Physics 321 Hour 29 Principal Axes

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 0 0 0 𝐼′ 𝑦𝑦 0 0 0 𝐼′ 𝑧𝑧 To do this we solve the eigenvalue problem. BA.C-CA.B

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 )

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

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 0 0 0 𝐽 𝑥𝑥 + 𝐽 𝑦𝑦 𝐽 𝑥𝑦 = 𝑥𝑦𝜎𝑑𝐴

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

Example Lamina a a 𝐈= 𝐽 𝑦𝑦 − 𝐽 𝑥𝑦 0 −𝐽 𝑥𝑦 𝐽 𝑥𝑥 0 0 0 𝐽 𝑥𝑥 + 𝐽 𝑦𝑦 = 1/3𝑚 𝑎 2 −1/4𝑚 𝑎 2 0 −1/4𝑚 𝑎 2 1/3𝑚 𝑎 2 0 0 0 2/3𝑚 𝑎 2 = 𝑚 𝑎 2 12 4 −3 0 −3 4 0 0 0 8

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

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.

Example 𝐈= 𝑚 𝑎 2 12 4 −3 0 −3 4 0 0 0 8 4−λ′ −3 0 −3 4−λ′ 0 0 0 8−λ′ =0 8−λ′=0 or 4−λ′ 2 −9=0 λ′=8 𝑜𝑟 4−λ′=3 𝑜𝑟 4−λ′=−3

Principal axis 1: 𝑧 axis with 𝐼 11 = 2𝑚 𝑎 2 3 Example 𝐈= 𝑚 𝑎 2 12 4 −3 0 −3 4 0 0 0 8 For λ ′ =8 λ= 8𝑚 𝑎 2 12 = 2𝑚 𝑎 2 3 4 −3 0 −3 4 0 0 0 8 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 = 0 0 1 Principal axis 1: 𝑧 axis with 𝐼 11 = 2𝑚 𝑎 2 3

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

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

Examples principal axes.nb