I'm trying to simulate a rotating sphere due to a torque on a specific point on the sphere.

Say the sphere is connected to a string (so variable length) on the bottom, on $(r, \theta, \phi)=(-R_{sphere},\pi,0)$. In which $\theta$ is the polar angle, and $\phi$ the azimuthal.

The torque at the point is $\vec{\tau}=\vec{r}\times\vec{F}$

The vector $\vec{r}$ is from the origin of the sphere to the point where the rod is connected to the sphere.

The force $\vec{F}$ is always in the direction of the origin.

The sphere is also being pulled on in positive $z$-direction, so the average net force over time is 0.

I have calculated the torque $\vec{\tau}$ and can now use $\vec{\tau}=I\vec{\alpha}$ to calculate the angular acceleration. For a solid sphere $I = \tfrac25 m \cdot r^2$.

Now I have the angular acceleration $\vec{\alpha}$, but how can I calculate the instantaneous angles $\theta,\phi$ from this vector? Such that I know the rotation of the sphere at any time. I can't simply transform the $\vec{\tau}$ to spherical coordinates, because the angular acceleration vector is not in the direction of movement.

I can try to integrate over time to get the angular velocity $\omega$, but I still wouldn't know how to update the new angular coordinates.

I made a simple illustration to clarify.

Illustration of situation

  • $\begingroup$ $\vec{L}=I\vec{\omega}$ (usually) and $\vec{\tau}=\frac{\mathrm{d}\vec{\vec{L}}}{\mathrm{d}t}$, and so usually $\vec{\tau}=I\vec{\alpha}$. $\endgroup$ – G. Paily Nov 5 '14 at 22:13
  • $\begingroup$ Is it really a rod, or is it a string? In other words - is the distance from the origin to the attachment point fixed? $\endgroup$ – Floris Nov 5 '14 at 22:18
  • $\begingroup$ @G.Paily I've corrected the notation $\endgroup$ – johnbaltis Nov 5 '14 at 22:18
  • 1
    $\begingroup$ @Floris it's a string, so the distance can vary. I updated the text. $\endgroup$ – johnbaltis Nov 5 '14 at 22:19

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