Equations relating changes between rotating and inertial frames

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Discussion Overview

The discussion revolves around the equations that relate changes in acceleration and velocity between inertial and rotating frames, particularly in the context of applications like Foucault's pendulum and the Coriolis effect on moving objects such as trains. Participants explore the conditions under which these equations are applicable and the implications of constant velocity in a rotating frame.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents the equation relating acceleration in inertial and rotating frames, noting its application to systems with acceleration, such as Foucault's pendulum.
  • Another participant argues that the Coriolis acceleration is present even when velocity is constant with respect to the Earth, suggesting that this implies a(R)=0.
  • A question is raised about the applicability of the velocity equation V(I) = V(R) + omega X (V(R)), leading to a clarification that this equation is not valid as initially presented due to unit inconsistencies.
  • A later reply corrects the velocity equation to v_I = v_R + omega X r, stating it applies when determining the velocity of an object from an inertial frame perspective.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the velocity equation and the implications of constant velocity in the context of the Coriolis effect. The discussion remains unresolved regarding the nuances of these equations and their conditions of use.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the applicability of the equations, particularly in relation to the definitions of acceleration and velocity in rotating frames. The discussion does not resolve the subtleties involved in these relationships.

bman!!
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i understand the reason and steps leading to the equation that relates acceleration in the inertial frame to acceleration in the rotating frame i.e.

a(I) = a(R) + 2(omega)Xv(R) + (omega)X(omega) X r

a(I) = acceleration in inertial frame
a(R) = acceleration in rotating frame
omega = rotation vector
X = cross product
v(R) = velocity in rotating frame

r = position vector


now i understand why this (or a rearrangement thereof) can be applied to a system like focaults pendulum, because you have acceleration going on, so obviously you would want to relate the changes between the two frames.

however, everyone remembers the simple problems where you simply use the coriolis term to calculate the direction and magnitude of the coriolis force on something like a moving train (i.e. 500 tonne train moving north at 100kph experiences a coriolis force of something liek 1500N eastwards)

however it struck me, that if the train is moving with constant velocity, then surely the above equation doesn't apply? and surely the equation linking the two vectors a together is simply the very original relation between the two frames for vector that is fixed in the rotating frame namely:

V(I) = V(R) + omega X (V(R))

or is it more subtle than this?
 
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Your original equation shows that there is a Coriolis acceleration even if v is constant wilth respect to the Earth. That just means a(R)=0.
 
pam said:
Your original equation shows that there is a Coriolis acceleration even if v is constant wilth respect to the Earth. That just means a(R)=0.

ah i see, that makes sense.

but when would

V(I) = V(R) + omega X (V(R))

be applicable? or is it just an intermediary?
 
bman! said:
but when would

V(I) = V(R) + omega X (V(R))

be applicable? or is it just an intermediary?

Never! Look at the units on the right-hand side. The first term has units of velocity, the second, acceleration. The equation you are looking for is

\mathbf v_I = \mathbf v_R + \mathbf \omega \times \mathbf r

So when does this equation apply? Simple: When you want to know the velocity of some object as seen by an observer fixed in the inertial frame.
 
ah i see now. cheers.
 

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