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I am a bit confused about what seems like it should be a perfectly straightforward problem in mechanics with a well defined solution.

A sphere approaches a wall from an angle and rebounds elastically. The coefficient of static friction between sphere and wall is large—no slipping occurs during the brief period of contact. Sphere and wall are both made from hard materials so that the contact patch is small. The initial velocity and rotation rate of the sphere are given (rotation is about the $z$ axis and motion takes place in the $xy$ plane). What are the final velocity (speed and direction) and rotation rate?

The source of my confusion is that at first glance, the solution seems under constrained: there are three unknowns ($[v_x, v_y, \omega]$, say), and only two obvious conserved properties (energy and angular momentum about the point of contact).

I suspect that the third constraint is that the magnitude of the velocity component perpendicular to the wall is the same before and after the collision—that $v_{y2}=-v_{y1}$ for a wall oriented in the $x$ direction—but I can’t immediately think of an argument to justify that suspicion with confidence.

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  • $\begingroup$ I answered a similar question a while back, physics.stackexchange.com/a/429834/197851 for the more general 3D case, but you can easily convert the equations into 2D. $\endgroup$
    – user197851
    Nov 11, 2018 at 3:09

2 Answers 2

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I'll give here the 2D version of my answer to this question since there's no guarantee that question (having been closed) won't disappear at some point.

If the wall is in the $x$ direction, as you suggested, let's assume that the sphere is approaching from the positive-$y$ side, so initially it has a velocity component $v_y<0$. The angular velocity $\omega$ is defined so that a positive value $\omega>0$ corresponds to anticlockwise rotation in the $xy$ plane.

Then the effects of the collision are completely determined by the two components of the impulse $(C_x,C_y)$ at the point of contact \begin{align*} mv_x' &= mv_x + C_x \\ mv_y' &= mv_y + C_y \\ I\omega' &= I\omega + RC_x \end{align*} where the primed quantities are those after the collision, $R$ is the radius of the sphere, $m$ is the mass, and $I$ is the moment of inertia. The third equation here represents the effect of an impulsive torque about the centre of the sphere. So there are two unknowns: $C_x$ and $C_y$.

If the collision is to conserve energy, there is one more condition to apply, but there are a couple of options that will satisfy this condition: a perfectly smooth wall and a perfectly rough wall. This is a bit subtle. The argument can be applied to a collision between two bodies of finite mass (as mentioned in the answer of @j-murray) and then (if it is appropriate) we can take the mass of one of them to infinity afterwards. But the argument applies equally well to two colliding bodies of finite mass. In the general case, the physical assumption is that the interaction must depend on the relative velocities of the two points that come into contact. It must also respect the symmetry of the collision, which is defined by the direction of that relative velocity vector and the normal to the surfaces. And of course it must be time reversible: if we reverse the post-collisional velocities, and apply the same collision rules, we should get the pre-collisional velocities again. Once those assumptions are made, there are really only these two possibilities which will conserve energy: smooth and rough collisions.

The smooth wall will correspond to $C_x=0$. This results in no change to the angular velocity, or to the component $v_x$; the component $v_y$ simply gets reversed $v_y'=-v_y$ so as to conserve energy.

In the case of the rough wall, the contact condition of roughness means that there is no slip, which in turn implies that the velocity of the point on the surface of the sphere, in contact with the wall, gets reversed. In the $x$ direction this means $(v_x'+\omega' R)=-(v_x+\omega R)$. In the $y$ direction we get $v_y'=-v_y$ again. A little bit of algebra shows that \begin{align*} C_x &= -2m\left(\frac{I}{I+mR^2}\right)(v_x+\omega R) \\ C_y &= -2m v_y \end{align*} Inserting these into the equations for $v_x'$, $v_y'$ and $\omega'$, it is possible to show that the total kinetic energy (translational plus rotational) is the same after the collision as before.


[EDIT following OP comment]

I'm assuming the rigid elastic rough hard sphere model which is described by S Chapman and TG Cowling, Mathematical Theory of Nonuniform Gases (3rd edition, Cambridge University Press, 1970). In this model, there are no internal degrees of freedom of the spheres: each sphere just has its linear and angular momenta, that's all. Also, I've assumed (as is common) that they behave kinematically as spherical tops, with a scalar moment of inertia. I can only give a simplified argument in the 2D case in favour of associating the energy conservation with the "perfectly rough" condition, i.e. reversal of the relative velocity of the point of impact. For the 3D case I recommend that book, or papers such as JW Lyklema Physica A, 96, 573 (1979).

