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Rolling, Both Slipping and Nonslipping

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A cylinder of radius r, mass m, and rotational inertia 1/2mr2 slides without rolling along a flat, frictionless surface with speed v0. At time t = 0 the object enters a region with friction (with coefficients μk and μs), as shown above. Initially the cylinder slips relative to the surface, but eventually it begins to roll. Set t = 0 when the object enters the region with friction.

    After the object enters the region with friction, but before it begins rolling without slipping, what is v(t), and what is ω(t)? When does the transition to rolling without slipping occur?

    2. Relevant equations

    F = ma
    Ff = μN
    [itex]\tau[/itex] = I[itex]\frac{d\omega}{dt}[/itex] = r x F
    v = r[itex]\omega[/itex]

    3. The attempt at a solution

    To find v(t),

    [itex]\Sigma F = ma[/itex]

    [itex] a = \frac{-F_f}{m} = \frac{-u_kN}{m} = \frac{-μ_kmg}{m} = -μ_kg[/itex]

    [itex] v = v_0 + \int a \ dt = v_0 - u_kgt[/itex]

    To find ω(t),

    [itex]\Sigma \tau = I \frac{dω}{dt} = -F_fr[/itex]

    [itex]\frac{1}{2}mr^2\frac{dω}{dt} = -μ_kmgr[/itex]

    [itex]\frac{dω}{dt} = \frac{-2μ_kg}{r}[/itex]

    [itex]ω(t) = \frac{-2μ_kg}{r}t[/itex]

    While my equation for v(t) agrees with the answer my book gives, the book has [tex]ω(t) = \frac{2μ_kg}{r}t[/tex] which omits the negative sign. I believe that this may be an incorrect omission on the book's part. It seems to me that if the force is happening in the [itex]-\hat{x}[/itex] direction and the radius is pointing in the [itex]-\hat{y}[/itex] direction, the torque should be in the [itex]-\hat{z}[/itex] direction. This implies that the angular velocity is increasing in the negative direction, so there should be a negative sign in the ω(t) function (how I did it). I would be happy to leave it there - case closed - except that it gives me a negative time for when the cylinder starts rolling without slipping (I set the v(t) and ω(t) functions equal, then then it gives me that [itex]t = \frac{v_o}{-μ_kg}[/itex]. What's the problem?
     
  2. jcsd
  3. May 5, 2013 #2

    ehild

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    Both correct. Usually the negative sign is ignored and only the magnitude of ω is considered in the textbooks, and also the rolling condition is established for the magnitude of ω and the speed of the CM.

    ω(t) is can not be equal to V(t) they have different dimensions :) So what exactly is the rolling condition?


    ehild
     
  4. May 5, 2013 #3
    Torque is defined as R×F and not F×R so the torque is in the +z direction. For an independent case like pure rotational motion the negative value for angular velocity is "acceptable" as it is just a sign convention. However here we are considering two motions and they should follow the same coordinate system. If +x direction is positive then clockwise direction is positive. Therefore torque is positive.
     
  5. May 5, 2013 #4
    Although he has written it wrongly has used the condition of rolling correctly but is getting the negative of the correct time.

    I agree that we should just use our common sense and take the magnitude. No need to think about the cross product for each problem.
     
  6. May 5, 2013 #5
    Didn't you state the condition already? v = rω, which is what I used, except that according to you the ω should be the magnitude (absolute value), which makes sense. I guess I didn't think about the fact that I was setting two different unit vectors equal to each other! Thanks for the help!

    I'm pretty sure not. Torque is certainly rXF, but if you examine the values, you do end up with a negative torque (assuming that [itex]\hat{x}[/itex] x [itex]\hat{y} = \hat{z}[/itex], which should always be true). I would submit a proof, except I'm not sure how to do 3x3 determinants in latex.
     
  7. May 5, 2013 #6

    ehild

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    When rolling, positive velocity of the CM involves clockwise rotation, that is, negative ω. So you have to use the v=-rω as rolling condition if you consider the angular velocity of the rolling body negative. But this equation is not a vector equation, better to use v=rω with the magnitudes.
    In vector notation, the velocity of the CM is v=ωxr with respect to the point of contact with the ground.
    And you are right, static friction produces negative torque with respect to the CM.
    ehild
     
    Last edited: May 6, 2013
  8. May 6, 2013 #7
    I see. This makes sense. Thanks once again!
     
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