Rolling ball and impulse (Too easy to be true?)

  • #1
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Homework Statement


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Homework Equations





The Attempt at a Solution



The change in velocity = a * change in ω

I think that makes sense since for pure rolling v = aω

296nluh.png


16 marks like that?
 

Answers and Replies

  • #2
haruspex
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You've assumed the impulse is horizontal, which I don't think is right. However, as far as I can see, correcting that does not change the answer.
 
  • #3
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You've assumed the impulse is horizontal, which I don't think is right. However, as far as I can see, correcting that does not change the answer.

which direction should J be pointing then?

For change in angular momentum, it's the component of J that is tangential to surface that contributes.

For change in linear momentum, it's the component of J that is perpendicular to the surface that contributes.

From this case, J has equal components on both, hence the angle should be 45 degrees with respect to the tangent at the point of contact. How do I show this?
 
  • #4
haruspex
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which direction should J be pointing then?
I don't know straight off, I just don't think you can assume it's horizontal. The ball strikes the cush obliquely. If there were no friction between the two you should take the impulse as perpendicular to the surface of the ball. Note that that would imply another impulse vertically from the table. But there's no slip between the ball and cush so it's not immediately clear which way the impulse will act. Because of the spin of the ball it could be above the horizontal, causing the ball to lift off the table slightly.
For change in angular momentum, it's the component of J that is tangential to surface that contributes.
You're taking moments about the point of contact of ball with table, right? And you're taking h as the distance from there to the line of the impulse. So you need the component of the impulse orthogonal to that h. I believe that will be the horizontal component of it.
For change in linear momentum, it's the component of J that is perpendicular to the surface that contributes.
Again, you only care about the horizontal component for your equation. That's why I concluded your answer turns out to be right anyway.
 
  • #5
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You're taking moments about the point of contact of ball with table, right? And you're taking h as the distance from there to the line of the impulse. So you need the component of the impulse orthogonal to that h. I believe that will be the horizontal component of it.

Since the ball is in pure-roll, the angular momentum L = Iω and that is with respect to the centre of the ball. So r x p, only the tangential component "spins" the ball and contributes to angular momentum.

Again, you only care about the horizontal component for your equation. That's why I concluded your answer turns out to be right anyway.

I'm not sure why we can just consider the horizontal component..
 
  • #6
haruspex
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Since the ball is in pure-roll, the angular momentum L = Iω and that is with respect to the centre of the ball. So r x p, only the tangential component "spins" the ball and contributes to angular momentum.
Angular momentum is always with regard to some point of reference. Only if the mass centre is stationary is it the same wrt all reference points.
I'm not sure why we can just consider the horizontal component..
Yes, it dawned on me yesterday after posting that that was wrong, but for some reason I've not been able to reach the site since then until just now.
I think that when there is unlimited friction the impulse will be so as to directly oppose the relative motion. In the OP case, that means it will be perpendicular to the line joining the point of contact with the cush and the point of contact with the table. This will likely give a different answer to the one you posted. But there's a problem then with the OP: if h < 2a, it implies the ball will become airborne briefly, not simply roll the other way. So now I tend to think the problem is not well posed.
 
  • #7
tiny-tim
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hi unscientific! :smile:
which direction should J be pointing then?

could be anywhere

if the table wasn't there, the ball would rotate down about the cushion, so clearly there's an impulse I from the table as well as the impulse J from the cushion

we know I must be vertical (because no friction at the table)

and we know Jy = I (because the ball doesn't rise, and because continuing forces like gravity can't be impulsive)

so try finding I and Jx :wink:
For change in angular momentum, it's the component of J that is tangential to surface that contributes.

for angular momentum about the centre, yes only Jtangent is relevant

but that's not necessarily the easiest point to take angular momentum about :wink:
For change in linear momentum, it's the component of J that is perpendicular to the surface that contributes.

if you mean the horizontal component of J (Jx), then yes
 
  • #8
haruspex
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could be anywhere
if the table wasn't there, the ball would rotate down about the cushion,
But the ball is initlially rolling, so as far as the cushion is concerned it is spinning. This will definitely change the angle of the impulse. It can't be just anywhere. It must be in a direction dictated by the known facts, and I see no alternative but to suppose it directly opposes the relative motion of the contacting surfaces.
 
