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Hypothesizing the KE lost in a number of collisons

  1. May 10, 2015 #1
    1. The problem statement, all variables and given/known data
    I suppose my question is a broad one. I am writing a Physics Extended Essay for the IB program (approx. 4000 words). My research question is "How does the angle of an inclined plane impact the horizontal translational kinetic energy of a solid cylinder [after it impacts and bounces across a horizontal surface, probably a wooden table, given no slipping]?", though this is not the question I need answered.
    upload_2015-5-10_22-14-46.png
    I am struggling to figure out how one might figure out theoretically how much KE the cylinder will lose due to the collisions (experimentally I can do this with a high speed camera).
    Is the KE lost to the collisions solely based on that of the vertical component, or is some of the horizontal component lost as well, or even the angular? Does the number of collisions matter?
    I do not have an experiment done yet, nor should I for this essay at this point, thus I have no numbers.
    If someone could direct me to a source, that would be fantastic.

    2. Relevant equations
    Coefficient of Restitution
    cimg198.gif
    Vy = mg*sinθ, where θ is the angle of incline
    3. The attempt at a solution
    I am beginning to figure that the number of collisions is theoretically insignificant, though that is if the KE lost in the collision is solely that of the vertical component, where KE lost = .5m*(Vy)^2 = .5m(mg*sinθ)^2
    Intuitively (which is a dangerous way to look at physics), I do not see the collisions having an effect on the horizontal component at any point.
     

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  2. jcsd
  3. May 11, 2015 #2

    CWatters

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    I can't really help but..

    Is it possible to have no slipping? During the contact phase of a bounce the normal component will increase from zero (just before impact) to some maximum value and then reduce to zero again (as it takes off again). I might be wrong but it seems to me that it must slip at some point in that process but perhaps it can be ignored?
     
  4. May 11, 2015 #3
    If we ignore friction and rolling resistance then i believe the answer is easy, as you say, it is only the vertical component that will lose kinetic energy and not the horizontal.

    However if there is friction and/or rolling resistance then during the contact time window (which will be small but not zero) some of the rotational kinetic energy and horizontal kinetic energy might be lost as well.

    Rolling resistance can be neglected if the cylinder and the horizontal plane are made from material that doesnt deform during the contact.

    So if we are left to deal only with friction, then what happens is that some of the total rotational kinetic energy (right before the first collision) will be converted to horizontal translational kinetic energy, so the horizontal kinetic energy will increase afterall.
     
    Last edited: May 11, 2015
  5. May 11, 2015 #4

    jbriggs444

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    The slipping cannot be ignored, but it can be quantified. For any gain (or loss) in linear momentum due to friction there will be a corresponding loss (or gain) in angular momentum. The final state of motion will be that unique state of motion where angular momentum and linear momentum match up for rolling without slipping.
     
  6. May 11, 2015 #5
    Thanks guys! I should have a lot of meat on my plate with this one, but your responses help with guiding me what I ought to be looking for.
     
  7. May 11, 2015 #6

    haruspex

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    In rolling contact at a steady speed on horizontal ground, there is no friction. There is drag and rolling resistance.
    Putting those aside for the moment, consider the rate of rotation as it reaches the bottom of the ramp. As you note, the vertical component of velocity will be lost during the bounces. What does that tell you about the rate of rotation when rolling contact is eventually restored?
    So what will the effect of friction be?
     
  8. May 11, 2015 #7
    Are you pointing out that the rate of rotation (omega) should equal the (eventual horizontal velocity component)/radius as soon as the rolling contact is restored? If that's the case, I don't see how friction plays into that. I do understand that dynamic friction is the way of the land when it is slipping and static friction is the force required to cause slipping (since there is no motion between the cylinder's contact point and the surface during perfect rotation), yet again, I don't understand how that knowledge applies into the angular and horizontal components.
    Regarding the change in angular momentum (from which change in kinetic energy can be then found), I don't see how one can predict either the time of contact or the angular acceleration in ΔL=I*Δt*α, which is my biggest problem with my experiment. Is that even possible? I can see how to figure it with the proper measurements taken, but I do not see a way to model it.
     
  9. May 11, 2015 #8

    haruspex

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    Yes, both at the approach to the bottom of the ramp, and again later. Which rotation rate will be the greater?
    Friction is the only force available to change the rotation rate.
    Static friction does not cause slipping. It's exceeding the limit of static friction that leads to slipping.
    You don't need to worry about time or acceleration. Use conservation laws. (For conservation of angular momentum, you can often eliminate unknown forces by careful selection of the reference axis.)
     
  10. May 11, 2015 #9

    jbriggs444

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    As I tried (and perhaps failed) to convey in post #4 above, you do not need to know the friction and you do not need to know the change in kinetic energy. You can and should know the initial values for horizontal linear momentum and angular momentum. You can compute the ratio between how much linear momentum can be gained due to friction per unit of angular momentum that is lost due to friction. If you wish, you can do that the obvious way -- figure out how much change in each quantity will result from a frictional impulse of 1 kg meter/second.

    Depending on exactly how you set up your variable naming conventions, that should lead you to two [or more] simultaneous linear equations in two [or more] unknowns.

    One equation will assert the relationship that you calculated above between angular momentum and linear momentum. It may be convenient to write that equation using starting linear momentum and starting angular momentum as named parameters. The other equation will assert the condition that must apply for rolling without slipping.

    Solve for final linear momentum and divide by mass. Or solve for final angular momentum, divide by moment of inertia and multiply by radius.
     
  11. May 11, 2015 #10
    Thanks. I will probably come back to this in a week or so and try to make sense of it all a bit more, but right now I've got other finals to study for. You guys put a lot of work into helping students, and should give yourselves hearty pats on the back.
     
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