Register to reply

Action of co-circular springs on a point mass

by omission9
Tags: action, cocircular, mass, point, springs
Share this thread:
omission9
#1
Apr16-12, 10:21 PM
P: 5
This problem is actually a simplification of something seen in a data visualization tool.
I think it is interesting in that it can be "translated" to a Hooke's Law problem.
Unfortunately, I am not sure how to proceed...
Suppose I have a point mass tethered to several springs. The springs are attached in a co-circular fashion to a rigid metal ring. The springs are equidistant from each other on the circular ring. The point mass is at some equilibrium position inside the ring.
Now, suppose we are able to slide one (and only one!) of the springs around the metal ring.
All the way around the circle.
How would you describe the movement of the point mass inside the ring? At first it seems that the point mass would move strictly in a circle of some varying radius. Instead what I am seeing from my software (in some cases) is that the point mass traces an elliptical path.
Why is that? I assume I haven't made some sort of coding error!
The position of the point mass when we start is for M springs
([itex]\sum_{j=0}^{M-1}[/itex] cos [itex]\Theta_j[/itex] * k[itex]_{j}[/itex], [itex]\sum_{j=0}^{M-1}[/itex] sin [itex]\Theta_{j}[/itex] * k[itex]_{j}[/itex] )
If we then move one of the springs to some new position (say it is the first spring)
The position is now
([itex]\sum_{j=1}^{M-1}[/itex] cos [itex]\Theta_j[/itex] * k[itex]_{j}[/itex] + [itex]\sum_{j=0}^{0}[/itex] cos [itex]\Theta_j[/itex] * k[itex]_{j}[/itex], [itex]\sum_{j=1}^{M-1}[/itex] sin [itex]\Theta_{j}[/itex] * k[itex]_{j}[/itex] + [itex]\sum_{j=0}^{0}[/itex] sin [itex]\Theta_{j}[/itex] * k[itex]_{j}[/itex])
This still seems to describe a circle to me. If we pick any spring (not just the first as in the example) and move it a full 360 I still only intuit a circular path.
Any advice and corrections would be appreciated!
Phys.Org News Partner Physics news on Phys.org
Vibrational motion of a single molecule measured in real time
Researchers demonstrate ultra low-field nuclear magnetic resonance using Earth's magnetic field
Bubbling down: Discovery suggests surprising uses for common bubbles
OldEngr63
#2
Apr16-12, 11:01 PM
P: 343
With this sort of a description, how do you think an ellipse would look?
omission9
#3
Apr17-12, 07:19 AM
P: 5
Quote Quote by OldEngr63 View Post
With this sort of a description, how do you think an ellipse would look?
Oh, I assumed people in this forum were familiar with basic analytic geometry. Let me explain.
You see, in general, an ellipse has parametric form
x=a cos [itex]\Theta[/itex]
y=b sin [itex]\Theta[/itex]
In my example problem we see that with some simplification we have a similar expression.
In my software experiments I am seeing the point mass trace an elliptical (non-circular) path. I am essentially 100% sure I do not have a software error and so I am wondering if I made an algebraic mistake which is leading me to an incorrect assumption?

mfb
#4
Apr17-12, 09:44 AM
Mentor
P: 11,831
Action of co-circular springs on a point mass

How do you get your expression for the position of the point mass? I would expect the solution of some quadratic equation (as the spring length is related to the euclidian distance) as the position.

Do you use the same equation in your simulation?
Maybe your formula is wrong - if the simulation uses a different method, this could explain the difference.

Instead what I am seeing from my software (in some cases) is that the point mass traces an elliptical path.
I assume that you mean an elliptical path with non-zero eccentricity, otherwise it would be a nice joke ;).
omission9
#5
Apr17-12, 10:10 AM
P: 5
Quote Quote by mfb View Post
How do you get your expression for the position of the point mass? I would expect the solution of some quadratic equation (as the spring length is related to the euclidian distance) as the position.
its just Hooke's law, basically, F=-kx. We have multiple springs so we have a sum of the forces.
The position x is given with the trig functions because the springs are arranged in a co-circular way.
mfb
#6
Apr17-12, 10:21 AM
Mentor
P: 11,831
x in F=-kx is the length deviation from equilibrium. You cannot assume that all your springs are connected in the coordinate origin and have a length of 0 in the equilibrium (which is what you do, if I read the formulas correctly).

Edit: After writing down some formulas, I think that you really underestimate the complexity of an exact solution. Are there any additional things you can use? Do you know the approximate number of springs? Do you know how large the movement of the mass is, compared to the ring radius? Much smaller than the radius?


Register to reply

Related Discussions
Total torque exerted by a mass that falls in a circular pattern around a point. Mechanical Engineering 2
Mass spring system where springs have mass Classical Physics 2
Spring Mass system where springs also have mass Introductory Physics Homework 1
Springs & circular motion Introductory Physics Homework 3
Effect of mass and springs on the damping of mass spring system Classical Physics 5