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Centripetal acceleration, incorporating a change in acceleration around the circle

  1. Jun 22, 2007 #1

    exi

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    1. The problem statement, all variables and given/known data

    A hawk flies in a horizontal arc of radius 13.5 m with a speed of 3.9 m/s.

    What is its centripetal acceleration? (I correctly found this to be 1.1267 m/s².)

    Next question is: If it continues to fly along the same horizontal arc but increases its speed at a rate of 0.54 m/s², what is the magnitude of acceleration under these new conditions?

    2. Relevant equations

    Centripetal acceleration = v² / r; possibly another one or two.

    3. The attempt at a solution

    Finding the first half of this was easy enough, as 3.9² / 13.5 = 1.1267. But I have tried three different ways of incorporating this acceleration and have gotten it wrong each time.

    The only other thing I can think of trying is using the circumference of this path as displacement, finding its velocity (10.335 after one "lap"), and using that to find a new acceleration, but I'd rather not blow any more answer submissions on this one question. By the above, I'm coming up with 7.913 m/s². Am I on the right track, or am I making this way too difficult?
     
  2. jcsd
  3. Jun 22, 2007 #2

    nrqed

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    Have you learned about tangential acceleration?

    If the speed is constant, there is only a centripetal acceleration directed toward the center of the circle with a magnitude given by [itex] a_c = \frac{v^2}{R} [/itex] as you know.

    If the speed is not uniform, there is an addition a tangential acceleration (tangent to the circle) given by [itex] a_t = \frac{dv}{dt} [/itex] i.e. it is equal to the rate of change of speed (notice that this is directly the value provided to you in the question).

    Now, notice that the two accelerations are perpendicular to each other. To find the total acceleration you have to do a vector sum of those two accelerations. The magnitude of the total acceleration is easy to find.

    Hope this helps.
     
  4. Jun 22, 2007 #3

    Doc Al

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    Staff: Mentor

    I'm not sure what you calculated there, but you are making things more difficult than necessary. You figured out the radial (centripetal) acceleration; They gave you the tangential acceleration. What's the resultant total acceleration?


    Oops... nrqed beat me too it!
     
  5. Jun 22, 2007 #4

    exi

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    No, we haven't really talked about that in class. Thanks for the run-down on it.

    And yep, 1.2494 m/s² it is.
     
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