  • #9
tiny-tim
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… I see no alternative but to suppose it directly opposes the relative motion of the contacting surfaces.

stop looking for a shortcut! :rolleyes:

the alternative is to use good ol' Newton's second law

the total impulse (I plus J) equals the change in momentum

and the total moment of impulse equals the change in angular momentum :smile:
 
  • #10
haruspex
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stop looking for a shortcut! :rolleyes:
I'm not looking for a shortcut. I'm looking for a logical basis for determining the direction of the impulse. What you suggest doesn't provide one. There's no way to determine the changes in momentum and angular momentum without knowing which way the impulse acts.
 
  • #11
TSny
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So far, I've only been able to deduce a range of possible heights, h (with unscientifics's value of 7##a##/5 as one of the extreme values). I believe I've made complete use of the angular and linear impulses. I don't see what else can be used. Does tiny-tim have something up his sleave? :smile:
 
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  • #12
haruspex
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Thanks TSny. What do you think of my argument that for no slippage on contact with the cushion the impulse must directly oppose the relative motion at point of contact? If I think of it as an elastic collision over time, with the cushion becoming compressed, it seems to me that that will be the direction of the elastic force.
In my youth, there was a popular toy called a superball. Are these still around? One trick with them was to throw them forwards with backspin. The ball would bounce back towards you, of course, but then bounce away again, then towards you again... The spin was reversed by each bounce, implying a large horizontal component to the impulse.
 
  • #13
TSny
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Thanks TSny. What do you think of my argument that for no slippage on contact with the cushion the impulse must directly oppose the relative motion at point of contact?
You're saying that just before the collision with the cushion, the point of the ball that makes contact with the cushion has a velocity ##\textbf{v}_0## that is perpendicular to the line connecting the point of contact with the cushion and the point of contact of the ball with the table. That makes sense, since the ball can be thought of as instantaneously rotating about the point of contact with the table.

And then you are arguing that the impulse should be in the opposite direction to ##\textbf{v}_0##. I'm going to have to think about that. Right now I don't see why there couldn't be a component of impulse along the line connecting the two points of contact. Maybe you can help me see that.

If I think of it as an elastic collision over time, with the cushion becoming compressed, it seems to me that that will be the direction of the elastic force.
In my youth, there was a popular toy called a superball. Are these still around? One trick with them was to throw them forwards with backspin. The ball would bounce back towards you, of course, but then bounce away again, then towards you again... The spin was reversed by each bounce, implying a large horizontal component to the impulse.

Yes, the superball! I was a teenager when they first came out. Hours of fun for less than a buck (back then). We used to throw it at an angle onto the floor so it would then bounce up under the seat of a chair or table and the darn thing would bounce right back to you. I think http://www.wham-o.com/product/SuperBall.html [Broken].

That reminds me, there have been several papers on the physics of superballs in the American Journal of Physics. I'll look them up and see if I can find anything that might help with this problem.
 
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  • #14
tiny-tim
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Does tiny-tim have something up his sleave? :smile:

i'm just waiting for the OP to show us how far he's got o:)
 
  • #15
haruspex
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I don't see why there couldn't be a component of impulse along the line connecting the two points of contact.
At the point of contact of ball with cushion, you can think of it as linear motion. It might as well be a rectangular block landing flat (but travelling obliquely) on the ground. The surfaces compress, without slipping, until instantaneously at rest. The force therefore directly opposes the motion; if there were any lateral force rest would not be achieved. Decompression should be like time reversal.
 
  • #16
TSny
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At the point of contact of ball with cushion, you can think of it as linear motion. It might as well be a rectangular block landing flat (but travelling obliquely) on the ground. The surfaces compress, without slipping, until instantaneously at rest. The force therefore directly opposes the motion; if there were any lateral force rest would not be achieved. Decompression should be like time reversal.

I can see how this argument would apply to a particle striking the cushion. But I don't see (yet) why the argument is valid for the sphere striking the cushion. If you consider a small volume element of the ball at the ball's surface where contact is made with the cushion, then that small element is going to experience not only the force from the cushion but also complicated forces from neighboring volume elements of the ball. It's the sum of all of these forces which brings the volume element momentarily to rest.
 
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  • #17
haruspex
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I can see how this argument would apply to a particle striking the cushion. But I don't see (yet) why the argument is valid for the sphere striking the cushion. If you consider a small volume element of the ball at the ball's surface where contact is made with the cushion, then that small element is going to experience not only the force from the cushion but also complicated forces from neighboring volume elements of the ball. It's the sum of all of these forces which brings the volume element momentarily to rest.
We can represent the motion of the ball as a sum of a linear motion and a rotation in any number of ways. If we centre on the part that contacts the cushion, the rotation about that point is immaterial to the forces in the impact (except as second-order effects over a longer time). It's like a rolling contact.
 

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