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Simple Question, Where am I going wrong ?

  1. Oct 5, 2006 #1
    Simple Question, Where am I going wrong!!?!

    The Question:

    A computer is reading data from a rotating CD-ROM. At a point that is 0.0251 m from the center of the disk, the centripetal acceleration is 250 m/s2. What is the centripetal acceleration at a point that is 0.0856 m from the center of the disc?


    My Work:

    To find V

    Ac = V^2/r

    250 m/s^2 = V^2/.0251

    V^2 = 6.275

    V = 2.50499501

    To Find Ac with the new radius:

    Ac = 2.50499501^2/.0856

    Ac = 73.30607477 m/s^2 <------------ This solution is wrong, is there anything wrong with my calculations?
     
  2. jcsd
  3. Oct 5, 2006 #2

    berkeman

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    Where did Ac = V^2/r come from? That's not the equation for centripital acceleration. Centripital acceleration ratios directly with the radius, and depends on the square of the angular velocity, I believe.
     
  4. Oct 5, 2006 #3

    arildno

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    Berkeman:
    V^2/r is perfectly correct; rewriting [itex]V=r\omega[/itex] we get:
    [tex]\frac{V^{2}}{r}=r\omega^{2}[/tex]
    V^2/r is the general expression for centripetal acceleration.
     
  5. Oct 5, 2006 #4

    arildno

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    mmiller:
    You need to determine the ANGULAR velocity in order to compute the linear velocity at some other radius.
    you've been given a major hint in the previous post.
     
  6. Oct 5, 2006 #5

    berkeman

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    Ah, thanks arildno. I think more in terms of omega, but I see your point. In that case OP, I think the error is in your second equation. And maybe think about using the omega form of the equation instead -- it should give you a more intuitive answer.
     
  7. Oct 5, 2006 #6
    i ahve question about this problem... would it also be possible to solve using conservation of angular momentum L?
     
  8. Oct 5, 2006 #7

    berkeman

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    I don't see how, especially considering how easy the problem is to solve with the angular velocity form of the centripital acceleration equation. How would you use Cons. of angular momentum to solve this? Do you get the same answer using the angular velocity form of the centripital acceleration equation?
     
  9. Oct 5, 2006 #8

    Thanks guys, but at this point we have not covered angular velocity in terms of centripetal acceleration, the problem should be able to be solved using these simple terms.

    Is there anything that I have done wrong? Based on my reasoning, the CD is rotating at a constant velocity, so finding that velocity and applying it in the formula with the given values should provide the correct answer.
     
  10. Oct 5, 2006 #9
    Since you're using v^2/r, you're using the tangential velocity at that point. Think about a carnival ride. If you were close to the center, you could probably run around in circles as fast as the ride is going, but if you're at the outer edge of the ride, you wouldn't be able to keep up. The speed you're running at is the velocity you're using in this problem.

    At the point closer to the center, the path it takes during one revolution is much shorter than the path a point would take along the outer edge. They both travel a distance equal to the circumference at each particular radius in the *same* amount of time. Thus, the point further away from the center must be moving faster.

    When you said the "CD is rotating at a constant velocity", I'm thinking that you meant "constant RPM's" (revolutions per minute) or per second... (whatever units you're using.)

    Hopefully this explanation helps, since you "haven't covered angular velocity." Good luck.
     
    Last edited: Oct 5, 2006
  11. Oct 5, 2006 #10

    Doc Al

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    But the velocity is not constant! The angular speed is constant, but a point on the disk far from the center must be moving faster than a point close to the center. (Both points go in a complete circle in the same time, but the outer circle is obviously larger.)

    Velocity is proportional to the distance from the center. You found the velocity at one point--use a ratio to find the velocity at the new point.
     
  12. Oct 5, 2006 #11

    I guess that's why they call you the Doc! Thanks so much, crystal clear explanation, I got the correct answer!

    :smile:
     
  13. Oct 6, 2006 #12

    arildno

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    He calls HIMSELF Doc, but you are right, he gave an admirable explanation nonetheless! :smile:
    (Probably because he IS a Doc)
     
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