Mechanics: acceleration & curvature relationship

In summary, you can solve for the driving force of an object following a trajectory by using information about the initial and final velocities and potential along the path. However, finding a unique solution is not possible without knowing more information about the trajectory and time taken.
  • #1
elegysix
406
15
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
 
Physics news on Phys.org
  • #2
Hi elegysix! :smile:

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
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
 

Attachments

  • description of force.pdf
    633.7 KB · Views: 175
  • #4
Thank you for the work, this is good.
in your equation,

[itex] m \ddot{q} = \frac{d\vec{r}}{dq}\cdot\vec{f}[/itex]

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

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
Suppose the particle has traveled 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
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
 

Attachments

  • q.pdf
    282.7 KB · Views: 194
  • #7
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
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
K^2 said:
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.

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
zsviben said:
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

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
elegysix said:
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.
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
Thanks for the advice!
 

1. What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration related to curvature?

Acceleration and curvature are closely related in the field of mechanics. In general, as an object travels along a curved path, it is constantly changing its direction of motion, which results in a non-zero acceleration. The greater the curvature of the path, the greater the acceleration.

3. What is the formula for calculating acceleration?

The formula for acceleration is a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time.

4. How does acceleration affect an object's motion?

Acceleration affects an object's motion by changing its velocity. If an object is experiencing a positive acceleration, its velocity will increase over time. If an object is experiencing a negative acceleration (also known as deceleration), its velocity will decrease over time.

5. Can an object have acceleration without changing its speed?

Yes, an object can have acceleration without changing its speed. This occurs when the object changes its direction of motion but maintains a constant speed. In this case, the object's velocity is changing, but its speed remains the same.

Similar threads

Replies
1
Views
788
Replies
9
Views
888
Replies
3
Views
693
Replies
2
Views
773
Replies
8
Views
400
Replies
22
Views
2K
Replies
17
Views
1K
  • Special and General Relativity
Replies
8
Views
490
Replies
3
Views
485
Back
Top