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Classic spinning symmetric top

  1. Apr 22, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Derive Euler's equation of motion for a rigid body: $$\dot{\vec{L}} + \vec{\omega} \times \vec{L} = \vec{G},$$ where ##\vec{L}## is the angular momentum in the body frame, ##\omega## is the instantaneous velocity of the body's rotation and ##\vec{G}## is the external torque.

    Subsequent parts are written in an attachment.

    2. Relevant equations
    Expressions for angular momentum and kinetic energy in rotating frame or body frame fixed in the body.

    3. The attempt at a solution
    I believe that equation is saying the rate of change of angular momentum in some inertial frame is the rate of change of angular momentum in the body frame + another term due to rotation of axes relative to the inertial frame. Is that right? I found a derivation of the result for a general vector A but I could not understand this equation fully: $$[\dot{A}]_{S_o} = [\dot{A}]_S + \omega \times [A]_S$$ (##S_o## denotes the inertial frame and S the non inertial one and they coincide instantaneously. ##[A]_x## denotes A measured in frame x.)

    The term on the left is the rate of change of A in the inertial frame. The first term on the RHS is the rate of change of A in S. Why is it then ##\omega \times [A]_S##? If we consider a turntable of radius R, then the velocity of a point on the circumference is ω x R relative to the centre of the turntable. But the centre of the turntable is fixed (so inertial) and a frame co-moving with a point on the circumference rotates relative to the centre so should it not be ##\omega \times [A]_{S_o}?##
    Many thanks.
     

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  3. Apr 23, 2014 #2

    CAF123

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    Or perhaps it is the case that given that the frames coincided instantaneously ##[A]_{S_o} = [A]_{S}##? But then this cannot be true for the whole motion.
     
  4. Apr 23, 2014 #3

    D H

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    Well, that's one way to express the kinematics transport theorem (not to be confused with the Reynold's transport theorem, or a number of other things called the transport theorem).

    First it's important to distinguish between what I'll call a reference system and a coordinate system. (Note well: The literature is entirely inconsistent. Some groups use reference frame for what I'm calling a reference system, some groups use reference frame for what I'm calling a coordinate system.)

    I'll use your turntable to distinguish the concepts, except I'll use a merry-go-round instead. Imagine two observers on the merry-go-round and another two on the ground. Each observer sets up a coordinate system based on how they see things. While the two observers on the merry-go-round might pick different basis vectors, those two rotating coordinate systems are embedded in the same rotating reference system. The same goes for the two observers on the ground. They can define two different non-rotating ("inertial") coordinate systems, but those two inertial coordinate systems are just different ways of expressing things from the perspective of the same inertial reference system.

    At a specific point in time, we'll have all four observers describe the displacement vector from the center of the merry-go-round to the nose of the spotted pony on the merry-go-round. All four observers see exactly the same vector. How they represent that vector in their observer-specific coordinate systems will differ, but it's still the same vector.

    It's time derivative is not the same to all observers. The two on the merry-go-round will say the derivative is zero. The pony isn't moving from the perspective of those rotating observers. The two on the ground will say the derivative is non-zero. Those two inertial observers see the same derivative vector. It's only the representations that differ.

    We can do the same for any other vector ##\vec a## that is substantively the same to all four observers. While the vector ##\vec a## is the same vector to all observers, it's time derivative is not. The relationship between the time derivative of ##\vec a## from the perspective of our rotating and non-rotating observers is
    [tex] \left(\frac {d\vec a}{dt}\right)_R + \vec \omega \times \vec a = \left(\frac {d\vec a}{dt}\right)_I [/tex]
    That's the coordinate-free version of the transport theorem. The subscript R on the left denotes the time derivative as observed by a rotating observer, the subscript I on the right denotes the time derivative as observed by a non-rotating (inertial) observer. Deriving this is non-trivial. Just take it as a given.

    Now let's make this specific to two different cartesian coordinate systems. Using a subscript R to denote one of the rotating observer's coordinate system and a subscript I to denote one of the inertial observer's coordinate system, the transport theorem becomes
    [tex] \frac {d\vec a_R}{dt} + \vec \omega_R \times \vec a_R = T_{I\to R} \frac {d\vec a_I}{dt} [/tex]
    where ##T_{I\to R}## is the time-varying transformation matrix from the inertial frame I to the rotating frame R.
     
  5. Apr 23, 2014 #4

    CAF123

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    Hi D H,
    If a is the vector from the middle of the merry-go round to the pony, in the frame co-moving with the pony, da/dt_R is zero, so da/dt_I = w x a. And the a on the right hand side there corresponds to both the representation in the rotating frame and the representation in the inertial frame?
     
