Question

Show that 2D potential flows satisfy Euler’s equations even if μ ≠ 0.

Answer 1

we learn that when an external torque acts on a body its angular momentum changes (and if no external torques act on a body its angular momentum does not change.) We learn that the rate of change of angular momentum is equal to the applied torque. In the first simple examples that we typically meet, a symmetrical body is rotating about an axis of symmetry, and the torque is also applied about this same axis. The angular momentum is just IωIω, and so the statement that torque equals rate of change of angular momentum is merely τ=Iω˙τ=Iω˙ and that’s all there is to it.

Later, we learn that LL = IωIω, where ll is a tensor, and LL and ωω are not parallel. There are three principal moments of inertia, and LL, ωω and the applied torque ττ each have three components, and the statement “torque equals rate of change of angular momentum” somehow becomes much less easy.

Euler’s Equations sort this out, and give us a relation between the components of the ττ , ll and ωω.

For Figure IV.5, I have just reproduced, with some small modifications, Figure III.19 from my notes on this Web site on Celestial Mechanics, where I defined Eulerian angles. Again it is suggested that those who are unfamiliar with Eulerian angles consult Chapter III of Celestial Mechanics.

In Figure IV.5, OxyzOxyz are space-fixed axes, and Ox0y0z0Ox0y0z0 are the body-fixed principal axes. The axis Oy0Oy0 is behind the plane of your screen; you will have to look inside your monitor to find it.

I suppose an external torque ττ acts on the body, and I have drawn the components τ1τ1 and τ3τ3 . Now let’s suppose that the body rotates in such a manner that the Eulerian angle ψψ were to increase by δψδψ . I think it will be readily agreed that the work done on the body is τ3δψτ3δψ . This means, following our definition of generalized force in Section 4.4, that τ3τ3 is the generalized force associated with the generalized coordinate ψψ. Having established that, we can now apply the Lagrangian Equation 4.4.1:

ddt(∂T∂ψ˙)−∂T∂ψ=τ3(4.5.1)(4.5.1) ddt(∂T∂ψ˙)−∂T∂ψ=τ3

Here the kinetic energy is the expression that we have already established in Equation 4.3.6. In spite of the somewhat fearsome aspect of Equation 4.3.6, it is quite easy to apply Equation 4.5.14.5.1 to it. Thus

∂T∂ψ˙=I3(ϕ˙cosθ+ψ)=I3ω(4.5.2)(4.5.2) ∂T∂ψ˙=I3(ϕ˙cosθ+ψ)=I3ω

where I have made use of Equation 4.2.3.

Therefore

ddt(∂T∂ψ˙)=I3ω3˙(4.5.3)(4.5.3) ddt(∂T∂ψ˙)=I3ω3˙

And, if we make use of Equations 4.2.1,2,3, it is easy to obtain

∂T∂ψ)=I1ω1ω2−I2ω2ω2=ω1ω2(I1−I2)(4.5.4)(4.5.4) ∂T∂ψ)=I1ω1ω2−I2ω2ω2=ω1ω2(I1−I2)

Thus Equation 4.5.14.5.1 becomes:

I3ω3˙−(I1−I2)ω1ω1=τ3(4.5.5)(4.5.5) I3ω3˙−(I1−I2)ω1ω1=τ3

This is one of the Eulerian Equations of motion.

Now, although we saw that τ3τ3 is the generalized force associated with the coordinate y, it will we equally clear that τ1τ1 is not the generalized force associated with q, nor is τ2τ2 the generalized force associated with ϕϕ. However, we do not have to think about what the generalized forces associated with these two coordinates are; it is much easier than that. To obtain the remaining two Eulerian Equations, all that is necessary is to carry out a cyclic permutation of the subscripts in Equation 4.5.54.5.5. Thus the three Eulerian Equation are:

I1ω1˙−(I2−I2)ω2ω3=τ1,(4.5.6)(4.5.6) I1ω1˙−(I2−I2)ω2ω3=τ1,

I2ω2˙−(I3−I1)ω3ω1=τ2,(4.5.7)(4.5.7) I2ω2˙−(I3−I1)ω3ω1=τ2,

I3ω3˙−(I1−I2)ω1ω2=τ3.(4.5.8)(4.5.8) I3ω3˙−(I1−I2)ω1ω2=τ3.

These take the place of τ=Iω˙τ=Iω˙ which we are more familiar with in elementary problems in which a body is rotating about a principal axis and a torque is applied around that principal axis.

If there are no external torques acting on the body, then we have Euler’s Equations of free rotation of a rigid body:

I1ω1˙=(I2−I3)ω2ω3,(4.5.9)(4.5.9) I1ω1˙=(I2−I3)ω2ω3,

I1ω2˙=(I3−I1)ω3ω1,(4.5.10)(4.5.10) I1ω2˙=(I3−I1)ω3ω1,

I3ω3˙=(I1−I2)ω1ω2.(4.5.11)(4.5.11) I3ω3˙=(I1−I2)ω1ω2.

