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Physics LibreTexts

13.18: Lagrange equations of motion for rigid-body rotation

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The Euler equations of motion were derived using Newtonian concepts of torque and angular momentum. It is of interest to derive the equations of motion using Lagrangian mechanics. It is convenient to use a generalized torque N and assume that U=0 in the Lagrange-Euler equations. Note that the generalized force is a torque since the corresponding generalized coordinate is an angle, and the conjugate momentum is angular momentum. If the body-fixed frame of reference is chosen to be the principal axes system, then, since the inertia tensor is diagonal in the principal axis frame, the kinetic energy is given in terms of the principal moments of inertia as

T=12iIiω2i

Using the Euler angles as generalized coordinates, then the Lagrange equation for the specific case of the ψ coordinate and including a generalized force Nψ gives

ddtT˙ψTψ=Nψ

which can be expressed as

ddt3iTωiωi˙ψ3iTωiωiψ=Nψ

Equation ??? gives

Tωi=Iiωi

Differentiating the angular velocity components in the body-fixed frame, equations (13.14.113.14.3) give

ω1ψ=˙ϕsinθcosψ˙θsinψ=ω2 ω1˙ψ=ω2˙ψ=0
ω2ψ=˙ϕsinθsinψ˙θcosψ=ω1 ω1˙ψ=ω2˙ψ=0
ω3ψ=0 ω3˙ψ=1
Table 13.18.1

Substituting these into the Lagrange Equation ??? gives

ddtI3ω3I1ω1ω2+I2ω2(ω1)=N3

since the ψ and ^e3 axes are colinear. This can be rewritten as

I3˙ω3(I1I2)ω1ω2=N3

Any axis could have been designated the ^e3 axis, thus the above equation can be generalized to all three axes to give

I1˙ω1(I2I3)ω2ω3=N1I2˙ω2(I3I1)ω3ω1=N2I3˙ω3(I1I2)ω1ω2=N3

These are the Euler’s equations given previously in (13.17.6). Note that although ˙ω3 is the equation of motion for the ψ coordinate, this is not true for the ϕ and θ rotations which are not along the body-fixed x1 and x2 axes as given in table 13.14.1.

Example 13.18.1: Rotation of a dumbbell

11.18.1.PNG
Figure 13.18.1: Rotation of a dumbbell.

Consider the motion of the symmetric dumbbell shown in the adjacent figure. Let |r1|=|r2|=b. Let the body-fixed coordinate system have its origin at O and symmetry axis ^e3 be along the weightless shaft toward m1 and vα=vαˆe1. The angular momentum is given by

L=imir×v

Because L is perpendicular to the shaft, and L rotates around ω as the shaft rotates, let ^e2 be along L.

L=L2^e2

If α is the angle between ω and the shaft, the components of ω are

ω1=0ω2=ωsinαω3=ωcosα

Assume that the principal moments of the dumbbell are

I1=(m1+m2)b2I2=(m1+m2)b2I3=0

Thus the angular momentum is given by

L1=I1ω1=0L2=I2ω2=(m1+m2)b2ωsinαL3=I3ω3=0

which is consistent with the angular momentum being along the ^e2 axis.

Using Euler’s equations, and assuming that the angular velocity is constant, i.e. ˙ω=0, then the components of the torque required to satisfy this motion are

N1=(m1+m2)b2ω2sinαcosαN2=0N3=0

That is, this motion can only occur in the presence of the above applied torque which is in the direction ^e1, that is, mutually perpendicular to ^e2 and ^e3. This torque can be written as N=ω×L.


This page titled 13.18: Lagrange equations of motion for rigid-body rotation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform.

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