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Center Punch Set
What is the minimum height of the center of mass as the water drains out?
A can in the shape of a symmetrical cylinder with mass M and height H is filled with water. The initial mass of the water is M, the same mass as the can. A small hole is punched in the bottom of the can, and the water drains out.
I have the correct equation for the center of mass: (x^2+H^2)/2(x+H)
I know that to find the minimum height of the center of mass i take the derivative of the above equation and set it equal 0 and solve for x, but i'm having a hard time doing that. the derivative i got was 0= (x/(x+H)) - ((x^2 + H^2)/(2(x+H)^2)). But every time I go to solve for x i get stuck. any help is great!
Your equation is correct,now let
Z=(x^2+H^2)/2(x+H)
Now differentiate we get
dZ=[2x(x+H)-(x^2+H^2)]/2(x+H)^2
For maxi ma and mini ma we have
dZ=0
we get 2x^2+2xH-x^2-H^2=0
or x^2+2xH-H^2=0
(x+H)^2=2H^2
x+H=sqrt(2).H or - sqrt(2)H
so x=sqrt(2)H-H or -sqrt(2).H-H
rejecting the second sol. we put x into eqn for Z
Z=(x^2+H^2)/2(x+H)
=[sqrt(2)-1]^2H^2+H^2/2(sqrt(2) H+H))
=([sqrt(2)-1]^2+1) H^2/2(sqrt(2)+1)) H
=H.[1+(sqrt(2)-1)^2]/2.(sqrt(2)+1)
=0.243 H
IVAN
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