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Rolling motion interrupted by a step (shorter than the radius)

  1. Mar 31, 2012 #1
    Hi,

    I have another query related to rolling motion.

    Lets say a disc is rolling purely with a velocity vo (this being the vcom).
    It encounters a step.lets say that the step has a height h, where h < r, i.e., the step has a height less than that of the height of the center of the disc. SO, only a corner of that step will touch that disc.
    Let the friction of the wall also be high enough that it resists slipping of the sphere upon contact. Then what happens to the velocity of the disc afterwards?
     
  2. jcsd
  3. Mar 31, 2012 #2

    K^2

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    If absolutely no slipping occurs, use conservation of energy. The potential energy of your disk/sphere went up by mgh, so subtract mgh from rotational kinetic energy and recalculate new speed from remaining energy.
     
  4. Apr 1, 2012 #3

    haruspex

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    There's been an impact - energy is NOT conserved.

    In general, need more info. E.g. disc could bounce.

    If we assume disc and step are inelastic then we can use conservation of linear and angular momenta, but it's tricky, and I don't guarantee to have got this right.

    Just prior to impact, the disc is instantaneously rotating about its point of contact with the floor.
    Let F be that point, S the tip of the step, and O the centre of the disc.
    The impulse is orthogonal to the line FS, and will bring the disc to a halt in that direction.
    S will be the new instantaneous centre of rotation.

    Moment of inertia of disc is m r^2/2.
    Prior to impact, moment of disc about its centre is m.r.v/2.
    Moment of disc about F = mvr/2 + mv.r = 3mvr/2
    Vertical distance from S to O = r - h
    Moment of disc about S = mvr/2 + mv(r-h)
    If just after impact disc has rotational speed w (and is in rolling contact with S) then by conservation of angular momentum about S:
    mvr/2 + mv(r-h) = 3mwrr/2
    So wr = v(1-2h/3r)

    Thereafter it still has to climb the step (energy being conserved from here).
     
  5. Apr 1, 2012 #4

    K^2

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    If h<r, it can't bounce without slipping. And for a rigid disk, slipping is the only way to lose energy. Ergo, no slipping = no loss.

    If the disk can deform, then you can lose energy that way, and yes, you'd need more info, but neither the linear nor angular momentum are conserved. As much momentum as needed can be transferred into or from the ground. And rolling up the step involves tangential forces on the wheel, which will again exchange angular momentum with ground.

    You aren't hoping to include the whole planet in computations, are you? In that case, you have your two constraints. Energy and rolling constraint. If energy is lost, you need to know how much.
     
  6. Apr 1, 2012 #5

    haruspex

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    1. Why couldn't it bounce without slipping? Roll a rubber tyre at a low step and see what happens. I'm not saying it would bounce back - but it would bounce.

    2. Mechanics questions commonly invoke inelastic strings and bodies in impulse questions. They do not imply there's no energy lost in the impact. Of course, there are no perfectly inelastic bodies in the real world - it's a limit condition. In reality, there are very large forces acting for very short times over minute distances. Impulse analysis glosses over that and concerns itself with the integral F.dt, i.e. the change in momentum.

    3. You're right about the tangential force from friction - forgot that - but I didn't use that so I don't think it affects my algebra.

    4. You can treat the step as simply a huge rigid mass, so it can be the whole planet if you like! Momentum that was in the radial direction OS is effectively transferred completely to that. The only moment left to consider is rotationally around S.

    So for now I stand by my formula.
     
  7. Apr 1, 2012 #6

    K^2

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    1. Because without losing the rolling constraint, back is the only direction it can bounce. And to bounce back without slipping, the collision point must be located above center point. This has to do with the fact that linear and angular momentum for a rolling body have a fixed ratio, while linear and angular impulse have ratio that depends on impact point.

    2. Fact that OP called it a disk implies rigid body. A perfectly rigid body is perfectly elastic by definition, as no work can be done against it. But even if it's not perfectly elastic, we can assume the loss to be small.

    3. Fact that you didn't use it is part of what's wrong with your algebra. At the moment of impact, you'll end up with forces generating torque, so angular momentum of your wheel changes.

    4. Yes, and an "infinite" mass is an infinite sink or source for momentum. Unless you actually take the body into account, give it a finite mass, and then take that mass as limit to infinity, you can't use conservation laws.


    What actually happens is that yes, the moment just before impact with S, the angular momentum around S is lower. However, during impact, tangential forces are present on the wheel which bring the angular momentum around S to the full value of mrv/2. Condition that the wheel doesn't slip require it.

