Physics 321 Hour 30 Euler’s Equations
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Physics 321 Hour 30 Euler’s Equations Bottom Line • When there are no external torques, we can describe the motion of rigid bodies with Euler’s equations. • One way of analyzing the system without solving the equations of motion is to use the “Method of Ellipsoids.” Space and Body Coordinates π3 π§ π2 π¦ π₯ π1 • Body coordinates are on principal axes • If possible, use c.m. as origin in both frames • If not possible, c.m. motion is easy Relating Coordinates • Start with πΏ in body coordinates πΏ1 π1 π2 π3 πΏ2 πΏ3 πΌ11 0 0 = π1 π2 π3 0 πΌ22 0 0 0 πΌ33 πΌ11 π1 = π1 π2 π3 πΌ22 π2 πΌ33 π3 π1 π2 π3 Relating Coordinates • Transform ππΏ/ππ‘ to space coordinates ππΏ ππΏ = + π × πΏ ππππ¦ ππ‘ π ππππ ππ‘ ππππ¦ π₯ π¦ π§ πΏπ₯ πΏπ¦ = π1 πΏπ§ π1 + π1 πΏ1 π2 π2 πΏ2 π2 π3 π3 πΏ3 π3 πΏ1 πΏ2 πΏ3 Relating Coordinates π₯ π¦ π§ Γπ₯ Γπ¦ = π1 Γπ§ Real external torque In space coordinates π2 π3 πΏ1 π1 πΏ2 + π1 πΏ1 πΏ3 “Perceived” body torque π2 π3 π2 π3 πΏ 2 πΏ3 Relating Coordinates π1 π2 π3 Γ1 Γ2 = π1 Γ3 Real external torque In body coordinates π2 π3 πΏ1 π1 πΏ2 + π1 πΏ1 πΏ3 “Perceived” body torque Γ1 = πΌ11 π1 − πΌ22 − πΌ33 π2 π3 π2 π3 π2 π3 πΏ2 πΏ3 Euler’s Equations Γ1 = πΌ11 π1 − πΌ22 − πΌ33 π2 π3 Γ2 = πΌ22 π2 − πΌ33 − πΌ11 π3 π1 Γ3 = πΌ33 π3 − πΌ11 − πΌ22 π1 π2 It’s usually very hard to find the actual torques in terms of the rotating body coordinates! Euler’s Equations – No Torques πΌ11 π1 = πΌ22 − πΌ33 π2 π3 πΌ22 π2 = πΌ33 − πΌ11 π3 π1 πΌ33 π3 = πΌ11 − πΌ22 π1 π2 We can do these! Example Rotating a book - eulerseqs.nb The Method of Ellipsoids πΌ11 = π 12 2 2 π + π , πΌ22 = πΌ33 = π 12 π 12 2 π2 + π 2 , π2 + π Letting π > π > π, then πΌ11 < πΌ22 < πΌ33 2 2 2 2 2 2 πΏ = πΌ11 π1 + πΌ22 π2 + πΌ33 π3 2 1 1 1 2 2 π = πΌ11 π1 + πΌ22 π2 + πΌ33 π3 2 2 2 2 The Method of Ellipsoids These are two ellipsoids with equations π1 2 π2 2 π3 2 2+ 2+ 2 =1 πΏ πΏ πΏ πΌ11 πΌ22 πΌ33 π1 2 π2 2 π3 2 2+ 2+ 2 =1 2π 2π 2π πΌ11 πΌ22 πΌ33 Note that the ellipsoids are in ω-space. They don’t predict motion. Possible values of ω are given by the intersection of the ellipsoids. Example ellipsoids.nb Euler’s Equations – No Torques, I11=I22 πΌ11 π1 = πΌ11 − πΌ33 π2 π3 πΌ22 π2 = πΌ33 − πΌ11 π3 π1 πΌ33 π3 = 0 π3 is a constant πΌ11 − πΌ33 π1 = π2 π3 ≡ Ωπ π2 πΌ11 πΌ11 − πΌ33 π2 = − π3 π1 ≡ −Ωπ π1 πΌ11 2 π2 = −Ωπ π1 = −Ωπ π2 Euler’s Equations – No Torques, I11=I22 π0 cos Ωπ π‘ π = −π0 sin Ωπ π‘ π3 πΌ11 π0 cos Ωπ π‘ πΏ = −πΌ11 π0 sin Ωπ π‘ πΌ33 π3 In the body axes: πΏ Ωπ = πΌ11 π0 sin α cos Ωπ π‘ π = π0 sin πΌ sin Ωπ π‘ π0 cos πΌ sin π cos Ωπ π‘ π3 = sin π sin Ωπ π‘ cos π Prolate object: Ωb<0, Ωs>0 Example football.nb
