1. Physics 321
Hour 29
Principal Axes

2. Bottom Line We can write 𝐿 =𝐈 𝜔 where 𝐈 is the inertia tensor:
𝐈= 𝐼 𝑥𝑥 𝐼 𝑥𝑦 𝐼 𝑥𝑧 𝐼 𝑥𝑦 𝐼 𝑦𝑦 𝐼 𝑦𝑧 𝐼 𝑥𝑧 𝐼 𝑦𝑧 𝐼 𝑧𝑧 = 𝐽 𝑦𝑦 + 𝐽 𝑧𝑧 − 𝐽 𝑥𝑦 − 𝐽 𝑥𝑧 −𝐽 𝑥𝑦 𝐽 𝑧𝑧 + 𝐽 𝑥𝑥 − 𝐽 𝑦𝑧 −𝐽 𝑥𝑧 −𝐽 𝑦𝑧 𝐽 𝑥𝑥 + 𝐽 𝑦𝑦
𝐽 𝑥𝑦 = 𝑥𝑦𝜌𝑑𝑉
Things are simpler with “principal axes”:
𝐈= 𝐼′ 𝑥𝑥 𝐼′ 𝑦𝑦 𝐼′ 𝑧𝑧
To do this we solve the eigenvalue problem.
BA.C-CA.B

3. The Inertia Tensor of a Point Mass
𝐿 =𝐈 𝜔
𝐈= 𝑚( 𝑟 2 − 𝑥 2 ) −𝑚𝑥𝑦 −𝑚𝑥𝑧 −𝑚𝑦𝑥 𝑚( 𝑟 2 − 𝑦 2 ) −𝑚𝑦𝑧 −𝑚𝑧𝑥 −𝑚𝑧𝑦 𝑚( 𝑟 2 − 𝑧 2 )

4. The Inertia Tensor of an Extended Object
𝐿 =𝐈 𝜔
𝐽 𝑥𝑦 = 𝑥𝑦𝜌𝑑𝑉
𝐈= 𝐼 𝑥𝑥 𝐼 𝑥𝑦 𝐼 𝑥𝑧 𝐼 𝑥𝑦 𝐼 𝑦𝑦 𝐼 𝑦𝑧 𝐼 𝑥𝑧 𝐼 𝑦𝑧 𝐼 𝑧𝑧 = 𝐽 𝑦𝑦 + 𝐽 𝑧𝑧 − 𝐽 𝑥𝑦 − 𝐽 𝑥𝑧 −𝐽 𝑥𝑦 𝐽 𝑧𝑧 + 𝐽 𝑥𝑥 − 𝐽 𝑦𝑧 −𝐽 𝑥𝑧 −𝐽 𝑦𝑧 𝐽 𝑥𝑥 + 𝐽 𝑦𝑦
𝑇= 1 2 𝜔 ∙𝐈∙ 𝜔

5. Example Lamina 𝐈= 𝐽 𝑦𝑦 − 𝐽 𝑥𝑦 0 −𝐽 𝑥𝑦 𝐽 𝑥𝑥 0 0 0 𝐽 𝑥𝑥 + 𝐽 𝑦𝑦
𝐈= 𝐽 𝑦𝑦 − 𝐽 𝑥𝑦 0 −𝐽 𝑥𝑦 𝐽 𝑥𝑥 𝐽 𝑥𝑥 + 𝐽 𝑦𝑦
𝐽 𝑥𝑦 = 𝑥𝑦𝜎𝑑𝐴

6. Example
Lamina a
𝐽 𝑥𝑥 = 𝑚 𝑎 𝑎 0 𝑎 𝑥 2 𝑑𝑥𝑑𝑦 = 𝑚 𝑎 2 𝑎 4 3 = 1 3 𝑚 𝑎 2 = 𝐽 𝑦𝑦
𝐽 𝑥𝑦 = 𝑚 𝑎 𝑎 0 𝑎 𝑥𝑦𝑑𝑥𝑑𝑦 = 𝑚 𝑎 2 𝑎 4 4 = 1 4 𝑚 𝑎 2

