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Rolling motion of a cylinder

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A solid homogeneous cylinder of mass M and radius R is moving on a surface with a coefficient of kinetic friction μk. At t=0 the motion f the cylinder is purely translational with a velocity v0 that is parallel to the surface and perpendicular to the central axis of the cylinder.
    Determine the time tR after which the cylinder performs pure rolling motion.

    2. Relevant equations

    None were given on the paper but I assume I'll be needing this:
    When rolling: ƩE = K = 0.5mv2 + 0.5Icmω2

    3. The attempt at a solution

    Here's what I've worked through so far

    When the cylinder is in pure translational motion:
    ƩE = 0.5Mv02 - fkd
    ƩE = 0.5Mv02knd

    When the cylinder is in pure rolling motion:
    ƩE = K = 0.5MvR2 + 0.5Icmω2

    Due to conservation of energy:
    0.5Mv02 - μknd = 0.5MvR2 + 0.5Icmω2

    As n = -Mg and Icm = 0.5mr2

    0.5Mv02 + μkMgd = 0.5MvR2 + 0.25MR2ω2

    Simplifying:
    0.5v02 + 9.8μkd = 0.5vR2 + 0.25R2ω2

    As vcm = rω

    0.5v02 + 9.8μkd = 0.5vR2 + 0.25vR2

    0.5v02 + 9.8μkd = 0.75vR2

    After this I can't really figure out how to go on apart from substituting d for:
    0.5(v0 + vR)t
    Which would add a time variable in.

    This gives: 0.5v02 + 4.9μk(v0 + vR)t = 0.75vR2

    The main things I'm stuck on are whether or not the vcm at t=t0 (v0) and the vcm at t=tR (vR) are the same or not and how to get rid of the coefficient of kinetic friction which I am not given a value for.
     
    Last edited: Apr 11, 2012
  2. jcsd
  3. Apr 12, 2012 #2

    ehild

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    You need time so it is better to use equations where time is included.
    The motion of the cylinder consist of translation of the CM and rotation about the CM. The force of friction decreases linear momentum and the torque of friction increases angular momentum. Write up both of them as functions of time and use the rolling condition v=Rω to find tR when pure rolling is established.

    ehild
     
  4. Apr 12, 2012 #3
    So that would make something like:

    Δp = Mv0 - MvR = -fkt

    and

    Δp = Icmω = fkRt

    so

    vR = v0 - μkgt

    and

    ω = (2μkgt)/R

    so

    vR = 2μkgt = v0 - μkgt

    v0 = 3μkgt

    I always seem to have too any unknowns left over
     
  5. Apr 12, 2012 #4

    ehild

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    Eliminate μkgt, you get VR. Substitute back to get tR in terms of the given data v0, R, μk.

    ehild
     
  6. Apr 12, 2012 #5
    Ok so:

    v0 = 3/2*vR
    vR = 2/3*v0

    But I don't think I can go on any further because I haven't actually been given any values for μk, v0 or R

    So the closest I can come is:

    t = v0/(3*μkg)
    t = v0/(29.4*μk)
     
  7. Apr 12, 2012 #6

    ehild

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    OK, it is correct. No numerical data are given, you have to express the time with vo, g and μk
    .

    ehild
     
  8. Apr 12, 2012 #7
    Ok thanks for your help it was much appreciated.
     
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