0
$\begingroup$

I'm a first year physics student and i've just learnt this equation for angular velocity in spherical polar coordinates:

$\omega=\dot{\phi}\mathbf{e_z}+\dot{\theta}\mathbf{e_\phi}$

The diagram i am using is on the RHS of this link:

https://en.wikipedia.org/wiki/Spherical_coordinate_system

I don't understand why this eqn is correct. Let's take the case where we imagine a vector going around in purely the xy-plane, a horizontal circle. Therefore $\theta=constant$ so in the above equation, $\dot{\theta}\mathbf{e_\phi}=0$. Now $\omega$ should be a vector that only has $\mathbf{e_x}$ and/or $\mathbf{e_y}$ component since it is only moving in that xy plane. But, from the equation, we're left with $\omega=\dot{\phi}\mathbf{e_z}$. But $\mathbf{e_z}$ is perpendicular to xy plane, so $\omega$ which is always pointing in a direction on this horizontal circle should have no $\mathbf{e_z}$ component.

I feel like i'm missing something very key. Anyone help me get out?

Much appreciated.

Improve this question
$\endgroup$
5
  • $\begingroup$ Rotations are definedabout an axis. So the rotation that you’re describing in the $x,y$ plane is about the $z$- axis. $\endgroup$
    – Sharat V Chandrasekhar
    Commented Apr 25, 2019 at 20:08
  • $\begingroup$ Your question is directly related to physics.stackexchange.com/questions/223217/… $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Apr 25, 2019 at 20:12
  • $\begingroup$ @GiorgioP that's very helpful. Thankyou very much $\endgroup$
    – Young's Modulus
    Commented Apr 25, 2019 at 20:17
  • $\begingroup$ Have look also to the link indicated as possible duplicate. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Apr 25, 2019 at 20:25
  • $\begingroup$ Yeah, i'm now convinced, thanks again @GiorgioP $\endgroup$
    – Young's Modulus
    Commented Apr 25, 2019 at 20:36

1 Answer 1

Reset to default
0
$\begingroup$

$\omega$ is not in the xy plane, it is perpendicular to the plane, so the equation is correct. $\omega$ always point in the direction of the axis around which the system instantaneously rotates. If you rotates a disk around its center of mass, then $\omega$ is perpendicular to the plane of the disk.

Improve this answer
$\endgroup$
4
  • $\begingroup$ is this a convention? If so, why? Because, when doing plane polars you have $v=r\hat{r}+r\dot{\theta}\hat{\theta}$ and in the case when r is constant, you're left with $r\dot{\theta}\hat{\theta}$ and in this case, the velocity of a particle is in the direction of $\hat{\theta}$, as opposed to perpendicular. $\endgroup$
    – Young's Modulus
    Commented Apr 25, 2019 at 20:14
  • $\begingroup$ Sorry to change up unit vector notation, i just quickly copied it from another post $\endgroup$
    – Young's Modulus
    Commented Apr 25, 2019 at 20:14
  • $\begingroup$ It is not by convention. Suppose that $\omega$ pointed in the plane of disk. By the symmetry of the disk, $\omega$ could point in any possible direction in the plane of the disk, which would make $\omega$ not well defined. The velocity that you have written is, as you have said yourself, the velocity of the particle, not its angular velocity. You can relate those two by $v = w \times r$, which if you work it out, gives you the correct answer. $\endgroup$
    – Shamaz
    Commented Apr 25, 2019 at 20:23
  • $\begingroup$ yeah, its making sense now, thanks. $\endgroup$
    – Young's Modulus
    Commented Apr 25, 2019 at 20:38

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