Question

Slippery and non-slippery thoughts.

I am trying to understand the difference in behaviors when two identical perfectly rigid spheres collide with some tangential velocity if a) their surface can slide against each other without restriction, and b) their surface cannot slide at all.

Assuming perfect rigidity and restitution, the first scenario is simple to describe: both spheres exchange the normal components of their relative velocity, resulting in a perfectly elastic collision. However, on the second, how will tangential velocity influence the resulting linear and angular velocities?
Thanks

Answer 1

When the balls cannot slide, the tangential component at the point of contact will create spin. I believe the normal component will be elastic, but the tangential will not be like in the case when they do slide. The momentum of the tangential will be conserved in both linear and rotational forms.

Answer 2

Thanks, eashmore, it also seems to me that would be the expected behavior. Furthermore, my intuition -which might of course be wrong- tells me that the spinning can be transformed back into linear motion if the rotating spheres collide again.

Think, for instance, of two fairly rough wooden spinning tops on a slightly parabolic surface, so that they are both pulled by gravity towards the same location, causing them to collide. If they are rotating in opposite directions at the exact same rate, then their collision will be very similar to a head-on collision because each top's tangential velocity will be cancelled out by the other top's tangential velocity. However, if both tops spin in the same direction, then their collision will be superelastic: part of the rotational motion will be transformed into linear motion upon collision and the tops will separate at a speed greater than they approached.

So linear and rotational motion seem to be able to transform into each other through the tangential collision of non-sliding surfaces. However, how could we model this mathematically? How could we calculate the expected linear and angular velocities after a collision?

Thanks for your patience, take care.

Answer 3

User69. Your intuition is correct. We know that energy is conserved, always. If we were able to have the tops collide in an elastic manner, we could use the Euler-Lagrange EOM.

www.people.fas.harvard.edu/~djmorin/chap6.pdf

E-L is modeled as such: \(L = T - V\), where \(L\) is the Lagrangian, \(T\) is the kinetic energy, and \(V\) is the potential energy. If the parabolic table were sufficiently smooth and had a sufficiently small slope, we could ignore potential energy losses and gains as the tops move and frictional losses as the tops spin. (Note: these conditions and assumptions can be modelled very closely in a laboratory.) With these assumptions, the Lagrangian reduces to \(L = T\). Let's note that kinetic energy, \(T\), has both linear and rotational forms, therefore, \(T = TKE + RKE\), where \(TKE\) is the translational kinetic energy and \(RKE\) is the rotational kinetic energy. You are familiar with translational kinetic energy. It is given mathematically as: \(TKE = {1 \over 2}mv^2\)
You may not be as familiar with rotational kinetic energy, it is expressed as: \(RKE = {1 \over 2} I \omega ^2\), where \(I\) is the moment of inertia, and \(\omega\) is the angular velocity.

Now, let's apply these equations to the actual collision. As established earlier, the collision will be elastic, and the potential energy is ignored. Therefore, before the collision:
\(L_{1i} = TKE_{1i} + RKE_{1i}\) and \(L_{2i} = TKE_{2i} + RKE_{2i}\)
If we want the collision to be "superelastic," with respect to the linear motion of the top, we know that \(TKE_i < TKE_f \) and \(RKE_i > RKE_f\)

This can be seen by applying conservation of energy to the Lagrangian: \(L_i = L_f\)

Therefore: \(TKE_i + RKE_i = TKE_f + RKE_f\)

I won't go into the kinematics of how to produce this result (I could be here all day). The direction each top is spinning is not important. The key here is to ensure that the difference in angular velocities between the two tops is not of such a great magnitude that slipping occurs. Therefore, this increase in TKE can be achieved regardless of spin direction. Geometry, materials, and vectors of the collision are all important. I might grab some tops at the dollar store and run some simple experiments at home.

Answer 4

Thanks again, eashmore. I will need some time to digest all this information, but I really appreciate your guidance. Best regards.

Answer 5

Additionally. Review the PDF file I linked to. It explains how the E-L EOM can be manipulated into F=ma form. This form might give you a better idea of what happens at the point of collision. N.B.: This requires some knowledge of matrices, calculus, and differential equations.

In the mean time, I'll work on an example problem. The more I think about this, the more I become curious as to how we could emulate a "superelastic" collision like this.

Answer 6

Do you know what I find interesting? In the suggested setup, two tops with exact opposite spins will collide in an elastic manner. Provided their kinetic energies before the collision are low, so will be the kinetic energies after the collision and the tops will remain within the same region in space. However, two tops with equal spins will collide superelastically and will therefore tend to separate from each other, even if their kinetic energies before the collision were low.

Opposite spins allow two objects to remain within small regions of space, whereas equal spins cause them to move away from each other. That's what I find interesting; it reminds me of the behavior of electrons within atomic orbitals.

I am going to read the Lagrangian method; I'll be looking forward to your example problem. May I suggest that you please consider your example particles spherical, perfectly rigid, and their surfaces completely incapable of sliding against each other, with no energy lost to friction? This is a little part of a big thought, and it would really help me a lot if I could view the behavior of such ideal objects on a computer screen.

Thanks in any case!

Answer 7

Your analogy to electrons is dead on. Remember that electrons with the same principle, azimuthal, and magnetic quantum numbers must have opposite spin numbers.

Also observe that if two tops have opposite spin directions and different speeds, the tops will bounce away from each other, ever so slightly, and continue to do so until finally their rotational speeds reach the same speed. When this happens, the tops mesh. This leads one to assume, that the RKE is being converted to TKE as to allow the tops to reach equilibrium.

Answer 8

Can I please abuse your kindness once more? This might be better on a different thread altogether, but I will post it here since I will still refer to our imaginary spinning-tops setup.

I have been warned many times not to think of electrons and other subatomic particles as spinning things, and not to think of their "spin" as an actual spin. However, if the spinning tops analogy can help me understand why electrons with equal spin cannot coexist in the same atomic orbital, maybe they can help me understand the possible values this "spin" property can take.

I suppose that spin=0 means no rotation. If our spinning tops could stay vertical without spinning, they would be spin 0 tops. From here on, values can be either negative or positive; I can visualize that as the two possible opposing directions of rotation. It is the numerical value what I have a little more trouble visualizing. Could an integer spin be viewed as a rotation around an axis parallel to the direction of motion, while a half-integer spin would be a rotation around an axis perpendicular to the direction of motion? Or should I just stop trying to view subatomic particles as rotating things, before my brain explodes? :-)

Answer 9

I would advise not viewing them as such. Take a look here if you wish to see how far you can go before your brain explodes. :-P http://math.ucr.edu/home/baez/spin/spin.html

Answer 10

I better say thanks now, before it really explodes :D Thanks a lot, you're very helpful. Gotta go now.