0
$\begingroup$

I am modeling a magnetic moment, $m$, and am using the sum of torques to do it. The torques are due to a time changing magnetic field, $B$, (that changes in direction and magnitude) and a dissipative term, $b$, i.e. a drag term related to velocity, $v$ (don't worry if the units don't look right, $b$ has a lot absorbed into it. Also, if it helps think of the dissipative torque the same way as a real damped pendulum). My sum of torques are as follows:

$$ \sum\tau = -b l\times v - m\times B $$

From this I can look at the rotation around a single axis (for example the $z$ axis with a moment of inertia $I$:

$$ \sum \tau_z = I \alpha_z $$

I am now left with three equations of motion, $\alpha_x, \alpha_y,$ and $\alpha_z$ each describing the rotation around it's axis. From here I can see how much it rotates in say the $xy$ plane by integrating $\alpha_z$ twice. I have no problems up to this point, but now my question is how do I convert this coordinate system (of describing the rotation around each axis) to conventional spherical coordinates to describe the rotation, i.e. the magnetic moment moved an inclination angle of $\theta$ and an azimuth angle of $\phi$. For some reason I'm really getting hung up on this.

$\endgroup$
  • $\begingroup$ BTW Rotational law of motion is $$ \vec{\tau}_C = \mathrm{I}_{C} \vec{\alpha} + \vec{\omega} \times \mathrm{I}_C \vec{\omega}$$ with ${\rm I}$ the 3x3 MMOI rotated to the global coordinate system. Point C is the center of mass. $\endgroup$ – ja72 Mar 23 '17 at 20:55
  • $\begingroup$ Rotational accelerations are vectors and transform as such. Why can't you use a regular cartesian to spherical transform? Or are you trying to express the equations of motion in spherical coordinates? $\endgroup$ – ja72 Mar 24 '17 at 12:55
1
$\begingroup$

I think you want to do this

$$ \begin{pmatrix} \alpha_x \\ \alpha_y \\ \alpha_z \end{pmatrix} = \alpha \begin{pmatrix} \cos \phi \cos \theta \\ \sin \phi \cos\theta \\ \sin \theta \end{pmatrix} $$

This is done with

$$\begin{align} \phi & = \tan^{-1} \left( \frac{\alpha_y}{\alpha_x} \right) \\ \theta & = \tan^{-1} \left( \frac{ \alpha_z}{\sqrt{\alpha_x^2+\alpha_y^2}} \right) \\ \alpha & = \sqrt{ \alpha_x^2+\alpha_y^2+\alpha_z^2} \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.