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Find Maximum Angular Speed

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data

    In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 1.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as

    θ =2.50t2 - 0.600t3

    where θ is in radians and t is in seconds. (a) Find the maximum angular speed of the roller. (b) What is the maximum tangential speed of a point on the rim of the roller? (c) At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation? (d) Through how many rotations has the roller turned between t=0 and the time found in part (c)?

    2. Relevant equations

    I think this has to do with translational and angular quantities. ac=v2/r=rω² might be useful.

    For part b, at=rα


    3. The attempt at a solution

    I took the derivative of the rotational position to get angular speed in terms of t. I know the radius is .5 m. I don't understand how a max speed can be reached, as it would increase indefinitely with time. I don't think I'm grasping the problem. I also don't understand how the roller could reverse its direction. Any help is much appreciated.
     
    Last edited: Jun 23, 2011
  2. jcsd
  3. Jun 23, 2011 #2

    I like Serena

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    Welcome to PF, menglish20! :smile:

    In the manufacturing process, the roller would be first accelerated to a max speed and then decelerated to a stop.

    You said you calculated the derivative. So what did you get?
    You did solve it for being equal to zero?
    Note that you're only interested in solutions where t > 0 and the where t is smaller than the time where the angular velocity becomes zero again.
     
  4. Jun 23, 2011 #3
    Glad to be here :smile:

    Right so the derivative would be
    ω=5t - 1.8t2
    So I solve for that set to zero, and i get t = 0, 2.78.

    So I know the time where the velocity peaks is between those times. I'm going to guess that it acts parabolic, so the peak must be the midpoint, so t=1.389 s.
    With that, I find vmax=3.47 rad/s

    For part c, the driving force should be removed at 2.78s correct? If not I guess I still don't understand what's going on in terms of the manufacturing process.
     
  5. Jun 23, 2011 #4

    I like Serena

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    Good! :smile:

    Actually, you calculated the max angular velocity here.
    This is not the tangential velocity.

    Edit: Do you know the relation between these two?
    Btw, you may have found by now, that the relevant equations you mentioned are not needed in this problem.

    Yes! :wink:
    At this time the angular velocity is zero, so if the angular acceleration is set to zero, it will remain zero, which is intended.
     
  6. Jun 23, 2011 #5
    v=rw

    So, I'd take .5 * 3.47 = 1.74 m/s.

    Then for part d,

    θ = 6.43, so the number of rotations would be 1.02.

    All these answers match the given solutions, thanks for clarifying this problem!
     
  7. Jun 23, 2011 #6

    I like Serena

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    You're welcome! :smile:

    And thanks for taking the time to finish this thread and say thanks.
     
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