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Period/Frequency of a Rotational Object

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data

    An object moves uniformly around a circular path of radius .20 meters, making one complete revolution every 2.00 sec. What are (a) the translational speed of the object, (b) the frequency of the object, and (C) the angular speed of the object?

    2. Relevant equations

    x = A Cos (angular speed * t)

    Angular speed = 2 pi frequency

    T = 1/f

    3. The attempt at a solution

    I am lost as to how to approach this one... I think I might need to solve for the angular speed first?
  2. jcsd
  3. Dec 3, 2007 #2
    Looks to me like you could easily find the frequency from the given info.

    making one complete revolution every 2.00 sec...

  4. Dec 3, 2007 #3
    Frequency = 1 / period correct? So, 1 / 2 sec = .5 rev/sec?
  5. Dec 3, 2007 #4
  6. Dec 3, 2007 #5
    Would this be suffice to solve it?

    r = .2m and T = 2s
    v = 2pi(r) / T
    a) so just plug in.

    b) f = 1/T so just plug in again

    c) omega = v/r so just plug in.

    On part c) how can I get the final answer to be in rad/sec???
  7. Dec 4, 2007 #6

    D H

    Staff: Mentor

    [itex]\omega = v/r[/itex] is in radians per second.
  8. Dec 4, 2007 #7
    Oh okay, so in part a) the r is reffering to 2 Pi Radians or not 2 pi radius???
  9. Dec 4, 2007 #8

    D H

    Staff: Mentor

    You have the right answer for part (a), [itex]v = 2\pi\, r/T[/itex]. Here is how this expression arises. The circumference of the circle is [itex]2\pi r[/itex]. This is the arc length the object travels in In one revolution. Dividing by the time gives the average speed. Since the speed is constant in uniform circular motion, the average speed is the speed, period. Given [itex]v = 2\pi\, r/T[/itex], what is [itex]\omega=v/r[/itex]?
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