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Drunken student (circula motion)

  1. Sep 3, 2006 #1
    Hello there!
    First, I really want to thank you guys for helping me out with my physic problems. I would not have been able to make such a good progress I have done in the last couples of days!
    I started a new chapter, it is about circula motion and I am having a hard time with this one:

    A drunken student fell as sleep on a bench. When he is finally waking up, he can see a street light which is 50 m away from him. He is getting up and is orbiting the street light with a constant angular speed of [tex] 0,03 s^{-1} [/tex] on a helix curve which is getting tighter and tighter. The helix curve has the form of
    [tex] s_\varphi=(R-k\varphi)(\cos\varphi \\ \sin\varphi) [/tex] (the cos and sin should be vektor components for x and y but I do not know how to do that with latex)

    Where R is the distance between the bench and the street light, [tex] \varphi [/tex] is the angle between the direct distance of the student and the street light and k is a constant of 2m.

    How great is the distance between the student and the bench after 6 min he stood up ?


    Here are my thoughts:
    To get the angle I multiply the angular speed with the time:
    [tex] \varphi= \omegat = 0,03(6*60) = 10.8 [/tex]

    I thought using that angle to get the x and y components of the student's position (zero of the coordinate system is in the center of the street light) with which I can get the modulus would be the solution but I never got the 62,116 m.
    I am aware that I have to consider the distance of the bench to the street light as well to get the over all distance between the bench and the student.

    Does anyone out there have a hint for me?
    Many thanks!
     
    Last edited: Sep 3, 2006
  2. jcsd
  3. Sep 4, 2006 #2

    andrevdh

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    With your suggestion I get 28.4 meters (which seems reasonable since he has completed only 1.7 revolutions). 62.116 meters is farther away than the initial 50 meters, which does not make sense!
     
    Last edited: Sep 4, 2006
  4. Sep 4, 2006 #3

    Kurdt

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    If the student is revolving round the lamp post then he can be further than 50 meters away when hes on the far side of a revolution. As to how to solve this I'm not sure at the minute. What is that angular speed? Radians per second or degrees per second?
     
  5. Sep 4, 2006 #4

    andrevdh

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    The helix is spiralling inwards as he circles the lamp post. This is also demonstrated by the equation. Fara's suggestion as to how to solve it works quite fine and I am quite sure that is how it was intended.
     
    Last edited: Sep 4, 2006
  6. Sep 4, 2006 #5

    Kurdt

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    I got it now. ok you know the guy has done ~1.7 revolutions so you know the angle from his starting position. You also know that after 1.7 revolutions he is 24.8 m from the lamp post. so using the lamp and the bench and the position after 6 minutes as reference points construct a triangle. You can work out one angle and you know that two sides are 24.8m and 50m in length and then using the cosine rule its easy to deduce the guys distance from the bench.

    Here are some useful cosine rule identities.

    http://www.acts.tinet.ie/cosinerule_673.html
     
  7. Sep 4, 2006 #6
    wow !
    With your help Kurdt I finally figured it out!
    The angular speed was in radians per second by the way. One mistake I did was that I totally forgot to calculate with radians instead of degrees. Another mistake I did was that I had R as a negative distance, which actually has to be + 50m. And thanks for the link to that page! It helped me to refresh my knowledge of geometry a lot.

    Have a good labor day!
     
  8. Sep 4, 2006 #7

    Kurdt

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    Glad I could help. It had me confused at first because I thought there was probably a more mathematical andmuch more complex way of doing it but once I dropped that idea it clicked. No labour day in the UK unfortunately but I'll have a good one none the less.
     
  9. Sep 4, 2006 #8

    andrevdh

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    The x-coordinate of the spiral is given by

    [tex]x = (R-k\varphi)\cos(\varphi)[/tex]

    and the y-coordinate

    [tex]y = (R-k\varphi)\sin(\varphi)[/tex]

    from which the distance the student is from the bench can be calculated with pythagoras as fara suggested.

    It is actually great fun to plot the spiral in Excel.
     
  10. Sep 4, 2006 #9
    @ andrevdh

    how can you plot that in Excel ? I did not know that I can to mathematical functions as well
     
  11. Sep 5, 2006 #10

    andrevdh

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