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Homework Help: Sphere sliding up a step - Inelastic Collision

  1. Jun 10, 2017 #1
    1. The problem statement, all variables and given/known data
    A sphere of radius R is rolling without slipping with a velocity v and collides inelastically with a step of height h < R. ¿What is the minimum velocity for which the sphere will be over the step?


    2. Relevant equations
    Total kinetic energy (maybe):
    [tex] \frac{1}{2}I\omega^{2}+\frac{1}{2}mv^2 [/tex]
    Gravitational potential energy:
    [tex] mgh [/tex]
    Moment of inertia about a point a distance d from the CM from de center of mass:
    [tex] I=I_{cm}+md^{2} [/tex]
    Rolling without slliping
    [tex]v=\omega R[/tex]
    3. A discussion of the problem
    This problem is confusing to me: I know the collision is inelastic so there is no conservation of kinetic energy, there is torque when the sphere pivots the step so the angular momentum is not conserved either, the linear momentum doesn't give a lot information and even though I tried to see what condition needs the torque about the pivot point (to give a net torque in the direction of rotation and thus passing the step), I'm not able to see what torque can compensate the one from gravity which doesn't let the sphere slide up:
    The sphere has an angular velocity, but I'm missing something conceptual because I can't figure out how to formally state that the angular momentum changed the right amount so that the sphere made it past the step.
    4. The attempt at a solution
    My (possible wrong) attempt was to say that the sphere lost kinetic energy because of the change on potential energy:
    [tex] \frac{1}{2}I\omega^{2}+\frac{1}{2}mv^2 +mgR = mg(R+h)\\
    v=\sqrt{\frac{10}{7}gh} [/tex]

    I hope you can help me :)
    Last edited: Jun 10, 2017
  2. jcsd
  3. Jun 10, 2017 #2
    It's [tex]v=\sqrt{\frac{10}{7}gh} [\tex]
  4. Jun 10, 2017 #3
    You are correct in that you are wrong, because as you said, additional energy is lost in the collision.

    You say angular momentum changes, but "the angular momentum of an object" does not exist. Only "the angular momentum of an object with respect to an origin" makes sense. Are you sure the angular momentum changes for every choice of origin?
  5. Jun 10, 2017 #4
    You are suggesting a reference frame in which there's no torque involved? I really don't see it
  6. Jun 10, 2017 #5
    Well, not exactly. We typically assume the collision lasts a very short time, such that the gravitational impulse, and the normal impulse from the ground, are ignorable. The torque due to gravity and the lower ground are justifiably ignored in the limit of a small collision time, because the corner force becomes more and more like a dirac delta spike and is thus the dominant effect in that limit.

    With that said, yes.
  7. Jun 10, 2017 #6
    Ok, I think I get your argument. If a choose my origin on the pivot I can omit the set up from before and think in two different moments: before the collision I have the sphere with some velocity [tex]v[/tex] and argular momentum [tex]\vec{r} x \vec{p}[/tex], after the collision I have other velocity and another r and the angular momentum about the pivot is conserved. So I can use conservation of both linear and angular momentum to solve the problem?
  8. Jun 10, 2017 #7
    What is r? Maybe the position vector of the center of mass with respect to the pivot point? You're saying this changes?
  9. Jun 10, 2017 #8
    I was trying to say rx thinking two dimensional. Otherwise I feel like out of information.
  10. Jun 10, 2017 #9
    Or should I think more about the components of that contact force. I'm kind of lost getting information or "critical" relations that satisfy that condition
  11. Jun 10, 2017 #10
    Well one key relation is that the collision is perfectly inelastic, which we can interpret as meaning that after the collision, the contact point is stuck, meaning it purely rotates about that point after the collision. So we have a ball rolling towards a corner, we know the exact motion of this rolling, and then we know that it purely rotates about that corner after they collide, we just don't know with what angular speed. Well you've pointed out that angular momentum is conserved about that pivot, so, if you are careful, you should be able to determine the angular speed which gives the same angular momentum immediately after the collision as there was before the collision.

    Then your energy considerations should help in determining if that angular speed is sufficient to reach the top.
  12. Jun 11, 2017 #11


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    That is actually a very hard question. Maybe it will rise high enough to be over the step, but peak too soon, and hit the edge of the step again. If it does, and there is friction on impact, will the rotational inertia be enough to complete the ascent in a second trajectory? Or a third??...

    In fact, it is not clear whether there is (sufficient) friction on impact with the step the first time to ensure no slipping then. It only says it had been rolling without slipping, which is a tautology.
  13. Jun 11, 2017 #12
    Assume the ball never slips. What about a reaction force from the floor (at the lowest point of the ball) at the moment of collision?
    If the ball just rests on the floor (without motion) and touches the edge of the stair then it is already a statically indeterminate system.
    Last edited: Jun 11, 2017
  14. Jun 11, 2017 #13


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    I do not see how that could be relevant. As Hiero notes in post #6, all strictly limited forces can be ignored during (what is presumed to be) the very brief impact.
  15. Jun 11, 2017 #14
    why do you think that this reaction must be limited?
  16. Jun 11, 2017 #15


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    Because there is no downward unlimited force threatening to penetrate the floor.
  17. Jun 11, 2017 #16
  18. Jun 11, 2017 #17
    horizontal forces also play a role in angular momentum equation
    and you can never deduce from the equations of impact that these forces are limited ; above I have explained why. If you propose that as a hypothesis -- no problem

  19. Jun 11, 2017 #18
    Man, thank you very much. I asked in other sites and that seems to be a satisfying answer, I was trying to think of it as an statics problem and got really confused. I guess the word collision is key to all the considerations you pointed out.

    It was a hard question for me too, because yes, it is open to suggest the right conditions for the sphere to go up, not only it's velocity and it happens in a lot of these classical mechanics problems from old Russian books.
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