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Math Help - How to calculate the volume of partial cylinder?

  1. #1
    Newbie JDwg's Avatar
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    How to calculate the volume of partial cylinder?

    Hello from Finland!
    I have a problem how to calculate the volume and the mass centroid of a partial cylinder. Please look at the image below. I have listed the given information there. Variable for a is: a=0...2R

    Ive tried to solve it, but I dont know how to do it... All help is appreciated! Thanks!
    -Janne
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  2. #2
    Newbie JDwg's Avatar
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    Solution when a=R

    I found that the solution for a=R is:
    V = 2/3*R^2*b
    c = R-3/16*PI*R
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  3. #3
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    Hyvää on! Miten selvisi?
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  4. #4
    Newbie JDwg's Avatar
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    Quote Originally Posted by TKHunny View Post
    Hyvää on! Miten selvisi?
    Solution for a=R I found from a Finnish tablebook called "Rakentajain Kalenteri 2004".
    But I need to also find solutions when a is not R...
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  5. #5
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    I don't understand the drawing.

    Is the height 'b' or something greater than 'b'?

    Are we using Calculus or only Geometry?
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  6. #6
    Newbie JDwg's Avatar
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    Clarification for b

    Quote Originally Posted by TKHunny View Post
    I don't understand the drawing.

    Is the height 'b' or something greater than 'b'?

    Are we using Calculus or only Geometry?
    Hi Tk

    Height is b. I clarified the drawing below.

    Basically Im hoping for an solution for V(olume) and mass (c)entroid in terms of b, a and R.
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  7. #7
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    Very good. Unfirtunately, I am not at home and have only limited PC access. Perhaps another able source can help you with this before I get around to in on Thursday. Sorry.
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  8. #8
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    Definitions and Orrientation:

    I set it up so that:
    1) The y-axis is parallel to the cut.
    2) The x-axis is the perpendicular bisector of the cut.
    3) The z-axis is parallel to 'b'.
    4) The entire structure is in the four quadrants were z > 0.
    5) The Origin of the x-y plane is at the center of the circle from the top view.

    This has two side effects.
    1) The centroid for the y-axis is obviously zero (0), due to symmetry.
    2) The equation of the cut-plane is (z-0) = (-b/a)(x-R)
    3) The centroid for the x-axis should be < 0. Intuitively clear.

    The volume, then, would be:

    2*\int_{-R}^{R-a}\int_{0}^{\sqrt{R^{2}-x^{2}}}\int_{0}^{b}\;dz\;dy\;dx + 2*\int_{R-a}^{R}\int_{0}^{\sqrt{R^{2}-x^{2}}}\int_{0}^{\frac{-b}{a}(x-R)}\;dz\;dy\;dx

    Let's see if you can get that far or if we need a different kind of solution.
    Last edited by TKHunny; October 29th 2009 at 07:50 PM. Reason: I came to my senses.
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  9. #9
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    Reality Check:

    The Volume is easier than that, ceratinly for a = R, but we'll need this machinery for the Centroid.

    For a = R

    1) Uncut semi-circular region x < 0

    \frac{1}{2}\cdot \pi R^{2}b

    2) Cut region, x > 0

    \frac{1}{3}\cdot \frac{1}{2}\cdot \pi R^{2} b

    Sure enough, this matches your answer from the book.
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  10. #10
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    Are you still with me?

    In THIS case, since we have symmetry in 'y', we can ignore it and just work with z and x. This makes it rather simple, since the x-z cross section is a simple trapezoid.

    M = \frac{R+(R-a) + 2R}{2} \cdot b

    M_{x} = \int_{-R}^{R-a}\int_{0}^{b}\;x\;dz\;dx + \int_{R-a}^{R}\int_{0}^{-\frac{b}{R-a}\cdot (x-R)}\;x\;dz\;dx

    Can you do M_{z}?
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