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Circular Motion and angle alpha

  1. Dec 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A plumb bob is deflected from the vertical by an angle alpha due to a massive mountain nearby. (A) Find an approximate formula for alpha in terms of the mass of the mountain the distance to its center and the radius and mass of the earth
    (B) Make a rough estimate of the mass of mt everest assuming it has the shape say of an equilateral pyramid or cone 4k m above its base and then (c) estimate the angle alpha of the pendulum bob if it is 5 km from the center of mt everest

    2. Relevant equations

    F= G*m1*m2/r^2

    G = 6.67*10^-11

    shoot idk throw some vector equations in there too

    3. The attempt at a solution

    well...i said
    Mm = mass of mountain
    m = mass of bob
    Dm = Distance to mountain from bob
    Fm = force exerted by mountain on bob
    Me = Mass of earth
    Re = radius of earth

    Fm = G*Mm*m/(Dm)^2
    g = G*Me/(Re)^2

    i made a right triangle with vector mg, vector Fm and hypotnuse of the deflected bob string

    using trig i got alpha = 90 - theta where tan(theta) = mg/Fm

    I know this can't be anywhere near to right...and i dont want to attempt other parts before i get this...PLEASE HELP!!!!!!
    Last edited: Dec 10, 2007
  2. jcsd
  3. Dec 10, 2007 #2


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    That's right, if you really mean alpha=90-theta where tan(theta)=mg/Fm. Then tan(alpha)=Fm/mg.
  4. Dec 10, 2007 #3
    Ok now can anyone help me figure out B....that is if i got A correct. I really honestly have no idea what they want me to do for B. Idk if im supposed to look up values for this or what (i can find a rough volume of it i suppose but its not like i know the base or anything) I am certain there is NO more info on the question

    PLZ HELP!!!
  5. Dec 10, 2007 #4


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    It says 'rough estimate'. This means 'order of magnitude'. You could say the base is a square 4km on a side, you could say it's a circle 3.5km in radius, it doesn't matter. They will only pay attention to the exponent in the angle (which will be REALLY small). You'll also need to know the density but you don't know the density either. Guess. Probably about the same density as any rock you can think of. Or the density of the earth, or even the density of water. You'll still be in the same ballpark.
  6. Dec 10, 2007 #5
    well that problem was officially boring. Thank you very much Dick!
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