Center of mass

Question

Initially, a cylindrical bottle of bottle of oil-and-vinegar salad dressing whose volume is 1/3 vinegar (p=1.01 g/cm^3) and 2/3 oil (p=0.910 g/cm^3) is shaken so that the oil and the vinegar are mixed uniformly. The bottle is then returned to the table and the oil and the vinegar become separated, with the oil floating on top of the vinegar. How has the height of the center of mass of the salad dressing changed as a result of the separation of the oil and the vinegar?

Answer 1

Let the height of the bottle be h and the area of the bottom be A.   In the analysis the area A can be taken as 1 without loss of generality, so I will take A =1.    The volume of the bottle is then h.  

 

Before the separation, the center of mass will be  at a height h/2 above the bottom.  

 

Afterwards,  there is a volume of (1/3) h water under a volume of (2/3) h oil.     The location of the center of mass of the water mass is  h/6 above the bottom,  the mass of the water is  (1/3)h x 1.01 .   The location of the center of mass of the oil is (2h/3)  above the bottom, the mass of the oil is  (2/3) h x .910.   

 

The standard formula for the height of the center of mass above the bottom then becomes:

 

        H  =  [  (h/6) x (1/3)h x 1.01 + (2h/3) x (2/3) h x .91 ] / [(1/3) h x 1.01 + (2/3) h x .91 ]

 

   This works out be   H = .489 h. 

 

  So the center of mass is lower after the separation by   h/2 - .489 h =   0.011 h.    This is a drop of 1.1%