Parallel-Axis and Perpendicular-Axis Theorems

Parallel-Axis and Perpendicular-Axis Theorems

Direct evaluation of

(1)

is the fundamental way to compute a moment of inertia.

Once one moment of inertia is known, however, two geometric theorems can often generate others immediately.

The parallel-axis theorem relates moments of inertia about two parallel axes:

(2)

Here:

• I_CM is the moment of inertia about an axis through the center of mass,
• the new axis is parallel to the center of mass axis,
• d is the perpendicular distance between the two axes,
• M is the total mass.

The perpendicular-axis theorem applies to a planar lamina:

(3)

where the x and y axes lie in the plane of the lamina and the z axis is perpendicular to the plane, with all three axes passing through the same point.

These theorems solve different geometric problems and have different conditions of validity.

1 Why axis changes matter

Moment of inertia is

(4)

Changing the axis changes the perpendicular distance r_⊥ of essentially every mass element.

Therefore the moment of inertia generally changes.

The parallel-axis theorem answers:

(5)

The perpendicular-axis theorem answers:

(6)

The two theorems should not be interchanged.

2 Geometry of two parallel axes

Let an axis through the center of mass be parallel to a second axis displaced by perpendicular vector d.

For a mass element dm, let ρ denote its position measured from the center of mass.

Only components perpendicular to the two axes contribute to the distance from either axis.

Let

(7)

be the perpendicular component of ρ.

Relative to the shifted axis, the perpendicular displacement is

(8)

Its squared distance is

(9)

3 Derivation of the parallel-axis theorem

Start with the moment of inertia about the shifted axis:

(10)

Expand the square:

(11)

Therefore

I

= ∫ ρ⊥2 dm + ∫ d2 dm − 2d ⋅∫ ρ⊥dm.

(12)

The first term is the center of mass axis moment:

(13)

The second term is

(14)

The final term vanishes because ρ is measured from the center of mass:

(15)

Therefore its perpendicular projection also vanishes:

(16)

Hence

(17)

This is the parallel-axis theorem.

4 Why the center of mass is essential

Suppose the reference axis is parallel to the final axis but does not pass through the center of mass.

Then

(18)

in general.

The cross term does not vanish, so the simple formula

(19)

is not generally valid.

The theorem has a very specific form:

(20)

The known axis must pass through the center of mass and be parallel to the desired axis.

5 The center of mass axis is the minimum parallel-axis inertia

Since

(21)

the theorem immediately implies

(22)

for every axis parallel to a given center of mass axis.

Equality occurs only when

(23)

Thus, among a family of parallel axes, the axis through the center of mass has the smallest moment of inertia.

6 Example 1: thin rod about one end

From M05-04, a uniform thin rod of length L and mass M has moment of inertia about a perpendicular axis through its center:

(24)

The parallel axis through one end is displaced by

(25)

Therefore

Iend	= ICM + Md2	(26)
	= ML2 + M2	(27)
	= ML2 + ML2	(28)
	= ML2.	(29)

Hence

(30)

This reproduces the direct integration result from M05-04.

7 Example 2: disk about a tangent axis perpendicular to its plane

A uniform disk has

(31)

about its symmetry axis through the center.

A tangent axis perpendicular to the disk is parallel to the symmetry axis and displaced by

(32)

Thus

Itangent,z	= MR2 + MR2	(33)
	= MR2.	(34)

8 Example 3: thin ring about a tangent axis perpendicular to its plane

For a thin ring,

(35)

A parallel tangent axis lies a distance

(36)

away.

Therefore

Itangent,z	= MR2 + MR2	(37)
	= 2MR2.	(38)

9 Parallel-axis theorem for discrete masses

The theorem is equally valid for a discrete collection of particles.

Let

(39)

and let

(40)

be each particle position relative to the center of mass.

Then

(41)

For a parallel axis displaced by d,

I	= ∑ imi|ρi⊥− d|2	(42)
	= ICM + Md2.	(43)

The continuous theorem is therefore the integral form of the same center of mass cancellation.

10 Inverse use of the parallel-axis theorem

Sometimes a moment of inertia about an offset axis is known and the center of mass value is desired.

Rearrange:

(44)

Because

(45)

the result must remain nonnegative.

A negative answer signals an incorrect axis, distance, mass, or input value.

11 Perpendicular-axis theorem

Now consider a thin planar lamina lying in the xy plane.

Let the x and y axes lie in the plane and intersect at point O.

Let the z axis pass through the same point and be perpendicular to the lamina.

