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Tangential Acceleration of uniform motion

  1. Jun 15, 2008 #1
    How do you find the Magnitude of tangential acceleration if you have uniform circular motion? I know the formula for Tangential Acceleration; however I have no clue how to apply it to determine the Magnitude?
     
  2. jcsd
  3. Jun 15, 2008 #2
    Its just the magnitude of the vector. It should just be V^2/r.
     
  4. Jun 16, 2008 #3
    The formula in the previous post is incorrect (that's the magnitude of the *radial* component of the acceleration). What formula are you using for tangential acc?
     
  5. Jun 16, 2008 #4

    Doc Al

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    Staff: Mentor

    Do you mean centripetal acceleration? If something is performing uniform circular motion, its tangential acceleration is zero. Or do you mean non-uniform circular motion, which will have a tangential component of acceleration?
     
  6. Jun 16, 2008 #5
    No, I mean tangential acceleration. That's probably the answer I'm looking for I just have to know how to express that the acceleration would be zero if it was uniform circular motion using words and one equation.
     
  7. Jun 16, 2008 #6
    Im suppose to use At=[dv/dt].
     
  8. Jun 16, 2008 #7

    Doc Al

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    Staff: Mentor

    OK, where v is the speed, not the velocity vector. For uniform circular motion, dv/dt = 0.
     
  9. Jun 16, 2008 #8

    rcgldr

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    Tangental acceleration can still exist on a object traveling in a circular path. The centripetal force just needs to change with respect to speed2, so it always equals m |v|2 / r.

    The magnitude of tangental acceleration would be the magnitude of angular acceleration times r = |angular acceleration| x r.
     
    Last edited: Jun 16, 2008
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