truesearch said:
I could not fully understand post 3, does your post agree with post 3?
I have to explain these things to 16 and 17 year old students.
In order to explain something, you need to understand it yourself at least one level higher, so that's a serious problem right there.
So first, let's try to explain it to you on a level above of what you'll need to explain to students.
Suppose, I have an object in an inertial frame of reference. For inertial frame of reference, we have Newton's Second Law.
F = ma
Now, acceleration is the second derivative on position. Furthermore, it is a vector. For a rotation problem 2D is sufficient. So let me rewrite it.
F_x = m \frac{d^2x}{dt^2}
F_y = m \frac{d^2y}{dt^2}
In a rotating reference frame, it's much easier to use polar coordinates.
x = r cos(\theta)
y = r sin(\theta)
Lets rewrite the equations for forces again.
F_x = m \frac{d^2}{dt^2}\left(r cos(\theta)\right)
F_y = m \frac{d^2}{dt^2}\left(r sin(\theta)\right)
Taking derivative once.
F_x = m \frac{d}{dt}\left(\frac{dr}{dt} cos(\theta) - r sin(\theta) \frac{d\theta}{dt}\right)
F_y = m \frac{d}{dt}\left(\frac{dr}{dt} sin(\theta) + r cos(\theta) \frac{d\theta}{dt}\right)
Taking derivative twice.
F_x = m \left(\frac{d^2r}{dt^2} cos(\theta) - 2 \frac{dr}{dt} sin(\theta)\frac{d\theta}{dt} - r cos(\theta) \left(\frac{d\theta}{dt}\right)^2 - r sin(\theta)\frac{d^2\theta}{dt^2}\right)
F_y = m \left(\frac{d^2r}{dt^2} sin(\theta) + 2 \frac{dr}{dt} cos(\theta)\frac{d\theta}{dt} - r sin(\theta) \left(\frac{d\theta}{dt}\right)^2 + r cos(\theta)\frac{d^2\theta}{dt^2}\right)
Suppose, now, that you are standing on the station rotating around r=0 with angular velocity ω. Suppose that you measure all angles and x/y relative to yourself. I'll call these θ', x', and y' coordinates. Clearly θ=θ' + ωt. (To within a choice of origin for θ) The equations for x' and y' are same polar equations as before.
x' = r' cos(\theta ')
y' = r' sin(\theta ')
The r'=r, of course, since we chose the coordinate system with the same center, and the center of rotation does not move.
Let us substitute θ=θ' + ωt into force equations.
F_x = m \left(\frac{d^2r}{dt^2} cos(\theta' + \omega t) - 2 \frac{dr}{dt} sin(\theta' + \omega t)\frac{d\theta}{dt} - r cos(\theta' + \omega t) \left(\frac{d\theta}{dt}\right)^2 - r sin(\theta' + \omega t)\frac{d^2\theta}{dt^2}\right)
F_y = m \left(\frac{d^2r}{dt^2} sin(\theta' + \omega t) + 2 \frac{dr}{dt} cos(\theta' + \omega t)\frac{d\theta}{dt} - r sin(\theta' + \omega t) \left(\frac{d\theta}{dt}\right)^2 + r cos(\theta' + \omega t)\frac{d^2\theta}{dt^2}\right)
Of course, we can also expand dθ/dt the same way.
F_x = m \left(\frac{d^2r}{dt^2} cos(\theta' + \omega t) - 2 \frac{dr}{dt} sin(\theta' + \omega t)\left(\frac{d\theta'}{dt}+\omega\right) - r cos(\theta' + \omega t) \left(\frac{d\theta'}{dt} + \omega\right)^2 - r sin(\theta' + \omega t) \frac{d^2\theta'}{dt^2}\right)
F_y = m \left(\frac{d^2r}{dt^2} sin(\theta' + \omega t) + 2 \frac{dr}{dt} cos(\theta' + \omega t)\left(\frac{d\theta'}{dt}+\omega\right) - r sin(\theta' + \omega t) \left(\frac{d\theta'}{dt} + \omega\right)^2 + r cos(\theta' + \omega t) \frac{d^2\theta'}{dt^2}\right)
Now, I'm going to simplify a bit, using the fact that x/r = cos(θ).
F_x = m \left(\frac{d^2r}{dt^2}\frac{x}{r} - 2 \frac{dr}{dt}\frac{y}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{x}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 - r^2 \frac{y}{r} \frac{d^2\theta'}{dt^2}\right)
F_y = m \left(\frac{d^2r}{dt^2}\frac{y}{r} + 2 \frac{dr}{dt}\frac{x}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{y}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 + r^2 \frac{x}{r} \frac{d^2\theta'}{dt^2}\right)
Since x/r and y/r simply give us directions, we go over completely to primed coordinates.
F_x' = m \left(\frac{d^2r}{dt^2}\frac{x'}{r} - 2 \frac{dr}{dt}\frac{y'}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{x'}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 - r^2 \frac{y'}{r} \frac{d^2\theta'}{dt^2}\right)
F_y' = m \left(\frac{d^2r}{dt^2}\frac{y'}{r} + 2 \frac{dr}{dt}\frac{x'}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{y'}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 + r^2 \frac{x'}{r} \frac{d^2\theta'}{dt^2}\right)
And we extract from it the parts that come directly from differentiating x' and y'.
F_x' = m \left(\frac{d^2 x'}{dt^2} - 2 \frac{dr}{dt} \frac{y'}{r} \omega - 2 r \frac{x'}{r} \frac{d \theta '}{dt}\omega - r \frac{x'}{r}\omega^2\right)
F_y' = m \left(\frac{d^2 x'}{dt^2} + 2 \frac{dr}{dt} \frac{y'}{r} \omega - 2 r \frac{x'}{r} \frac{d \theta '}{dt}\omega - r \frac{x'}{r}\omega^2\right)
The above already contains the centrifugal and Coriolis terms. In vector form, it can be equivalently written like this.
F = m(a' + 2 \omega \times v - \omega^2 r \hat{r})
So long as you define vector ω as vector perpendicular to x/y plane.
Finally, suppose you want to continue using Newton's 2nd in this coordinate system. Then you would have to define.
F' = F - 2m\omega \times v + m\omega^2 r \hat{r}
Where F is the real forces acting on the object, and the two additional terms are Coriolis and centrifugal fictitious forces respectively.
Obviously, this is not how you are going to explain the subject to high school kids. But this is level of understanding you should have before teaching centrifugal forces.
)
