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Rotational dynamics

  1. Mar 19, 2007 #1


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    1. The problem statement, all variables and given/known data

    Does anyone know of a treatment of rotational dynamics especially the heavy top and precession of the equinoxes which uses only vectors and tensors. I've got treatments in terms of Lagrange's equations, but I wanted something using only torques etc.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 19, 2007 #2


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    Does this help?
    http://theory.phy.umist.ac.uk/~mikeb/lecture/pc167/rigidbody/gyro.html [Broken]

    There's a good reason why general theory is not usually done using vectors and tensors: adding up large rotations is not commutative.

    A 90 degree rotation about X followed by a 90 degree rotation about Y is not the same as rotation about Y and then about X.

    The consequence is that arbitrary large rotations are not vector quantities! The "easy" way to get over that hurdle is to use Euler angles and a Lagrangian formulation instead.
    Last edited by a moderator: May 2, 2017
  4. Mar 19, 2007 #3


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    Many thanks, that's just what I'm looking for.
  5. Mar 19, 2007 #4

    D H

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    Don't do that! http://www.google.com/search?client=safari&rls=en&q="Euler+angles+are+evil"&ie=UTF-8&oe=UTF-8".

    This first result is covered in most upper-level mechanics course: The relationship between time derivative of a vector quantuty in a rotating versus non-rotating frame.

    Suppose we have two reference frames that share the same origin but one has inertial axes while the other is rotating at some rate [itex]\vect \omega[/itex] with respect to this inertial frame. The time derivative of some vector quantity [itex]\vect q[/itex] depends on the observer's reference frame:

    \left(\frac {d\vect q}{dt}\right)_I =
    \left(\frac {d\vect q}{dt}\right)_R + \vect \omega \times \vect q

    This can be applied to the problem of rigid body rotational dynamics to get a tensor/vector based version of Euler's equations for a rigid body.

    \mathbf I &\text{\ be the inertia matrix for some body} \\
    \vect \omega &\text{\ be the rotation rate of the body with respect to inertial}

    where both [itex]\mathbf I[/itex] and [itex]\vect \omega[/itex] are represented in the coordinates of the rotating body (body frame coordinates).

    The angular momentum of the body with respect to inertial represented in body frame coordinates is

    [tex]\vect L = \mathbf I\;\vect \omega[/itex]

    Differentiating with respect to time,

    [tex]\left(\frac {d\vect L}{dt}\right)_R =
    \frac {d\mathbf I}{dt}\;\vect \omega +
    \mathbf I\;\frac {d\vect \omega}{dt}[/itex]

    Using the generic relation for the time derivative of a vector quantity,

    [tex]\left(\frac {d\vect L}{dt}\right)_I =
    \frac {d\mathbf I}{dt}\;\vect \omega +
    \mathbf I\;\frac {d\vect \omega}{dt} +
    \vect\omega\times(\mathbf I\;\vect \omega)

    The rotational equivalent of Newton's second Law is

    [tex]\left(\frac {d\vect L}{dt}\right)_I = \vect N[/tex]

    where [itex]\vect N[/itex] os the net external torque acting on the body.

    Combining the above,

    \frac {d\mathbf I}{dt}\;\vect \omega +
    \mathbf I\;\frac {d\vect \omega}{dt} +
    \vect\omega\times(\mathbf I\;\vect \omega) = \vect N

    Note that if [itex]\mathbf I[/itex] is constant, the above reduces to

    \mathbf I\;\frac {d\vect \omega}{dt}
    = \vect N - \vect\omega\times(\mathbf I\;\vect \omega)

    The term [itex]\vect\omega\times(\mathbf I\;\vect \omega)[/itex] is the rotational analog of the Coriolis force.

    Finally, Euler's equations result in the special case of [itex]\mathbf I[/itex] being a diagonal matrix.
    Last edited by a moderator: Apr 22, 2017
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