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Circular Motion Problem

  1. Sep 18, 2016 #1
    [moderator's note: Thread moved from forum General Physics, so no template shown]

    The angular velocity of a process control motor is (10−1/2t^2) rad/s, where t is in seconds.

    a. At what time does the motor reverse direction?
    I got 4.5 s which is correct

    b. Through what angle does the motor turn between t =0 s and the instant at which it reverses direction?

    I do not know how to solve part b
     
  2. jcsd
  3. Sep 18, 2016 #2
    That was all that I was given. I am looking for the Delta theta
     
  4. Sep 18, 2016 #3
    How do you get rads out of rad/s and time?
     
  5. Sep 18, 2016 #4
    Multiply by time?
     
  6. Sep 18, 2016 #5

    kuruman

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    Do you know how angular velocity is related to angular displacement?
     
  7. Sep 18, 2016 #6
    velocity is the derivative of displacement
     
  8. Sep 18, 2016 #7

    kuruman

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    Right, and displacement is the integral of velocity. So, ...
     
  9. Sep 18, 2016 #8
    displacement = 10t-1/6t^3
     
  10. Sep 18, 2016 #9

    kuruman

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    Right. I would write this as, displacement = 10t-(1/6) t^3 otherwise you might think that t^3 is multiplied by 6. Now reread question b. Can you answer it?
     
  11. Sep 18, 2016 #10
    Can I just plug in time? so 45 - (1/6)*4.5^3 = 29.81?
     
  12. Sep 18, 2016 #11

    kuruman

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    Sure. You derived an expression that gives you the angular displacement at any time t and you are looking for the angular displacement at the specific time t = 4.5 s.
     
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