Euler's Equations for Rigid Body Rotation

Euler’s Equations for Rigid Body Rotation

The rotational dynamics law in an inertial frame is

(1)

For a general three-dimensional rigid body,

(2)

need not be parallel to

(3)

M05-09 introduced

(4)

and M05-10 introduced principal axes, in which

(5)

The principal axes are fixed in the body, so they rotate relative to an inertial frame.

This creates the key subtlety of three-dimensional rigid body dynamics:

(6)

Euler’s equations resolve this by combining angular-momentum dynamics with the derivative rule for a rotating frame.

The result is

τ1 = I1ω1 + (I3 − I2)ω2ω3,		(7)
τ2 = I2ω2 + (I1 − I3)ω3ω1,		(8)
τ3 = I3ω3 + (I2 − I1)ω1ω2.		(9)

These are Euler’s equations for rigid body rotation in principal-axis body coordinates.

1 Inertial frame and body frame

Let

(10)

be a fixed inertial basis.

Let

(11)

be a body-fixed principal-axis basis.

The body basis rotates with angular velocity

(12)

relative to the inertial basis.

The angular velocity vector can be written in body components as

(13)

The components

(14)

can change even while the body basis itself rotates.

2 Derivative of a vector in a rotating frame

For any physical vector

(15)

the inertial and rotating-frame derivatives are related by

(16)

To see why, write

(17)

Differentiate in the inertial frame:

I	= Ȧ1e1 + Ȧ2e2 + Ȧ3e3	(18)
	+ A1e1 + A2e2 + A3e3.	(19)

For a basis rotating with angular velocity

(20)

(21)

Therefore

(22)

The remaining component derivatives form

(23)

Thus the rotating derivative theorem follows.

3 Angular momentum in principal-axis coordinates

Choose the body axes to be principal axes through the center of mass.

Then

(24)

Therefore

(25)

Because the body is rigid,

(26)

are constant in the body frame.

4 Derivation of Euler’s vector equation

Start with

(27)

Apply the rotating-frame derivative theorem:

(28)

In the principal body frame,

(29)

and the matrix

(30)

is constant.

Thus

(31)

Therefore

(32)

This is Euler’s equation in compact vector form.

The overdot on the body components means differentiation of the components

(33)

with respect to time.

5 Derivation of the component equations

Write

(34)

and

(35)

The cross product is

ω × L

= .

(36)

Its first component is

(37)

The second is

(38)

The third is

(39)

Therefore

τ1 = I1ω1 + (I3 − I2)ω2ω3,		(40)
τ2 = I2ω2 + (I1 − I3)ω3ω1,		(41)
τ3 = I3ω3 + (I2 − I1)ω1ω2.		(42)

6 What the coupling terms mean

Consider the first equation:

(43)

The first term,

(44)

resembles the fixed-axis law

(45)

The second term,

(46)

arises because the principal-axis basis rotates while angular momentum has components along that rotating basis.

This term is often called a gyroscopic coupling term.

It is not an additional physical torque.

It appears because the angular-momentum components are being differentiated in a rotating body frame.

7 Why diagonal inertia does not remove dynamic coupling

In principal coordinates the inertia tensor is diagonal:

(47)

Nevertheless Euler’s equations remain coupled through products such as

(48)

Diagonalization removes algebraic coupling in

(49)

It does not remove the kinematic coupling caused by the rotating basis.

This distinction is fundamental.

8 Special case: rotation about one principal axis

Suppose

(50)

Then Euler’s equations reduce to

(51)

The other two equations become

(52)

Thus the familiar scalar law

(53)

is recovered for rotation exactly about a principal axis.

9 Example 1: principal-axis spin-up

A rotor has

(54)

It initially rotates only about principal axis 3 and experiences torque

(55)

Because

(56)

Euler’s third equation becomes

(57)

Therefore

(58)

This is exactly the fixed-axis result.

10 Torque-free Euler equations

For torque-free motion,

(59)

Euler’s equations become

I1ω1 + (I3 − I2)ω2ω3 = 0,		(60)
I2ω2 + (I1 − I3)ω3ω1 = 0,		(61)
I3ω3 + (I2 − I1)ω1ω2 = 0.		(62)

Even with zero external torque, the body-frame angular-velocity components can vary.

This does not violate angular-momentum conservation.

The physical vector

(63)

is constant in inertial space, while the body axes rotate beneath it.

11 Conservation of angular-momentum magnitude

For torque-free motion,

(64)

Therefore

(65)

Its magnitude is also constant:

(66)

This gives one conserved quadratic surface in angular-velocity component space.

12 Conservation of rotational kinetic energy

For a rigid body,

(67)

Power delivered by external torque is

(68)

Therefore

(69)

For torque-free motion,

(70)

Thus torque-free motion conserves both:

• angular-momentum magnitude,
• rotational kinetic energy.

