1. Apr 25, 2005

### physicsfun

Consider a collision between two spheres where we have both elastic force and dissipative force.

Let x is a variable where it is let to be the sum of their radii when they are just touching, and that distance decreasing during the collision. After the collision, x is irrelevant.

Given, u*(dv/dt)= F_elastic+F_dissipative = -h*x - k*v*x^(1/2) ---------(1)

where v= dx/dt, and u=m1*m2/(m1+m2)

QUESTION: Corresponding to the two forces are impulses, both during compression and during restitution. Impulses corresponding to the dissipative force F_dis are: P_dis-c and P_dis-r.
PROVE that ---------> P_dis-c= - P_dis-r ----------------(2)

Note: I was also told that the above results hold for a more general solution, and that the x does not matter, all that is necessary is remembering that the F_dissipative is proportional to velocity, i.e. can treat the two forces as a function in that:
u*(dv/dt) = -f(x) - g(x)*v ---------(3)

Last edited by a moderator: Apr 26, 2005
2. Apr 25, 2005

### OlderDan

I'm sort of getting the picture here, but I don't understand the separation of P_dis-c and P_dis-r. Can you say what these are?

3. Apr 25, 2005

### physicsfun

Thanks Dan for looking at this thread... :]

P_dis-c and P_dis-r are the impulses during compression and restitution respectively. And I was told that they corresponds to the dissipative force F_dissipative (or F_dis).

But I'm not sure if I understand that... like are those P_dis-c and P_dis-r just corresponding to F_dis? What about F_elastic? Cos then what would happen to each of the two terms in the integral of the forces in equation (1)?

... cos I know that integral of force over time is P (impulse)... but I'm not sure what does each of the integral of F_el and F_dis become..?

Thanks!!

Last edited by a moderator: Apr 25, 2005
4. Apr 25, 2005

### OlderDan

OK. This is beginning to make some sense. The model is that the forces between the spheres are separable into an elastic force and a dissipative force with their associated impulses. Since the elastic forces are functions only of x, and x goes from r1+r2 to some minimum value and then restores to r1+r2, the work done by those forces will integrate to zero regardless of impact time. On the other hand, the dissipative forces are proportional to velocity as well as a function of |x|, and since velocity changes direction. they are going to do work against the spheres at all times. That is why they dissipate energy.

During compression, the velocity will start out higher than it will be at the end of restitution because of the dissipation. The average velocity during compression will be higher than during restitution, but both involve the same change in |x|, so the restitution process will involve a smaller dissipative force (lower velocity) but will take more time. It is plausible that the compression and restitution impulses will have the same magnitude. The hint that you have been given suggests to me that the scaling down of the dissipative force by the reduction of velocity will be exactly offset by the scaling up of the time interval over which the integral must be done. It may require nothing more than a change of integration variable to prove the impulses equal.

I think I have the general approach in hand, but obviously there is still work to be done to complete the proof. I assume a head on collision from a center of mass perspective will do since everything is cast in terms of the x-variable. I need to so some other things, but I hope you will follow up with your progress. If this is of any help, and I can clarify anything I have said I'll be checking back.

5. Apr 25, 2005

### physicsfun

Thanks Dan! This is great help!... Now I'm more clear on the scenario and in fact, I think I almost got it! :):):)

Just One thing though...

Can you please elaborate more on this? How is it that it is the "work done"? I thought the integral of our force integral here, by taking the dt to the right hand side of equation (1), would be integrating w.r.t. time, and that would be the impulse... Can you please show me more on how can I prove that the elastic force is zero?

Thanks!

6. Apr 25, 2005

### physicsfun

Actually... another thing.... I think my answer may be a little off... so I want to ask a little more of what you mean here:

What do you mean by "scaling down the dissipative force etcetc.. offset by the scaling up ....etc."??
I remember my professor said something about the x does not matter,that we can just take it as a function as a whole... so is it true that the integral of the second term of force from Eqt. (1) is just the sum of P_r and P_c?
Where does the v in the second term come into place? doesnt matter? How bout the fact that v=dx/dt? Does that affect any part of the integral of eqt.(1)?

