# Homework Help: Angular Speed

1. Dec 1, 2008

### mburt3

1. The problem statement, all variables and given/known data
A uniform disk is set into rotation with an initial angular speed about its axis through its center. While still rotating at this speed, the disk is placed in contact with a horizontal surface and released. What is the angular speed of the disk once pure rolling takes place (in terms of omega final and initial)?

2. Relevant equations
1/2mv(int)^2 + 1/2Iw(int)^2 = 1/2mv(final)^2 + 1/2Iw(final)^2
Torque=d(L)/d(t)
Normal Force=mg
I=1/2mr^2
L=Iw

3. The attempt at a solution
I knew that the only once a disk rolls without slipping does w=v/r
I simplified the first equation as follows:
v(int)^2 + 1/2r^2w(int)^2 = v(final)^2 + 1/2v(final)^2
v(int)^2 + 1/2r^2w(int)^2 = 3/2v(final)^2

I also knew that the only force acting on the disk was the surface. So I knew that
Fr= d(L)/d(t)
mgr=[w(final)I-w(int)I]/d(t)
from there I got that
g=[r(w(final)-w(int))]/2d(t)

I really have no idea where to go from here though. Any push in the right direction would be appreciated!
Thanks!

2. Dec 1, 2008

### Staff: Mentor

The only horizontal force acting on the disk is friction from the surface. That force does two things: (1) It accelerates the center of mass, (2) It creates a torque that slows down the initial rotation.

At some point the translational speed will be just enough compared to the rotational speed so that the condition for rolling without slipping will be met. Solve for that point.

3. Dec 1, 2008

### mburt3

So do I use those equations....or is it a lot easier than I am thinking? I was thinking you could do something with work being equal to final ke - initial ke and also that
torque x change in time= change in angular momentum.
I guess I am still really lost.

4. Dec 2, 2008

### Staff: Mentor

I suggest that you use Newton's 2nd law for translation and rotation, and a bit of kinematics: Δv = at; Δω = αt.

5. Dec 2, 2008

### mburt3

Thank you for your help so far by the way!
I set up an equation for the rotation, but I'm still a little confused on the rotation.

I(alpha)=F(friction)r
(1/2mr^2)a/R=F(frict)r
ma/2=F(frict)

F(frict)=

I guess I really don't know how to relate them?

I was also thinking that F(frict)rt=I(w(fin)-w(int))

I'm sorry I have to keep asking questions about it, this rotational stuff is killing me!

6. Dec 2, 2008

### Staff: Mentor

Do this for rotation:
Fr = Iα = (1/2 mr²)α
Thus:
α = 2F/mr (note that the disk rotation is slowing down)

Now use kinematics to find ω as a function of time:
ω = ω0 - αt

Do the same thing for translation--find the velocity as a function of time.

Then solve for the time when the condition for rolling without slipping is met.