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Can anyone help my start this problem?

  1. Nov 28, 2005 #1
    A 0.150-m-radius grinding wheel, starting at rest, develops an angular speed of 12.0 rad/s in a time interval of 4.00 s. What is the centripetal acceleration of a point 0.100 m from the center when the wheel is moving at an angular speed of 12.0 rad/s?
  2. jcsd
  3. Nov 28, 2005 #2

    Tom Mattson

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    First state the formula for centripetal acceleration. Then look for those quantities in the problem statement.

    Give it a try. If you get stuck then post what you have done and where you got stuck. But you must show an attempt at the problem in order to receive help here.
  4. Nov 28, 2005 #3
    Ac= v^2 / r

    12^2 / .100 which I got 1440. and the answer is 14.4
  5. Nov 28, 2005 #4


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    Homework Helper

    It is true that centripetal acceleration = v^2/r, where v is the tangential velocity.

    However, the 12 you are using is angular velocity. You may convert this [tex]\omega = \frac{v}{r}[/tex] or use the alternative formula: [tex]a_c = \omega^2r[/tex]
  6. Nov 28, 2005 #5
    :blushing: ooh....thanks for the help :smile:
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