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Drawing the angular velocity graph from the acceleration vs. time graph?

  1. Sep 16, 2010 #1
    1. So this is one of my homework problems and I just can't seem to get it, and my attempts are verging on pathetic. I feel like this shouldn't be this hard, maybe I'm just missing something. This is the acceleration vs. time graph: http://session.masteringphysics.com/problemAsset/1070314/6/04.EX33.jpg and the question it is asking is :The figure shows angular acceleration versus time. Draw the corresponding graph of angular velocity versus time. Assume omega _0 = 0
    Please help, I never took physics in high school and right now physics is making my head feel like somebody hit me with a bat.

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2010 #2

    rock.freak667

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    If you are not able to use calculus, then the fact that the area under the angular acceleration vs. time graph gives the angular velocity.

    So if you find the area from t=0 to t=1, you will get the angular velocity at t=1. So just find the areas for different ranges and then plot the graph. Note how the slope of the angular acceleration graph changes at t=2.
     
  4. Sep 16, 2010 #3
    I tried doing that, but on the y axis, which is angular velocity I can only plot whole numbers, and finding the area under the curve from t=0-t=1 gives me an area of 2.5
     
  5. Sep 16, 2010 #4

    rock.freak667

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    Why do you need to plot whole number for? Although the calculus approach would be much easier, are you allowed to use calculus to draw the graph?
     
  6. Sep 16, 2010 #5
    I am allowed to use calculus, what I was saying is that the velocity i find for t=1 is equal to 2.5 m/s and the online graph that they have given to me to plot my results only allows me to plot (1,2) or (1,3), not (1,2.5) like i need
     
  7. Sep 16, 2010 #6

    rock.freak667

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    In that case you might need to round up to 3.

    But if you can use calculus, then for t=0 to t=2, you can see that gradient of the ang. acc. vs. t graph is 2, meaning that the equation of that section is α=2t and since ω=∫α dt, then you can get how 'ω' will look for that range.
     
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