A cylindrical soda can has a radius of 4 cm and a height of 12 cm. When the can

Question

A cylindrical soda can has a radius of 4 cm and a height of 12 cm. When the can is full the center of mass is 6 cm above the base of the can. When soda is removed from the can the center of mass drops. However, when the can is empty the mass is also 6 cm above the bottom of the can so at some point, the center of mass would have to rise. Find the depth of the soda in the can for which the center of mass is the lowest. Neglect the mass of the can, and assume the density of the soda is 1 g /cm3 and the density of air is 0.001 g/cm3. Do by Integration please

Explanation / Answer

From symmetry, we can ignore the radius of the can. We also know from symmetry that the average height of the drink is halfway up and the average height of air is halfwa y from the top of the liquid to the top of the can. Thus, total mass when fluid is at height h(ignoring radius, which will multiply everything by 4^2* pi = 16 pi) is 1h + .001(12-h) = .012 + .999h Meanwhile, the mass times the average height is 1h(1/2h) + .001(12-h)(12+h)/2 = 1/2h^2 + .0005(144-h^2) = .072 + .4995h^2 Thus, the center of mass is at (.072 + .4995h^2)/(.012 + .999h) We find the minimum height of the center of mass by taking the derivative. .999h(.012 + .999h) - .999(.072 + .4995h^2)/(.012 + .999h)^2 Note that the denominator is irrelevant. Let's multiply through by 1,000,000 999h(12+999h)-999(72+499.5h^2) = 499000.5h^2 + 11988h-71928 = 999(499.5h^2+12h-72) We can ignore 999 and solve using the quadratic formula Ignoring the negative solution, we get h = (-12+sqrt(12^2+4*499.5*72))/999 = 0.367841160623577 The center of mass at this point (this question is not asked) is (.072 + .4995h^2)/(.012 + .999h) = 0.36784116 Interestingly, this is the height of the fluid in the can.

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