• Support PF! Buy your school textbooks, materials and every day products Here!

Collision with wall

  • Thread starter Tusike
  • Start date
  • #1
139
0

Homework Statement


A hockey-disk with mass m, radius "r" collides with a wall in an angle "alpha" with velocity "v". There is friction, but the place where the ball touches the wall doesn't slide. How does the velocity component that is parallel to the wall change? The collision is completely elastic


Homework Equations


Fdt = mdx
Frdt = (theta)*w


The Attempt at a Solution


I actually have a problem after the solution, with the no sliding part. First, here's my solution:

Let's the parallel be the x axis, the other one y. Since the collision is elastic, the y component will only become the opposite of what it was.
Let the collision take place during a dt amount of time, in which the average friction force upon the ball is F. The x component of the velocity was v*sin(alpha) before the collision, and is changed by dvx. So we can write:
F*dt = mdvx
Also, during this time a torque causes the hockey-disk to rotate, and if the angular momentum will be w, we can write:
F*r*dt = (1/2)mr^2 *w
From the above two equations, we get that dvx = r*w/2
Since no other forces except F are applied in the x direction, the energy is conserved:
m(vsinalpha)^2 = m(vsinalpha - dvx)^2 + (theta)*w^2
m*dvx^2 + 1/2*m*r^2*w^2 = 2mvsin(alpha)*dvx ,substitute in dvx:
m*(1/4)*r^2*w^2 + (1/2)*m*r^2*w^2 = mvrwsin(alpha) , divide by mrw:
rw/4 + rw/2 = vsin(alpha)
3/4*r*w = vsin(alpha)
w=4vsin(alpha)/(3r)
and so dvx = 2vsin(alpha)/3.

That would be great and the problem is supposed to be done, but here's my real problem. Since the point where the ball and the wall touch isn't supposed to slide, it's velocity should be zero during and at the end of the collision process. Let this velocity be ux, we know that at the end:
ux = vsin(alpha) - dvx - r*w = vsin(alpha) -2vsin(alpha)/3 - 4vsin(alpha)/3 = -vsin(alpha)

Now that isn't zero! What did I do wrong?
 

Answers and Replies

  • #2
139
0
anyone?
 
  • #3
ehild
Homework Helper
15,477
1,854
I do not think your equations for the change of momentum and angular momentum are valid. Remember rolling friction. There is a force and a torque acting on the rolling body and the torque is not equal to the force of friction times R.
The disk does not slide which means that the point where the disk touches the wall is in rest. Let be vx' the x component of the velocity of the CM after collision, and the angular momentum ω. v'xCM+ωR=0, that is ω=-v'x/R.

ehild
 
  • #4
139
0
I'm not sure I have to take into account the rolling friction. Then there would be energy loss, so I couldn't write the energy conservation and solve the equations.

And if there was rolling friction, it would act as a force and a torque as the normal (unsliding) friction. So you could just say "F" represent the average of both, and my equation dvx = r*w/2 would still be true...

Perhaps the answer is simply that the no-sliding can only happen if energy is not conserved, and so the that equation isn't true?

(In my equation your v'x = vsin(alpha)-dvx, so:)
If so, from my equation dvx=r*w/2, and your equation w=(vsin(alpha) - dvx) / r I get the result that dux = vsin(alpha)/3, which is half of the "original"!
So the only real question remains: is it true, that the energy is NOT conserved? If that's true, then we're done:)

EDIT: and if the energy is not conserved, why the not?
 
  • #5
139
0
OK I'm summarizing the possibilities and the results from them. Either the momentum and angular momentum equation, the energy equation, or the no-sliding equation is incorrect.
Let the resulting x velocity be u.

1) The no sliding equation is incorrect, we get what I wrote down first:
dvx = 2vsin(alpha)/3
u = vsin(alpha) - dvx = vsin(alpha) / 3

2) If the energy equation is incorrect, we get what I just wrote down in the above post:
dvx = vsin(alpha)/3
u = vsin(alpha) - dvx = vsin(alpha)* 2/3

3) If the momentum equations are incorrect, from the energy & no sliding we get:
(dvx I didn't calculate)
u = vsin(alpha) * sqr(2/3)

So, which one is the real answer, and why? :) thanks!
 
  • #6
ehild
Homework Helper
15,477
1,854
The problem states that the collision is completely elastic. So the energy is conserved. I did not mean that there was rolling friction, I just wanted to show that interaction between the wall and disk was equivalent with a force F and a torque M, but M was not necessarily equal to F*R, just in the case of rolling friction.

ehild
 
  • #7
139
0
I still don't see how that is possible. Looking at only the x components:
Any time during the collision, if any dF force acts on the disk, it acts on the outer edge, so it's R away from the center. So it's torque would be dF*R... So the angular momentum can only be changed by the sum of these dF*R torques, which results in an average F*R torque during the whole process. And during this time, the average force upon the disk was F, so my equations should be true.

What I'm trying to say, is that the force that can act as a torque acts on the disk itself as well, I don't see how the two can be separated so that M wouldn't equal F*r, in case of any friction. Could you give an example of where this wouldn't be true?
 