Last edited by a moderator: Apr 25, 2005
7. Apr 25, 2005

### OlderDan

From the original force equation

F_elastic+F_dissipative = -h*x^(3/2) - k*v*x^(1/2)

we see that the elastic force is F_elastic = -h*x^(3/2). This tells us that there is no explicit time dependence in the elastic force. It only depends on the compression of the spheres. Although this is a different exponent on x, this is directly anaoagous to the compression of a spring where F = -kx. The work done to compress the spring is (1/2)kx^2 regardless of the time it takes to do the compressing. An ideal spring stores the work done to compress it as potential energy, and delivers that energy back upon restoration. The same thing will happen here. There will be a potential energy stored in the compressed spheres that will be converted back to kinetic energy upon restitution.

Upon looking at this again though I am disturbed by the expression of the forces in terms of x and the definition of x. It seems to me that the defining equation is incorrect. It should say F_elastic = -h*(Deltax)^(3/2). Otherwise, the force would exist for all x, being greatest whene the spheres first touch and decrease as x gets smaller; that makes no sense. The x in the dissipative force should also be a Deltax. Alternatively, the variable x should represent the deviation of the separation of the centers from the sum of the radii.

8. Apr 25, 2005

### OlderDan

Consider an analogous situation. The normal distribution function is of the form sqrt(1/2pi)(1/sigma)exp((x-mu)/sigma)^2. Integrated over x from -inf to inf you always get 1. But if you look at the function you see that as sigma gets larger the height of the peak is scaled down. There is a corresponding and proportional spreading of the peak that maintains the integral at 1, regardless of the value of sigma.

What I am suggesting is that the dissipation force that is proportional to velocity is going to be lower during restitution than it was during compression (analogous to a larger sigma), but that the time it takes for restitiution to be completed is going to be extended proportionately leaving the value of the integral (the impulse) unchaged.

I think what your instructor is telling you is pretty much what I am telling you. When you do the time integral, however F_elastic varies with time you can change the integration to be over x instead, and by symmetry since x goes one way and the goes back to where it started that integral will be zero. Then my hunch is that the velocity factor multiplying the dissipative force will be offset by the increased time interval. You can probably change the time
to a spacial integral. After all (dx/dt)dt is dx, right? I think that's it... I think that is all you need!! Just change the integrals to integrals over dx and they will be symmetric and equal for compression and restitution.

9. Apr 26, 2005

### physicsfun

Sorry... stilll don't understand...

Thanks!! but I'm sorry.. I still need a little more guidance...

so, when we take the integral of the original equation:

I'm not sure I got the math part... or how to write that out... so when I take the integral of the F_elastic in eqt. (1), how do I "change integration to be over x instead"? Do you mean the rate of change is proportional, i.e. dx=dt??

As for the second term... int[F_dis]dt would become: int[ g(x)dx] then?
If I really do out the integration it would become (2/3)*k*x^(3/2)....
I don't understand what you mean... so isn't the F_dissipative the whole of g(x)*v ?
You mentioned that something about symmetry for compression and restitution...Do you mean that the x in the equation shows symmetry? x^3/2?? Do I think of it as a graph of, for example, y=x^3/2? cos that isn't a symmetrical graph... or is it?

Also, int[g(x)dt ] = P_r + P_c , is that right? So what we're doing here is try to show that the rest of Eqt. (1) to be zero in order to prove P_r+P_c=0?

So then, for the LHS of the original equation is intdv... which is u*(change in velocity)... is the change in velocity = the relative final velocities of both spheres minus their initial relative velocities?
And so, just from, say plotting the graph out, we know that this term is also zero? ... or is it not?

I'm really sorry.. I wish I had gotten it sooner... but I hope you can still explain this a little further... thank you!...

