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28.2: Free Rotation of a Symmetric Top Using Euler’s Equations

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This is a problem we’ve already solved, using Lagrangian methods and Euler angles, but it’s worth seeing just how easy it is using Euler’s equations.

For I1=I2, the third equation gives immediately Ω3= constant. .

Then, writing for convenience

Ω3(I3I1)/I1=ω

the first two equations are

˙Ω1=ωΩ2,˙Ω2=ωΩ1

These equations can be combined to give

d(Ω1+iΩ2)/dt=iω(Ω1+iΩ2), sо (Ω1+iΩ2)=Aeiωt

That is, (Ω1,Ω2) moves around a circle centered at the origin with constant angular velocity. So Ω21+Ω22=|A|2 stays constant, and Ω3 is constant, the angular velocity vector has constant length and rotates steadily about the axis x3.

From

L1=I1Ω1,L2=I2Ω2,L3=I3Ω3

it follows that the angular momentum vector also precesses at a steady rate about x3. This is, remember, in the body frame—we know that in the fixed space frame, the angular momentum vector is constant! It follows that, as viewed from the outside, the x3 axis precesses around the fixed angular momentum vector at a steady rate.

Of course, the rate is the same as that found using Euler’s angles, recall from the previous lecture that

L=(I1Ω1,I1Ω2,I3Ω3)=(I1˙θ,I1˙ϕsinθ,I3(˙ϕcosθ+˙ψ))

so in precession

L3=Lcosθ=I3(˙ϕcosθ+˙ψ) and ˙ϕ=L/I1

so

˙ψ=Lcosθ(1I31I1)=Ω3(I3I1)/I1


This page titled 28.2: Free Rotation of a Symmetric Top Using Euler’s Equations is shared under a not declared license and was authored, remixed, and/or curated by Michael Fowler.

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