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Particle going in a circle with speed U problem

  1. Jun 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello all,
    I am having some trouble with answering the problem below, mostly because I do not know what the letters stand for and what kind of graph is meant to be drawn. Any help on this would be greatly appreciated. Thanks

    For a particle going in a circle with speed U, draw a graph showing:
    x(t)
    y(t)
    u(t)=dx/dt
    v(t)=dy/dt
    du/dt
    dv/dt


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 26, 2013 #2

    gneill

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    Staff: Mentor

    In your diagram you've indicated x and y axes. x(t) will be the x position with respect to time, y(t) the y position with respect to time. So to begin, draw two more graphs with t as the horizontal axis and place x on the vertical axis of one and y on the vertical axis of the other.

    Clearly u(t) and v(t) are meant to be the x-velocity and y-velocity. du/dt is the derivative of y-velocity with respect to time (what do we usually call the derivative of a velocity?). You'll want to draw similar graphs for these items.
     
  4. Jun 26, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Are you serious? You don't know what the letters stand for? That should have been the first thing you learned! x and y are the coordinates of that particle in some coordinate system. u and v, as is said in the problem are the derivatives (do you know what a derivative is?) of x and y and so are the x and y components of the velocity vector. du/dt and dv/dt are the x and y components of acceleration.

    It helps to know that a circle, with center at (0, 0) and radius R has equation [itex]x^2+ y^2= R^2[/itex]. And, of course, [itex]sin^2(x)+ cos^2(x)= 1[/itex]. That means that parametric equations for the circle are [itex]x= Rcos(\theta)[/itex], [itex]y= R sin(\theta)[/itex]. Now, the circumference of the circle is [itex]2\pi R[/itex] and if the particle is moving at constant speed U, it will complete the circle (moving through an angle of [itex]2\pi[/itex] radians) in time [itex]\frac{2\pi R}{U}[/itex] so has angular speed [itex]\frac{U}{R}[/itex]. That means the angle, at time t, is [itex]\theta= \frac{U}{R}t[/itex].
     
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