A cylindrical soda can has a radius of 5cm and a height of 15cm. When the can is

Question

A cylindrical soda can has a radius of 5cm and a height of 15cm. When the can is full of soda, the center of mass of the contents of the can is 7.5cm 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 7.5 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 point. Neglect the mass of the can, and assume the density of the soda is 1g/cm3 and the density of air is 0.0001g/cm3.

Explanation / Answer

From the given question,

radius(r)=5cm

height(H)=15cm

density of soda(d1)=1g/m3

density of air(d2)=0.0001g/m3

mass = density x volume

volume = area x height

m= a h d

h is taken from base.

formula for centre of mass is x=( m1 x1 + m2 x2)/(m1+m2)

x= [a h1 d1 (h1/2) + a h2 d2 (h1 + h2/2)]/(a h1 d1 + a h2 d2)

let h1=h, h2=15-h

x= [a h d1 (h/2) + a (15-h) d2 (h + (15-h)/2)]/(a h d1 + a(15- h) d2 )

=[h2d1+(15-h)(15+h)/2 d2]/(h d1 + (15- h) d2 )

=[h2d1+(225-h2)/2 d2]/(h d1 + (15- h) d2 )

for maxima or minima dx/dh=0

dx/dh=[(h d1 + (15- h) d2 )(2hd1 + (-2h/2)d2) - (h2d1+(225-h2)/2 d2)(d1 -d2)]/(h d1 + (15- h) d2 )2=0

[h (d1-d2) +15d2 )(2hd1 - hd2) - (h2d1+(225-h2)/2 d2)(d1 -d2)]=0

we can take d1-d2=1-.00001=1 (approx)

(h +0.0015)(2h -0.0001h)-(h2+5000(225-h2)=0

2h2-0.0001h2+0.0030h-0.00000015h-h2-1125000+5000h2=0

0.00000015=0 ( approximation)

h2(2-0.0001-1+5000) +h(0.003)-1125000=0

5001h2+0.003h-1125000=0

solving this we can get depth for lowest point of centre of mass.

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