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.

  • $\begingroup$ 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. $\endgroup$ Mar 23, 2017 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? $\endgroup$ Mar 24, 2017 at 12:55

1 Answer 1


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

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


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