Last edited by a moderator: Apr 26, 2005
10. Apr 26, 2005

### OlderDan

What I now think is that you can forget about the elastic force for a minute and concentrate on the dissipative force

$$\int_{t_1}^{t_2} -kvx^{1/2} dt =-k\int_{t_1}^{t_2} x^{1/2}\frac{dx}{dt} dt =-k\int_{x_i}^{x_{min}} x^{1/2} dx +k \int_{x_{min}}^{x_i} x^{1/2} dx = P_{dis-c} + P_{dis-r}$$

$$P_{dis-c} = -k\int_{x_i}^{x_{min}} x^{1/2} dx$$

$$P_{dis-r} = k \int_{x_{min}}^{x_i} x^{1/2} dx = -k \int_{x_i}^{x_{min}} x^{1/2} dx = P_{dis-c}$$

t1 is the time of first contact and t2 is the time of separation. There is a time in between when the compression stops and the restitution begins. Call it tmin for time at minimum x. tmin is not centered between t1 and t2 because the velocity is greater during compression so it will happen faster than restitution. However, by changing the variable of integration to x, the integral from the initial x to the minimum x applies to both of these intervals. The signs of the integrals reflect the change in velocity. The two integrals are the same. Thats the best I can do for now. I'm not that great a mathematician, so there may be some technical flaw in this, but I believe it is the right idea.

11. Apr 26, 2005

### Andrew Mason

I haven't had time to study this but it looks like you would have to solve this non-homogeneous differential equation.

There may be an easier approach if you view the problem in the frame of reference of the centre of mass. In that frame, the forces of compression exerted each sphere are equal and opposite at all times during compression. The forces of restitution exerted by each sphere are equal and opposite at all times during restitution.

I'll have another look at it tomorrow.

AM

12. Apr 26, 2005

### physicsfun

Thanks Dan!! That is totally the right idea! The equations made it so much clearer!
(just one thing about the sign... but we can fix that no problem.) Great! this helps out a lot, thank you!

13. Apr 26, 2005

### physicsfun

Thanks Andrew!

14. Apr 26, 2005

### Andrew Mason

That's it in a nutshell. So long as the spheres separate, the restorative impulse has to be equal (and opposite) to the compressive impulse because the impulses are a function of x not t:

$$p_{dis-C} =\int_{t1}^{t2}F_{c}dt = \int_{t1}^{t2} \frac{dx}{dt}g(x)dt = \int_{x_0}^{x_{min}}g(x)dx = G(x_{min}) + C - (G(x_0) + C)$$

$$p_{dis-R} = \int_{t2}^{t3} \frac{dx}{dt}g(x)dt = \int_{x_{min}}^{x_0}g(x)dx = G(x_0) + C - (G(x_{min}) + C)$$

$$p_{dis-R} + p_{dis-C} = (G(x_0) + C) - (G(x_{min}) + C) + (G(x_{min}) + C) - (G(x_0) + C) = 0$$

AM

15. May 12, 2005

Collision time

I have a similar question related to this problem...

Let's say we just take this same scenario, but this time we just have elastic force, but No dissipative force. F only equals to the first term, and velocity is still v=dx/dt, u=reduced mass, and h is constant as above.

So let's take F=u*dv/dt= -h*x^(3/2) this time, how do we find the collision time (write it as an integral over position in terms of initial velocity v_i, and Maximum compression X_m)?

Last edited by a moderator: May 12, 2005
16. May 12, 2005

### OlderDan

That's the idea. Have you tried to do the integral?

17. May 12, 2005

I didn't know how to do it.... cos the answer I need to find is some complicated thing:

t=2(X_m)/(v_i)* [Integral from 0 to 1 of (1-k^5/2)^(-1/2) dk]

---- how can I get to this?
oh, and sorry I shouldve mentioned this: I need to replace the resulting integral over x by a dimensionless variable k--- and I have no idea what does that--replacing something by a dimensionless variable---even means. Ideas?

Thanks.

Last edited by a moderator: May 12, 2005
18. May 12, 2005