Acceleration in rotating frame understanding

[Awesomesauce]

Homework Statement

Not a homework problem but:
I am revising rotating frames of reference right now and

I know that: a_{inertial}=(\frac{dv}{dt}_{inertial})_{rot}+(w x v_{inertial})

But I cannot seem to understand what (\frac{dv}{dt}_{inertial})_{rot} means; more so the 'rotational' bit outside the brackets of the true acceleration term.

Thank you for your help, and sorry about the formating; I have no idea how to use latex.

Homework Equations

n/a

The Attempt at a Solution

Inertial acceleration viewed from the rotating frame? That makes no sense whatsoever. I honestly can't physically understand this.

 

[TSny]

This can certainly be confusing.

First, it should be clear that if we have some vector V, then that vector will have certain components in the inertial coordinate system and at the same time certain components with respect to the rotating system and the components will of course generally be different in the two frames. In some sense, a vector does not “belong” to any coordinate system; it may be expressed relative to the inertial frame or it may be expressed relative to the rotating frame. It is not hard to derive equations for transforming the components of V in the inertial frame to the components of V in the rotating frame. The transformation equations depend on time due to rotation of one frame relative to the other. Thus, if V is any old vector that has, say, components (2, -3, 7) at some instant of time in the inertial frame, we could use the transformation equations to find the components of the vector at that instant of time in the rotating frame.

Next, suppose we have a time dependent vector V(t). At each instant of time we can construct the components of V(t) in each coordinate system. We would generally find that the components of V(t) change with time in each frame and that the rate of change of the components would be different in each frame. So, (dV/dt)_inert ≠ (dV/dt)_rot. Rather, (dV/dt)_inert = (dV/dt)_rot + ω x V(t). This relates the rate of change of a vector according to the inertial frame to the rate of change of the same vector according to the rotating frame.

Now for the confusing part. Suppose V(t) happens to be the velocity of a particle as measured in the inertial frame. To emphasize this, write the vector as V_inert(t). Likewise, we can let V_rot(t) denote the velocity of the particle as measured in the rotating frame. It is crucial to understand that these are two very different vectors. Like any vector, you could express V_inert(t) in terms of components in the inertial frame or in terms of components in the rotating frame. Again, just because V_inert(t) is the velocity of the particle as measured in the inertial frame, it does not mean that V_inert(t) “belongs” to the inertial frame. V_inert(t) is a vector which, like all vectors, can be expressed in terms of components relative to either frame. Likewise for V_rot(t). Moreover, the components of V_inert(t) in the rotating frame are not the same as the components of V_rot(t) in the rotating frame. (Think about a particle at rest relative to the inertial frame. V_inert(t) = 0 and will have all components = 0 in both frames. But V_rot(t) for this particle will not be zero and will have nonzero components in both frames.)

Since the relation (dA/dt)_inert = (dA/dt)_rot + ω x A(t) holds for any vector function A(t), we have

(dV_inert/dt)_inert = (dV_inert/dt)_rot + ω x V_inert(t)

By definition, the vector on the left is the acceleration of the particle as measured in the inertial frame. The first term on the right does not represent the acceleration of the particle as measured in either frame. It represents the rate at which the vector V_inert(t) is changing relative to the rotating frame. In words, it’s the rate at which the velocity vector as measured in the inertial frame is changing relative to the rotating frame. It has little physical significance, but it nevertheless is a well-defined quantity.

 

[Awesomesauce]

Wow. That was the best explanation I have ever read! Thank you so much for your time, that makes complete sense