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Thread: Volume of falling sphere below horizontal line as sphere falls past line

  1. #1

    Volume of falling sphere below horizontal line as sphere falls past line

    A sphere of radius 10 cm sits on a straight horizontal line. It then drops at a constant rate of 1 cm/s until the entire sphere is below the line, after 20s. Could anybody tell me a formula for how to calculate the volume of the sphere that is below the line at any time < 10s? Volume of a sphere = (4/3) × pi × r3

    I'd be grateful for any help.

  2. #2
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    Quote Originally Posted by Gorgiewave View Post
    A sphere of radius 10 cm sits on a straight horizontal line. It then drops at a constant rate of 1 cm/s until the entire sphere is below the line, after 20s. Could anybody tell me a formula for how to calculate the volume of the sphere that is below the line at any time < 10s? Volume of a sphere = (4/3) × pi × r3

    I'd be grateful for any help.
    What are your thoughts?

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  3. #3
    My thoughts are that there will need to be an integral with an infinite number of infinitely fine horizontal sections. The curve of the sphere is constant.

    The graph of the second derivative of volume under line against time will have a maximum point when the sphere is halfway down.

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    You need some preliminary thoughts.

    "sits on a straight line" -- So 100% of the sphere is above the line?

    "below the line" -- Is the sphere allowed to droop below ground level?

    Perhaps a drawing and a better description? You will need dr/dV, I would think.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Gorgiewave View Post
    A sphere of radius 10 cm sits on a straight horizontal line. It then drops at a constant rate of 1 cm/s until the entire sphere is below the line, after 20s.

    Could anybody tell me a formula for how to calculate the volume of the sphere that is below the line at any time < 10s? Volume of a sphere = (4/3) × pi × r3
    I think the search term for the sort of partial sphere thing you're considering is "spherical cap". Perhaps relate that sort of formula to the height, noting that dh/dt = -1cm/sec...?

  6. #6
    Quote Originally Posted by tkhunny View Post
    You need some preliminary thoughts.

    "sits on a straight line" -- So 100% of the sphere is above the line?

    "below the line" -- Is the sphere allowed to droop below ground level?

    Perhaps a drawing and a better description? You will need dr/dV, I would think.
    The sphere sits dead on the line, with zero space between it and the line and no part of the sphere below the line. The sphere then falls one diameter, i.e., 20cm. Now, the topmost point of the sphere is dead on the line, while the rest is below. There are now two parallel horizontal lines, 20cm apart.

    When the sphere has dropped 10cm, i.e., halfway, half of the volume is below the first line. I'd like to know how much is below at, for example, t=3s or t=8.34292s.

  7. #7
    Quote Originally Posted by stapel View Post
    I think the search term for the sort of partial sphere thing you're considering is "spherical cap". Perhaps relate that sort of formula to the height, noting that dh/dt = -1cm/sec...?
    Thanks very much. And the formula for the volume of a spherical cap is , so I should be able to continue from there.

    Thanks stapel.

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