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Using Triple integrals to solve torque around a point.

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    A cylindrical coffee cup (8 cm in diameter and 10 cm tall) is filled to the brim
    with coffee. Neglecting the weight of the cup, determine the torque at the handle
    (2 cm from edge of cup 5 cm up from bottom of cup).

    The easy way would be to just use the center of mass of the cup but that isnt the way its told to be done. So triple integrals it is.

    density of water(ρ) = 1g/cm3


    2. Relevant equations
    T = r x F

    if R is from


    3. The attempt at a solution
    So to set up my coordinate's I put the origin in the center of the cylinder with x going out of the page y to the right and z up.


    T = r x F
    = r x mFg (Fg = force of gravity)
    dT = r x dmFg (dm = ρdxdydz)

    [itex]\int[/itex] [itex]\int[/itex][itex]\int[/itex] |r|ρgdzdydx

    I think (tell me if I'm wrong) the bounds of the first integral (from left to right):
    -4 to 4, -[itex]\sqrt{}[/itex]4-x2 to [itex]\sqrt{}[/itex]4-x2, -5 to 5.

    but where I'm having problem is finding an equation that relates r ( the vector from the handle to the cup) for each dm. If someone could help me out that would be great.
     
    Last edited: Oct 4, 2011
  2. jcsd
  3. Oct 5, 2011 #2
    I don't think you have to integrate anything (but then I have been drinking). You are told to ignore the mass of the cup. Assume the total mass of the water acts at it geometric center.

    Torque = r X F = [8/2 +2]*m*g where m = ρ*V

    ??
     
  4. Oct 5, 2011 #3

    ehild

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    The force of gravity is mg. Do not multiply with the mass again.

    As g is vertical (parallel to the z axis), the vector product [itex]\vec r \times dm \vec{g} = -g (y \hat x -x \hat y)dm[/itex]

    ehild
     
  5. Oct 5, 2011 #4
    Now how am I supposed to work with that. It makes sense but now I'm integrating a vector field and the only thing i can think to do is take the magnitude of that. Shouldn't it be different because the of symmetry.

    Plus i tried taking the magnitude of that field and it was not fun by the time i had to do the second integral.
     
    Last edited: Oct 6, 2011
  6. Oct 6, 2011 #5

    ehild

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    The torque is a vector, has both x and y components, depending on the orientation of the coordinate system. You can find its magnitude after having the components, but the magnitudes do not add.

    ehild
     
  7. Oct 6, 2011 #6
    So how would i go about solving the problem from this point on. I mean i get a torque vector and obviously there is an answer (0.296 N * m) but I don't see how we can get from the vector field to this answer.
     
  8. Oct 6, 2011 #7

    ehild

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    How do you set up your coordinate system? Show a picture. How do you calculate the torque vector and what is the result? If you have the torgue vector with components Tx, Ty, the magnitude is sqrt(Tx2+Ty2).

    ehild
     
    Last edited: Oct 6, 2011
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