GRE Physics Companion: Euler’s Equations for Rigid Body Rotation
In a principal-axis body frame,
| τ1 = I1ω1 + (I3 − I2)ω2ω3, | | (1)
|
| τ2 = I2ω2 + (I1 − I3)ω3ω1, | | (2)
|
| τ3 = I3ω3 + (I2 − I1)ω1ω2. | | (3) |
Figure 1. A compact strategy for Euler-equation problems: use principal axes, identify torque
components, retain gyroscopic coupling terms, then apply conservation laws or linearization where
appropriate.
1 High-value GRE facts
- Euler’s equations are most compact in a body-fixed principal-axis frame.
- The rotating derivative theorem produces the ω × L term.
- A diagonal inertia tensor does not eliminate dynamic coupling.
- Rotation exactly about one principal axis reduces to τi = Iiωi.
- Zero external torque does not imply constant body-frame angular-velocity components.
- Torque-free motion conserves angular momentum in inertial space.
- torque-free rigid body motion also conserves rotational kinetic energy.
- For an axisymmetric torque-free body, the symmetry-axis angular-velocity component
is constant.
- Minimum- and maximum-moment principal-axis rotations are stable.
- Intermediate-axis rotation is unstable.
Part I: Original GRE-style problems
Problem 1: rotating derivative
For a vector A expressed in a frame rotating with angular velocity ω,
- (dA∕dt)I = (dA∕dt)B
- (dA∕dt)I = (dA∕dt)B + ω × A
- (dA∕dt)I = ω ⋅ A
- (dA∕dt)I = A ×ω only
- (dA∕dt)I = 0
Problem 2: first Euler equation
Which is Euler’s first principal-axis equation?
- τ1 = I1ω1
- τ1 = I1ω1 + (I3 − I2)ω2ω3
- τ1 = I2ω2 + (I1 − I3)ω3ω1
- τ1 = I1ω12
- τ1 = I2I3ω2ω3
Problem 3: principal-axis spin
A body rotates only about principal axis 3. Then Euler’s third equation becomes
- τ3 = I3ω3
- τ3 = I1ω1
- τ3 = (I2 − I1)ω1ω2 only
- τ3 = 0 necessarily
- τ3 = I3ω32
Problem 4: torque-free body
For a torque-free rigid body,
- all body-frame angular-velocity components must be constant
- angular momentum is constant in inertial space
- rotational kinetic energy must decrease
- angular momentum must vanish
- the body must rotate about the intermediate axis
Problem 5: axisymmetric body
For a torque-free axisymmetric body with I1 = I2, which angular-velocity component is
constant?
- ω1 only
- ω2 only
- ω3
- all three necessarily
- none
Problem 6: body-frame precession frequency
For I1 = I2 = I⊥, I3 = I∥, the body-frame transverse frequency is
- ω3
- (I∥− I⊥)ω3∕I⊥
- I⊥ω3∕I∥
- (I∥ + I⊥)ω3
- zero always
Problem 7: stable axes
For I1 < I2 < I3, which torque-free principal-axis rotations are linearly stable?
- only axis 1
- only axis 2
- only axis 3
- axes 1 and 3
- all three
Problem 8: intermediate-axis theorem
For I1 < I2 < I3, rotation mainly about axis 2 is
- linearly stable
- unstable to small perturbations
- impossible
- possible only with I1 = I3
- always static
Problem 9: energy
For a torque-free rigid body,
is
- constant
- always zero
- always increasing
- always decreasing
- equal to L
Problem 10: gyroscopic coupling
If τ1 = 0 but ω2ω3≠0 and I2≠I3, then
- ω1 must be zero
- ω1 can be nonzero
- ω1 is undefined
- angular momentum vanishes
- the body is in pure translation
Problem 11: numerical Euler equation
Let
and
Then ω1 is
- −24
- −12
- 0
- 12
- 24
Problem 12: attitude propagation
Euler’s equations alone determine
- the complete orientation history without any additional equation
- only translational position
- body-frame angular-velocity dynamics, while attitude requires separate kinematics
- only center of mass motion
- only the inertia tensor
Part II: Complete worked solutions
Solution 1
The rotating derivative theorem is
Answer: (B).
Solution 2
Euler’s first principal-axis equation is
Answer: (B).
Solution 3
With
the coupling term in the third equation vanishes:
Answer: (A).
Solution 4
Zero external torque implies
Thus angular momentum is constant as an inertial vector. Answer: (B).
Solution 5
For
the third torque-free Euler equation becomes
Hence
Answer: (C).
Solution 6
The transverse equations reduce to
where
Answer: (B).
Solution 7
For ordered moments
small perturbations oscillate about axes 1 and 3. Both are stable. Answer: (D).
Solution 8
Linearization about axis 2 produces an equation of the form
The growing exponential solution makes the intermediate-axis rotation unstable. Answer:
(B).
Solution 9
For torque-free motion,
Thus
Answer: (A).
Solution 10
From
a nonzero coupling term requires a compensating
Answer: (B).
Solution 11
| 0 | = 2ω1 + (5 − 3)(4)(6) | (24)
|
| = 2ω1 + 48. | (25) |
Therefore
Answer: (A).
Solution 12
Euler’s equations propagate angular-velocity components. Orientation requires separate attitude
kinematics. Answer: (C).
2 GRE checklist
- Use Euler’s component equations only in a principal-axis body frame unless using the
more general tensor form.
- Keep the ω × L term when converting inertial derivatives to body derivatives.
- Remember that zero torque preserves inertial angular momentum, not necessarily body
components of angular velocity.
- Use energy and angular-momentum invariants as fast checks in torque-free problems.
- For axisymmetric bodies, exploit I1 = I2 immediately.
- Minimum- and maximum-moment spins are stable; intermediate-moment spin is
unstable.
- gyroscopic coupling can change one body component even when the matching torque
component is zero.
- Separate rotational dynamics from attitude kinematics.
References
References
[1] PhysicsLibrary, M05-10, Principal Axes and Principal Moments of Inertia.
[2] J. R. Taylor, Classical Mechanics, University Science Books, 2005.
[3] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley,
2002.
[4] K. R. Symon, Mechanics, 3rd ed., Addison-Wesley, 1971.