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PHYS 705: Classical Mechanics Euler’s Equations 2
Euler’s Equations (set up)
We have seen how to describe the kinematic properties of a rigid body. Now, we would like to get equations of motion for it.
1. We will follow the Lagrangian Formalism that we have developed.
2. For generalized coordinates, we will use the Euler’s angles with one point of the rigid body being fixed (no translation, just rotation)
3. As we have seen previously, the rotational kinetic energy is given by 1 1 TωIω I 2 2 i ij j 4. Choose the body axes to coincide with the Principal axes, then
1 2 T Iii i ( no sum; I is diagonalized! ) 2 ij 3
Euler’s Equations (set up)
Note:
-We still have the freedom to align xˆ 3 z ˆ (from the body axes) to any one of the 3 Principal axes. - The three Euler’s angles ,, give the orientation of the Principal axes of the body axes relative to the fixed axes.
5. A general rotation (an inf. one here) d Ω along a given axis in the body frame can be decomposed into three rotations along the Euler’s angles. Then, the time rate of change of this rotation ω d Ω dt can also be written as, ωω ω ω (we write this as a sum since the angular changes are infinitesimal) - These three different pieces correspond to time rate of change of the individual rotations along each of the three Euler’s angles. 4
Euler’s Equations (set up)
-Now, our task is to express each of these in terms of the body coordinate
xx1,, 2 x 3 We will go through the three individual Euler steps now:
a) ω : We are in the fixed axes and we do a rotation along the x3 zˆ 0 In the fixed axes, we have ω 0 fixed To express it in the body axes, we apply the Euler rotations BCD
T 0 sin sin Note: Since 0,0, is ω BCD 0 cos sin already in the z ˆ body cos direction, 0 0 D0 0 5
Euler’s Equations (set up)
nd ˆ b) ω : This is the (2 ) rotation along the “line of nodes” ( x 1 x in the intermediate ,, coordinate system) In the intermediate axes, we have ω 0 ξ 0 To express it in the body axes, we apply the Euler rotations BC
T cos Note: Since ,0,0 is ω BC 0 sin body already in the x ˆ 0 0 direction, C0 0 0 0 6
Euler’s Equations (set up)
c) ω : Finally, the last rotation is along the x 3 zˆ of the ', ', '
0 In the ', ', ' axes, we have ω 0 ' To express it in the body axes, we apply the Euler rotations B
T 0 0 Note: Since 0,0, is ω B 0 0 already in the z ˆ body direction, 0 0 B0 0 7
Euler’s Equations (set up)
Putting all three pieces together, we have
sin sin cos ωω ω ω cos sin sin cos
These are the components of ω expressed in the “body” frame using the Euler’s angles. 8
Euler’s Equations (set up)
Alternatively, one can think of how the vector ω being ' “projected” along different set of bases:
- basis along the Euler directions e1,, e 2 e 3 e 1 ω - basis along the body frame xyˆ ,,ˆ z ˆ e 3 ωe1 e 2 e 3 ' ωx xˆ y y ˆ z z ˆ e 2
' Note: the basis set e 1 ,, e 2 e 3 defining an infinitesimal rotation along the Euler angles is NOT an orthogonal set of vectors.
(Note: Here and onward, space frame is primed and body frame is unprimed.) 9
Euler’s Equations (derivation)
Now, we will continue with our equation of motion for a rotating rigid body.
I is diagonalized since we’ve chosen the body 1 T I 2 2 i i axes to lay along the principal axes and we will
call the nonzero diagonal elements, Iii I i
Without further assuming the nature of the applied forces acting on this system, we will use the following general form of the E-L equation:
dT T Qi is the generalized force (including Qi dt qi q i forces derivable from conservative and non-conservative sources) 10
Euler’s Equations (derivation)
Let calculate the equation of motion explicitly for :
1 2 T Ii i T T i I i (E’s sum) 2 i i i sin sin cos 1 2 3 ω cos sin sin I1 1 I 2 2 I 3 3 cos
I1 10 I 2 2 0 I 3 3 1
I3 3
d T I3 3 dt 11
sin sin cos Euler’s Equations (derivation) ω cos sin sin cos Now, we need
T T i 1 2 3 I1 1 I 2 2 I 3 3 i
Note that: 1 cos sin sin 3 0 2 2 sin sin cos 1
Thus, we have,
T I I 0 II 1 12 2 2 1 1 12 2 2 1 12
Euler’s Equations (derivation)
Now, we need to calculate the generalized force with respect to :
Since the Euler angle is associated with a rotation about the z ˆ axis in the “body” frame, we have,
r zˆ i ˆ Q Fi Fi z ri ri i i
zˆ N N3 r r i i
used ABC BCA ri r i sin r i zˆ r i 13
Euler’s Equations (derivation)
Finally, putting everything together, the E-L equation gives,
dT T Q dt
II3 3 1 1 2 I 2 1 2 N 3 I3 3 II 1 2 1 2 N 3
- One can calculate the E-L equation for , but (they are ugly) we are not doing them here !
