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PHYS 705: Classical Euler’s Equations 2

Euler’s Equations (set up)

We have seen how to describe the kinematic properties of a . Now, we would like to get equations of for it.

1. We will follow the Lagrangian Formalism that we have developed.

2. For , we will use the Euler’s with one point of the rigid body being fixed (no translation, just )

3. As we have seen previously, the rotational kinetic is given by 1 1 TωIω   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 along the Euler’s angles. Then, the 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       D0  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  ,, )        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,   C0  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        B0  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 ωe1    e 2   e 3 '   ωx 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, 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 acting on this system, we will use the following general form of the E-L equation:

dT   T Qi is the generalized (including     Qi dt qi   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 10  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  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)

 dL  dL  N      ω L dt  fixed  dt body Writing this vector equation out in the components of the body axes,

dL Ni    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)

di  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 ( 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,

I2I3   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 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

1tA cos  t  0  and 2tA 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 Tω  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 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  2I 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,

ω1 xˆ 1t 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 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

2I1  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.