Posted by **Karl Haxell** on Tuesday, May 17, 2011 at 7:16am.

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 -
**MathMate**, Tuesday, May 17, 2011 at 3:54pm
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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