13 Checking energy conservation directly from Euler’s equations

Multiply the three torque-free equations by

(71)

respectively, and add.

The derivative terms give

(72)

The coupling terms sum to

	ω1ω2ω3	(73)
	= 0.	(74)

Hence

(75)

The nonlinear coupling redistributes rotational motion among body components without changing total kinetic energy.

14 Symmetric rigid body

Consider an axisymmetric body with

(76)

For torque-free motion, Euler’s third equation becomes

(77)

Therefore

(78)

The first two equations become

I⊥ω1 + (I∥− I⊥)ω2ω3	= 0,	(79)
I⊥ω2 − (I∥− I⊥)ω3ω1	= 0.	(80)

Define

(81)

Then

ω1	= −ΩBω2,	(82)
ω2	= ΩBω1.	(83)

15 Body-frame precession of angular velocity

Differentiate the first symmetric-body equation:

(84)

Using

(85)

(86)

Similarly,

(87)

Therefore

(88)

The transverse angular-velocity vector rotates around the symmetry axis in the body frame at angular frequency

(89)

16 Example 2: torque-free symmetric body

Let

I⊥	= 2.0 kg m2,	(90)
I∥	= 3.0 kg m2,	(91)
ω3	= 10 rad∕s.	(92)

Then

ΩB	= (10)	(93)
	= 5.0 rad∕s.	(94)

Thus the transverse body-frame angular-velocity components rotate with frequency

(95)

If initially

(96)

one convenient solution is

ω1(t)	= cos(5t),	(97)
ω2(t)	= sin(5t),	(98)

in radians per second.

17 Steady torque-free rotation about a principal axis

Set two angular-velocity components to zero.

For example,

(99)

The torque-free Euler equations then give

(100)

Therefore rotation exactly about any principal axis is a torque-free steady solution.

But not every such solution has the same stability.

18 Stability about the minimum principal axis

Order the principal moments as

(101)

Consider dominant rotation about axis 1:

(102)

with small perturbations

(103)

To first order, the second and third Euler equations give

I2ω2 + (I1 − I3)Ωω3	= 0,	(104)
I3ω3 + (I2 − I1)Ωω2	= 0.	(105)

Differentiate the first and substitute the second:

(106)

The coefficient is positive.

Therefore the perturbation oscillates rather than grows exponentially.

Rotation about the minimum-moment principal axis is linearly stable.

19 Stability about the maximum principal axis

Now consider dominant rotation about axis 3:

(107)

Linearization gives an equation of the form

(108)

Again the coefficient is positive.

Thus rotation about the maximum-moment principal axis is also linearly stable.

20 Intermediate-axis instability

Now consider dominant rotation about axis 2:

(109)

The linearized first and third equations give

I1ω1 + (I3 − I2)Ωω3	= 0,	(110)
I3ω3 + (I2 − I1)Ωω1	= 0.	(111)

Differentiate the first equation and substitute the second:

(112)

The coefficient multiplying

(113)

appears with a minus sign.

Solutions therefore contain exponentially growing and decaying terms.

Rotation about the intermediate principal axis is unstable.

This is the intermediate-axis theorem, also called the tennis-racket theorem.

21 Example 3: growth rate near the intermediate axis

Let

(114)

in consistent inertia units.

Suppose the dominant intermediate-axis spin is

(115)

The perturbation equation is

(116)

where

λ	= Ω	(117)
	= 6	(118)
	= 2 s−1.	(119)

Thus

(120)

Small perturbations can therefore grow rapidly.

22 Applied torque and gyroscopic response

When

(121)

Euler’s equations show that an applied torque component does not necessarily change only the corresponding angular-velocity component.

For example,

(122)

Even if

(123)

the component

(124)

can change if

(125)

and

(126)

This cross-component response is central to gyroscopes and precession.

23 Example 4: instantaneous coupling

Suppose

I1	= 2 kg m2,	(127)
I2	= 3 kg m2,	(128)
I3	= 5 kg m2,	(129)

and at one instant

ω1	= 0,	(130)
ω2	= 4 rad∕s,	(131)
ω3	= 6 rad∕s.	(132)

Let

(133)

Euler’s first equation gives

0	= 2ω1 + (5 − 3)(4)(6)	(134)
	= 2ω1 + 48.	(135)

Therefore

(136)

A body-axis angular-velocity component changes even though no external torque acts along that body axis.

24 Power in body coordinates

The instantaneous rotational power is

(137)

In principal-axis components,

(138)

Euler’s equations are consistent with

(139)

The gyroscopic coupling terms do not create or destroy energy by themselves.

Their net contribution to

(140)

cancels.

