Physics 321 Hour 31 Euler’s Angles

Space and Body Coordinates 𝑒 3 𝑧 𝑒 2 𝑦 𝑥 𝑒 1 Body coordinates are on principal axes If possible, use c.m. as origin in both frames If not possible, c.m. motion is easy

Euler’s Equations – No Torques 𝐼 11 𝜔 1 = 𝐼 22 − 𝐼 33 𝜔 2 𝜔 3 𝐼 22 𝜔 2 = 𝐼 33 − 𝐼 11 𝜔 3 𝜔 1 𝐼 33 𝜔 3 = 𝐼 11 − 𝐼 22 𝜔 1 𝜔 2

Euler’s Equations – No Torques, I11=I22 𝐼 11 𝜔 1 = 𝐼 11 − 𝐼 33 𝜔 2 𝜔 3 𝐼 22 𝜔 2 = 𝐼 33 − 𝐼 11 𝜔 3 𝜔 1 𝐼 33 𝜔 3 =0 𝜔 3 is a constant 𝜔 1 = 𝐼 11 − 𝐼 33 𝐼 11 𝜔 2 𝜔 3 ≡ Ω 𝑏 𝜔 2 𝜔 2 =− 𝐼 11 − 𝐼 33 𝐼 11 𝜔 3 𝜔 1 ≡ −Ω 𝑏 𝜔 1 𝜔 2 =− Ω 𝑏 𝜔 1 =− Ω 𝑏 2 𝜔 2

Euler’s Equations – No Torques, I11=I22 In the body axes: 𝜔 = 𝜔 0 cos Ω 𝑏 𝑡 − 𝜔 0 sin Ω 𝑏 𝑡 𝜔 3 𝐿 = 𝐼 11 𝜔 0 cos Ω 𝑏 𝑡 − 𝐼 11 𝜔 0 sin Ω 𝑏 𝑡 𝐼 33 𝜔 3

In the space axes: Ω 𝑠 = 𝐿 𝐼 11 Ω 𝑠 = 𝐿 𝐼 11 𝜔 = 𝜔 0 sin α cos Ω 𝑠 𝑡 𝜔 0 sin 𝛼 sin Ω 𝑠 𝑡 𝜔 0 cos 𝛼 𝑒 3 = sin 𝜃 cos Ω 𝑠 𝑡 sin 𝜃 sin Ω 𝑠 𝑡 cos 𝜃 Prolate object: Ωb<0, Ωs>0

Example football.nb

Constants of the Motion, No Torque 𝐿 = 𝐼 11 𝜔 0 cos Ω 𝑏 𝑡 − 𝐼 11 𝜔 0 sin Ω 𝑏 𝑡 𝐼 33 𝜔 3 in body, has constant length Ω 𝑏 = 𝐼 11 − 𝐼 33 𝐼 11 𝜔 3 𝜔 3 is constant 𝐿 in space is constant 𝐿 𝑧 is constant cos 𝜃= 𝐿 𝑧 /𝐿 so θ is constant cos 𝛼= 𝜔 𝑧 /𝜔 so ω is constant

Example HW31 Answers.nb

Unit Vectors 𝑒 ′ 3 = 𝑒 3 = sin 𝜃 cos 𝜑 𝑥 + sin 𝜃 sin 𝜑 𝑦 + cos 𝜃 𝑧 𝑒 ′ 1 = cos 𝜃 cos 𝜑 𝑥 + cos 𝜃 sin 𝜑 𝑦 − sin 𝜃 𝑧

Angular Velocities 𝜓 is spin about the body 3-axis 𝜃 is tipping of the body 3-axis 𝜑 is precession about the space z-axis 𝜔 = 𝜑 𝑧 + 𝜃 𝑒 ′ 2 + ψ 𝑒 3 𝑧 = 𝑒 3 cos 𝜃 − 𝑒 ′ 1 sin 𝜃 𝜔 =− 𝜑 sin 𝜃 𝑒 ′ 1 + 𝜃 𝑒 ′ 2 +( ψ + 𝜑 cos 𝜃 ) 𝑒 ′ 3

Angular Momentum 𝐿 =− 𝐼 11 𝜑 sin 𝜃 𝑒 ′ 1 + 𝐼 22 𝜃 𝑒 ′ 2 + 𝐼 33 ( ψ + 𝜑 cos 𝜃 ) 𝑒 ′ 3 𝐿 3 = 𝐼 33 ψ + 𝜑 cos 𝜃 𝐿 𝑧 = 𝐿 ∙ 𝑧 = 𝐼 11 𝜑 sin 2 𝜃+ 𝐿 3 cos 𝜃 → 𝜑 = 𝐿 𝑧 − 𝐿 3 cos 𝜃 𝐼 11 sin 2 𝜃 For torque-free systems, Lz and L3 are constants, so 𝜑 is also constant.

Kinetic Energy 𝑇= 1 2 𝐼 11 𝜔 1 2 + 1 2 𝐼 22 𝜔 2 2 + 1 2 𝐼 33 𝜔 3 2 𝑇= 1 2 𝐼 11 𝜔 1 2 + 1 2 𝐼 22 𝜔 2 2 + 1 2 𝐼 33 𝜔 3 2 = 1 2 𝐼 11 𝜑 2 sin 2 𝜃+ 𝜃 2 + 1 2 𝐼 33 ψ + 𝜑 cos 𝜃 2