- There is a smarter way to get EOM for the other two dofs…
Since nothing required our choice of to lay along . Then, by a 3 zˆ(x3 ) symmetry argument, the other components of ω should have a SIMILAR form. 14
Euler’s Equations (derivation)
This then gives,
I1 1 II 2 3 2 3 N 1 (same cyclic symmetry as the equation for ) I2 2 II 3 1 3 1 N 2 3
I3 3 II 1 2 1 2 N 3
In principle, one can get out the 2 and 3 equation by solving for 2 and 3 simultaneously from the , Euler-Lagrange equations.
These are called the Euler’s Equations and the motion is described in terms of the Principal Moments ! 15
Euler’s Equations (derivation)
The following is Goldstein’s (Newtonian) derivation. We start with,
dL dL N ω L dt fixed dt body Writing this vector equation out in the components of the body axes,
dL Ni L idt ijkjk Choose the body axes to coincide with the Principal axes, so that
Li I ii (no sum, just writing out the components)
di NIi i ijkjkk I dt (no sum) 16
Euler’s Equations (derivation)
Writing this index equation out explicitly for i 1,2,3 , we have,
I11 123233 I 132322 IIII 11 2 3 23 N 1
I2 2 231 3 I 1 1 213 1 IIII 3 3 2 2 3 1 3 1 N 2
I33 312122 I 321211 IIII 33 1 2 12 N 3
So, this gives us the same set of Euler’s equations as previously.
The Euler’s Equations describes motion in the body frame. ω and N are vectors expressed in the body frame. 17
Torque Free Motion of a Symmetric Top
A symmetric top means that: I1 I 2 I 3
(example will be a long cigar-like For concreteness, let I1 I 2 I 3 objects such as a juggling pin) Euler equations (torque free) are: dL I1 1 II 2 3 2 3 ω L dt body I2 2 II 3 1 3 1 or
I3 3 II 1 2 1 2 0 L Iω ωL
Trivial case ( ω is along one of the principal axes):
ω is along one of the eigendirection of I and L ω L Iω ω L 0 ω 0 18
Torque Free Motion of a Symmetric Top
Interesting case ( ω is NOT along one of the principal axes):
We still have, 3 0 since I1 I 2
3 const Note: x ˆ 3 is along the body’s symmetry axis.
And, the rest of the Euler equations give,
I2I3 I 1 I 3 1 2 3 2 3 I1 I 1 Note: I1 I 2
II3 1 II 3 1 2 3 1 3 1 I2 I 1 19
Torque Free Motion of a Symmetric Top
I3 I 1 Let 3 const I1 Then, the remaining two Euler’s equations reduce simply to,
1 2
2 1 Taking the derivative of the top equation and substitute the bottom on the right, we have,
2 1 2 1 1
2 Since, 0 we have the solution: A,0 will be determined by ICs
1tA cos t 0 and 2tA sin t 0 20
Torque Free Motion of a Symmetric Top
Looking at this deeper… First in the “body” frame,
- We know that 3 is a constant and 1 & 2 oscillates harmonically in a circle.
2 2 2 So, 1 2 3 const ω const
In the “body” axes, this description for ω can be visualized as ω
precessing about xˆ 3 . x 3 -The projection of ω onto the x 3 axis is fixed. ω -The projection of ω onto the x 1 x 2 plane rotates as a parametric circle with a rate of I I x2 3 1 3 const x1 I1 (This is called the “body” cone) 21
Torque Free Motion of a Symmetric Top
Now, let look at this same situation in the “fixed” frame,
Observations: 1 1. Energy is conserved in this problem so that Tω L const rot 2 2. The situation is torque free so that L is fixed in space.
