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Rotational dynamics

  • Thread starter Ray
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Ray
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Homework Statement



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.

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
AlephZero
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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.
 
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  • #3
Ray
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Many thanks, that's just what I'm looking for.
 
  • #4
D H
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The "easy" way to get over that hurdle is to use Euler angles and a Lagrangian formulation instead.
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:

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

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.

Let
[tex]
\begin{matrix}
\mathbf I &\text{\ be the inertia matrix for some body} \\
\vect \omega &\text{\ be the rotation rate of the body with respect to inertial}
\end{matrix}
[/tex]

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)
[/tex]

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,

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

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

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

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.
 
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