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Angular and Tangential Variables

  1. Nov 6, 2004 #1
    I have one last homework question that I am stuck on. I am attaching the picture.

    The earth has a radius of 6.38x10^6m and turns on its axis once every 23.9 hours. a) what is the tangential speed (in m/s) of a person living in Ecuador, a country that lies on the equator? b) at what latitude (i.e. the angle(theta) in the drawing and in degrees) is the tangential speed one-third of a person living in Ecuador

    I solved part a, no problem, but part b??????
    part a:
    converted time from hours to seconds 86040s
    omega(angular speed)= 6.28rad/86040s = 7.3x10^-5 rad/s
    Vt=r(omega)= 466m/s

    but I am lost on b and I have been working on it for hours!!!

    Attached Files:

  2. jcsd
  3. Nov 6, 2004 #2
    The centripetal force is the same no matter where you are on the earth.
    Now what has changed from the equator to this new latitude? The radius of the path travelled by the person. On the equator, it is simply the radius of the earth, but since the earth is spinning on a vertical axis as shown in the diagram, the new radius of the person's path is now his distance from the vertical axis (not the radius of the earth). Substitute into the equations to find that distance, which should allow you to find the angle theta.
  4. Nov 6, 2004 #3
    I am not sure I understand what you mean. Why are you using a Force equation when it is asking for tangential speed. Do I need to take 1/3 from the intial tangential speed I found?
  5. Nov 6, 2004 #4
    I get almost 90°. That can't be right...hmm, I think I assumed incorrectly that centripetal force is constant. Allow me to rethink. Maybe someone else can chime in and help out in the mean time.
  6. Nov 6, 2004 #5
    The answer in the back of the book is 70.6 degrees and I still am not sure why we are using centripetal force equations?
  7. Nov 6, 2004 #6


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    Sirus is suggesting that you look at the size of the circular path a person travels in a day as a function of the latitude. I think it's a great suggestion! :-)
  8. Nov 6, 2004 #7


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    You only need to consider the speed and latitude. The question has nothing to do with forces.
  9. Nov 6, 2004 #8
    I'm sorry, I am not sure why I don't understand. I have 12 pieces of scratch paper and I keep going around and circles! I just can't figure out what equations to use!
  10. Nov 6, 2004 #9


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    Take a deep breath. You basically have a uniformly rotating sphere. A point on the surface of the sphere travels in a circle of the course of a single rotation. The size of the cirlce the point travels depends on the latitude of the point. A point at the equator travels in a large circle and points near the poles travel in small circles.

    Use basic geometry/trigonometry to find the radius of the circle as a function of latitude. If you know the radius then you can calculate the circumference of the circle which is the distance the point travels in a full day. Divide the distance travelled by the time and you have the speed.
  11. Nov 6, 2004 #10
    Try to figure out the (horizontal, in the picture) distance from the axis of rotation to any point on the Earth's surface by using trig. Then, since the Earth is rotating with a certain angular velocity w, you can get v from v=wr, where r- distance from the axis of rotation. (The angular velocity w is the number of radians the Earth turns through in a second; you can calculate it from the period of the Earth and the fact that it's going to turn though 360 deg, or 2pi radians, in one period).
  12. Nov 6, 2004 #11
    I just don't understand what you mean by "as a function of the latitude" isn't that what I am looking for? Where does the 1/3 part come in? I have been trying a^2+b^2=c^2, but I don't think I am using the right numbers or if that is even what I should be using
  13. Nov 6, 2004 #12
    well, you can express the distance from the axis of rotation of a point on the surface of Earth as a function of the lattitude. then you can express the tangential velocity in terms of this distance. then you can set the velocity to one third of that in ecuador and see what value lattitude must have in order for that equation to be satisfied. suggestion: use trigonometry instead of pythagoras' theorem. (edit: that is, use sines and cosines. pythagoras' theorem is, of course, also trigonometry in a sense :redface: )
  14. Nov 6, 2004 #13
    Yes, that seems like the correct approach. Pinky2468, I apologize for having confused you earlier. I was completely wrong in my approach. Forget about forces.
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