For a mass element at coordinates

(46)

the squared distance to the z axis is

(47)

The squared distance to the x axis is

(48)

and to the y axis,

(49)

Therefore

(50)

Multiply by dm and integrate:

Iz	= ∫ (x2 + y2) dm	(51)
	= ∫ y2 dm + ∫ x2 dm.	(52)

Thus

(53)

This is the perpendicular-axis theorem.

12 Conditions for the perpendicular-axis theorem

The formula

(54)

requires:

1. the body is planar, or treated as an ideal thin lamina,
2. the x and y axes lie in the plane,
3. the z axis is perpendicular to the plane,
4. all three axes intersect at the same point.

It is not generally valid for a three-dimensional body of finite thickness.

For example, it should not be applied directly to a solid sphere.

13 Why the theorem fails for a general three-dimensional body

For a three-dimensional mass element at

(55)

the moments about the coordinate axes are

Ix	= ∫ (y2 + z2) dm,	(56)
Iy	= ∫ (x2 + z2) dm,	(57)
Iz	= ∫ (x2 + y2) dm.	(58)

Then

Ix + Iy

= ∫ (x2 + y2 + 2z2) dm.

(59)

Only when

(60)

for every mass element does this reduce to

(61)

That is precisely the planar-lamina condition.

14 Example 4: disk about a diameter

A uniform thin disk has

(62)

about its central symmetry axis perpendicular to the disk.

By rotational symmetry in the plane,

(63)

The perpendicular-axis theorem gives

(64)

Therefore

(65)

Thus the moment of inertia of a thin disk about any diameter is

(66)

15 Example 5: thin ring about a diameter

For a thin ring,

(67)

By symmetry,

(68)

Therefore

(69)

so

(70)

16 Example 6: rectangular plate

Consider a uniform rectangular plate with side length a along x and side length b along y.

About the central x axis,

(71)

About the central y axis,

(72)

Therefore the perpendicular-axis theorem gives

Iz	= Ix + Iy	(73)
	= M(a2 + b2).	(74)

Hence

(75)

matching the direct two-dimensional integration from M05-04.

17 Combining the two theorems

The two theorems can be used sequentially.

For example, find the moment of inertia of a thin disk about a tangent line lying in the plane of the disk.

First use the perpendicular-axis theorem to find the moment about a central diameter:

(76)

Then use the parallel-axis theorem to shift that diameter to the parallel tangent line:

Itangent,in plane	= Idiameter + MR2	(77)
	= MR2 + MR2.	(78)

Thus

(79)

This is a useful example because neither theorem alone completes the problem.

18 Example 7: rectangular plate about an edge in its plane

Take the rectangular plate of side lengths a and b.

Suppose the desired axis is parallel to the x axis and lies along an edge at

(80)

The central parallel-axis value is

(81)

The displacement is

(82)

Therefore

Iedge	= Mb2 + M	(83)
	= Mb2.	(84)

Notice that the dimension a does not appear because distance to this axis depends only on the y coordinate.

19 Composite bodies

For a body composed of several pieces,

(85)

provided each I_k is evaluated about the same final axis.

A useful workflow is:

1. identify the final axis,
2. take each component’s known center of mass inertia about a parallel axis,
3. shift it to the final axis using

   (86)

4. add all component contributions.

For holes or cavities, treat removed material as a negative contribution.

20 Offset cavity

Suppose a circular hole is cut from a thin plate.

Let the full plate have inertia

(87)

about the desired axis.

Let the removed circular disk have mass

(88)

and center of mass moment

(89)

If the hole center is displaced by distance d from the desired axis, its contribution about that axis would have been

(90)

Therefore the remaining body’s moment of inertia is

(91)

21 Example 8: numerical composite rotor

A rotor contains:

• a uniform disk of mass 4.0 kg and radius 0.30 m centered on the axis,
• a point mass of 1.0 kg attached at radius 0.50 m.

The disk contributes

Id	= MR2	(92)
	= (4.0)(0.30)2	(93)
	= 0.180 kg m2.	(94)

The point mass contributes

(95)

Thus

(96)

This example does not require a parallel-axis shift for the disk because its center of mass axis already coincides with the final axis.

22 Example 9: offset disk component

A small uniform disk of mass m and radius a is mounted so that its symmetry axis is parallel to the final rotation axis but displaced by distance d.

Its center of mass inertia is

(97)

Therefore

(98)

This pattern appears frequently in gears, flywheels, eccentric rotors, and composite mechanisms.

