Rotational Dynamics: Vectors & Tensors for Heavy Top & Equinox Precession

[Ray]

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

 

[AlephZero]

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

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.

 

[Ray]

Many thanks, that's just what I'm looking for.

 

[D H]

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 \vect \omega with respect to this inertial frame. The time derivative of some vector quantity \vect q depends on the observer's reference frame:

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

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
<br /> \begin{matrix}<br /> \mathbf I &amp;\text{\ be the inertia matrix for some body} \\<br /> \vect \omega &amp;\text{\ be the rotation rate of the body with respect to inertial}<br /> \end{matrix}<br />

where both \mathbf I and \vect \omega 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

\vect L = \mathbf I\;\vect \omega[/itex]<br /> <br /> Differentiating with respect to time,<br /> <br /> \left(\frac {d\vect L}{dt}\right)_R =&lt;br /&gt; \frac {d\mathbf I}{dt}\;\vect \omega +&lt;br /&gt; \mathbf I\;\frac {d\vect \omega}{dt}[/itex]&lt;br /&gt; &lt;br /&gt; Using the generic relation for the time derivative of a vector quantity,&lt;br /&gt; &lt;br /&gt; \left(\frac {d\vect L}{dt}\right)_I =&amp;lt;br /&amp;gt; \frac {d\mathbf I}{dt}\;\vect \omega +&amp;lt;br /&amp;gt; \mathbf I\;\frac {d\vect \omega}{dt} +&amp;lt;br /&amp;gt; \vect\omega\times(\mathbf I\;\vect \omega)&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; The rotational equivalent of Newton&amp;#039;s second Law is&lt;br /&gt; &lt;br /&gt; \left(\frac {d\vect L}{dt}\right)_I = \vect N&lt;br /&gt; &lt;br /&gt; where \vect N os the net external torque acting on the body.&lt;br /&gt; &lt;br /&gt; Combining the above,&lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; \frac {d\mathbf I}{dt}\;\vect \omega +&amp;lt;br /&amp;gt; \mathbf I\;\frac {d\vect \omega}{dt} +&amp;lt;br /&amp;gt; \vect\omega\times(\mathbf I\;\vect \omega) = \vect N&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; Note that if \mathbf I is constant, the above reduces to&lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; \mathbf I\;\frac {d\vect \omega}{dt}&amp;lt;br /&amp;gt; = \vect N - \vect\omega\times(\mathbf I\;\vect \omega)&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; The term \vect\omega\times(\mathbf I\;\vect \omega) is the rotational analog of the Coriolis force.&lt;br /&gt; &lt;br /&gt; Finally, Euler&amp;#039;s equations result in the special case of \mathbf I being a diagonal matrix.