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Find the Friction (Rotational Problem)

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data
    One afternoon, you are playing on a wooden basketball court with a hula‐hoop to amuse yourself. The hoop has a mass of 500g and a radius of 55 cm. After practicing a bit, you find that you can launch the hoop so that its plane is vertical, it isn’t rotating before it hits the floor, and it doesn’t bounce after hitting the floor. For these launch conditions, the hoop’s CM moves initially with a speed of 3 m/s along the floor. You notice that as the hoop slips along the floor it begins to rotate, and that just before it hits the far wall 20 m away from the launch point, it “catches” and begins to roll without slipping. What is the coefficient of kinetic friction between the hoop and the floor?


    2. Relevant equations
    τ=Iα=f*r where f is frictional force
    f=μmg
    f=ma
    I = MR^2

    3. The attempt at a solution
    Tried various method to solve for μ, but it ends up canceling itself. i.e.:
    f*r=MR^2*(a/r)
    a=f/M
    f =f/M
     
  2. jcsd
  3. Nov 9, 2013 #2

    Simon Bridge

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    Please show your reasoning.
    Notice that the hoop slides for a while before it "catches" - what is happening in this part of the motion?
    What is the condition for rolling without slipping?
     
  4. Nov 9, 2013 #3

    haruspex

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    It's rather hard to follow if you keep using the same symbol (f) to mean different things. Please define all your variables and name them uniquely.
     
  5. Nov 11, 2013 #4
    Sorry for the confusion but here is what I was trying to do
    At the bottom of the hoop, as the hoop starts to rotate, that must mean that friction is causing the rotation. In my free body diagram I had. friction to left of point of contact and Normal up and weight down. Friction acting on the object cause it rotate so it will have angular acceleration. Friction cause the hoop to have acceleration since it should be slowing down. so f (friction ) = ma. By torque eqn. τ=f*l=Iα=MR^2(a/r) -> which arrives to the same equation of f = ma. And from here I am lost.
     
  6. Nov 11, 2013 #5

    haruspex

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    You got the same equation because you assumed αr = a. But you can only be sure of that once it is rolling. So these equations are telling you something you didn't know, that αr = a holds even while it is slipping. But it might not be terribly useful.
    Think about velocity. You are given the initial linear and rotational velocities, and a relationship between the final velocities.
     
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