#1 and #2 means that ω must also precesses around L in the fixed frame
x3 ' - Assume L lies along the x 3 ' axis in the fixed frame. L ω - L is a constant vector - The dot product ω L must remain constant.
x2 ' (This is called the “space” cone) x1 ' 22
Torque Free Motion of a Symmetric Top
Observations (in the fixed axes) cont:
3. The three vectors ω ,,( L xˆ 3 body ) always lie on a plane. Consider the following product:
L ω xˆ 3 where x ˆ 3 is in the zˆ direction in the body axes
L 2 xˆˆ 1 1 x 2 2 L x ˆ 1 1 L x ˆ 2 Since the body axes are chosen to lie along the principal axes, we have
Li I ii ( no sum )
L ω xˆ 3 2I 1 1 1 I 2 2 0
(for a symmetric top) I1 I 2 23
Torque Free Motion of a Symmetric Top
Observations (in the fixed axes) cont:
This means that all three vectors ω ,, L xˆ 3 always lie on a plane.
ˆ ω x3 ω xˆ 3 L ω xˆ 0 (for a symmetric top with L 3 or without torque)
http://demonstrations.wolfram.com/Angu Summary: larMomentumOfARotatingRigidBody/
-ω precesses around the “body” cone -ω also precesses around the “space” cone
- All three vectors ω ,, L xˆ 3 always lie on a plane
- L is chosen to align with xˆ 3 ' in the space axes 24
Torque Free Motion of a Symmetric Top
This can be visualized as the body cone rolling either inside or outside of the space cone !
Precession Rate
I3 I 1 3 I1
Case 1: I1 I 3 Case 2: I1 I 3 25
Motion of a Symmetric Top with Constant
All three vectors ω ,, L xˆ 3 remains to lie on a plane,
ˆ ω x3 ω xˆ 3 L ω xˆ 0 L 3
In fixed frame, set of body axes and L precesses around ω
http://demonstrations.wolfram.com/AngularMomentumOfARotatingRigid Body/ 26
Stability of General Torque Free Motion
Consider torque-free motion for a rigid body with I1 I 2 I 3 Again, we have chosen the body axes to align with the principal axes.
As an example, we will consider rotation near the x 1 axis (similar analysis can be done near the other two principal axes).
this means that we have,
ω1 xˆ 1t x ˆ 2 t x ˆ 3
where t , t are small time-dependent perturbation to the motion
For stability analysis, we wish to analyze the time evolution of these two quantities to see if they remain small or will they blow up. 27
Stability of General Torque Free Motion
Plugging our perturbation into the Euler’s equations, we have
III1 1 2 3 2 3 0 III 1 1 2 3 0
III2 2 3 1 3 1 0 III 2 3 1 1 0
III3 3 1 2 1 2 0 III 3 1 2 1 0
Assume small perturbations and drops higher order terms ( ), the first equation gives,
1 0 1 const 28
Stability of General Torque Free Motion
And, the other two equations reduces to,
I3 I 1 1 0 I2
I1 I 2 1 0 I3 Taking the derivative of the top equation and substitute the bottom into it,
II3 1 II 3 1 I1 I 2 1 1 1 I2 I 2 I3
I3 I 1 I 1 I 2 2 1 I2 I 3 29
Stability of General Torque Free Motion
Since we have chosen to have I 1 I 2 I 3 , the constant
2I1 I3 I1 I 2 2 1 0 (Note: we have switch the order I2 I 3 2 And, we can write of I 1 , I 3 so that is explicitly positive.) 2
The solution to this ODE is oscillatory, i.e., t Aeit Be it
it it and t Ae' Be ' where A, B, A’, & B’ depends on ICs
Thus, both of the small perturbations are oscillatory and the
rotation about the x 1 axis is stable ! 30
Stability of General Torque Free Motion
With a similar calculation for rotation near the x 3 , one can show again
that small perturbations are oscillatory and motion about the x 3 axis is stable.
However, a similar analysis will show that the oscillatory motion for the
perturbations will become exponential if we consider rotation near the x 2 axis.
Summary: Without any applied torque, motion around the principle axes with the largest and the smallest principal moments are stable while motion around the intermediate axis is unstable.
http://www.youtube.com/watch?v=XALe27bnUm8