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Engineering Maths

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A perfectly cylindrical container standing upright with a top and bottom (an oil drum
or coke can for example) has an empty mass of M and a height of H. It is filled with
a liquid of uniform density of variable mass m up to height h. When the container is
full the centre of gravity is in the centre (H/2). As the container is emptied and m
reduces the centre of gravity moves down. Once the container is empty the centre
of gravity is again at the centre (height H/2). Using calculus calculate the value of h
in terms of H, m and M when the centre of gravity is at its lowest position.

  • Engineering Maths - ,

    m is a function of height h, namely
    m(h) = ρAh

    The centre of gravity is at a distance x from the bottom of the cylinder, given by:

    x(h)=(M*H/2+m*h/2)/(M+m)
    substitute m=m(h)
    x(h)=(MH/2+ρAh²/2)/(M+ρAh)

    1. Differentiate with respect to h,
    2. equate derivative to zero for maximum
    3. Solve for h when dx/dh=0 to get
    hmax=(sqrt(M^2+ρAHM)-M)/(ρA)
    the negative root is rejected.

    Check the calculations.

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