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Angular velocity (graph)

  1. Mar 27, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-3-27_1-45-42.png

    2. Relevant equations


    3. The attempt at a solution
    the displacement should be the area under the line , it was traveling backwards for 3 seconds, then it would take an additional 3 seconds to get back to the starting position. From there, the wheel rotates forward for 4 more seconds for a total of 10 seconds.

    So do I need to find the area of the entire triangle (from seconds 0 to 10) then subtract the smaller triangle (from seconds 0 to 6) to be left with only the area of (seconds 6 to 10)?
     
  2. jcsd
  3. Mar 27, 2015 #2
    I tried starting from second 6 (because it would be at position = 0)

    then the wheel rotates for 4 seconds , and that should be the displacement, is that correct?
     
  4. Mar 27, 2015 #3

    Simon Bridge

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    You mean it was rotating in the negative direction ... maybe the negative direction was "clockwise", we are not told.
    You can do it either way - if you do it the second way, notice that the angular velocity at t=6s will not be zero.
    The area inside 6s<t<10s will give you the total displacement.

    You can also do it by figuring the equation of the line and/or using suvat equations.
    Basically pick one ... if you are uncertain, try two methods and see if they agree.
     
  5. Mar 27, 2015 #4
    I did the first way:

    area of whole triangle = 10 x 30 x .5 = 150

    area of small triangle = 6 x 18 x .5 = 54

    whole triangle minus small triangle = 150 - 54 = 96
    is this what I'm suppose to get?
    is that 96 already in radians?
     
  6. Mar 27, 2015 #5

    NascentOxygen

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    You won't get the right answer.

    I think it is better to go:
    triangle of positive area - triangle of negative area
    7 x 21 x 0.5 - 3 x 9 x 0.5
     
  7. Mar 27, 2015 #6
    yes, your way is correct but i can't figure why the way I calculated mine came out to be wrong.
     
  8. Mar 27, 2015 #7

    NascentOxygen

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    Sketch the graph and attach it here. Geometry will show where you are wrong.
     
  9. Mar 27, 2015 #8
    tried my best to draw out the graphs with only ms paint.

    ioFlMZV.png

    so this is what I was suppose to do, the green triangle should be above the x axis. then subtract the red triangle from the green.

    LWSo7Ai.png

    here is what I did, which was wrong, I took the area of the whole triangle, and subtracted the red triangle. As you can see, the area of green is much bigger.
     
  10. Mar 27, 2015 #9

    jbriggs444

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    Note that the red triangle that you evaluated the area of is the wrong one. You want the area above the slanted line and below the x axis.

    That is a cosmetic error, however, because the area of both those triangles will be equal.
     
  11. Mar 27, 2015 #10

    NascentOxygen

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    Very good.

    So, starting with the area of that monster triangle, subtract the red triangle, then subtract it again, then deduct the rectangular area of green that you don't want, should, in a roundabout way, give the answer, too!

    Much better to sketch the graph right from the start so that you can see what you are doing.
     
  12. Mar 28, 2015 #11

    Simon Bridge

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    Just to be clear...
    The area in "the displacement is the area under the graph" is between the graph and the time axis.
    Thus the triangle inside 0<t<3s has height -9rad/s and base 3s, so it's area is -9(rad/s)x3(s)/2 = -13.9 rad.
    Notice how the units multiply just the same as the numbers do, so (rad/s) times (s) = (rad)?

    Note: since you were not given "the whole triangle" as part of the problem statement, you had to do some working to derive/deduce it.
    You should show that part of your working in your written answer too.
     
  13. Mar 28, 2015 #12

    BvU

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    Dear king of goons,

    I think it's good to remind you of Simon's post #3: once all the artwork is done and you have an answer you feel confident about, you should take the extra step and find the same answer "figuring the equation of the line and/or using suvat equations". It'll give you a good experience to feel the power of abstraction.
     
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