1 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). As we have seen previously, the rotational kinetic energy is given by T 1 1 2 2 i ij j 4. Choose the body axes to coincide with the Principal axes, then T 1 2 2 ii i ( no sum; ij is diagonized! )
Euler s Equations (set up) Note: -We still have the freedom to align (from the body axes) to any one of the Principal axes. - The three Euler s angles give the orientation of the Principal axes of the body relative to the fixed axes.,, zˆ 5. A general rotation (an inf. one here) along a given axis in the body frame can be decomposed into three rotations along the Euler s angles. Then, the time dω dt rate of change of this rotation can also be written as, xˆ - These three different pieces correspond to time rate of change of the individual rotations along each of the three Euler s angles. dω (we write this as a sum since the angular changes are infinitesimal)
4 Euler s Equations (set up) -Now, our task is to express each of these in terms of the body coordinate x, x, x 1 2 We will go through the three individual Euler steps now: a) : We are in the fixed axes and we do a rotation along the n the fixed axes, we have z ˆ To express it in the body axes, we apply the Euler rotations BCD body fixed 0 0 0 sin sin BCD 0 cos sin cos x Note: Since is already in the direction, 0,0, ẑ 0 0 D 0 0 T
5 Euler s Equations (set up) b) : This is the (2 nd ) rotation along the line of nodes ( in the intermediate coordinate system) 0 n the intermediate axes, we have 0 x 1 ˆ To express it in the body axes, we apply the Euler rotations BC,, body cos BC 0 sin 0 0 ξ x Note: Since is already in the direction, ˆx,0,0 T C0 0 0 0
6 Euler s Equations (set up) z c) : Finally, the last rotation is along the of the ˆ 0 n the axes, we have ' ', ', ' To express it in the body axes, we apply the Euler rotations B x 0 ', ', ' body 0 0 B 0 0 Note: Since is already in the direction, 0,0, ẑ 0 0 B 0 0 T
7 Euler s Equations (set up) Putting all three pieces together, we have These are the components of expressed in the body frame using the Euler s angles. sin sin cos cos sin sin cos
8 Euler s Equations (set up) ' Alternatively, one can geometrically project defined along the Euler directions onto the body axes (red frame) e1, e 2, e e ' e 1 e 2 ' e e e Note: the basis set 1 2 xˆ yˆ zˆ x y z defining an infinitesimal rotation e1, e 2, e 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. T 1 2 2 i i is diagonalized since we ve chosen the body axes to lay along the principal axes and we will call the nonzero diagonal elements, ii 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: d T T dt q i qi Q i Q i is the generalized force (including forces derivable from conservative and non-conservative sources)
10 Euler s Equations (derivation) Let calculate the equation of motion explicitly for : T T i i (E s sum) i i i 1 2 11 22 0 0 1 1 1 2 2 T 1 2 2 i i sin sin cos cos sin sin cos d dt T
11 Euler s Equations (derivation) Now, we need T T i 1 2 1 1 2 2 i sin sin cos cos sin sin cos Note that: 1 cos sin 2 sin sin sin cos 2 1 0 Thus, we have, T 1122210 1 1 2 221
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 axis in the body frame, we have, ẑ Q i Fi r i zˆ N N F ˆ i z ri i ẑ r i ri r i used ABCBCA r i ri r i zˆ r sin i
1 Euler s Equations (derivation) Finally, putting everything together, the E-L equation gives, d T T dt Q N 1 1 2 2 1 2 N 1 2 1 2, - 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 zˆ( x ) Since nothing required our choice of to lay along. Then, by a symmetry argument, the other components of should have a SMLAR form.