We can write an equation for the change of kinetic energy $\Delta K$ $$ 2\Delta K = m({v_x'}^2-v_x^2) + m({v_y'}^2-v_y^2) + I({\omega'}^2-\omega^2) $$ Now, I'm just going to assert (as being physically reasonable) that $v_y'=-v_y$, as in a standard 1D elastic collision, unaffected by the details of the impulse in the $x$ direction. This means that the change in $v_y$ has no effect on $\Delta K$, and we can write the remaining terms as \begin{align*} 2\Delta K &= \underbrace{m(v_x'-v_x)}_{C_x}(v_x'+v_x) + \underbrace{I(\omega'-\omega)}_{RC_x}(\omega'+\omega) \\ &= C_x [(v_x'+v_x)+R(\omega'+\omega)] \\ &= C_x [g_x' + g_x] \end{align*} where I have inserted the equations for the change in $v_x$ and $\omega$ in terms of $C_x$, as well as defining $g_x=v_x+R\omega$ as the $x$-velocity of the point on the sphere which is in contact with the wall.

In order for $\Delta K$ to be zero, there are only two possibilities: either $C_x=0$, in which case we have a smooth hard collision, or $g_x'=-g_x$, in which case the velocity at the point of impact is reversed (and we can derive the equation for $C_x$ given earlier).

In Lyklema's paper (if you can get hold of it) you'll see that the analogous equation in 3D is $|\mathbf{g}_\perp'|^2=|\mathbf{g}_\perp|^2$, where $\mathbf{g}_\perp$ is that part of the vector $\mathbf{g}$ which is perpendicular to the normal to the surfaces in contact. (The derivation is for two colliding spheres, but it can easily be adapted to one sphere colliding with another massive, huge, sphere, which acts like a wall). This leaves open the possibility of conserving energy by rotating $\mathbf{g}_\perp$ through some arbitrary angle in the plane of contact to give $\mathbf{g}_\perp'$. However, it is hard to argue that this makes physical sense, without ascribing some exotic properties to the surfaces (chirality). The non-chiral option is to make the angle $180^\circ$ giving $\mathbf{g}_\perp'=-\mathbf{g}_\perp$.

To be honest, in many places this reversal of the velocities at contact is often taken to be the defining property of the perfectly rough sphere model; the detailed justification involves a fair bit of work.

[2nd, hopefully last, EDIT following chat]

I'm going to stick with my previous point: the hard, elastic, rough sphere model is fully specified by this collision rule, that the relative velocity of the points of impact on the two bodies is reversed. The quotations below (going back to 1894!) explain the physical thinking of those who proposed the model, which (in my opinion) largely matches the description of the collision given in the question.

I accept that it is possible to extend the model so as to have different collision rules, while still satisfying the conservation laws: Lyklema's paper (cited above), I think, goes into this in some detail. So one can argue that the mechanical problem is not fully specified (although I am still happy with the assumption made in my answer, that the reversal of velocity perpendicular to the wall is not affected by the impulse tangential to the wall). There are plenty of other papers around which attempt similar extensions of the model, but beware, it's easy to inadvertently make a mistake.

Finally, having established the collision rule for two spheres of different radius, mass, and moment of inertia, respecting all the conservation laws, one can carefully take the limit that one of them becomes very large and heavy, and initially motionless, hence acting as a wall.

Quoting S Chapman and TG Cowling 3rd ed (full ref above) p217 $\S$11.6,

These results are first applied to the rough elastic spherical molecules of Bryan and Pidduck, these being by definition such that at a collision the relative velocity of the spheres at their point of impact is exactly reversed.

Quoting GH Bryan Rep Brit Assoc Advan Sci (1894), p83:

The cases of perfectly rough spheres or circular discs having both their normal and tangential coefficients of restitution unity furnish interesting examples for solution. We may imagine the spheres and discs covered over with perfectly elastic fine teeth or minute projections by whose action the tangential components of the relative velocity are reversed at impact ....

Quoting FB Pidduck Proc Roy Soc A, 101, 101 (1922)

Imagine two spheres to collide and grip each other, so as to bring the points of contact to relative rest. A small elastic deformation is produced, which we suppose to be released immediately afterwards, the force during release being equal to that at the corresponding stage of compression. Thus the relative velocity of the points of contact is reversed by collision.

