# Homework Help: Hoop with backspin - Conservation of Angular Momentum

1. Apr 13, 2014

### mintsnapple

1. The problem statement, all variables and given/known data

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. Apr 13, 2014

### TSny

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".