GRE Physics Companion: Principal Axes and Principal Moments of Inertia
Principal axes satisfy
In the principal-axis basis,
Figure 1. A compact strategy for principal-axis problems: exploit symmetry first, solve the
eigenvalue equation, normalize eigenvectors, verify orthogonality, and check the diagonalization
with tensor invariants.
1 High-value GRE facts
- Principal axes are eigenvectors of the inertia tensor.
- Principal moments are its eigenvalues.
- A real symmetric inertia tensor has real eigenvalues and orthogonal eigenvectors.
- products of inertia vanish in principal coordinates.
- Distinct principal moments imply orthogonal principal axes.
- Repeated principal moments make the corresponding principal directions nonunique.
- Rotating exactly about a principal axis gives L ∥ ω.
- Principal moments are stationary values of the scalar moment of inertia over axis
direction.
- trace and determinant are unchanged by orthogonal diagonalization.
- In principal coordinates, rotational kinetic energy contains no cross terms.
Part I: Original GRE-style problems
Problem 1: eigenvector meaning
A unit vector e is a principal axis of inertia if
- Ie = 0
- Ie is perpendicular to e
- Ie = Ipe
- det I = 0
- all tensor entries are equal
Problem 2: diagonal principal basis
In a principal-axis basis, the inertia tensor
- is antisymmetric
- is diagonal
- has only off-diagonal entries
- must equal the identity matrix
- has zero trace
Problem 3: orthogonality
Two principal moments are distinct. Their principal-axis eigenvectors are
- parallel
- orthogonal
- necessarily equal
- both zero
- complex
Problem 4: two-dimensional eigenvalues
For
the principal moments are
- 3 and 10
- 4 and 9
- 5 and 8
- 6 and 7
- −4 and 9
Problem 5: invariant check
For the tensor in Problem 4, the determinant equals
- 13
- 20
- 32
- 36
- 40
Problem 6: angular momentum
If a body rotates exactly about principal axis 2 with angular speed ω, then
- L = I2ωe2
- L = I1ωe1
- L is perpendicular to ω
- L = 0
- L has arbitrary direction
Problem 7: kinetic energy
In principal coordinates, the rotational kinetic energy is
(I1 + I2 + I3)(ω1 + ω2 + ω3)2
(I1ω12 + I
2ω22 + I
3ω32)
- I1I2I3
|L|2
- I1 + I2 + I3
Problem 8: axisymmetric body
An axisymmetric body has
Which statement is correct?
- no principal axes exist
- only one principal direction exists
- every direction in the 1-2 eigenspace is principal
- every direction in three-dimensional space is principal
- I3 must be zero
Problem 9: spherical inertia
If
then
- no direction is principal
- only coordinate axes are principal
- every direction is principal
- angular momentum is always zero
- kinetic energy is always zero
Problem 10: extremum property
The stationary values of
subject to
occur when
- n is an eigenvector of I
- n = 0
- I = 0
- det I = 0 only
- the tensor is antisymmetric
Problem 11: thin disk
For a thin uniform disk about its center,
Each diameter has principal moment
MR2
MR2
MR2
- MR2
- 2MR2
Problem 12: numerical 3D tensor
For
the three principal moments are
- 2, 6, 10
- 4, 8, 10
- 6, 6, 10
- 4, 6, 12
- 2, 8, 12
Part II: Complete worked solutions
Solution 1
A principal axis is an eigenvector direction:
Answer: (C).
Solution 2
Principal-axis coordinates diagonalize the symmetric inertia tensor. Answer: (B).
Solution 3
Distinct eigenvalues of a real symmetric matrix have orthogonal eigenvectors. Answer:
(B).
Solution 4
The characteristic equation is
| 0 | = (5 − Ip)(8 − Ip) − 4 | (11)
|
| = Ip2 − 13I
p + 36 | (12)
|
| = (Ip − 4)(Ip − 9). | (13) |
Thus the principal moments are
Answer: (B).
Solution 5
Answer: (D).
Solution 6
If
then
Answer: (A).
Solution 7
In principal coordinates the tensor is diagonal, so no cross terms appear:
Answer: (B).
Solution 8
A repeated eigenvalue has a multidimensional eigenspace. Every direction in that eigenspace is an
eigenvector with the same principal moment. Answer: (C).
Solution 9
If
then
for every direction. Answer: (C).
Solution 10
Using a Lagrange multiplier gives
Thus stationary directions are eigenvector directions. Answer: (A).
Solution 11
For a planar disk,
Rotational symmetry gives
Hence
Answer: (B).
Solution 12
The z direction gives eigenvalue 10.
For the in-plane block,
the eigenvalues are
Thus the principal moments are
Answer: (B).
2 GRE checklist
- Treat principal-axis problems as symmetric-matrix eigenvalue problems.
- Use geometry and symmetry before doing algebra.
- Distinct principal moments give orthogonal principal axes.
- Normalize eigenvectors before building the transformation matrix.
- Check trace and determinant before accepting the result.
- Remember that repeated moments make principal directions nonunique within the
degenerate eigenspace.
- Rotation exactly about a principal axis gives L ∥ ω.
- In principal coordinates, energy and angular momentum component formulas become
diagonal.
References
References
[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005.
[2] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley,
2002.
[3] K. R. Symon, Mechanics, 3rd ed., Addison-Wesley, 1971.