1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cylinder sliding down a track

  1. Oct 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A uniform solid cylinder (m=0.230 kg, of small radius) is at the top of a similar ramp, which has friction. The cylinder starts from rest and rolls down the ramp without sliding and goes around the loop. Find the speed of the cylinder at the top of the loop.

    [​IMG]
    2. Relevant equations

    PE - potential energy
    KE - kinetic energy
    KEr - rotational kinetic energy
    I - moment of inertia
    w - omega

    KE0 + PE0 + KEr0 = KEf + PEf + KErf

    3. The attempt at a solution

    At first I was tripped up over the small radius part, until I realized that the radius would cancel out with final height of the potential energy and I deduced that the initial rotational and kinetic energies would cancel out to get something to this fashion:

    mgh0 = 1/2mv^2 + mg2R + 1/2Iw^2
    mgh0 = 1/2mv^2 + mg2R + 1/2I(v^2/r^2)
    mgh0 = 1/2mv^2 + mg + 1/2Iv^2

    I plugged and chugged my numbers to get a velocity of 7.18, but then I realized I forgot to incorporate the friction force, so how would that fit in to the mathematical equation?
     
  2. jcsd
  3. Oct 26, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    OK.
    Not sure what you did here. What happend to R and r?

    I didn't do the calculation myself, but you need to double check your formulas. What did you use for I?

    Since the cylinder rolls without slipping, no energy is lost to friction: Mechanical energy is conserved.
     
  4. Oct 26, 2007 #3
    sorry, R=r so I should have had them in the same caps. Since they're the same I canceled them out

    I = 1/2MR^2 <---- I looked it up in my book and it said the moment of inertia for solid cylinders is that
    so does that mean the fact that there is friction is negligible in this situation?
     
  5. Oct 26, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    R is the radius of the loop; r is the radius of the rolling cylinder. They are not the same. (And even if they were, how would they cancel?)

    Rewrite the final equation that you used to get your answer.


    Good.


    It's not that the friction is negligibly small, and thus can be ignored. Friction is essential--it's what makes the cylinder roll instead of slide. But from an energy point of view, the friction does no work so energy is still conserved. (So it has no effect on your energy equation.)
     
  6. Oct 26, 2007 #5
    looks like goofed here. Ok so R stays, but r isn't given so somehow it drops out, but I'm not sure how it would.
     
  7. Oct 26, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Small r appears in the formula for I.
     
  8. Oct 26, 2007 #7
    oh, no duh, thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?