Soren4
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I'm trying to understand the relations between the existence of Coriolis force and the conservation of angular momentum. I found this explanation on Morin.
I do not understand the two highlighted parts. In particular it seems that Coriolis force is there to change the angular momentum of the person in the lab frame. But this cannot be true since Coriolis force is a fictitious force, existing only in the rotating frame. I don't see clearly the link between Coriolis force and conservation of angular momentum in this case. Can anyone give some further explanations about this?A carousel rotates counterclockwise with constant angular speed ω.Consider someone walking radially inward on the carousel (imagine a radial line painted on the carousel; the person walks along this line), at speed v with respect to the carousel, at radius r. [...]
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Take d/dt of L = mr^2ω, where ω is the person’s angular speed withrespect to the lab frame, which is also the carousel’s angular speed. Using dr/dt =− v, we have
$$dL/dt =− 2mrωv+mr^2(dω/dt)\tag{1}$$
What if the person doesn’t apply a tangential friction force at his feet? Then the Coriolis force of 2mωv produces a tangential acceleration of 2ωv in the rotating frame, and hence also in the lab frame (initially, before the direction of the motion in the rotating frame has a chance to change), because the frames are related by a constant ω. This acceleration exists essentially to keep the person’s angular momentum (with respect to the lab frame) constant.
[...] To see that this tangential acceleration is consistent with conservation of angular momentum, set dL/dt = 0 in Eq. (1) to obtain 2ωv = r(dω/dt) (this is the person’sω here, which is changing).
The right-hand side of this is by definition the tangential acceleration. Therefore, saying that L is conserved is the same as saying that 2ωv is the tangential acceleration (for this situation where the inward radial speed is v).