  6. Apr 23, 2014 #5

    D H

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    No. It's just a vector. How that vector is represented is not an issue in that first perspective. There's a big difference between what that vector represents and how that vector is represented. In this case, the vector ##\vec a## from the center of the merry-go round to the pony's nose represents the exact same thing to all of our observers. They all agree on it's length, they all agree on what other objects that vector point to.

    When you take the time derivative that's no longer the case. The two inertial observers agree that their time derivatives are substantively the same vector, even if their coordinate representations differ. The two rotating observers similarly agree that their time derivatives are substantively the same vector. An inertial and rotating observer will not agree. To one the pony is moving and to the other, it's stationary.

    Now what happens when you substitute angular momentum ##\vec L## for that vector ##\vec a##? Keep in mind that angular momentum in the rotating frame and angular momentum in the inertial frame are the same vectors. They just have different coordinate representations.
     
  7. Apr 23, 2014 #6
    Consider a rotating unit vector. Because it is unit, its length is constant, so ##0 = {d \over dt} \left( \vec u \cdot \vec u\right) = 2 \dot {\vec u} \cdot \vec u##, i.e., the derivative of a unit vector is orthogonal with the vector itself. This should also be intuitively clear. Because of that, there exists vector ##\vec \omega## such that ##\dot {\vec u} = \vec \omega \times \vec u##. Vector ##\vec \omega## is the "angular velocity" of vector ##\vec u##.

    Now let's take any vector ##\vec v(t) ## and represent it in a rotating frame. A rotating frame is specified by three rotating unit vectors ##\vec i, \vec j, \vec k## whose derivatives because of the above are given by ##\vec \omega \times \vec i, \vec \omega \times \vec j, \vec \omega \times \vec k## respectively. Because ## \vec v(t) = x(t) \vec i + y(t) \vec j + z(t) \vec k ##, ## \dot {\vec v} = \dot x \vec i + \dot y \vec j + \dot z \vec k + x \vec \omega \times \vec i + y \vec \omega \times \vec j + z \vec \omega \times \vec k = \vec p + \omega \times \vec v ##, where ## \vec p = \dot x \vec i + \dot y \vec j + \dot z \vec k ##. Observe that ##\vec p## looks like the derivative of ##\vec v## with the frame unit vectors fixed, or the "derivative in the co-rotating frame". This is what is meant by that funny bracket notation.
     
  8. Apr 23, 2014 #7

    CAF123

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    Hi voko,
    So the ##\vec v## in that formula is the coordinate representation of a point relative to the co-moving frame. ##\vec \omega \times \vec v## encapsulates the fact that the co-moving coordinate frame is rotating relative to a fixed frame, so the quantity gives the rate of rotation of the co-moving frame relative to this fixed frame. ##\vec p## gives the rate of change of a vector describing a point relative to the co-moving frame. Yes?

    How do you know, in your derivation, that ##\dot{\vec{v}}## is the rate of change of the vector v in the fixed frame?

    Going back to the merry-go round example, and ##\vec a## describes a vector from the centre to a pony. In a frame co-moving with the riders, this vector remains fixed so ##\vec p = 0## and therefore ##\dot{\vec a} = \vec \omega \times \vec a##
     
  9. Apr 23, 2014 #8

    D H

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    If by "fixed frame" you mean the rotating frame, he didn't. voko's result is the rate of change of the vector v in the inertial frame. The unit vectors are constant in the rotating frame, so the derivative in the rotating frame is ##\dot v(t)_R = \dot x(t) \hat \imath + \dot y(t) \hat \jmath + \dot z(t) \hat k##. It's when you compute the derivative in the inertial frame that you get those extra derivative terms.

    That ##\dot{\hat u} = \vec \omega \times \hat u## is a non-trivial exercise. It's usually done in terms of "bad physics math" -- a big handwaving exercise. For a decent derivation you'll either have to look at a graduate level aerospace engineering text (and even there, most authors still wave their hands rather vigorously), a graduate level classical physics text (ditto on most authors using vigorous hand waving), or a decent upper level undergrad or graduate level math text on Lie groups (but here you might have to read between the lines).
     
  10. Apr 23, 2014 #9

    CAF123

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    Sorry, I meant the inertial frame. What I don't understand is if the basis ##\left\{i,j,k\right\}## is used to describe some point in the rotating frame (if i interpreted voko correctly) then why would the time derivative of some vector in that frame ##v(t)##(I.e v(t) had components specified in the rotating frame) give the rate of change of that vector in the inertial frame?
     
