A right cylindrical can with mass M, height H,and uniform density is initially filled with soda of mass m.  We punch small holes in the top and bottom to drain the soda, we then, consider the height h of the centre of mass of the can and any soda in it.  What is h a)initially and b)when all the soda is drained?

(c)What happens to h during draining of the soda?(d)If x is the height of the remaining soda at any given instant find x (in terms of M, H and m) when the centre of mass reaches the lowest point.(See figure 1).

 

Please explain the answer also.

 

hi friend its simple

initially when no soda was filled the centre of mass will be at the centre of the cylinder of the cylinder

now considering the case (a)

now consider the cylinder to be made up of two parts

one with height h (lower one )

and other with height H-h(upper one)

now calculate the centre of mass of lower part it will be at h/2(from symmetry)

 from centre of cylinder and will have mass M/2+m

now calculate the centre of mass of the upper part

again it will be at height (H-h)/2 from the centre of the cylinder and

will have mass=M

consider the two centre of masses as point and calculate their centre of mass

it will come by taking proper sign convention taking origine to be centre of

cyliner.you will get the ans easily of other also

plzz rate me

there is a correction in 11th line it is M/2

bye

 

Hi Aman how do you attach a mass of M/2 to both the parts. It should also be a function of soda in the can.

 

Ayush here comes my answer. I am answering just part (d) and all the answers will come from there. Let at any time the cylinder contains soda upto a height x.

Mass of can=M

center of mass of can = H/2

Mass of soda upto level x = mx/H

center of mass os soda = x/2

Center of mass of whole system (h)=(MH/2+mx^2/(2H))/(M+mx/H)

 

for part a) and b) put x=H and x=0 and you get h+H/2

for c) minimize x by diffrentiating h