# Need a little help!

1. Jun 7, 2007

### scw287

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 0.030 m from the center of the disc, the centripetal acceleration is 118 m/s2. What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?

2. Relevant equations

Ac=v^2/r-centripual acceleration
possibly v=2pier/T

3. The attempt at a solution
Aa=Va^2/Ra
Ab=Vb^2/Rb
substituted Va=1.667*r*a, 1.667*ra<<<1.667 I got by dividing .050m/.030m to get the ratio of the radius'

I set up the equations like this:

118^2/.5^2=(1.667*118 m/s^2*V)^2/(1.667*.30)
>>crossed multiplied .25(196.706V)^2=13924(.5001)
solved for V and got V=.720<<<this is wrong

I think I'm on the right track I think theres an error on the way I set up my equations before substituting in the numbers.

Last edited: Jun 7, 2007
2. Jun 7, 2007

### Staff: Mentor

What is Va?

How does V ratio with radius?

3. Jun 7, 2007

### Staff: Mentor

Not quite sure what you are doing, but here's a hint that might help: Rewrite the formula for centripetal acceleration in terms of angular velocity (omega) instead of tangential velocity (v). (How are those two quantities related?)

4. Jun 7, 2007

### scw287

Va means velocity times the acceleration

5. Jun 7, 2007

### scw287

I haven't learned angular velocity yet so I'm not sure how to relate the two. I was trying to use the formula for centrepitual acceleration A=v^2/r
I then set up the equation for the first distance from the center of the CD and then set up a second equation for the second distance from the center. I used the ratio .05/.03=1.667 to tie the two distances together. I then subtitued all the known and solved for the velocity.

6. Jun 8, 2007

### Staff: Mentor

No it doesn't, not as you've used it:

Aa=Va^2/Ra
Ab=Vb^2/Rb

You've used a and b as the first and second points. Call them V1 and V2 if you like to be clearer.

Whatever notation you use, my hint was to calculate the linear velocity (Va or V1 or whatever), and then use what you know about how the velocity varies with radius to solve the problem.

7. Jun 8, 2007

### esalihm

use the expression for centripetal acceleration and see how it is related to the radius. thry to write it in terms of something that these two points have in common (like period of the motion)

then solve for the period. since the period is the same for both, equate the two get the answer.

8. Jun 8, 2007

### scw287

So I'm trying to learn how to do it using the angular velocity way. I'm not familiar with this method and I was hoping if you could check my work.

a = omega^2 r

So a is proportional to r.

a(r) = a(@ r') (r / r')

plug the numbers in and i get

a(.050)=118(.030)(.050/.030)

=118<<i don't think this answer is right, I have a feeling there is something wrong with the left side maybe I'm not supposed to multiply a*the radius???

9. Jun 8, 2007

### esalihm

a is proportional to r but not as "constant = a*r"

look at the equation you wrote first a=omega^2 r
then you get "omega^2(which is a constant) = ........)

10. Jun 8, 2007

### scw287

Its equal to the angular velocity so would it be 118*(.050/.030)

11. Jun 8, 2007

### Staff: Mentor

Exactly right.
Careful with your notation. Express it like this:
a_2 = a_1 (r_2/r_1)

See my note above.

12. Jun 8, 2007

### Staff: Mentor

Exactly.....

13. Jun 8, 2007

### scw287

that is much easier to understand!