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Circular Motion with time-dependent radial acceleration

  1. Nov 17, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle is traveling on a circle with a radius R. The particle's radial acceleration is given as:
    [itex] a_r=At^4 [/itex]
    At time [itex] t=0 [/itex] the particle is at [itex] (R,0) [/itex].

    A. Find the distance that the particle has traveled as a function of time [itex] S(t) [/itex].
    B. Display the particle's acceleration in polar coordinates.
    C. Display the particle's acceleration in Cartesian coordinates.

    2. Relevant equations
    ##a_r=\frac {v^2} {R}##
    ##x=rcos\theta, y=rsin\theta##

    3. The attempt at a solution
    I assumed at first that since the particle is moving in a uniform circular motion, I can use
    [itex] a_r=\frac {v^2} {R} [/itex]
    [itex] At^4=\frac {v^2} {R} [/itex]
    [itex] v=\sqrt{RAt^4} [/itex]
    Then I treated [itex] v [/itex] as [itex] \frac {ds} {dt} [/itex] and got
    [itex] S(t)=\sqrt{RA}*\int_0^t t'^2 \, dt' = \frac{\sqrt{RA}t^3} {3} [/itex]

    But then I realized that the radial acceleration changes according to time. Wouldn't that affect the solution? Also, I'm confused as to converting the acceleration from polar to Cartesian. I'm not sure if the same rules apply to acceleration and exactly what are my ##\hat r## and ##\hat \theta##.
  2. jcsd
  3. Nov 17, 2014 #2


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    There is nothing stating that the motion is uniform. In fact, the changing radial acceleration tells you that it is not. However, ##a = v^2/r## holds for any circular motion - otherwise the particle would not be moving in a circle.
  4. Nov 18, 2014 #3
    I see. Does that mean that there is necessarily tangential acceleration as well? If so, any hint on finding it?
  5. Nov 18, 2014 #4


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    Yes, there must be a tangential acceleration. You have already found the velocity, so finding the acceleration is a simple matter of differentiating.
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