Equations relating changes between rotating and inertial frames

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SUMMARY

The discussion focuses on the equations that relate acceleration in inertial frames to acceleration in rotating frames, specifically the equation a(I) = a(R) + 2(omega)Xv(R) + (omega)X(omega) X r. The conversation highlights the application of these equations in systems like Foucault's pendulum and the Coriolis effect experienced by a moving train. A key point established is that the equation V(I) = V(R) + omega X (V(R)) is not applicable in its original form, as it mixes units of velocity and acceleration. The correct formulation for velocity in inertial frames is v(I) = v(R) + omega X r.

PREREQUISITES
  • Understanding of inertial and rotating frames of reference
  • Familiarity with vector calculus, specifically cross products
  • Knowledge of Coriolis acceleration and its implications
  • Basic principles of classical mechanics, including forces and motion
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  • Study the implications of Coriolis acceleration in various physical systems
  • Learn about Foucault's pendulum and its demonstration of Earth's rotation
  • Explore the mathematical derivation of the equations relating inertial and rotating frames
  • Investigate practical applications of these concepts in engineering and physics
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Students of physics, engineers working with rotating systems, and anyone interested in the dynamics of motion in different reference frames.

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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