Solving a Spinning Disk Physics Exam Problem

[discoverer02]

Here's the problem and I'm close to the answer, but I guess close isn't good enough on a Physics exam.

A spinning solid disk, rotating with angular velocity Wo, is put down on a level surface. It slides and rolls until it reaches an angular velocity W at which it rolls without sliding. Show that W = Wo/3

I place my origin on the ground so that angular momentum is conserved.

The disk spins clockwise
1 = right when the disk is placed on the floor.
2 = right when the disk begins rolling without spinning.
L = angular momentum.
cm = center of mass.
I = moment of inertia about cm
R = radius
M = Mass
V = Velocity of cm

L1spin + L1cm = L2spin + L2cm

IWo + 0 = IW - RMV ==> with no slip V = RW
(2/5)MWoR^2 = (2/5)MWR^2 - MWR^2
-2Wo/3 = W

Besides probably messing up the signs, why do I end up with twice what I need?

 

[gnome]

hmmm...

I've never done anything like this one & I don't have an answer, but I have some questions that may help...

It looks like you're using the moment of inertia for a solid sphere rather than a cylinder. Is that part of the problem?

Clearly friction is involved here, otherwise the disk would just keep spinning. Doesn't that mean that neither energy nor momentum are conserved?

Can this be solved without knowing μ_s? Can it be that regardless of what the initial angular velocity is, it will always scrub off exactly 2/3 of its speed before it stops slipping, independent of the coefficient of static friction? That's certainly counter-intuitive.

On the other hand, maybe it's μ_k that's relevant, since the disk is sliding initially. But once it stops slipping, static friction must take over, right?

 

[gnome]

This is amazing!

Let
Δt = time elapsed until disk stops slipping
f = force of friction while slipping (assumed this to be constant)

Then the linear momentum gained by the disk is equal to the impulse provided by the friction so
fΔt = Mv_cm ....Equation 1

and the change in angular momentum is equal to the angular impulse which (I hope ) is given by
TΔt = Iω₀ - Iω

and T = fR so
fRΔt = (1/2)MR²(ω₀ - ω)
fΔt = (1/2)MR(ω₀ - ω)

Now, substuting from Equation 1:
Mv_cm = (1/2)MR(ω₀ - ω)
v_cm = (1/2)R(ω₀ - ω)

and v_cm = Rω so
Rω = (1/2)R(ω₀ - ω)
ω = (1/2)ω₀ - (1/2)ω
(1/2)ω₀ = (3/2)ω
ω₀ = 3ω

Whoda thunk that?

...

By the way, go to the Mathematics forum and see Greg Bernhardt's announcement about Making math symbols and the thread Additional math notation for tips on formatting.

 

[discoverer02]

You're right. I was taking the moment of inertial of a sphere rather than a disk and I also messed up this signs in the equation.

Yours is a very interesting approach. I'm impressed. I tend to stay within the mechanics of what was presented in class and not think about the problem enough. It obviously pays to think about the problem.

Thanks very much for the help.