rotations........

no one trying................isn't it a good question??

i did it jus roughly...dont blame me for wrong results...
on the left side of the symettry line, we have depth of the water surface 'h' as a function of 'x' as

h=\frac{x}{2g}[(\omega _{1}^{2}+\omega _{2}^{2})x-2\omega _{2}^{2}R]

ill do it properly now...

im sorry 'x' here is the distance from the left end of the vessel..

ok gordo..u are doing theory job...thats good.........

but i wanted to know how fluid shape will look like........surely it will be irregular.....but rough idea.....

luk trusting that what i have put down is even roughly rite,
we have h=0 at x=2Rw₂² /(w₁² +w₂²)
now the cases that uve put down,
if w1>w2
we wud have this point where it comes back to normal height at a point in the left side of the symettry line of the container..
if w1=w2
we have this happening on the symettry line itself, i mean the liquid surface wud first go down and come up again at the symettry line itself...
for the third case i need to reluk at the trajectory on the right side of the symettry line...

i have a strong feeling motion of container doesnt affect shape in any way as long as liquid bulk has a particular ω

any objections to abv statement ??????