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Dynamics - Normal and Tangential Coordinates

  1. Sep 12, 2007 #1
    1. The problem statement, all variables and given/known data
    An outdoor track is full circle of diameter 130 meters. A runner starts from rest and reaches her maximum speed in 4 seconds with constant tangential acceleration and then maintains that speed until she completes the circle with a total time of 54 seconds. Determine the magnitude of the maximum total acceleration of the runner


    2. Relevant equations
    [tex]a_t = \frac{dv}{dt}[/tex]
    [tex]a_n = \frac{(v_t)^2}{r}[/tex]
    [tex]a = \sqrt{(a_t)^2+(a_n)^2}[/tex]


    3. The attempt at a solution
    Ugh.. I attempted this problem for about thirty minutes and realized I needed some help. Can someone give me some guidelines? If I am able to find the maximum tangential velocity somehow, then I think I can do the problem, but I just cannot find it.
     
    Last edited: Sep 13, 2007
  2. jcsd
  3. Sep 13, 2007 #2

    Dick

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    The runner accelerates at constant rate for 4 sec, hence speed at the end of that time is V=a*(4 sec). Then travels at speed V for 50 sec. How far did the runner travel during the 4 sec, acceleration period (in terms of a) and how far during the 50 sec constant speed period (in terms of a). if you add them the total is 130m. Solve for a.
     
  4. Sep 13, 2007 #3
    Ok so for the first 4 seconds

    [tex]
    s = \frac{1}{2}at^2 = 16a
    [/tex]

    is the distance.

    and from 4 seconds to 54 seconds

    [tex]
    s = s_0 + v_0t + \frac{1}{2}at^2
    [/tex]

    Acceleration is equal to 0 at this point so

    [tex]
    s = 16a + 4a * 50 = 216a
    [/tex]

    [tex]
    216a = 130m
    [/tex]

    a = 0.602 m/s^2

    Is this really the total acceleration? What about the normal component of the acceleration? That needs to be accounted as well, doesn't it? If that is the case, does this acceleration represent the tangential acceleration?
     
  5. Sep 13, 2007 #4

    Dick

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    (1/2)*a*t^2 with t=4 doesn't give you 16a. And, yes, everything so far is tangential. Use the formulas you've quoted to handle the normal part. Notice that the magnitude of the normal acceleration varies as the runner speeds up.
     
  6. Sep 13, 2007 #5
    The equation used to find the normal acceleration is

    [tex]
    a_n(t) = \frac{(v_t(t))^2}{r}
    [/tex]

    But the velocity is increasing only until 4 seconds. Until then, the equation is

    [tex]
    v_t(t) = a_t*t
    [/tex]

    After then, the velocity remains constant.

    How do I associate this fact with the normal acceleration equation? I can't possibly just plug it in. Do I define a step function?
     
    Last edited: Sep 13, 2007
  7. Sep 13, 2007 #6

    Dick

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    The normal component of the acceleration increases in magnitude for the first 4 sec, then remains constant. The tangential component remains constant for the first 4 sec, and then drops to zero. At what point in time do you think the total magnitude will be greatest? Attention to the error in computing (1/2*a*t^2!!!!
     
  8. Sep 13, 2007 #7
    Wait, if that is the case, we didn't even need tangential velocity in terms of time, since the greatest normal acceleration occurs on the greatest tangential velocity, which is t = 4 seconds. Can you check this work for me?


    [tex]
    s = \frac{1}{2}at^2 = 8a
    [/tex]

    [tex]
    s = s_0 + v_0t + \frac{1}{2}at^2
    [/tex]

    [tex]
    s = 8a + 4a * 50 = 208a
    [/tex]

    [tex]
    208a = 130m
    [/tex]

    [tex]
    a_t = 0.625 m/s^2
    [/tex]

    [tex]
    a_n = \frac{v^2_t(t)}{r}
    [/tex]


    [tex]
    a_n = \frac{(0.625 * 4)^2}{130}
    [/tex]

    [tex]
    a_n = 0.048 m/s^2
    [/tex]

    [tex]
    a = \sqrt{(0.625^2)+(0.048^2)}
    [/tex]

    [tex]
    a= 0.626 m/s^2
    [/tex]
     
  9. Sep 13, 2007 #8

    Dick

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    OOOPS. Hold it. 130m is the diameter, so it's NOT the circumference and it's NOT the radius. I've been being sloppy. Can you correct those numbers???? Sorry.
     
  10. Sep 13, 2007 #9
    Ah noticed it too


    [tex]
    s = \frac{1}{2}at^2 = 8a
    [/tex]

    [tex]
    s = s_0 + v_0t + \frac{1}{2}at^2
    [/tex]

    [tex]
    s = 8a + 4a * 50 = 208a
    [/tex]

    [tex]
    208a = 130 * 3.14
    [/tex]

    [tex]
    a_t = 1.9625 m/s^2
    [/tex]

    [tex]
    a_n = \frac{v^2_t(t)}{r}
    [/tex]


    [tex]
    a_n = \frac{(1.9625 * 4)^2}{65}
    [/tex]

    [tex]
    a_n = 0.483m/s^2
    [/tex]

    [tex]
    a = \sqrt{(1.9625^2)+(0.483^2)}
    [/tex]

    [tex]
    a= 1.964m/s^2
    [/tex]
     
  11. Sep 13, 2007 #10

    Dick

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    I don't like the a_n calculation. It looks like you should get a number near 1.
     
  12. Sep 13, 2007 #11
    Ahhh I don't like it either. I will never trust windows calculators from now on. Microsoft should just burn in hell.



    [tex]
    s = \frac{1}{2}at^2 = 8a
    [/tex]

    [tex]
    s = s_0 + v_0t + \frac{1}{2}at^2
    [/tex]

    [tex]
    s = 8a + 4a * 50 = 208a
    [/tex]

    [tex]
    208a = 130 * 3.14
    [/tex]

    [tex]
    a_t = 1.9625 m/s^2
    [/tex]

    [tex]
    a_n = \frac{v^2_t(t)}{r}
    [/tex]


    [tex]
    a_n = \frac{(1.9625 * 4)^2}{65}
    [/tex]

    [tex]
    a_n = 0.948m/s^2
    [/tex]

    [tex]
    a = \sqrt{(1.9625^2)+(0.948^2)}
    [/tex]

    [tex]
    a= 2.18m/s^2
    [/tex]
     
  13. Sep 13, 2007 #12

    Dick

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    Looks ok now. Finally.
     
  14. Sep 13, 2007 #13
    PROBLEM SOLVED! OMFGWTFBBQ! THANKS!
     
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