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Angular Velocity

  1. May 23, 2017 #1
    The question:

    A wheel was rotating at 12.5 rad/s when a torque was applied for 15.7 s. The angular velocity increased to 54.0 rad/s. What angle did the wheel turn through in that time?

    My Solution:

    theta = Wo*t+1/2*a*t^2
    = (12.5 rad/s)(15.7)+(1/2)(41.5)(15.7)^2

    therefore,

    theta = 5310.9 rad.

    Comment:

    This answer doesn't seem right to me and I am stuck on what to do.

    Thank you in advance!
     
  2. jcsd
  3. May 23, 2017 #2

    Orodruin

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    Check the physical dimensions of the numbers you have entered!
     
  4. May 23, 2017 #3
    Welcome to PF.
    It looks like you found α = 41.5. That would have been true if the specified velocity change occurred within 1 second. But it took 15.7 seconds.
     
  5. May 23, 2017 #4
    Thank you for your responses, so I should do α = 41.5/15.7 to break it down to seconds and then use α = 2.66 rad/s as my value for acceleration?

    That then gives me an answer of 524 rad. This seems more reasonable!
     
    Last edited: May 23, 2017
  6. May 24, 2017 #5

    haruspex

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    Near enough - I get 522.
    But angular movements at constant acceleration are analogous to linear ones. All the usual SUVAT equations carry over. If you remember those, for each of the five variables (initial speed, final speed, acceleration, distance, time) there is an equation that omits one and connects the other four. In this case you have initial speed, final speed and time, and you want to find distance. So there is no need to find acceleration if you pick the right equation.
     
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