That's pretty much all I can contribute here!

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  • $\begingroup$ +1 This comes close to resolving the issue for me--I especially appreciate the insight that the collision must be time reversible--but there is still one logical step that I don't completely follow. Why does no slip imply that the velocity of the point of contact gets reversed? Given that that point can exchange internal forces with other parts of the sphere, that fact isn't obvious to me. $\endgroup$
    – Ben51
    Nov 11, 2018 at 12:00
  • $\begingroup$ I've made an attempt at explaining this by editing my answer. As I say at the end, a formal proof needs a fair bit of work, but I've given a couple of references in which you can find the details. $\endgroup$
    – user197851
    Nov 11, 2018 at 14:01
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    $\begingroup$ Following our useful chat yesterday with @j-murray I have made a final edit to my answer. The Lyklema paper gives at least some useful insight, and I don't feel able to add much more. Perhaps you'll get another answer which clarifies things a bit more. $\endgroup$
    – user197851
    Nov 12, 2018 at 13:32
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You're right. The reason for your confusion is that you've inserted an unphysical object (an immobile wall) into the problem without laying out the rules governing how it interacts with the rest of the system, i.e. the ball.

Consider a 1D problem as a toy example. A ball of mass $m$ with velocity $v$ collides elastically with a moveable wall of mass $M$. What are the final velocities $u_b$ and $u_w$ of the ball and wall, respectively?

Demanding the conservation of momentum and energy, we have $$ m v = mu_b + Mu_w$$ $$\frac{1}{2}mv^2 = \frac{1}{2}mu_b^2 + \frac{1}{2}Mu_w^2$$

After a wee bit of algebra, we find that $$ u_b = \frac{m - M}{m+M} v$$ $$ u_w = \frac{2m}{m+M} v $$

Now we consider the question of an immovable wall. Such a wall does not actually obey Newton's laws, so to model one, we consider a moveable wall as above and take the limit as $M\rightarrow \infty$. We now find that $$ u_b = -v$$ $$ u_w = 0$$ in accordance with what we'd expect.


So to conclude, when we talk about immoveable walls, what we mean is moveable walls in the limit as the walls become infinitely massive. Including such a wall (which again, does not obey Newton's laws) in a physical problem requires us to add in the reflection of the velocity vector by hand.

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  • $\begingroup$ I would not agree that a wall attached to the earth does not obey Newton's laws. You transfer momentum to the wall during the collision, but it is so massive that the velocity associated with that momentum is vanishingly small. So conservation of (linear) momentum, is valid, just not useful. In your 1D example with no rotation, conservation of energy leads immediately to the result. When energy can be partitioned between translation and rotation, it becomes less obvious. $\endgroup$
    – Ben51
    Nov 11, 2018 at 12:05
  • $\begingroup$ I didn't say a wall attached to the earth doesn't obey Newton's laws, I said an immobile wall doesn't obey Newton's laws. In the latter case, momentum is explicitly not conserved because the initial and final momentum of the wall are zero. If you account for the (tiny but nonzero) movement of the wall, then you can use the conservation of momentum to obtain the last constraint you're looking for; if you take the limit of an infinitely heavy wall, however, you need to add that constraint (the flipping of the velocity vector) in by hand. $\endgroup$
    – J. Murray
    Nov 11, 2018 at 12:43
  • $\begingroup$ It is, again, conservation of energy and not momentum that requires that the rebound speed in an elastic collision between a ball and a very heavy object is equal to the speed before the collision. And when there is an unknown rotation rate, energy conservation is not sufficient to determine translational velocity. $\endgroup$
    – Ben51
    Nov 11, 2018 at 13:14
  • $\begingroup$ I'm talking about the fact that $v_y' = - v_y$, I'm not referring to the rebound speed. This comes from the conservation of momentum, taken in the $M\rightarrow \infty$ limit. $\endgroup$
    – J. Murray
    Nov 11, 2018 at 13:48
  • $\begingroup$ You have (appropriately) used conservation of energy in your answer. This leads to an unambiguous result in the case with no rotation. I have no difficulty understanding this case. Change in rotational energy cannot be ignored in the conservation of energy in the case I'm asking about. I fail to see how the 1-D case with no rotation adds illumination. $\endgroup$
    – Ben51
    Nov 11, 2018 at 13:56

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