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Just to ruin the punchline upfront: With this question, I'm trying to do a sanity check about the logic of insisting on a trajectory and only then thinking about forces.

I took Classical Mechanics with Thornton and Marion's 5th edition "Classical Dynamics of Particles and Systems" many years ago and I'm going through it again. I'm on page 32 looking at equation 1.98 which is

$$\vec a = [\frac{d^2 r}{dt^2}-r(\frac{d \theta}{dt})^2]\vec e_r + [r\frac{d^2 \theta}{dt^2}+2\frac{dr}{dt}\frac{d\theta}{dt}]\vec e_\theta .$$

It took me a bit of time on Math StackExchange (with help from @npojo) but I relearned that $\vec e_\theta$ is a true unit vector and $not$ an angle of any kind whatsoever. But the direction of $\vec e_\theta$, and $\vec e_r$ for that matter, are determined by $\theta$, but never mind.

Anyways, I am now able to derive the above equation for acceleration using the definition of $\vec e_r$ and $\vec e_\theta$ and the product rule... starting from the general curve:

$\vec r(t)=r(t)\vec e_r$

But this all got me thinking about geometry and the ratio of mass to net force... which brings me to my question. Since, for a real situation with a point mass, $\frac{\vec F_{net}}{m}$ must match the above $\vec a$ which I derived purely from the geometry of $\vec r(t)$, is it fair to say that the geometry of the trajectory can be realized by the particle's movement if and only if we are able to create a net force and mass ratio that matches exactly with the above acceleration for all $t$? I'm assuming the answer is yes but I just want to check.

Any thoughts would be much appreciated!

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Don't forget what the equation $\mathbf{a}=\ddot{\mathbf r} = (\ddot{r} - r \dot{\theta}^2) \mathbf{\hat r} + (r\ddot{\theta} - 2\dot{r}\dot{\theta}) \mathbf{\hat θ}$ tells us. It relates the components of the acceleration to the rate of change of the coordinates, and the force is just the acceleration multiplied by the mass. So, you should be correct to say that the geometry of the trajectory must correspond to the force applied, since the acceleration is in the same direction as the net force.

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