  • #8
ehild
Homework Helper
15,477
1,854
I have understood what you stated, but I think you are wrong. I suggest that the disk and the wall touch each other with an extended area. In this case, the effect of the wall on the disk is equivalent with a force F and a torque τ, but τ≠ r*F. An example is rolling friction (see in Budó's Kisérleti fizika I). You do not know anything about the interaction between the wall and the disk. But it is known that the point on the rim of the disk did not slide along the wall during this interaction, so the friction did not do any work, the energy was conserved, and as a result of this interaction, both the x component of the velocity of the CM and the angular velocity around the CM have changed.

ehild
 
  • #9
139
0
I see now; that explanation was exactly what I was looking for. Thanks again!
 
  • #10
139
0
Hi! I was perfectly content with the explanation to my problem in the OP, until today when I came across the 1994 IPhO, Theoretical Problem 3 (http://www.jyu.fi/ipho/). The problem assumes an elastic collision, but doesn't assume that the point of contact does not slide during the entire process, so I can understand that energy dissipates because of the sliding. That's why in the answer given at the end, it can be checked that the energy of the system has indeed lowered.
However, in the solution they assume the torque M to be equal to F*r, where F is the frictional force (see their equation (4)). I may be overlooking something completely, but I don't see the difference in the usage of this equation in that problem and the one I mentioned originally. What's the reason they can use the concept of the torque being F*r throughout the collision, and I couldn't? Without it (and without the energy conservation) the problem couldn't be solved, but still that shouldn't allow the usage of it.

Thanks!
-Tusike
 
  • #11
798
0
1/ How come an elastic collision allows sliding? As long as there is sliding, there is kinetic friction (I will clarify the difference between static/kinetic friction and rolling friction later). As kinetic friction occurs when there is relative motion between the points of contact of the 2 bodies (i.e. the disk and the wall), it always dissipates energy as heat, and so, kinetic energy is nor conserved, i.e. the collision is not elastic.

2/ The torque that ehild mentions, I guess, is the rolling friction. Rolling friction is due to the deformation of the contact surface during collision. In the case of a wheel rolling on a horizontal plane for example, it is a torque and not a force, and thus, is not associated with any "d" in the formula M = Fd (as there is no F here!). Making a conjecture to your problem, rolling friction is a moment (not really a torque), effectively associated with some "d" but not the radius r of the disk.
On the other hand, static/kinetic friction is a force, and thus, effectively associated with an "d", which is radius r in the problem. I see no problem in saying M = Fr, if F is kinetic friction.
(Ahh, I feel pretty clumsy here, not so good at explaining things :biggrin:)

Anyway, when it comes to deformation of bodies, it becomes complicated. I don't think problems at this level require that much understanding and knowledge.

3/ The solution of the IPho problem is correct. And they didn't assume an elastic collision: they didn't have any energy conservation equation! Besides, most of the times, IPho problems are of high caliber (though the problems may occasionally have mistakes), so you should trust them.

4/ The problem in #1 is wrong I think. For instance, right when collision starts, the disk has the parallel component velocity, while the wall is at rest, so there is relative motion between the point of contact of the disk and the wall, i.e. the disk slides.
Or perhaps the disk initially also rotates with a specific angular velocity that we have to figure out???
 
Last edited:
  • #12
139
0
Here's what I think sums this all up:
If the point of contact doesn't slide, F cannot be kinetic friction. Also, if the "point" of contact doesn't slide, we must still think of the collision as one without deformation, otherwise the problem would say "point". If so, there isn't any rolling friction either, which leaves us with static friction. This way the torque will in fact be F*r, with F being the same force causing the disks momentum to change. This leaves us with the energy-conservation law to be untrue in problem of the OP, which can be explained by what you stated in 4/, that it is impossible for the point of contact not to slide right at the beginning when it has no angular velocity but a parallel velocity component.

In the IPhO problem we don't know whether there is kinetic, static, or rolling friction (probably there's more than one), but still, we assume that the deformation is so small that the F average force of the frictional forces can be thought of as an F*r torque causing the rotation, which explains why their equation (4) is correct.

Do I see it correctly?
 
  • #13
798
0
It's generally correct, but I think I need to clarify some points:

Rolling friction is inevitable, howsoever small deformation is. Only in the case of ideal rigid body (or rigid body for short), where it's a point of contact, not a surface, is rolling friction is zero.
See picture attached. I draw 2 cases:
_ Case 1: a wheel rolling on a horizontal plane. The normal force N = weight P. Due to deformation, N is located away from P by a distance r. The torque by the couple N & P is T = Nr = Pr, which is rolling friction. So in this case, if deformation is small, r is small and thus T is small.
_ Case 2: the disk collides with the wall on a horizontal plane. In collision, generally the force N is very large, so even if deformation is small or r is small, we cannot say anything about T = Nr, which is also rolling friction.
_ Ideally, when there is no deformation, r = 0, howsoever large N is (N cannot be infinitely large), rolling friction is 0.

Usually, the problems at IPhO level will assume ideal rigid bodies, except when there is further clarification. This is why we can safely remove the factor of rolling friction.
 

Attachments

Related Threads on Collision with wall

  • Last Post
Replies
1
Views
8K
Replies
17
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
559
  • Last Post
Replies
24
Views
4K
  • Last Post
Replies
3
Views
853
Replies
5
Views
633
Replies
5
Views
11K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
17
Views
2K
Top