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Two cylinders rolling without slipping

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Two cylinders (radius R) start from rest and roll without slipping down ramps. They both have the same masses and moments of inertia. The ramps are both at the same angle. One cylinder (A) is directly on the ramp. The other (B) is hanging from two smaller cylindrical bits (radius r) which stick out from either side and are in contact with the ramp. Which reaches the bottom first?

    2. Relevant equations

    τ = I α
    τ = F r
    mgh
    1/2 mv^2
    1/2 Iω^2

    3. The attempt at a solution

    I used torque to figure out which would have the greater ω. The force providing torque is friction, which must be the same for each cylinder (same mass, same normal force). The difference is in r; so the cylinder with the smaller r-value must have the smaller torque and thus the smaller ω at any moment in time. I.e.:

    τ = F_f r = I α = I Δω/Δt
    ω(t) = (F_f t / I) r

    So it would seem that cylinder B, with the smaller r, must also have a smaller ω.

    I know that the following is true for both cylinders:
    mgh = 1/2 mv^2 + 1/2 Iω^2

    So it would seem that cylinder B must have a greater v, because it has a smaller ω.

    But I'm confused for a few reasons:
    1. For rolling without slipping, v_cm = r ω. So if ω is smaller, so is v. This (obviously) contradicts the conclusion that smaller ω means greater v.
    2. If friction is providing torque, it is increasing the rotational kinetic energy of the cylinders. Does that then mean that it is doing work? That would make the energy equation above incorrect (since it assumes zero work is done).
    3. When I imagine the problem in my head, cylinder B seems like it should spin faster than cylinder A.

    Any insight into this question is very much appreciated!
     
  2. jcsd
  3. Dec 10, 2011 #2
    Mmm... do the cilindrical bits have mass? If they don't, then they would not affect the moment of inertia... If you have some diagram or picture you can upload, if would greatly help in understanding the problem :)
     
  4. Dec 10, 2011 #3
    I don't have a scanner, so I had to make the image myself:

    Wx4Ym.png
    In case that doesn't work: http://imgur.com/Wx4Ym

    The mass of the extra bits on cylinder B can be ignored because both objects have the same total mass and same moment of inertia...
     
  5. Dec 10, 2011 #4
    If both cilinders have the same normal force (which is evenly distributed in the bits), then the friction force is the same, and they come to the floor at the same time. Cilinder B could be lower than A, but I think we can ignore this.
     
  6. Dec 10, 2011 #5
    Thanks, but the prof who asked this question claimed that one of them reached the ground first. Unfortunately I can't remember which and am unable to ask him right now.

    They both start with the same potential energy, but once they're rolling, one of them spins faster and so has less translational kinetic energy. Just can't decide which.
     
  7. Dec 10, 2011 #6
    But, if the inner surfaces that are in contact to the two sides (bases) of the cilinder are not frictionless, there would be a reduction of the torque... And they would also reduce kinetic energy...
    Anyway, all seems to indicate that they come at the same time.
     
  8. Dec 10, 2011 #7
    But, if the inner surfaces that are in contact to the two sides (bases) of the cilinder are not frictionless, there would be a reduction of the torque... And it would also reduce kinetic energy...
    Anyway, ignoring that, all seems to indicate that they come at the same time.
     
  9. Dec 10, 2011 #8
    What indicates that they arrive at the same time? I don't follow your reasoning. You said it's because the friction forces are the same? They are, I think, but they must also provide different torques if they're applied at different distances from the axis of rotation...
     
  10. Dec 10, 2011 #9
    I got it. The difference in the radius of cilinder A and the radius of the bits of cilinder B should account for the difference in angular speed...
     
  11. Dec 10, 2011 #10
    Exactly. (And yes, you can ignore friction between the sides of cylinder B and the inside of the track.)

    Any ideas on how to proceed?
     
  12. Dec 10, 2011 #11
    Can you put what you have together with moment of force? That'll help you know which one reaches first.
     
  13. Dec 10, 2011 #12
    Yes, friction forces are applied at different distances from the axis. That said, the torque is greater for cilinder A, and it comes first.
    One question: is the lower part of cilinder B contributing to normal force? If it isn't, the problem is now solved.
     
  14. Dec 10, 2011 #13
    I'm guessing it doesn't contribute to anything because B is spinning on an axe Δ which is it itself.
     
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