Hoop with backspin - Conservation of Angular Momentum

[mintsnapple]

Homework Statement

Homework Equations

L = Iw
v = wr

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?

 

[TSny]

Hello.

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

That looks good.

a. In the beginning, there is only rotational angular momentum. L_i = Iw_i

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".