CD spinning - Centripetal Acceleration

In summary, the conversation is about finding the centripetal acceleration at a point on a rotating CD-ROM, given the acceleration and distance at another point. The equation used is Ac = v^2/r, with v representing angular velocity. The solution involves finding the value of w (angular velocity) and plugging it into the equation, giving a final answer of 992.8197 m/s^2, which may seem counterintuitive but is correct due to the changing velocity with distance.
  • #1
thatgirlyouknow
58
0

Homework Statement



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?

Homework Equations



Ac = v^2/r


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!
 
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  • #2
thatgirlyouknow said:
Since v is constant...
V is not constant, but [itex]\omega[/itex] (angular speed) is. Find or derive an alternate expression for centripetal acceleration in terms of [itex]\omega[/itex].
 
  • #3
We haven't dealt with w in my class. What is it?
 
  • #4
thatgirlyouknow said:
We haven't dealt with w in my class. What is it?

angular velocity. v = Rw
 
  • #5
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
thatgirlyouknow said:
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.

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
Ah, gotcha. Thanks so much!
 

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration of an object moving in a circular path. It is always directed towards the center of the circle and its magnitude is equal to v²/r, where v is the velocity of the object and r is the radius of the circle.

2. How is centripetal acceleration related to CD spinning?

In the case of CD spinning, the centripetal acceleration is caused by the friction between the CD and the player's spindle. As the CD spins, the centripetal acceleration keeps it moving in a circular path.

3. What factors affect the centripetal acceleration of a CD spinning?

The centripetal acceleration of a CD spinning is affected by its angular velocity, which is determined by the speed at which the CD is spinning, and the radius of the CD. The larger the radius or the faster the spinning speed, the greater the centripetal acceleration.

4. How does centripetal acceleration affect the sound quality of a CD?

Centripetal acceleration has a direct impact on the sound quality of a CD. If the centripetal acceleration is too high, it can cause the CD to warp or deform, leading to skipping or distortion of the sound. This is why it is important to maintain a proper balance between the centripetal acceleration and the speed of the CD spinning.

5. Can centripetal acceleration be harmful to a CD?

Yes, a high centripetal acceleration can be harmful to a CD and can damage it. This is why it is important to handle CDs with care and ensure that they are spinning at a proper speed to avoid damage. It is also recommended to use a CD player with a speed control feature to adjust the centripetal acceleration according to the specific CD being played.

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