 # Cud not solve the following problem, plzz help.Q. A uniform soda can of mass 0.140 kg is 12.0 cm tall and filled with 1.31 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 COM of the can and contents :-(a) Initially(b) After the can loses all the soda(c) What happens to h as the soda drains out?(d) If x is the height of the soda at any given instant, find COM when x reaches its lowest point. 12 years ago

Dear Raghu,

in (a) and (b) it is clear from symmetry that h=6.0 cm,(c) as the soda drains out, the com of the can stays constant whereas that of the soda falls. but the mass of soda also falls and therefore, so does its weightage in the com expression. Therefore the com initially falls, reaches a minimum and then rises up to its starting value.

h= (1.31(x/12)(x/2)+(6*0.14))/(1.31(x/12)+0.14), differentiating this expression w.r.t. x and equating the derivative to zero we get minimum value of h=2.8cm