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Time From Slipping to Rolling Motion

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    EDIT: A solid homegeneous cylinder of mass M and radius R is moving on a surface with a coefficient of kinetic friction μ(k).
    At t=0 the motion of the cylinder is purely translational with a velocity v(0) that is parallel to the surface and perpendicular to the central axis of the cylinder.
    Determine the time t after which the cylinder performs pure rolling motion.

    I've tried with energy equations to determine this and got a solution with the right units, but it seems fundamentally wrong in that for friction to do the work in converting translational to angular velocity there must be a loss of energy, right?
    Will detail as "Attempt 1"

    Also tried with simple kinematics and got another solution with the right units, but after doing it I realised that kinematics make all sorts of assumptions too, if mapping the trajectory of a projectile as a parabola ignores frction, then is this really any more valid?
    will detail as "Attempt 2"

    All thoughts and comments welcomed.

    2. Relevant equations
    E(kt) = 1/2 mv^2
    E(ka) = 1/2 Iω^2

    I = 1/2 MR^2

    3. The attempts at a solution

    Attempt 1:
    E(kr i) + E(kt i) = E(kt f) + E(kr f)
    0 + (1/2)mv(o)^2 = (1/2)mv(f)^2 + (1/2)Iω(f)^2
    mv(o)^2 = mv(f)^2 + Iω(f)^2

    No slipping at t (by definition) so v(f) = Rω(f)
    mv(o)^2 = m(R^2)ω(f)^2 + Iω(f)^2

    I = (1/2)mR^2
    mv(o)^2 = m(R^2)ω(f)^2 + (1/2)m(R^2)ω(f)^2
    v(o)^2 = (R^2)ω(f)^2 + (1/2)(R^2)ω(f)^2
    v(o)^2 = (3/2)(R^2)ω(f)^2
    ω(f) = √(2/3)v(o)/R

    And ω(f) = ω(i) + αt
    √(2/3)v(o)/R = ω(i) + αt

    ω(i) = 0
    t = √(2/3)v(o)/(Rα)

    τ = Iα, α = τ/I
    t = √(2/3)v(o)I/(Rτ)

    τ = F(friction)R
    F(friction) = μ(k)F(normal)
    F(normal) = mg
    τ = μ(k)mgR
    t = √(2/3)v(o)I/(μ(k)mgR^2)

    Again, I = (1/2)mR^2
    t = (1/2)√(2/3)v(o)/(μ(k)g)
    (1/2)√(2/3) = √(1/4)√(2/3) = √(1/6)

    t = √(1/6)v(o)/(μ(k)g)


    Attempt 2:

    v = v(i) + at
    ω = ω(i) + αt

    Determining variables:
    v(i) = v(0)

    a = F(f)/M = F(n)μ(k)/M = μ(k)g
    But it's acting in the negative direction as velocity must be decreasing, so -μ(k)g

    ω(i) = 0

    α = τ/I
    τ = F(f)R = μ(k)MgR
    I = (1/2)MR^2
    So α = (μ(k)MgR)/((1/2)MR^2) = 2μ(k)g/R

    Plugging them in:
    v = v(o) - μ(k)gt
    ω = 2μ(k)gt/R

    Condition of rolling: v = ωR -> v/R = ω
    Equating and solving:
    (v(o) - μ(k)gt)/R = 2μ(k)gt/R
    v(o) - μ(k)gt = 2μ(k)gt
    v(o) = 3μ(k)gt

    t = v(o) / ( 3 μ(k) g )
     
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 23, 2012 #2
    Hi, Im not sure I understand your problem statement exactly but this is not necessarily the case. For example a drum pure rolling down an incline. The translational energy loss from friction is exactly made up by the rotational kinetic energy gained from the torque, so that the system is not dissipative.
     
  4. Apr 24, 2012 #3
    Thanks for responding.
    And sorry, I didn't have enough time to word it perfectly on account of needing to attend a lecture, I'll find the original wording and edit the first post.

    As for energy loss, I'd treat that as a special case.
    With this there is definitely slipping, and non-zero relative velocity between the surfaces + friction --> heat generation.
     
  5. Apr 24, 2012 #4

    ehild

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    Homework Helper
    Gold Member

    If you want to solve the problem with work-energy theorem, then the work done by friction equals the change of KE: both translational and rotational. The energy is not conserved. You get a relation between final speed and distance travelled, but you need the time.

    The second approach is correct if the rolling resistance can be ignored.

    ehild
     
  6. Apr 24, 2012 #5
    From a quick wiki, rolling resistance would rely on deformation of the cylinder or the surface.
    Based on the low level at which we've studied friction so far, I'll venture it can be ignored in this problem.
    Will go with the second method, cheers. :]
     
    Last edited: Apr 24, 2012
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