Physics 321 Hour 29 Principal Axes
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Physics 321 Hour 29 Principal Axes Bottom Line • We can write πΏ = ππ where π is the inertia tensor: πΌπ₯π₯ πΌπ₯π¦ πΌπ₯π§ π½π¦π¦ + π½π§π§ −π½π₯π¦ −π½π₯π§ −π½π₯π¦ π½π§π§ + π½π₯π₯ −π½π¦π§ π = πΌπ₯π¦ πΌπ¦π¦ πΌπ¦π§ = πΌπ₯π§ πΌπ¦π§ πΌπ§π§ −π½π₯π§ −π½π¦π§ π½π₯π₯ + π½π¦π¦ π½π₯π¦ = π₯π¦πππ • Things are simpler with “principal axes”: πΌ′π₯π₯ 0 0 πΌ′π¦π¦ 0 π= 0 0 0 πΌ′π§π§ To do this we solve the eigenvalue problem. The Inertia Tensor of a Point Mass πΏ = ππ π(π 2 − π₯ 2 ) π= −ππ¦π₯ −ππ§π₯ −ππ₯π¦ π(π 2 − π¦ 2 ) −ππ§π¦ −ππ₯π§ −ππ¦π§ π(π 2 − π§ 2 ) The Inertia Tensor of an Extended Object πΏ = ππ π½π₯π¦ = πΌπ₯π₯ π = πΌπ₯π¦ πΌπ₯π§ πΌπ₯π¦ πΌπ¦π¦ πΌπ¦π§ π₯π¦πππ πΌπ₯π§ π½π¦π¦ + π½π§π§ πΌπ¦π§ = −π½π₯π¦ πΌπ§π§ −π½π₯π§ 1 π = πβπβπ 2 −π½π₯π¦ π½π§π§ + π½π₯π₯ −π½π¦π§ −π½π₯π§ −π½π¦π§ π½π₯π₯ + π½π¦π¦ Lamina Example a a π½π¦π¦ π = −π½π₯π¦ 0 −π½π₯π¦ π½π₯π₯ 0 π½π₯π¦ = 0 0 π½π₯π₯ + π½π¦π¦ π₯π¦πππ΄ Example Lamina a a π½π₯π₯ π½π₯π¦ π π 4 π π 1 2 π₯ ππ₯ππ¦ = 2 = ππ2 = π½π¦π¦ π 3 3 0 0 π π π π π4 1 = 2 π₯π¦ππ₯ππ¦ = 2 = ππ2 π 0 0 π 4 4 π = 2 π Example Lamina a a π½π¦π¦ π = −π½π₯π¦ 0 −π½π₯π¦ π½π₯π₯ 0 0 1/3ππ2 −1/4ππ2 0 = −1/4ππ2 1/3ππ2 π½π₯π₯ + π½π¦π¦ 0 0 ππ2 4 −3 0 = −3 4 0 12 0 0 8 0 0 2/3ππ2 Diagonalizing the Inertia Tensor πΏ = ππ πΌ′π₯π₯ 0 0 πΌ′π₯π₯ ππ₯ πΌ′π¦π¦ 0 ? πΏ′ = πΌ′π¦π¦ ππ¦ What if π′ = 0 0 0 πΌ′π§π§ πΌ′π§π§ ππ§ 1 1 1 2 2 2 π = πΌ′π₯π₯ ππ₯ + πΌ′π¦π¦ ππ¦ + πΌ′π§π§ π′π§ 2 2 2 Furthermore, if π′π₯ π′ = 0 0 1 2 πΏ′ = πΌ′π₯π₯ π′ π = πΌ′π₯π₯ π′π₯ 2 Diagonalizing the Inertia Tensor • Find the eigenvalues: det π − λπ = 0 • For each λ, find the eigenvectors: πΌπ = λπ • 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 π by hand. Example ππ2 4 −3 0 π= −3 4 0 12 0 0 8 4 − λ′ −3 0 =0 −3 4 − λ′ 0 0 0 8 − λ′ 8 − λ′ = 0 or 4 − λ′ 2 −9=0 λ′ = 8 ππ 4 − λ′ = 3 ππ 4 − λ′ = −3 Example ππ2 4 −3 π= −3 4 12 0 0 For λ′ = 8 0 0 8 8ππ2 2ππ2 λ= = 12 3 π 4 −3 0 π −3 4 0 π = 8 π π 0 0 8 π π 4π − 3π 4π − 3π = 8 π π 8π −4π − 3π = 0 −4π − 3π = 0 8π = 8π Principal axis 1: π§ axis with πΌ11 = π 0 π = 0 π 1 2ππ2 3 Example a For λ′ = 1 a 4 −3 0 π −3 0 π 4 0 π = π π 0 8 π π 4π − 3π 4π − 3π = π π 8π Principal axis 2: ππ2 λ= 12 3π − 3π = 0 3π − 3π = 0 8π = π 1 2 π 1 1 π = 1 2 0 π π₯ + π¦ axis with πΌ22 = ππ2 12 Example a For λ′ = 7 a 4 −3 0 π −3 0 π 4 0 π =7 π π 0 8 π π 4π − 3π 4π − 3π = 7 π π 8π Principal axis 2: 7ππ2 λ= 12 −3π − 3π = 0 −3π − 3π = 0 8π = π 1 2 π 1 1 π = −1 2 0 π π₯ − π¦ axis with πΌ33 = 7ππ2 12 Examples principal axes.nb
