Circular motion when F=ma' I apologize in advance if this question is deemed too general or too similar to this and this question.
How would mechanics be different if $F=mx'''$ instead of $F=ma$? I feel like I have intuition about what kind of place an $F=mv$ world would be like, since, after all, this is part of our everyday experience dominated by friction. Let me make my question more specific.
If I pick a ball attached to the end of a string in a $F=mv$ world, I cannot make it move in uniform circular motion by swinging it around. I can separate the ball and the end of string by pulling the two apart, but then pulling on the string simply pulls the ball towards me. $F=mv$ is like living in a very viscous liquid. I've heard that at a very small scale (like that of bacteria), water is like this and this has consequences for the propulsion of microorganisms.
In reality, it is possible to give an initial velocity to the ball and then to tug on the string to rotate the velocity vector so that it is moving in uniform circular motion. 
Please correct me if any of the above is wrong.
But what happens if $F=mx'''$? I have no intuition about what would happen with the string and ball example in this case. You could give the ball a constant velocity and zero initial acceleration. But then how does it move constrained by a string that is applying a force which changes acceleration. Would it spiral? What kind of setup of forces could force a particle into circular motion? Is there any way to gain intuition about the consequences of $F=mx'''$?     
 A: Confine the motion to a plane, and for convenience make it the complex plane.  Describe the position of a particle in uniform circular motion by $z = Re^{i\omega t}$.  Then the jerk (derivative of acceleration} is $z''' = -i\omega^3 z$.  Compare that to velocity $z' = i\omega z$.  The jerk points opposite the velocity.  Thus, to move in uniform circular motion with $F = ma'$, you must constantly feel a force pushing you backwards, opposite your current direction of motion.  Additionally, you must have the correct initial conditions for $z(0)$, $z'(0)$, and $z''(0)$.
Presumably, a string could not provide  this force, because the force would still act in the direction of the string.  We would have $z''' \propto z$, the force in the direction of the motion, if the string followed Hooke's law.  The solutions are $z = e^{\alpha t}$ with $\alpha$ some constant times a cube root of unity.  The interesting solutions have both periodic and exponential behavior, so the particle would spiral out or spiral in with this force law.
Free particles in this universe move the same way particles with constant force move in our universe - on parabolas.
