let us discuss about pressure distribution
in a rotating tube filled with the liquid.
here we can see we are having a tube of length
l which is open at one end, containing some
fluid, closed at the other end is rotating
at an angular speed omega.
now due to rotation the fluid inside will
experience a centrifugal force and will have
a tendency to move out. and as the other end
is closed, a pressure distribution is developed
within the fluid. and say we are require to
find . pressure at a distance, x from.
the axis of rotation say this is a point,
ay. and we are required to find pressure,
what will be the value of pressure at a distance
x, lets calculate. in this situation the rotational
motion of the tube at every point of the,
tube it’ll experience an outward centrifugal
force, which is m omega square r, or this
centrifugal acceleration as omega square r
at a distance r from, the axis of rotation.
now to find out pressure lets consider, an
element of width d-r. at a radial distance
r. now at this point the centrifugal acceleration
will be omega square r, due to which the pressure
will, continuously increase . say if at one
end of this d-r pressure is p on the other
end it can be taken as p plus d-p. and this
pressure increment d-p. will be due to the
level of width d-r it can be written as d-p
as.
d-r, ro, and g. instead of g will take the
centrifugal acceleration which is omega square
r. so we can simply state . pressure between,
a level, which is this atmospheric level say
it is, at a depth . l-one, up to which air
column is there and beyond which fluid is
present so we can simply state.
at this point say pressure is p-atmospheric
or say this point is one. and this point is
two which is located at the point ay in the
tube, so we can simply state pressure at two
will be more then pressure at 1 so p-2 minus
p-1 can be given here as integration of d-p,
which can be written as rho omega square r,
d-r. which is integrated from, length of r,
which is varying from l-one to x, l-one is
the,
here column length above the , fluid , up
to the open end of the tube.
so it’ll be integrated, r will be integrated
from l-one to x. so if we just integrate the
values it is ro omega square and it’ll be
r-square by two with limits from l-one to
x, so it can be give as half ro omega square,
x square minus, l-one square.
so in this situation this is the pressure
difference p-two minus p-one, a pressure at
a point ay can be given as , at point 1 pressure
is p-atmospheric.
so it can be given as p-atmospheric plus,
half ro omega square, x square minus , l-one
square this is the relation, we can develop.
you just need to very carefully analyze the
whole process , how we have calculated the
pressure at point ay due to the rotation of
tube as this concept is going to be used in
different kind of applications.
