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 of soda, the center of mass of the contents of the can is 6cm above the                    base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty                    (filled only with air), the center of mass is once again 6 cm above the base on the axis of the can. Find the depth of soda in the can for which the center of                    mass is at its lowest poing. Neglect the mass of the can, and assume the density of the soda is 1 g/cm^3 and the density of air is 0.001 g/cm^3

Explanation / Answer

Let height of soda be y

Volume of soda = y*pi*4^2

mass of soda = 1*y*pi*4^2 = 50.272y g

mass of air = (12-y)*3.142*4^2*0.001 = 0.0502*(12-y)

Centre of mass of soda = y/2

Centre of mass of air = y +((12-y)/2) = 6+y/2

Centre of mass of can = (50.272y*y/2 + 0.05027*(12-y)*(6+ y/2))/(50.272y +0.0502*(12-y)) = Y

dY/dt = 0

y = 0.36785 cm = height of soda from bottom of can = lowest centre of mass

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