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Hoop with backspin - Conservation of Angular Momentum

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data
    73flg2.png


    2. Relevant equations

    L = Iw
    v = wr

    3. The attempt at a solution
    Friction acts on the ball while it is skidding, but goes away when the hoop starts to roll, because the velocity is 0 at a point on the ground. This is when v = wr.

    When skidding, friction decreases translational motion but the torque increases rotation.

    When the ball starts to roll without slipping, its angular momentum will be its rotational and translational angular momentum:
    L_f = Iw_f + mrv_f
    At this point, since w_f = v_f/R, and I for a hoop = MR^2

    L_f = MR^2*v_f/R + MRv_f = 2MRv_f

    a. In the beginning, there is only rotational angular momentum. L_i = Iw_i
    Since w_i = v_i/2R and I = MR^2,
    L_i = MR^2*v_i/2R = MRv_i/2
    Equating L_i = L_f

    MRv_i/2 = 2MRv_f
    v_f = v_i/4

    b. Similarly, L_i = MR^2 *v_i/R = MRv_i
    Equating angular momentums,
    MRV_i = 2MRv_f
    v_f = v_i/2

    c. Similarly, L_i = MR^2 *2v_i/R = 2MRv_i
    So 2MRv_i = 2MRv_f
    v_f = v_i

    So, my answer for a is correct, however for b and c, the final velocities are 0 and -v_i/2, respectively. Why?
     
  2. jcsd
  3. Apr 13, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello.

    That looks good.

    Is that true? What about the contribution of v0 to the initial angular momentum?

    Also, be sure to take into account the directions of the angular momenta. Note that ω0 is a "backspin".
     
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