Example 4.5.14.5.1

In the above drawing, a rectangular lamina is spinning with constant angular velocity ωω between two frictionless bearings. We are going to apply Euler’s Equations of motion to it. We shall find that the bearings are exerting a torque on the rectangle, and the rectangle is exerting a torque on the bearings. The angular momentum of the rectangle is not constant – at least it is not constant in direction. We shall calculate the torque (its magnitude and its direction) and see what is happening to the angular momentum.

We note that the principal (second) moments of inertia are

I1=13mb2I2=13ma2I3=13m(a2+b2)I1=13mb2I2=13ma2I3=13m(a2+b2)

and that the components of angular velocity are

ω1=ωcosθω2=ωsinθω3=0.ω1=ωcos⁡θω2=ωsin⁡θω3=0.

Also, ω˙ω˙ and all of its components are zero. We immediately obtain, from Euler’s Equations, that τ1τ1 and τ2τ2 are zero, and that the torque exerted on the rectangle by the bearings is

τ3=(I2−I1)ω1ω2=13m(a2−b2)ω2sinθcosθτ3=(I2−I1)ω1ω2=13m(a2−b2)ω2sinθcos⁡θ

And since

sinθ=ba2+b2√andcosθ=ba2+b2√,sin⁡θ=ba2+b2andcos⁡θ=ba2+b2,

we obtain

τ3=m(a2−b2)ab3(a2+b2)ω2τ3=m(a2−b2)ab3(a2+b2)ω2

Thus ττ, the torque exerted on the rectangle by the bearings is directed normal to the plane of the rectangle (out of the plane of the paper in the instantaneous snapshot above).

The angular momentum is given by L=lωL=lω. That is to say:

⎛⎝⎜L1L2L3⎞⎠⎟=13m⎛⎝⎜b2000a2000a2+b2⎞⎠⎟⎛⎝⎜ωcosθωsinθ0⎞⎠⎟(L1L2L3)=13m(b2000a2000a2+b2)(ωcos⁡θωsin⁡θ0)

L1=13mb2ωcosθ=13mab2a2+b2√ωL1=13mb2ωcos⁡θ=13mab2a2+b2ω

L2=13mb2ωsinθ=13mab2a2+b2√ωL2=13mb2ωsin⁡θ=13mab2a2+b2ω

L3=0L3=0

L=13mabωL=13mabω

L2/L1=a2sinθb2cosθ=cotθ=tan(90°−θ)L2/L1=a2sinθb2cosθ=cot⁡θ=tan(90°−θ)

This tells us that LL is in the plane of the rectangle, and makes an angle 90° - θθ with the xx-axis, or q with the yy-axis, and it rotates around the vector ττ. ττ is perpendicular to the plane of the rectangle, and of course the change in LL takes place in that direction. The torque does no work, and ωω and TT are constant. The reader might find an analogy in the situation of a planet in orbit around the Sun in a circular orbit.. The planet experiences a force that is always perpendicular to its velocity. The force does no work, and the speed and kinetic energy remain constant.

The torque on the plate can be represented as a couple of forces exerted by the bearings on the plate, each of magnitude τ32a2+b2√,τ32a2+b2, or m(a2−b2)6(a2−b2)32ω2m(a2−b2)6(a2−b2)32ω2 Forces exerted by the plate on the bearings are, of course, in the opposite direction.

Example 4.5.24.5.2

Figure IV.6 shows a disc of mass mm, radius aa, spinning at a constant angular speed ωω about at axle that is inclined at an angle θθ to the normal to the disc. I have drawn three body-fixed principal axes. The xx- and yy- axes are in the plane of the disc\boldsymbol; the direction of the xx-axis is chosen so that the axle (and hence the vector ωω ) is in the zxzx-plane. The disc is evidently unbalanced and there must be a torque on it to maintain the motion.

Since ωω is constant, all components of ω˙ω˙ are zero, so that Euler’s Equations are

τ1=(I3−I2)ω3ω2,τ1=(I3−I2)ω3ω2,

τ2=(I1−I3)ω1ω3,τ2=(I1−I3)ω1ω3,

τ3=(I2−I1)ω2ω1,τ3=(I2−I1)ω2ω1,

Now ω1=ωsinθ,ω2=ωcosθ,I1=14ma2,I2=14ma2,I3=11ma2ω1=ωsin⁡θ,ω2=ωcos⁡θ,I1=14ma2,I2=14ma2,I3=11ma2

Therefore τ1=τ3=0,andτ2=−14ma2ω2sinθcosθ=−18ma2ω2sin2θτ1=τ3=0,andτ2=−14ma2ω2sinθcosθ=−18ma2ω2sin2θ

(Check, as always, that this expression is dimensionally correct.) Thus the torque acting on the disc is in the negative yy-direction.

Can you reconcile the fact that there is a torque acting on the disc with the fact that is it moving with constant angular velocity? Yes, most decidedly! What is not constant is the angular momentum LL, which is moving around the axle in a cone such that L˙=−τ2jL˙=−τ2j, where jj is the unit vector along the yy-axis.