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Circular Motion of the earth

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data

    The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles. a.) Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees. b.) If the flight from Kampala to Singapore takes 12 hours, what is the plane's angular speed relative to the earth? (in rad/s)

    2. Relevant equations



    3. The attempt at a solution

    I tried drawing a right triangle to find the angle. I assumed that it would be the cos of 5000/4000, but that didn't work.
     
  2. jcsd
  3. Oct 5, 2007 #2

    learningphysics

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    You have 3 points... the center of the earth... and the two cities...

    So these 3 points form a triangle... what are the lengths of the 3 sides?

    which is the angle you need... draw a picture...
     
  4. Oct 5, 2007 #3
    The center of the earth is the point from which both lines exit. There would be a straight horizontal line coming from the earth, which would be 5000. Then there would be another line extending somewhat horizontal and somewhat vertical, which would be 4000.
     
  5. Oct 5, 2007 #4

    learningphysics

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    you should have 3 lines... what is the distance from the center to kampala? what is the distance from the center to singapore? what is the distance from kampala to singapore?
     
  6. Oct 5, 2007 #5
    The distance from Kampala to Singapore is 5000 miles. The distance from the center to Kampala is 4000 miles, and the distance from the center to Singapore is the same.
     
  7. Oct 5, 2007 #6

    learningphysics

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    cool. now you need the angle between the 2 4000 mile sides... you can use the cosine rule to get this angle.
     
  8. Oct 5, 2007 #7
    So it would be the cos(4000/4000)?
     
  9. Oct 5, 2007 #8

    learningphysics

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  10. Oct 5, 2007 #9
    I looked that up before and found cos(C) = a^2 + b^2 - c^2 / 2ab. I got .219 rad, which was wrong.
     
  11. Oct 5, 2007 #10

    learningphysics

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    I'm not getting that... just to be sure this is the formula you need to be using:

    [tex]cos(C) = \frac{a^2+b^2 -c^2}{2ab}[/tex]

    what did you use for a and b? what did you use for c?
     
  12. Oct 5, 2007 #11
    a and b would be 4000 and c would be 5000.
     
  13. Oct 5, 2007 #12

    learningphysics

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    I see what happened. cos(C) = 0.219. you forgot to take the arccos...
     
  14. Oct 5, 2007 #13
    So would it be the inverse of cos which is 1.35?
     
  15. Oct 5, 2007 #14

    learningphysics

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    yes. 1.35 radians.
     
  16. Oct 5, 2007 #15
    I tried that and it said it's wrong.
     
  17. Oct 5, 2007 #16

    learningphysics

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    Hmm... I don't understand it... did the question want radians or degrees?
     
  18. Oct 5, 2007 #17
    It wants both.
     
  19. Oct 5, 2007 #18

    learningphysics

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    Ah... I was considering the distance between the two places as a straight line... maybe the 5000 miles is supposed to be the curved distance... in which case 5000 = R*theta, so theta = 5000/4000 = 1.25 radians...
     
  20. Oct 5, 2007 #19
    Force acting on a Skier

    can you guys assist me solve this.
    Determine the normal force acting on the skier at point A, where his velocity is 20 m/s, weigth = 60kg and the radius of curvature, p is 30m.

    The available options as answers are:

    A. 1368N
    B. 1857N
    C. 1386N
    D. 1636N


    I have worked with the formula

    F = MV2/R

    i.e F = (60 * 20*20)/30 = 800N

    as u can see this doesn't correspond with any of the options above
     
  21. Oct 5, 2007 #20

    Astronuc

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    A separate thread would be appropriate for this problem.

    I presume there is a figure associated with this problem. The net force will depend on the orientation of the trajectory. In addition to a centripetal force, there is also the weight of the skier, W = mg. Weight and centripetal force are vectors and would be additive to obtain the net force.
     
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