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Tangential Speed at the Equator and Beyond

  1. Nov 24, 2007 #1
    [SOLVED] Tangential Speed at the Equator and Beyond

    1. The problem statement, all variables and given/known data
    The earth has a radius of 6.38 X 10^6m and turns on its axis once every 23.9 h. (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. theta) is the tangential speed one third that of a person living in Ecuador?
    (There is a picture with the earth and an axis of rotation through the center. The equator is labeled and an angle (theta) is shown from the axis of rotation and the equator up to a point above the equator making a triangle like the theta = rs triangle.)


    2. Relevant equations
    v=wr


    3. The attempt at a solution
    I solved the first part using v=wr as follows:
    w = 6.28/86040s = 7.3 X 10^-5 rad/s
    v = (7.3 X 10^-5 rad/s)(6.38 X 10^6)
    v = 466 m/s

    I know the speed will be 1/3 of 466m/s in the second part or 310.67m/s.
    I'm just confused how to find the angle theta. I tried using v=wr again and finding the radius at the second part but then that still doesn't help me find the angle.
    Any thoughts?
     
    Last edited: Nov 24, 2007
  2. jcsd
  3. Nov 24, 2007 #2
    Ok, so the cos of the angle I'm looking for would give me the new radius, right?
    So if I can find that radius with v=wr using the data from the first part...
    310.67=(7.3X10^-5)r
    r = 4255753.4 m

    Can I use that to find the angle?
     
  4. Nov 24, 2007 #3
    your r must be

    [tex]r=R\sin\theta[/tex] or [tex]r=R\cos\theta[/tex]

    Look at the angle and you'll see cos or sin.
     
  5. Nov 24, 2007 #4
    Gotcha,
    however I'm trying to find the angle itself. Could I use inverse tan to find the angle?
     
  6. Nov 24, 2007 #5
    Imagine the earth with its axis of rotation drawn in from top to bottom. There is a line for the equator and a line drawn from the axis of rotation on the equator up an angle ([tex]\theta[/tex]) to a point on the edge of the circle (earth). This point is the new area and point where the tangential speed is 1/3 that of the person on the equator. I am trying to find the angle [tex]\theta[/tex].
     
  7. Nov 24, 2007 #6
    It really gives you a triangle where one side is your radius of the earth, the angle[tex]\theta[/tex] and the new radius as the other side....

    The [tex]\theta[/tex] = [tex]\frac{s}{r}[/tex] triangle....

    But I dont know s or [tex]\theta[/tex]

    Ok, I hope someone has some ideas on this one because I'm completely stumped. I've been trying it and re-reading over it for two days now. I understand the connection I just can't understand how to get the angle.
     
    Last edited: Nov 24, 2007
  8. Nov 25, 2007 #7

    Shooting Star

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    Draw the diagram correctly. From the new point where you want to find speed, drop a perp on the axis of rotation. If that perp length is x, then x=(radius of earth)*cos(theta), from the triangle formed.

    Tangential speed = x*w.
     
  9. Nov 25, 2007 #8
    I'm not sure I understand what you mean by "drop a perp".
    I'm not really looking for speed as I know its 1/3 that of the speed at the equator and I have found that speed. What I'm looking for is theta, the angle between the equator at the axis of rotation and the new point further north of the equator (latitude).
     
  10. Nov 25, 2007 #9

    Shooting Star

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    It means, "draw a perpendicular from the point in question to the axis of rotation of the earth".

    From what I've given above, if you know tangential speed, then you can find x. Then you can find cos(theta). And then we'll try to find theta...
     
  11. Nov 25, 2007 #10
    Ok, so if I draw a perp from the new point to the axis of rotation I get a right triangle where x (the perp I drew) = radius of earth * cos(theta)
    Tangential speed = x*w
    Tangential Speed = 1/3 466m/s = 155.33m/s

    So, 155.33 = x * w
    w = angular velocity which I don't know.
     
  12. Nov 25, 2007 #11

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    Do you know how long the earth takes to complete one rotation about its axis?
     
  13. Nov 25, 2007 #12
    Ok, so then...

    [tex]\omega[/tex] = [tex]\frac{2 \pi}{86040sec}[/tex]

    [tex]\omega[/tex] = 7.3 x 10[tex]^{-5}[/tex]

    If so, from the equation before, x = 2127808.2

    Since, x = radius of the earth * cos(theta)
    2127808.2 = 6380000 * cos(theta) ?
     
  14. Nov 25, 2007 #13

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    So find theta. A day has 86400 secs, not 86040.

    Let me show you how this is actually supposed to be done. Let x be as above, that is, the dist of the pt from the axis. Let v be the speed at the equator, and v1 is the speed at latitude theta. You have to find theta for which v1 = v/3.

    v = rw and v1 = xw =>
    v1/v = x/r = cos(theta) [from diagram] =>
    1/3 = cos(theta) => theta = 70.5 deg.

    This is the same as you've got.
     
  15. Nov 25, 2007 #14
    I see that you got the right answer, but I'm still not sure how.
    From my last post, cos(theta) = .334

    I understand in your post where v = rw and v1 = xw
    I'm confused at v1/v = x/r = cos(theta).....
     
    Last edited: Nov 25, 2007
  16. Nov 27, 2007 #15

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    Your ans is good enough.

    If v = rw and v1 = xw, then dividing 2nd eqn by 1st, you get v1/v = x/r.
    The value of x/r is also cos(theta) from the diagram of the right angled triangle.
    It's given that v1/v = 1/3.

    So, cos(theta) = 1/3.
     
  17. Nov 27, 2007 #16
    Ok, I see....thank you again!
     
  18. Oct 13, 2008 #17
    Re: [SOLVED] Tangential Speed at the Equator and Beyond

    Where did that 86040seconds come from?
     
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