23 A useful theorem-selection strategy

Ask the following questions.

If the new axis is parallel to a known axis through the center of mass:

(99)

If the body is a planar lamina and three axes intersect at the same point, with two in the plane and one perpendicular:

(100)

If neither geometry applies:

(101)

24 Parallel-axis theorem and radius of gyration

Write

(102)

and

(103)

The parallel-axis theorem gives

(104)

Divide by M:

(105)

Thus shifting the axis increases the squared radius of gyration by exactly d².

25 Energy consequence of an axis shift

For the same rigid body rotating at the same angular speed ω about two parallel axes,

(106)

Using

(107)

the energy difference is

ΔK

= Md2ω2.

(108)

Therefore

(109)

This is not an independent theorem, but it makes the physical effect of the larger shifted-axis inertia clear.

26 Connection to the inertia tensor

The parallel-axis theorem has a three-dimensional tensor generalization.

The perpendicular-axis theorem does not become a general three-dimensional identity of the same form because it relies on every mass element having zero coordinate Normal to the lamina.

Later articles introduce the inertia tensor, principal axes, and the full three-dimensional relation between angular momentum and angular velocity.

For now, the scalar axis moments are sufficient.

27 Common mistakes

1. Using the parallel-axis theorem when the two axes are not parallel.
2. Using I = I₀ + Md² when the known axis does not pass through the center of mass.
3. Using the distance from the center of mass to a point instead of the perpendicular distance between the two parallel axes.
4. Subtracting Md² when shifting away from the center of mass axis.
5. Forgetting that the center of mass axis gives the minimum moment among parallel axes.
6. Using the perpendicular-axis theorem for a solid three-dimensional body.
7. Applying I_z = I_x + I_y when the three axes do not intersect at the same point.
8. Forgetting that the x and y axes must lie in the plane of the lamina.
9. Assuming I_x = I_y without an actual symmetry that justifies it.
10. Combining component moments that are referred to different final axes.
11. Subtracting a cavity’s center of mass inertia without first shifting it to the desired axis.
12. Confusing the parallel-axis theorem with the perpendicular-axis theorem because both involve axis geometry.
13. Using a standard table value without checking which axis that table value refers to.
14. Forgetting that a tangent axis in the plane of a disk and a tangent axis perpendicular to the disk have different moments of inertia.

28 Practice exercises

1. Starting from I = ∫ r_⊥² dm, derive the parallel-axis theorem in vector form.
2. Explain exactly why the cross term vanishes when the reference axis passes through the center of mass.
3. Use the parallel-axis theorem to derive the moment of inertia of a uniform rod about a perpendicular axis through one end.
4. A disk has mass M and radius R. Find its inertia about an axis perpendicular to the disk and tangent to its rim.
5. A thin ring has mass M and radius R. Find its inertia about an axis perpendicular to the ring and tangent to its rim.
6. Prove the perpendicular-axis theorem directly from the coordinates (x,y) of a planar mass element.
7. Explain why the perpendicular-axis theorem fails for a general three-dimensional body.
8. Use the perpendicular-axis theorem to derive the moment of inertia of a thin disk about a diameter.
9. Use the perpendicular-axis theorem to derive the moment of inertia of a thin ring about a diameter.
10. Use both the perpendicular-axis and parallel-axis theorems to find the moment of inertia of a thin disk about a tangent line in the plane of the disk.
11. Find the moment of inertia of a rectangular plate about an in-plane axis along one edge.
12. A body has I_CM = 3.0 kg m², M = 5.0 kg, and an offset parallel axis at d = 0.40 m. Find I.
13. A body has I = 8.0 kg m² about an offset axis, M = 10 kg, and d = 0.50 m. Find I _CM.
14. Derive k_g² = k _g,CM² + d².
15. Explain how to compute the inertia of a plate with an off-center circular hole using negative mass and the parallel-axis theorem.

29 Summary

For an axis parallel to a center of mass axis and displaced by perpendicular distance d,

(110)

The center of mass axis has the smallest moment of inertia among all parallel axes:

(111)

For a planar lamina with x and y axes in the plane and a perpendicular z axis through the same point,

(112)

The parallel-axis theorem shifts an axis.

The perpendicular-axis theorem relates orthogonal axis moments of a planar body.

When neither theorem applies, the fundamental definition remains

(113)

These tools complete the geometric foundation needed for M05-06, where moment of inertia enters rotational dynamics through torque and angular acceleration.

References

References