A uniform soda can of mass 0.143 kg is 11.9 cm tall and filled with 0.357 kg of

Question

A uniform soda can of mass 0.143 kg is 11.9 cm tall and filled with 0.357 kg of soda (figure below). Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. (Initially the soda can is full.) (a) What is the height h of the com of the can and contents initially? (b) What is the height h of the com of the can and contents after the can loses all the soda? (c) If x is the height of the remaining soda at any give instant, find x when the com reaches its lowest point. I just need the answer for C

Explanation / Answer

given that

m = mass of the can = 0.143 kg

h = height of the can = 11.9 cm

M = initial mass of soda = 0.357 kg

let actual height of the soda = x

let height of the center of mass = y

By definition of center of mass (COM)

y = (mass of can * height of can's COM + mass of liquid * height of liquid's COM) / (total mass)

height of the can's center of mass = h/2 = 11.9/2 = 5.95

mass of liquid

When x = 11.9, the mass of liquid is 0.357kg (full can).

When x=0, the mass of liquid is 0 (empty can).

So we can say that mass of liquid = (0.357/11.9)*x = 0.03*x

The height of the liquid's COM = x/2 = 0.5*x

So

y = (0.143 * 5.95 + 0.03x * 0.5x)/0.143 + 0.03x

y = (0.850 + 0.015 x^2) / (0.143 + 0.03x).......(eq1)

part(a)

whem x = 11.9 (can is full)

y = (0.850 + 0.015(11.9)^2) / (0.143 +0.03*11.9)

y = 5.95 m    (the COM is half)

part(b)

when x = 0 (can is empty)

y = 5.95 m ((the COM is half )

part(c)

differentiate the eq1with respect to x

dy/dx = ((0.143+0.03x)(0.03x) - (0.03x)(0.850+0.015x))/(0.143+0.03x)^2

dy/dx = ((0.0042x + 0.0009x ) - (0.0255x +0.00045x^2))/(0.020449+0.00429x+0.0009x^2)

dy/dx has to be zero .

dy/dx = 0

x = (-0.143 + sqrt(0.143(0.143+0.357)))/0.357

x =(-0.143 + 0.26)/0.357

x = 0.348 m

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