In this illustration we'll analyze the shape
of liquid surface in a rotating vessel. we
are given that in a cylindrical vessel. a
liquid is filled partially, and it is rotated
at constant angular speed omega. and it is
saying if liquid of also rotates along with
the vessel, it is seen that near to side walls
liquid rises up, as shown in the figure. and
we are required to find the shape y equal
to f of x of the surface with respect to origin
at the lowest point o. now, to analyze it
we can, consider a particle of liquid on the
surface in the rotating vessel. here say if
this is the shape and redrawing the figure
and we consider this as x direction, and this
as y direction. and the vessel is rotating
at angular speed omega. now if i consider
particle on the surface of this liquid. then
it'll be revolving in a circular path, say
of, radius x this point if, having coordinate
x comma y, then in this situation we can consider
this would be revolving in circle of radius
x. the particle would be experiencing an outward
centrifugal force in the frame of this container.
as m omega square x. and this will experience
a downward force m g which is balance by the
normal reaction. onto the particle due to
the, layer of liquid beneath it, now in this
situation, if we draw a tangent of, this point,
along this curve. then we can say if slope
of this, tangent is theta. here we can see
with the vertical normal reaction is also
making an angle theta. and we can write, if
this particle is p we can write as shown in
figure. for equilibrium. of particle p. in
frame of container. we use. here we can write
n sine theta is equal to m omega square x.
and n coz theta is equal to m g. and from
these relations. if this first equation this
second equation then 1 by 2 will result, tan
theta that is equal to omega square x by,
g. and in this curve we can write down the
slope tan theta is d y by d x for this curve
which is omega square x by g. and here we
can write d y is equal to omega square x by
g d x. and if we integrate it at, point o.
x was zero y was zero and at point p x and
y are the coordinates. so this will result
us, the integration y is equal to omega square
x square by, 2 gee, that is the equation of,
the shape of the curve, obtained due to rotation
of this fluid that is the result of this problem.
the same result you can also obtain by using
bernoulli's theorem for the rotating fluid
in ground frame. so i am leaving it as an
exercise for you to evaluate, the same result
by using bernoulli's theorem for the rotating
fluid in round frame.
