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Graphing rotation & linear quantites

  1. Oct 27, 2006 #1
    While working on her bike, Amanita turns it upside down and gives the front wheel a counterclockwise spin. It spins at approximately constant speed for a few seconds. During this portion of the motion, she records the x and y positions and velocities, as well as the angular position and angular velocity, for the point on the rim designated by the yellow-orange dot in the figure. View Figure Let the origin of the coordinate system be at the center of the wheel, the positive x direction to the right, the positive y position up, and the positive angular position counterclockwise. The graphs View Figure begin when the point is at the indicated position.
    [​IMG]

    Which of the graphs corresponds to x position versus time?
    -> would this be a [ C ] because the position doesn't move as time passes by? Or is this different for rotational?

    Which of the graphs corresponds to angular position versus time?
    Would this be [ E ] since it starts with position 0, then it goes to a positive position, then negatice, to posive etc?

    Which of the graphs corresponds to y velocity versus time?
    What do they mean by y velocity? (verticle velocity)
    would the verticle velocity do negative to positive and back to negative again? Is this possible? so graph [ G ] ?

    Which of the following graphs corresponds to angular velocity versus time?
    I have no idea...

    Thanks :approve:
     
    Last edited: Oct 27, 2006
  2. jcsd
  3. Oct 27, 2006 #2

    radou

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    The velocity vector is tangent to the circle, and you can decompose it to x and y components, so I guess that's what was meant ba x and y velocity.
     
  4. Oct 27, 2006 #3
    What about the other responses? I'm not too 100% sure... would there be a point when velocty y would be 0? so it must be one of the sin or cos graphcs?
     
  5. Oct 27, 2006 #4

    radou

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    Think about it this way: fix your eyes on a point on the rotating circle. The velocity vector will always be tangential to the circle, so, at which point will the y-component of velocity vanish? (Hint: the two points at which there is only the x-component.)
     
  6. Oct 27, 2006 #5
    Would it be [ E ]? the velocity can have a negative value? (since Y is negative) but why isn't it a straight line? isn't the velocity constant?

    Edit: E was wrong, but the wheel starts rotating at position 15 minutes on a clock.. if you can visualize that :)
     
    Last edited: Oct 27, 2006
  7. Oct 27, 2006 #6

    radou

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    Which of the graphs corresponds to angular position versus time?

    Okay, angular position is given (for constant angular velocity, as stated) , by [tex]\phi(t)=\omega\cdot t[/tex], where [tex]\omega[/tex] is angular velocity, so it's a linear function.

    Which of the following graphs corresponds to angular velocity versus time?

    Well, we said that angular velocity is constant, right? :smile:
     
  8. Oct 27, 2006 #7

    Since they're linear ('cause they're constant) is it the graph C? (positive, but what about the linear graphics that is negative, decreasing and increasing, those are confusing me)
     
  9. Oct 27, 2006 #8

    radou

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    Well, that depends if you defined counter clockwise (I think you said that was the direction of rotation) as positive or negative.
     
  10. Oct 27, 2006 #9
    since its counterclockwise it would go to positive (so increasing) but then in reaches right back down to negative.. so non of the line graphics work?
     
  11. Oct 27, 2006 #10

    radou

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    It does not reach anything negative. Angular speed is constant. If you accepted counter clockwise as positive, then it's the graph on the positive side.
     
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