25 Angular acceleration versus body-component derivatives

A subtle notation issue deserves emphasis.

The physical angular acceleration vector is

(141)

Using the rotating derivative theorem on

(142)

itself gives

(143)

But

(144)

Therefore

(145)

So the body-component derivatives of angular velocity are indeed the body components of angular acceleration.

This simplification is special to differentiating

(146)

itself.

26 Euler equations in matrix form

Let

(147)

Then

(148)

Therefore

(149)

This form is convenient for numerical rigid body simulation.

27 Time integration and attitude

Euler’s equations determine the angular-velocity components.

They do not by themselves determine the body’s orientation.

A complete rigid body simulation must also propagate attitude, for example with:

• a direction cosine matrix,
• Euler Angles,
• a unit quaternion.

Thus a three-dimensional simulation contains two coupled pieces:

1. rotational dynamics for

   (150)

2. attitude kinematics for the body’s orientation.

This distinction is important in spacecraft, aircraft, robotics, and inertial-navigation applications.

28 Numerical example: one Euler-equation step

Let

(151)

(152)

and

(153)

Euler’s equations give

ω1	=	(154)
	= −4 rad∕s2,	(155)
ω2	=	(156)
	= 3 rad∕s2,	(157)
ω3	=	(158)
	= −0.8 rad∕s2.	(159)

Therefore

(160)

For a short explicit Euler time step

(161)

(162)

Thus

(163)

A production simulation would normally use a higher-quality integrator and propagate attitude consistently.

29 Common mistakes

1. Writing τ = Iα as the complete law for arbitrary three-dimensional rotation.
2. Forgetting that the principal-axis basis rotates with the body.
3. Differentiating body components as though the body basis were inertially fixed.
4. Omitting the term ω × L.
5. Using Euler’s principal-axis component equations in a basis that does not diagonalize the inertia tensor.
6. Mixing inertial-frame components and body-frame components in one equation.
7. Treating gyroscopic coupling terms as extra external torques.
8. Assuming zero external torque implies constant body-frame angular-velocity components.
9. Assuming torque-free motion means the body rotates about one fixed body axis.
10. Forgetting that angular momentum is constant in inertial space during torque-free motion.
11. Forgetting that kinetic energy is also constant for a torque-free rigid body.
12. Assuming all three principal-axis rotations have the same stability.
13. Forgetting that the intermediate principal axis is unstable.
14. Confusing body-frame precession of ω with inertial precession of a symmetry axis.
15. Integrating angular velocity without separately propagating attitude.

30 Practice exercises

1. Derive the rotating-frame derivative theorem for an arbitrary vector.
2. Starting from τ = (dL∕dt)_I, derive the compact body-frame equation

   (164)

3. Expand the compact equation into Euler’s three principal-axis component equations.
4. Show that rotation exactly about one principal axis reduces Euler’s equations to τ_i = I_iω_i.
5. For torque-free motion, prove directly from Euler’s equations that rotational kinetic energy is constant.
6. Prove that L² = I ₁²ω ₁² + I ₂²ω ₂² + I ₃²ω ₃² is constant in torque-free motion.
7. Derive the torque-free symmetric-body equations for I₁ = I₂ = I_⊥.
8. Show that ω₁² + ω ₂² is constant for the torque-free symmetric body.
9. Derive the body-frame precession frequency Ω_B = (I_∥− I_⊥)ω₃∕I_⊥.
10. Linearize the torque-free equations about steady rotation around the minimum principal axis and show that the perturbations oscillate.
11. Repeat for the maximum principal axis.
12. Linearize about the intermediate principal axis and show that one perturbation mode grows exponentially.
13. For I₁ = 1, I₂ = 2, I₃ = 3, and Ω = 6 rad/s, reproduce the intermediate-axis growth rate.
14. For I = diag(2, 3, 5) kg m², ω = (1, 2, 3) rad/s, and τ = (4, 0,−2) N m, compute ω.
15. Explain why Euler’s equations must be combined with attitude kinematics in a full rigid body simulation.

31 Summary

The inertial angular-momentum law is

(165)

For a body-fixed rotating frame,

(166)

In principal-axis coordinates,

(167)

so

(168)

The component equations are

τ1 = I1ω1 + (I3 − I2)ω2ω3,		(169)
τ2 = I2ω2 + (I1 − I3)ω3ω1,		(170)
τ3 = I3ω3 + (I2 − I1)ω1ω2.		(171)

For torque-free motion:

• L is constant in inertial space,
• rotational kinetic energy is constant,
• body-frame angular-velocity components can still vary.

For

(172)

rotation about the minimum and maximum principal axes is stable to small perturbations, while rotation about the intermediate principal axis is unstable.

M05-12 uses these results to study gyroscopes, precession, and nutation.

References

References