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Homework Help: I Am STUCk! pls enlighten me

  1. Sep 10, 2004 #1
    Hi Guys,

    I tried drawing a free body diagram, But the term 34.5 north latitude is getting on my nerves... the closest feeling i got wat it meant was the angle made between the Weight of the bob(line of action drawn Diagonally) and the horizontal.

    Q. A plumb bob does not hang exactly along a line directed to the center of the Earth, because of the Earth's rotation. How much does the plumb bob deviate from a radial line at 34.5° north latitude? Assume that the Earth is perfectly spherical and of radius 6.40 x 106m.

    i understand there is a formula for the deviation but wifout the formula is there any other way to solve it? Shd i try resolving the F component using the angle 34.5.. Wat equations?? PLS help! :confused: :bugeye:
     
  2. jcsd
  3. Sep 10, 2004 #2

    Tide

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    If [itex]\hat i[/itex] is up and [itex]\hat j[/itex] is horizontal (pointed toward local south) then the effective acceleration due to gravity accounting for the rotation is [itex]\vec a = (-g + \Omega ^2 R \cos \lambda) \hat i + \Omega ^2 R \sin \lambda \hat j[/itex] where [itex]\lambda[/itex] is the latitude and [itex]\Omega[/itex] is the angular rotation rate.
     
  4. Sep 10, 2004 #3
    I understand that u have resolved the components into the horizontal and vertical axis, but still how do u apply this equation to get the deviation with tension unknown... although it would be canceled eventually... Any more takers?
     
  5. Sep 10, 2004 #4

    Tide

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    All you need to do is determine by what angle the total vector I gave you deviates from the ordinary grav acceleration vector.
     
  6. Sep 11, 2004 #5
    Hmmm sorry but how do u do that... i jus found out that the 2 equations to resolve the three vectors into their vertical and horizontal components are:

    [itex]\T \cos \phi + m \Omega ^2 R \cos^2 \theta = mg [/itex] ---> vertical forces

    [itex]\T \sin \phi = m \Omega ^2 R \cos \theta \sin \theta [/itex] ----> horizontally

    where phi is the deviation to be found....
    BUT i DUn understand why it is cos^2 and not cos, and for the bottom equation, there is an additional cos theta.
     
  7. Sep 11, 2004 #6

    Tide

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    You can find the angle between two vectors [itex]\vec A[/itex] and [itex]\vec B[/itex] from [itex]\vec A \cdot \vec B = A B \cos \phi[/itex].
     
  8. Sep 11, 2004 #7
    I am totally confused by you now that u have introduced the dot product and u oni started with an acceleration vector... Oh mAn... Sorry to bother u... can u Show me at least half of the full working towards getting the deviation...
     
  9. Sep 11, 2004 #8

    Tide

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    Sorry - didn't mean to confuse you. Here's a simpler way:

    Look at the vector I gave you in #2. Divide the [itex]\hat j[/itex] component by the [itex]\hat i[/itex] component. The result is the (negative) tangent of the deflection angle so:

    [tex]\tan \phi = \frac {\Omega ^2 R \sin \lambda}{g - \Omega ^2 R \cos \lambda}[/tex]
     
  10. Sep 11, 2004 #9
    Ok based on ur equation, when i subs in all the values of the question:

    [tex]\tan \phi = 0.001959 [/tex]
    [tex]\phi = 6.43 \deg [/tex] ---> which is still considerably bigger

    Actual ans is 0.0925 deg..
    Is there any link u can establish with ur equation and the 2 equations i gave you?
     
  11. Sep 11, 2004 #10

    Tide

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    I think you made an error in your calculations. I get 0.14 degrees from my equation but I realize that in my original equations R should really be the distance from the axis of your location which, in your example is the radius of the Earth divided by [itex]\sqrt 2[/itex] in which case I get 0.1 deg. We may have used slightly different numbers for the radius of the Earth and for g.
     
  12. Sep 12, 2004 #11
    Ehh yupz...i agree tat it is 0.11196 degrees, but in fact the ans shd be 0.0925 degrees, my mistake for that 6.43.... i found out that the angle of deviation is given by a formula:

    [tex] \phi= \frac {\Omega ^2 R \sin \lambda \cos \lambda}{g - \Omega ^2 R \cos^2 \lambda} [/tex] , where [itex]\lambda[/itex] is the latitude and [itex]\Omega[/itex]
    is the angular rotation rate. g=9.81, and Earth radius=6.4 x 10^6. and [itex] /phi [/itex] is in radian form.

    SomeHow the oni thing that puzzles me is the additional [itex] \cos \lambda [/itex] on the top and bottom.
     
  13. Sep 12, 2004 #12

    Tide

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    One [itex]\cos \lambda[/itex] is from breaking the centripetal acceleration into horizontal and vertical components and the other is from the fact that the point in question travels about a circle of radius [itex]R \cos \lambda[/itex] as the Earth rotates.

    I'm only guessing here but the number we got and "the answer" are sufficiently close that I suspect "the answer" was obtained using slightly different numbers than we used - e.g. someone might have used g = 10 m/s^2 instead of 9.81 m/s^2 or they might have ignored the small term in the denominator. If you find out do let us know!
     
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