GRE Physics Companion: The Inertia Tensor and Products of Inertia

GRE Physics Companion: The Inertia Tensor and Products of Inertia

The core relation is

(1)

with

(2)

For an arbitrary unit axis,

(3)

1 High-value GRE facts

1. The inertia tensor is real and symmetric.
2. Diagonal entries are scalar moments about coordinate axes.
3. In this article, P_xy = ∫ xy dm but I_xy = −P_xy.
4. Reflection symmetry often makes products of inertia vanish.
5. L need not be parallel to ω.
6. A diagonal inertia tensor does not imply equal principal moments.
7. The moment about unit axis n is n^T In.
8. rotational kinetic energy is ω^T Iω.
9. For a planar lamina, I_zz = I_xx + I_yy.
10. Principal axes diagonalize the inertia tensor.

Part I: Original GRE-style problems

Problem 1: tensor symmetry

For a mass inertia tensor in Cartesian coordinates,

1. I_xy = −I_yx always
2. I_xy = I_yx
3. all off-diagonal entries are positive
4. all diagonal entries vanish
5. the tensor must be antisymmetric

Problem 2: product sign

Using the convention P_xy = ∫ xy dm, the tensor component is

1. I_xy = P_xy
2. I_xy = −P_xy
3. I_xy = P_xy²
4. I_xy = 0 always
5. I_xy = I_xx + I_yy

Problem 3: reflection symmetry

A body is symmetric under x →−x. Which products must vanish?

1. P_xy and P_xz
2. P_xy and P_yz
3. P_xz and P_yz
4. only P_yz
5. none

Problem 4: planar lamina

For a lamina in the xy plane,

1. I_zz = I_xx − I_yy
2. I_zz = I_xx + I_yy
3. I_xx = I_yy + I_zz
4. I_zz = 0
5. all three moments are equal

Problem 5: angular momentum

If

(4)

and

(5)

then

(6)

is

1. (1, 1, 0)
2. (2, 2, 0)
3. (2, 4, 0)
4. (4, 2, 0)
5. (6, 6, 0)

Problem 6: parallelism

For the tensor and angular velocity in Problem 5,

1. L is parallel to ω
2. L is perpendicular to ω
3. L is not parallel to ω
4. L = 0
5. ω = 0

Problem 7: arbitrary-axis moment

For

(7)

and unit axis

(8)

the scalar moment is

1. 2
2. 3
3. 4
4. 6
5. 8

Problem 8: rotational energy

For the tensor in Problem 5 and

(9)

the rotational kinetic energy is

1. 1
2. 2
3. 3
4. 4
5. 6

Problem 9: trace

The trace of the inertia tensor equals

1. ∫ r² dm
2. 2 ∫ r² dm
3. 3 ∫ r² dm
4. total mass only
5. zero

Problem 10: coordinate rotation

If vector components transform as r′ = Qr with orthogonal Q, then the inertia tensor transforms as

1. I′ = QIQ^T
2. I′ = I + Q
3. I′ = Q^T Q
4. I′ = I⁻¹
5. I′ = QI

Problem 11: shift along an axis

A center of mass inertia tensor is diagonal. The origin is shifted distance d along the x axis. Which center of mass moment is unchanged by the tensor parallel-axis shift?

1. I_xx
2. I_yy
3. I_zz
4. all three increase equally
5. all three remain unchanged

Problem 12: principal axis

If e is an eigenvector of I with eigenvalue I_p, and

(10)

then

1. L is perpendicular to ω
2. L = I_pω
3. L = 0
4. L has arbitrary direction
5. I_p = 0 necessarily

Part II: Complete worked solutions

Solution 1

The inertia tensor is symmetric because

(11)

Thus

(12)

Answer: (B).

Solution 2

With

(13)

the off-diagonal tensor component is

(14)

Answer: (B).

Solution 3

Under x →−x, both xy and xz change sign. Their paired contributions cancel:

(15)

Answer: (A).

Solution 4

For z = 0,

Ixx	= ∫ y2 dm,	(16)
Iyy	= ∫ x2 dm,	(17)
Izz	= ∫ (x2 + y2) dm.	(18)

Therefore

(19)

Answer: (B).

Solution 5

Multiply:

(20)

Answer: (C).

Solution 6

The vectors

(21)

and

(22)

are not scalar multiples of each other. Answer: (C).

Solution 7

In	= nT In	(23)
	= (2) + (6)	(24)
	= 4.	(25)

Answer: (C).

Solution 8

K	= ωT Iω	(26)
	= (1, 1, 0) ⋅ (2, 4, 0)	(27)
	= 3.	(28)

Answer: (C).

Solution 9

Adding the diagonal entries gives

(29)

Answer: (B).

Solution 10

For an orthogonal coordinate transformation,

(30)

Answer: (A).

Solution 11

A shift along the x axis leaves all perpendicular distances to the x axis unchanged. Thus I_xx is unchanged. Answer: (A).

Solution 12

The eigenvector equation is

(31)

Therefore

L	= I(ωe)	(32)
	= Ip(ωe)	(33)
	= Ipω.	(34)

Answer: (B).

2 GRE checklist

1. State the origin and coordinate axes before writing an inertia tensor.
2. Check the sign convention used for products of inertia.
3. Use symmetry before performing integrals.
4. Do not assume L ∥ ω.
5. Use I_n = n^T In for arbitrary axes.
6. Use K = ω^T Iω for rotational energy.
7. Remember that tensor components change when the coordinate axes rotate.
8. Principal axes are eigenvector directions of the symmetric inertia tensor.

References

References