# Mechanics: acceleration & curvature relationship

1. Apr 20, 2012

### elegysix

Is there a way to take a known path, all the existing forces/fields along it, and solve for a driving force function that results in an object moving along that path? but solving it without knowing the speed along that path, only the path itself, and maybe the total time taken from A to B along the path.

the reason I ask: normally force fields are given as functions of position,
and F=ma - so if acceleration can be related to the curvature of a path as a function of position, the problem will get a lot simpler.

The idea is to say

Fdriving(r) + Fexisting fields(r) = ma(r)

but I don't know a(r) outright, because it depends on the driving force, which I want to solve for. so can I take a path S that a(r) will lie on, and change this equation

Fdriving(r) = ma(r) - Fexisting fields(r)

into this one

Fdriving(r) = kC(r) - Fexisting fields(r) ?

is there some relation between acceleration and Curvature of a path/trajectory, C?

or is this even possible?

thanks for any ideas

2. Apr 20, 2012

### I like Serena

Hi elegysix!

I believe the typical method to solve these kind of problems are numerical shooting methods that make use of Runge Kutta to solve the differential equation.
Usually it's not really possible to solve the equation mathematically.

3. Jul 11, 2012

### zsviben

Hi,

I just wanted to see if this helps somehow?
(I played a little, and I do not guarantee that it is 100% correct)

Bye

#### Attached Files:

• ###### description of force.pdf
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633.7 KB
Views:
64
4. Jul 12, 2012

### elegysix

Thank you for the work, this is good.

$m \ddot{q} = \frac{d\vec{r}}{dq}\cdot\vec{f}$

If f is given as f(r), and say the trajectory is a circle, is $\frac{d\vec{r}}{dq}$

then the rate of change of a circle? (like dy/dx or dx/dy of y^2 + x^2 = c)

at the same time, how exactly is q measured? a unit of length along a trajectory?

thanks

5. Jul 12, 2012

### haruspex

Suppose the particle has travelled distance x along the path and is moving with velocity v. The direction of v is so as to continue along the next δx, but the speed is unconstrained. Left to the whims of the external forces (which may be a function of v), the velocity v+δv at x+δx may not be what is required to keep v in the direction of the path. A lateral driving force is therefore prescribed to achieve the necessary change in direction. However, the force in direction of travel may be chosen arbitrarily. Repeating this at each infinitesimal δx drives the particle along the path.
Abrupt changes in direction of path can be handled without requiring infinite lateral forces by bringing the particle to rest instantaneously.
Along any smooth section of path, the initial speed and component of driving force in the direction of travel (as a function of x or time, say) can be specified arbitrarily, but the lateral driving force function is then fixed.

6. Jul 13, 2012

### zsviben

Hi,

is the force in the radial direction ?
I put one calculation where the force is in the radial direction and has the same origin as the circle (see q.pdf).

Yes, q is the length on a circle (q goes from 0 to 2πR and above).
q/R is then the angle in radians.

Bye

#### Attached Files:

• ###### q.pdf
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282.7 KB
Views:
76
7. Jul 13, 2012

### zsviben

Hi,

Do you mean under the rate of change of a circle the change of radius?
I do not mean such a thing, trajectory is fixed, and the derivative only means the unit vector in the direction of movement (velocity unit vector).

8. Jul 13, 2012

### K^2

Lets take a small mass and fire it in vacuum under constant gravity. The net force applied to the mass is just its weight. Now I take an identical mass, and slowly move it along the same curve. I will have to support the mass to do that, so the net force is not equal to the weight.

Therefore, I have two solutions for exactly the same trajectory with different forces. Constraining total time does not help, because I can carry the mass following the same ballistic trajectory at first faster, and then slower than the actual ballistic flight, giving me same total time, and again, different force applied.

So the answer is no. You cannot find a unique solution for driving force knowing only the trajectory and the time taken.

You can find the average driving force, however, knowing only the initial and final velocities and the potential along the path.

9. Jul 14, 2012

### elegysix

I realize that you need more information than just the total time and trajectory, I think though that I could minimize either total time or total work in order to find a unique solution - that might also require something like a maximum force, which is more realistic too.

I would like to get the starting point down first, given a curve and a mass, How do I go from f(r)=m*a(t) to something like f(r)+f(t)=m*C(r) where C is, or is related to, the trajectory and not time. C(r) is what I need to first figure out. f(t) is the family of solutions that I will later minimize or use to find a unique solution.

10. Jul 14, 2012

### elegysix

I think you might have it figured out, but will you do one more example because I am still somewhat confused?

Could you do this for a non-circular trajectory passing by a radial force origin? like say f(r)=1/r^2, and the trajectory is along y= x^4 - 1 or something similar. Would like to see the Force required to keep the mass on the line.

Or if it is easier, a non-radial force like f=-kx, and a trajectory of y=x for example.

thanks

11. Jul 14, 2012

### K^2

Yes, now you are getting into something that is known as Optimal Control Theory. The general problem of Optimal Control is stated thus. Given an initial state and final state, subject to certain constraints, such as specific path, find input, such as applied force, that would optimize a given functional of the trajectory. The functional is usually an integral along the trajectory.

This doesn't guarantee that you'll have unique solution, but it often does. You can optimize time, but without further constraints, the solution will diverge. Obviously, with infinite forces, you can traverse path infinitely fast, giving zero time. You can optimize time under certain constraints however, such as maximum acceleration at any given point.

Optimizing work done has the same problem, since in "perfect world" positive work of accelerating body will offset negative work to slow it down. But if you integrate over just positive work or over absolute value of work done, you can get something that way. If we are talking about a practical case of a rocket following a trajectory, as another example, you might want to optimize fuel consumption under certain time constraint, etc.

In general, optimal control problems are very difficult to solve and require numerical methods. However, if both reaction to input and functional are linear, you can use Linear Optimal Control, which produces a coupled set of linear differential equations for inputs, and can be solved exactly.

A good place to start looking for more information is the Wikipedia article on the topic. Optimal Control. But keep in mind that math associated with OCT is pretty serious.

12. Jul 14, 2012