We were discussing earlier about the stability
of floating bodies and we will continue with
that. So, if you recall the sketch which we
were discussing in the previous class. We
identified a point know as metacentre and
the motivation behind identifying this point
is that when it is a floating body and it
gets displaced and its submerged within the
water changes its configuration the center
of buoyancy does not remain fixed at its own
position.
So, referring to a fixed center of buoyancy
might not work as a stability criteria. Now,
we will try to find out as what we have done
for the case of totally submerged bodies that
how we can find out a stability criteria for
floating bodies. As you can see the presence
of or the existence of these metacentre is
something which is equivalent to the presence
or existence of the center of buoyancy for
a case when the center of buoyancy remains
at its own position.
So, the metacentre somehow reflects the location
of the center of buoyancy with respect to
the axis of symmetry of the body. So, it has
some relationship with the center of buoyancy
definitely. But by specifying the metacentre
or by specifying the center of buoyancy just
as their absolute locations. It is not just
possible to talk about the stability criteria.
So, we have to look into more details. To
look into more details we will set up some
coordinate axis.
Say we have x-axis like this say we have G
axis like and the y axis is perpendicular
to the plain of the figure. So, the y axis
is this dotted line which represents an axis
in the other view. Now, if consider that how
we calculate the submerged the volume the
volume of the solid which there within the
fluid. Let us say at a distance of x we take
a small element of width Tx and this element
has a depth of z.
So, this element if you look into the other
view 
will be like this which is basically located
at a distance x from the y axis. Let us say
that this shaded area is the dA. So, you can
say z*dA is the representation of the elemental
volume of this shaded portion. Now, when it
is tilted even when it is tilted we fix the
x and z axis with respect to the body and
we keep them as same. So, we still have this
as the x axis and may be this as the z axis.
These are fixed relative to the body. Although,
the body has tilted we are using a body fitted
set of axis. Now, let us say that the angle
of tilt is theta if the angle of tilt is theta
then what happens? We can find out that what
is now the displaced volume. So, if you consider
again say at a distance x from here some strip
width dx. Now, if you consider the displaced
volume the displaced volume is of course corresponding
to this part.
This part is like z same as this one because
it is the same axis that has got tilted so
if this is z, this displaced part is also
z. But you have the additional displaced part
which may be let us mark with the different
colour. So, this is an additional displaced
volume. So, this additional displaced part
will correspond to a particular length along
the z axis and what is this? Remember that
if this angle is theta then the angle of tilt
of the x axis with respect to the horizontal
is theta.
So, what is this dimension? You know that
this is x so that is x tan theta. So, if we
calculate what is the displacement in centroid
of the displaced volume? That is what is the
difference between the x coordinate of say
the original center of buoyancy let us give
it a name that there was original center of
buoyancy say CB and now the new center of
buoyancy CB prime. So, the difference in the
x coordinate of CB prime and CB.
That is what we are interested to find out.
What is that? Center of buoyancy you can calculate
by utilizing the formula for centroid of a
volume which is here the displaced volume.
So, this will be integral of x DB. So, this
x what is DB so you have this total height
as x ten theta + z dA, dA is the area in the
other view. So, the depth times the dA is
the volume. So, this divided by the volume
of the immersed part.
Because of the symmetry the volume of the
total immersed volume does not change whatever
comes down the same volume goes up above the
water. So, the total displace volume does
not change minus what was the original one?
Integral of x z dA/v. It may be easier if
we mark the original CB and the displaced
CB here. So, there has been a displacement
from CB to CB prime. The location of the center
of buoyancy has got shifted.
So, this is nothing but = CB, CB prime that
is the length that we are talking about. And
this is integral of x square tan theta dA/
. Theta being like a constant parameter tan
theta you can take out of the integral and
this area integral is carried out over the
shaded area dA. So, what does it represent
if you take tan theta out of the integral?
Second moment of area with respect of which
axis?
You now refer to this plan view, integral
of x square dA is the second movement of area
with respect to the y axis. So, it is tan
theta * Iyy/ v where what is Iyy? It is integral
of x square dA. Now referring to this figure
you can say that for a small angle theta CB,
CB prime this particular small distance is
a small arc of may be a circle. So, you can
write this as approximately CBM*theta, for
small theta. It is just like S =r theta for
a small part of an arc of a circle.