    How do you check that? Consider a perfectly rigid wheel for which energy is conserved. The only solution is for w to remain constant during collision, and that requires a change of angular momentum around S unless the wheel slips. Now, if the wheel does absorb some energy, then the wheel will slow down, but your analysis is off either way. If it doesn't work in limiting case of no energy loss, it doesn't work.
     
  8. Apr 1, 2012 #7

    haruspex

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    1. No, it could bounce up. In the real world, a bounce takes time. During that time it can be in "rolling" contact - i.e. not slipping, but deforming suitably. Have you ever played with a superball? These demonstrate it beautifully. If you spin one on a horizontal axis and drop it onto a hard floor it will reverse spin with each bounce and hop backwards and forwards!

    2. Try reading some of these:

    http://www.efm.leeds.ac.uk/CIVE/CIVE1140/section04/mechanics_sec04_full_notes02.html#_Toc466176305
    "Although momentum is conserved, it is important to realise that energy is always lost in an inelastic collision"

    http://www.physics.isu.edu/~hackmart/impulsemomentum.pdf [Broken]
    "Momentum is conserved in inelastic collisions but kinetic
    energy is not."

    http://web.me.com/dtrapp/ePhysics.f/WDmomentum.html

    3. I took moments about S. Forces through S, such as the tangential frictional force, have no moment about S.

    4. Likewise, the impulse inflicted on the step is force through S, so has no moment about S.
     
    Last edited by a moderator: May 5, 2017
  9. Apr 1, 2012 #8

    K^2

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    I'm telling you to treat this collision as elastic and see what happens. Yes, forces through S don't change moment about S. Problem is, there is a moment when body is in contact with S and F, and force through F most certainly generates moment around S.

    Look, lets start by taking a civilized choice of pivot - through the center of mass. Now we have 3 quantities to consider. ΔL, Δpx, and Δpy. And we have 2 constraint forces during collision, which allow me to fix these 3 quantities at any value I wish. Fortunately, we do have some constraints on the initial and final quantities.

    Before collision: py=0, px=mv, L=mvr/2

    After collision, the body rotates around S. We don't know |p|, but we can fix py/px, because it has to be perpendicular to SO line.

    [tex]\frac{p_y}{p_x}=\frac{h}{\sqrt{r^2-h^2}}[/tex]

    We also know that L=|p|r/2 both before and after collision. (Rolling constraint.)


    That's 3Δs and only 2 equations to constraint them. The 3rd equation can only come from energy conservation. Now, you can put in some damping at this stage. You can say Ef=ζEi, and you will get different answers depending on the value of that ζ, which can be anywhere in the [0,1] range. And nothing in the problem actually fixes the specific quantity.
     
  10. Apr 1, 2012 #9
    First of all, thank u both for pondering over this.

    And using haruspex's idea of taking angular momentum of the ball at the point of contact, I have got the correct answer.

    Thanks a lot man!!!
     
    Last edited: Apr 1, 2012
  11. Apr 1, 2012 #10

    haruspex

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    Yes, in general there will be a coefficient of restitution e between 0 and 1.

    But it's the inelastic case I found interesting. And it would appear that's what was wanted (he/she has the right answer now).
    (The term 'inelastic' is conventionally used, perhaps misleadingly, to imply the e = 0 case. I would not equate it with rigidity or stiffness, which relate to elastic modulus.)

    Btw, I'm pretty sure subsequent motion would be that the disc would lose contact with the step immediately after impact and become airborne.
     
  12. Apr 1, 2012 #11

    K^2

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    In a real case, certainly. But if the problem insists on rolling constraint, it can be implemented in theory without any trouble.

    It's not the right answer, though. I mean, I'm sure you can find the energy loss such that it is satisfied, but there is absolutely nothing in the problem to point that way.
     
  13. Apr 1, 2012 #12

    haruspex

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    As I read it, the rolling constraint applies up to and including the moment of impact.
    I doubt there's any way it could remain in contact thereafter.

    It might help to imagine something close to this in reality. Think of a steel ball rolled at a low step on a concrete floor. Such a ball dropped on such a floor won't bounce much, so we have something close to an inelastic collision. But I'm sure you'd predict that it would become airborne, however briefly, after hitting the step.

    To get the answer I got, all you have to do is assume the disc is inelastic and apply conservation of angular momentum. Since that gave the 'right' answer (I guess this is from a textbook), that appears to put it beyond dispute (unless you wish to take it up with the textbook author).

    Did you check out the online tutorials I listed?
     
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