  11. Apr 23, 2014 #10

    D H

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    Let's go back to my rotating and inertial observers. The rotating observers see those rotating frame unit vectors as constant, so the only parts of ##\vec v(t) = x(t)\hat i + y(t)\hat j + z(t)\hat k## that vary with time are the components ##\{x(t),y(t),z(t)\}##. The inertial observers see those rotating frame unit vectors as rotating, which means it's better to write that vector using the rotating frame basis vectors as ##\vec v(t) = x(t)\hat i(t) + y(t)\hat j(t) + z(t) \hat k(t)## when viewed from the perspective of an inertial observer.
     
  12. Apr 24, 2014 #11
    The "point in the rotating frame" bit is misleading. Note that vector ##\vec v## was originally given without referencing any frames, it was "just" a vector. Then I spoke of representing it in a particular frame. Any frame, rotating or not, can be associated with three mutually orthogonal unit vectors, and any vector can be represented in that frame using those vectors, giving rise to coordinates of that vector in that frame. It is the coordinates that are frame-dependent, but not (necessarily) the vector itself. That is what I did; the vector was frame-independent, but its coordinates x, y, z were. This frame-dependency of the coordinates is exactly what gives the two terms in the formula for the full derivative: if we just differentiate the three coordinate functions, we obtain three other functions, which we can interpret as some vector using the original basis vectors; this is ##\vec p## in my derivation, and is known via some other confusing notation and confusing names.
     
  13. Apr 24, 2014 #12
    I omitted the proof that ##\vec \omega## is uniquely defined and is the same for all the three basis vectors of the rotating frame. That, even though it may be almost obvious, gets to be rather lengthy to prove both correctly and lucidly. Overall, I like the exposition in Landau & Lifshitz Vol. 1, where it takes a few pages.

    I never cease to wonder how all that knowledge was derived before the notion of vectors was well established and the vectorial symbolism got good traction. That was truly non-trivial.
     
  14. Apr 25, 2014 #13

    CAF123

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    Thanks D H, voko.
    Is that correct? This means that as viewed from the observer attached to the inertial frame, the pony undergoes rotational motion only.

    My next question is more relevant to the problem at hand. In the body frame, we write that ##T=\frac{1}{2}(I_1 w_1^2, I_2 w_2^2, I_3 w_3^2)##. But in the body frame, we are rotating with the top, so why is the kinetic energy of this form? Is it because in the body frame we feel the centrifugal force and so an observer attached to the body frame knows he is rotating and therefore his kinetic energy is like so?
     
  15. Apr 25, 2014 #14
    Yes, that is correct. That is in fact the familiar equation relating linear and angular velocities: ## \vec v = \vec \omega \times \vec r ##.

    I do not understand what that equation means. If ##T## is kinetic energy, it must be scalar.

    In a frame co-moving with a rigid body, the kinetic energy of the body is zero. Pretty much by the definition of kinetic energy, which is based solely on the speed of an object.
     
  16. Apr 25, 2014 #15

    CAF123

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    Ok, so in my turntable example, viewed from the centre, the motion of a point on the circumference is purely rotational.
    :eek: Rather ##T = \frac{1}{2}(I_1 w_1^2 + I_2w_2^2 + I_3 w_3^2)##

    One part of the question is to show that kinetic energy is conserved. But why is the kinetic energy of the form above?
     
  17. Apr 25, 2014 #16
    Well, there is more than one approach to that. I believe you and I discussed a while ago how you can decompose the kinetic energy of a system into the kinetic energy of the centre of mass and the "internal" kinetic energy. That could be useful here.
     
  18. Apr 25, 2014 #17

    CAF123

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    Yes. The kinetic energy to an inertial frame is the kinetic energy in the CoM + the kinetic energy due to rotation about CoM. The kinetic energy due to rotation about CoM is then ##T## above and since the body is just rotating (i.e not translating and rotating), the speed of the CoM is zero. Yes?

    I can show that kinetic energy is conserved, by solving the Euler equations for ##w_1, w_2## and seeing that the sum of ##w_1^2 +w_2^2## is constant.
     
  19. Apr 25, 2014 #18
    That looks good.

    Assuming that you have already established the equation relating kinetic energy, angular velocity and the tensor of inertia, that should complete the proof.
     
  20. Apr 25, 2014 #19

    CAF123

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    That should just be ##T##. It is strange though because the next part of the question after showing the kinetic energy is conserved is to show that the sum ##w_1^2 + w_2^2## is constant. But I used this already in my derivation of the constancy of the kinetic energy.

    If in the energy relative to the inertial frame is ##T##, then what is the energy relative to the body frame, or frame co-moving with the body? My notes say that it is also ##T##.
     
  21. Apr 25, 2014 #20
    I have re-read the problem and it requires an explicit demonstration of energy conservation. Which probably means that you should start with the definition of kinetic energy and Euler's equations. Sorry that I misdirected you.
     
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