I'm trying to understan kinematic inertial explanation of the apparent deviation of objects due to fictitious forces in rotating Earth.

Take an object moving from the equator northwards, or southwards. The observed deflection in the rotating Earth frame can be explained inertially saying that the velocity of rotation (which is the velocity of the ball, at the beginning) is greater at the equator than at the poles, so the object goes faster than the points on the surface of the Earth, thus it seems deflected.

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But how can this be explained via the conservation of angular momentum? $$L=mrv_{\theta}$$

In this kind of motion the object gets nearer to the axis of rotation, but the vector $r$ which is the distance from the center of the Earth and the object does not change. Morevorer $r$ must be used because the angular momentum is conserved only if the pivot point is the center of the earth (otherwise gravity exerts a torque). So how can one use the conservation of angular momentum to explain correctly the phenomenon?

  • 2
    $\begingroup$ Your formula gives a scalar angular momentum, that is, the projection of the vector angular momentum onto the rotation axis. In this formula, the meaning of $r$ is: distance to the axis, and $v_θ$ is the part of velocity parallel to the equator. Cylindrical coordinates. $\endgroup$ – L. Levrel Apr 19 '16 at 13:49
  • $\begingroup$ Thanks for the answer! Is that component of the angular momentum conserved? Thus does $v_{\theta}$ increases (in the inertial frame) while the object goes northwards? $\endgroup$ – Sørën Apr 19 '16 at 17:46
  • $\begingroup$ If the object is only constrained to the surface and otherwise free to move, the only force acting on it is vertical, its direction crosses the Earth axis, so its torque projected on this axis is zero, and this component of angular momentum is conserved. See also physics.stackexchange.com/a/250615/113338 P.S.: you have quite a lot of questions with good answers but none accepted; you'd be nice to reward the efforts of the answerers. $\endgroup$ – L. Levrel Apr 20 '16 at 12:42

The vector $\mathbf r$ does change, even though its magnitude is nearly constant. Most importantly, the component of $\mathbf r$ which is perpendicular to the rotation axis is decreasing in this example. This fact leads to the explanation for which you are searching.

Another way of seeing this would be to use this definition of angular momentum, which allows you to use the vector $\mathbf r$ directly:

\begin{align}\mathbf{L}&=\mathbf{r}\times m\mathbf{v}\\ &= \mathbf{r}\times\mathbf{p}, \end{align}

  • $\begingroup$ Torque can be defined about a point as well! $\endgroup$ – George Smyridis Apr 18 '16 at 23:06
  • $\begingroup$ @ George Smyridis Okay, but given that he uses a definition of angular momentum that does not include a cross product, should I really get into that? I do see your point that my first sentence could be misleading to other readers, though. I will reword it. $\endgroup$ – Duncan Harris Apr 18 '16 at 23:14

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