Understanding Equations in Classical Physics: Force, Motion, and Acceleration

[PFuser1232]

In classical physics, when we say, for example:
$$\sum_{}^{} \vec{F}_r = -mr \omega^2 \hat{r}$$
are we saying that the force is what changes $\omega$ and keeps $r$ constant, which results in circular motion? Or are we saying that $\omega$ is what "causes the force"? Or are we just saying that if $\vec{F}_r$, $r$, and $\omega$ satisfy the above equation, then the motion in circular?
More generally, when we say $\sum_{}^{} \vec{F} = m \vec{a}$ are we saying that "force causes acceleration"? Or are we saying (in an inertial reference frame) "the particle has a nonzero acceleration, therefore, a force must have acted upon it".

 

[mfb]

"Force causes acceleration" is a convenient description because usually you control the force (e. g. by pushing something) and the acceleration is the result of your action then.
In terms of physics, it is probably better to say "this equation is satisfied" or "this equation is satisfied if and only if the motion is circular".

 

[A.T.]

Or are we saying...

For physics it doesn’t really matter how you say it. The quantitative prediction matters.

 

[PFuser1232]

"Force causes acceleration" is a convenient description because usually you control the force (e. g. by pushing something) and the acceleration is the result of your action then.
In terms of physics, it is probably better to say "this equation is satisfied" or "this equation is satisfied if and only if the motion is circular".

Suppose I want to explain why parabolic motion is not circular in terms of polar coordinates (not cartesian coordinates), here's how I think of it:
The radial component of the weight (net radial force) is NOT equal to $-mr\omega^2$, instead, it is equal to $m(\ddot{r} - r \omega^2)$. In other words, $\ddot{r}$ is nonzero which implies that $r$ is varying. I found the question of "why $r$ varies" impossible to answer in polar coordinates though, so I switched to cartesian coordinates (which are more suited for this problem) and found that $r$ varies.
Is my reasoning correct?

 

[nasu]

If you know that is parabolic, isn't this enough "argument" for not being circular?
I think that a better question will be "what will be the motion if the force is only gravity, near the surface of the Earth (constant direction and magnitude)".
Polar coordinates are indeed not so useful for this problem, as the force is not radial but it has constant direction.

 

[anachin6000]

In classical physics, when we say, for example:
$$\sum_{}^{} \vec{F}_r = -mr \omega^2 \hat{r}$$
are we saying that the force is what changes $\omega$ and keeps $r$ constant, which results in circular motion? Or are we saying that $\omega$ is what "causes the force"? Or are we just saying that if $\vec{F}_r$, $r$, and $\omega$ satisfy the above equation, then the motion in circular?
More generally, when we say $\sum_{}^{} \vec{F} = m \vec{a}$ are we saying that "force causes acceleration"? Or are we saying (in an inertial reference frame) "the particle has a nonzero acceleration, therefore, a force must have acted upon it".

Normaly, the force causes acceleration because of the first Newton's law: In an iertial sistem, if no external action exists on a body it mantains it's relative constant velocity motion or it's relative rest. You can better understand this after you study impulse and momentum.