7. Example
Lamina a a
𝐈= 𝐽 𝑦𝑦 − 𝐽 𝑥𝑦 0 −𝐽 𝑥𝑦 𝐽 𝑥𝑥 𝐽 𝑥𝑥 + 𝐽 𝑦𝑦 = 1/3𝑚 𝑎 2 −1/4𝑚 𝑎 2 0 −1/4𝑚 𝑎 2 1/3𝑚 𝑎 /3𝑚 𝑎 2
= 𝑚 𝑎 −3 0 −

8. Diagonalizing the Inertia Tensor
𝐿 =𝐈 𝜔
What if 𝐈′= 𝐼′ 𝑥𝑥 𝐼′ 𝑦𝑦 𝐼′ 𝑧𝑧 ? 𝐿 ′ = 𝐼′ 𝑥𝑥 𝜔 𝑥 𝐼′ 𝑦𝑦 𝜔 𝑦 𝐼′ 𝑧𝑧 𝜔 𝑧
𝑇= 1 2 𝐼′ 𝑥𝑥 𝜔 𝑥 𝐼′ 𝑦𝑦 𝜔 𝑦 𝐼′ 𝑧𝑧 𝜔′ 𝑧 2
Furthermore, if
𝜔 ′ = 𝜔′ 𝑥 0 0
𝐿′ = 𝐼′ 𝑥𝑥 𝜔 ′ 𝑇= 1 2 𝐼′ 𝑥𝑥 𝜔′ 𝑥 2

9. Diagonalizing the Inertia Tensor
Find the eigenvalues: det 𝐈−λ𝟏 =0
For each λ, find the eigenvectors: 𝐼 𝜔 =λ 𝜔
The three eigenvectors define the “principal” axes.
We’ll usually let Mathematica do that for us, but you should be able to diagonalize a lamina 𝐈 by hand.

10. Example
𝐈= 𝑚 𝑎 −3 0 −
4−λ′ −3 0 −3 4−λ′ −λ′ =0
8−λ′=0 or 4−λ′ 2 −9=0
λ′=8 𝑜𝑟 4−λ′=3 𝑜𝑟 4−λ′=−3

11. Principal axis 1: 𝑧 axis with 𝐼 11 = 2𝑚 𝑎 2 3
Example
𝐈= 𝑚 𝑎 −3 0 −
For λ ′ = λ= 8𝑚 𝑎 = 2𝑚 𝑎 2 3
4 −3 0 − 𝑎 𝑏 𝑐 =8 𝑎 𝑏 𝑐
4𝑎−3𝑏 4𝑏−3𝑎 8𝑐 =8 𝑎 𝑏 𝑐 −4𝑎−3𝑏=0 −4𝑏−3𝑎=0 8𝑐=8𝑐 𝑎 𝑏 𝑐 =
Principal axis 1: 𝑧 axis with 𝐼 11 = 2𝑚 𝑎 2 3

12. Principal axis 2: 1 2 𝑥 + 𝑦 axis with 𝐼 22 = 𝑚 𝑎 2 12
Example a
For λ ′ = λ= 𝑚 𝑎 2 12
4 −3 0 − 𝑎 𝑏 𝑐 = 𝑎 𝑏 𝑐
4𝑎−3𝑏 4𝑏−3𝑎 8𝑐 = 𝑎 𝑏 𝑐 𝑎−3𝑏=0 3𝑏−3𝑎=0 8𝑐=𝑐 𝑎 𝑏 𝑐 =
Principal axis 2: 𝑥 + 𝑦 axis with 𝐼 22 = 𝑚 𝑎 2 12

13. Principal axis 2: 1 2 𝑥 − 𝑦 axis with 𝐼 33 = 7𝑚 𝑎 2 12
Example a
For λ ′ = λ= 7𝑚 𝑎 2 12
4 −3 0 − 𝑎 𝑏 𝑐 =7 𝑎 𝑏 𝑐
4𝑎−3𝑏 4𝑏−3𝑎 8𝑐 =7 𝑎 𝑏 𝑐 −3𝑎−3𝑏=0 −3𝑏−3𝑎=0 8𝑐=𝑐 𝑎 𝑏 𝑐 = − 1 1 0
Principal axis 2: 𝑥 − 𝑦 axis with 𝐼 33 = 7𝑚 𝑎 2 12

14. Examples
principal axes.nb