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Angular motion

  1. Jan 6, 2005 #1
    A coin with a diameter of 2.30 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 18.4 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 2.17 rad/s2, how far does the coin roll before coming to rest?


    Can anyone help- I'm trying to study for a physics final and I'm getting nowhere!!! Thanks.
     
  2. jcsd
  3. Jan 6, 2005 #2
    Unless I'm missing something here, you can use this formula:

    vf2 = vi2 + 2aΔx

    Dividing both sides by R2, the coin's radius squared, we get:

    ωf2 = ωi2 + 2αΔθ

    Where α is the coin's angular acceleration, which is negative in this case. From here you can find Δθ and multiplying by R you can find the total distance travelled by the coin.
     
    Last edited: Jan 6, 2005
  4. Jan 6, 2005 #3
    Great.....thanks :biggrin:
     
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