14 Euler s Equations (derivation) This then gives, N 1 1 2 2 1 N 2 2 1 1 2 N 1 2 1 2 (same cyclic symmetry as the equation for ) and 2 and n principle, one can get out the equation by solving for, 2 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, N dl dt fixed dl L dt body Writing this vector equation out in the components of the body axes, N dl L dt i i ijk j k Choose the body axes to coincide with the Principal axes, so that L i i i (no sum, just writing out the components) di Ni i ijkjkk dt (no sum)
16 Euler s Equations (derivation) Writing this index equation out explicitly for, we have, N 1 1 12 2 12 2 2 1 1 2 2 1 N 2 2 21 1 1 21 1 2 2 1 1 2 N 12 1 2 2 21 2 1 1 1 2 1 2 i 1, 2, So, this gives us the same set of Euler s equations as previously. The Euler s Equations describes motion in the body frame. vectors expressed in the body frame. N and are
17 Torque Free Motion of a Symmetric Top A symmetric top means that: 1 2 For concreteness, let Euler equations (torque free) are: 1 1 2 2 2 2 1 1 1 2 1 2 0 1 2 (example will be a long cigar-like objects such as a juggling pin) Trivial case ( is along one of the principal axes): or dl dt body L L L is along one of the eigendirection of and L L 0 0 L
18 Torque Free Motion of a Symmetric Top nteresting case ( is NOT along one of the principal axes): We still have, since 0 1 2 const ˆx Note: is along the body s symmetry axis. And, the rest of the Euler equations give, 2 1 1 2 2 1 1 1 1 2 1 1 2 1 Note: 1 2
19 Torque Free Motion of a Symmetric Top Let 1 1 const 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 Cs t Acost 1 0 and t Asin t 2 0
20 Torque Free Motion of a Symmetric Top Looking at this deeper First in the body frame, 1 & 2 - We know that is a constant and oscillates harmonically in a circle. So, const 2 2 2 1 2 const n the body axes, this description for can be visualized as ˆx precessing about. x 1 x x 2 (This is called the body cone) -The projection of onto the axis is fixed. -The projection of onto the plane rotates x 1 2 as a parametric circle with a rate of 1 1 x x const
21 Torque Free Motion of a Symmetric Top Now, let look at this same situation in the fixed frame, Observations: 1. Energy is conserved in this problem so that Trot 2. The situation is torque free so that L is fixed in space. 1 L 2 const #1 and #2 means that must also precesses around L in the fixed frame x ' L - Assume L lies along the axis in the fixed frame. - L is a constant vector L x ' - The dot product must remain constant. x 1 ' x 2 ' (This is called the space cone)
22 Torque Free Motion of a Symmetric Top Observations (in the fixed axes) cont: ˆ Lx,, ( body). The three vectors always lie on a plane. Consider the following product: L ˆ x where ˆx is in the direction in the body axes ẑ ˆ ˆ ˆ ˆ L x x Lx Lx 2 1 1 2 2 1 1 2 Since the body axes are chosen to lie along the principal axes, we have L ( no sum) i i i ˆ 0 L x 2 1 1 1 2 2 (for a symmetric top) 1 2
2 Torque Free Motion of a Symmetric Top Observations (in the fixed axes) cont: This means that all three vectors always lie on a plane. ˆ Lx,, xˆ ˆx L L x ˆ 0 (for a symmetric top with or without torque) Summary: http://demonstrations.wolfram.com/angu larmomentumofarotatingrigidbody/ - precesses around the body cone - also precesses around the space cone - All three vectors always lie on a plane - L is chosen to align with in the space axes ˆ Lx,, xˆ '
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 1 1 Case 1: 1 Case 2: 1
25 Motion of a Symmetric Top with Constant ˆ Lx,, All three vectors remains to lie on a plane, xˆ ˆx L L xˆ 0 n fixed frame, set of body axes and precesses around L http://demonstrations.wolfram.com/angularmomentumofarotatingrigid Body/
26 Stability of General Torque Free Motion Consider torque-free motion for a rigid body with 1 2 Again, we have chosen the body axes to align with the principal axes. As an example, we will consider rotation near the axis (similar analysis can be done near the other two principal axes). this means that we have, t t t xˆ xˆ xˆ, t 1 1 2 where are small time-dependent perturbation to the motion x 1 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 0 0 1 1 2 2 1 1 2 0 0 2 2 1 1 2 1 1 0 0 1 2 1 2 1 2 1 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, 1 1 2 1 2 1 0 0 Taking the derivative of the top equation and substitute the bottom into it, 1 1 1 2 1 1 1 2 2 1 1 2 2 1 2
29 Stability of General Torque Free Motion Since we have chosen to have, the constant And, we can write 2 1 2 2 1 1 2 2 1 0 2 The solution to this ODE is oscillatory, i.e., (Note: we have switch the order, of so that is explicitly 1 positive.) 2 and it it t Ae Be t A' e B' e i t it where A, B, A, & B depends on Cs Thus, both of the small perturbations are oscillatory and the rotation about the x 1 axis is stable!
0 Stability of General Torque Free Motion With a similar calculation for rotation near the, one can show again that small perturbations are oscillatory and motion about the axis is stable. However, a similar analysis will show that the oscillatory motion for the x x 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