# CD spinning - Centripetal Acceleration

1. Sep 15, 2007

### thatgirlyouknow

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

A computer is reading data from a rotating CD-ROM. At a point that is 0.0244 m from the center of the disk, the centripetal acceleration is 283 m/s2. What is the centripetal acceleration at a point that is 0.0856 m from the center of the disc?

2. Relevant equations

Ac = v^2/r

3. The attempt at a solution

So for the first point:
283 = v^2/(.0244)
283 *.0244 = v^2
v^2 = 6.9052

Since v is constant, then plug it into the next equation:

6.9052/(.0856) = Ac
Ac = 80.668 m/s^2

However, this does not agree with the answer I should be getting. I don't understand where my math went wrong. This seems like a such an easy problem. Any suggestions? Thanks!

2. Sep 15, 2007

### Staff: Mentor

V is not constant, but $\omega$ (angular speed) is. Find or derive an alternate expression for centripetal acceleration in terms of $\omega$.

3. Sep 15, 2007

### thatgirlyouknow

We haven't dealt with w in my class. What is it?

4. Sep 15, 2007

### learningphysics

angular velocity. v = Rw

5. Sep 15, 2007

### thatgirlyouknow

Ok so solving for w gives me 107.696. Then:
v = Rw
v = .0856*107.696
v = 9.219
v^2/r = 992.8197 m/s^2

This can't be right, because as radius increases, the centripetal acceleration decreases. The second acceleration should be less than 283 m/s2.

6. Sep 15, 2007

### learningphysics

Your work looks good to me. As radius increases, cent. acc. decreases only for constant velocity... but velocity changes with R here...

cent acc = v^2/r = (rw)^2/r = rw^2, so as r increases cent. acc. increases.

7. Sep 15, 2007

### thatgirlyouknow

Ah, gotcha. Thanks so much!