5}\,R \approx 0.63R$ .

What to remember

1Parallel axes: $I = I_{cm} + Md^2$ ; I is least through the centre of mass.
2Perpendicular axes (flat bodies only): $I_z = I_x + I_y$ .
3Ring about a diameter MR²/2; disc about a diameter MR²/4; rod about an end ML²/3.
4Radius of gyration: $I = Mk^2$ .

Check yourself

Can the perpendicular axis theorem give a solid sphere's I about a diameter from its I about another diameter?

A body has I = 0.5 kg·m² about an axis through its centre of mass. Its mass is 2 kg. About a parallel axis 0.5 m away, I is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousCalculating Moment of Inertia Next Rolling: Rotation Plus Translation

The parallel axis theorem#

The perpendicular axis theorem#

Radius of gyration#

Check yourself

Keep exploring

The parallel axis theorem#

The perpendicular axis theorem#

Radius of gyration#

Check yourself

Keep exploring

The parallel axis theorem

The perpendicular axis theorem

Radius of gyration

The parallel axis theorem

The perpendicular axis theorem

Radius of gyration