A metal soda can of uniform composition has a mass of 0.122 kg and is 10.4 cm ta
ID: 1474029 • Letter: A
Question
A metal soda can of uniform composition has a mass of 0.122 kg and is 10.4 cm tall (see the figure). The can is filled with 1.36 kg of soda. Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is the height h of the center of mass of the can and contents (a) initially and (b) after the can loses all the soda? (c) If x is the height of the remaining soda at any given instant, find x when the center of mass reaches its lowest point. Give your answers in cm.
Explanation / Answer
m1 =1.36 kg , m2 =0.122 kg , H =10.4 cm
Since the can is of uniform construction, its COM is at the point in the center of the can (i.e., half its height and at the center of its area, assuming a cylindrical can). The COM of the liquid is going to be 1/2 the remaining height of liquid in the can, again at the center of the area.
h = H/2 =10.4 /2 =5.2 cm
a)The height of the COM in both cases is 5.2 cm cm. The height of the COM of the can itself is 6cm, no matter what the liquid level. When all the liquid is present, the height of its COM is 5.2 cm
(b) When there is no liquid, the height of its COM is 5.2 cm.
c) As the liquid drains, the height of the COM decreases.
Height of soda =1.36/10.4 = 0.131 kg/cm
If M is the mass of remaining soda and h is its height at any given time, the height of the COM (x) will be:
x = [5.2*0.122 + (h/2)*(0.131*h)] / (0.122 +0.131*h)
x= (0.6344+0.0655*h^2) / (0.122 +0.131*h)
To find the minimum value of x, we find derivative of the the function above.
dx/dt = [0.131h/ (0.122 +0.131*h) ] - [ (0.6344+0.0655*h^2)(0.131) / (0.122 +0.131*h)2] =0
By solving above equation we get
x= 2.317 cm