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:

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

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.

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

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

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