Because this is very small we are assuming
a small displacement we test stability not
with the large displacement but with the small
displacement. M is the metacentre. Now, for
the same smallness tan theta will be approximately
=theta of course you are expressing theta
in radian that is understood implicitly. So,
then if you equate the two parts what you
will get CBM=Iyy/v. Now, look into 
this figure. In this figure you see that M
is above G.
When M is above G the couple movement created
by the forces is trying to restore the body
to its original configuration. However if
G was above M it would have been just the
opposite case that you can clearly visualize.
So, what is important here is the location
of M relative to G just like for a totally
submerged body it was location of B relative
to or center of buoyancy relative to G. So,
what we are interested to find out is what
is the height GM?
So, GM you can write it as CBM –CBG=Tyy/v
–CBG. See why we have written the formula
in this way? In the right hand side you have
terms which are independent of the deformed
configuration or deflected condition. So,
if you see a Iyy V these can be calculated
based on what is the original configuration
and CB is the location of original center
of buoyancy relative to the body. G is the
location of center of gravity relative to
the body.
And distance between those two that does not
change with deflection. So, we are able to
express something which should be a function
of deflection in terms of certain things which
are not function of deflection. Therefor this
is also not a function of deflection and m
remains sort of fixed. But you have to keep
in mind that there are certain assumptions
that go behind this. What are the important
assumptions?
Theta is small so this analysis is valid only
when you are having a small theta not only
that we have implicitly assumed that the z
axis is the axis of symmetry for calculating
or for coming up with expression. So, we are
essentially dealing with the simplified case
with symmetric bodies that we have to also
keep in mind. Let us look into an example
where we illustrate the use the this expression
for finding out the stability criteria.
So what is the stability criteria here? Let
us summarize.
If M is above G it is stable. If M is below
G it is unstable and obviously if M coincides
with G it is neutral. So, you can clearly
understand that M plays the equivalent role
of center of buoyancy in this case. But it
is not exactly the center of buoyancy. We
consider a very simple example and we will
illustrate the use of this through that example.
Let us say that you have a body of this shape
rectangle parallel perpendicular shape and
you can give dimensions.
Let us say A, this is B and let us say this
is C. Say this is partially immersed in a
fluid so it is like a floating body. Now,
let us try to see that how we can apply this
stability criteria here and what are the important
issues. First important issue is that when
you apply this criteria what should be the
y axis with respect to which you are considering
the second moment of area. If you look into
this problem very critically you will see
that there are certain non-trivial issues.
Like you can consider may be this as one of
the frontal surfaces. You might consider this
as one of the frontal surface or even the
third one. So, accordingly it is not very
straightforward to say so when we drew the
picture of the boat see it could be this front
part. It would be the side part both are exposed
to the water. I mean both are within the water.
So, if you could say that what should be the
corresponding axis that you need to consider?
Or does the choice of the axis change if you
shift your attention from the frontal area
to the side area. At the end, what you are
bothered with? You are bothered with the plan
view so when you have the plan view which
is like the top view of this one. What is
the plan view? It is the intersection of the
body with the free water surface, free fluid
surface so that particular view it boils down
to the same when you consider this surface
or this surface.
But again when you look from the side and
when you look from this one may be in one
case you are considering this as the y axis.
In another case you are considering this as
the y axis. Both axis are relative to the
top. So, the question is with respect to which
axis you should evaluate this. See, this is
what? This is like evaluation of a safety
criteria that it will be stable. So, when
you want that it should be stable what should
be the guideline?
The guideline should be that like you want
this to be positive. That means this height
is greater than this height. Then like M is
above G so the way in which this is written
the right hand side has to be positive. So,
more positive means so to say it more stable
so to say. So, if we take the least of this
one, least possible value of this one and
still find that the right hand side is positive
then for all other cases it should be positive.
So, it is like a safety design. So, you look
into the most adverse condition make sure
that it is positive even for that. So, for
better conditions it would always be better.
So, out of these two axis for one axis, what
will be the I? It will be Ba cube/12 another
will be ab cube/ 12. So, you take the smaller
one that means if B is smaller than a as an
example you take this one. So, then when you
substitute that in this expression and still
you calculate this as positive.
You are assured that it is safe. Because with
respect to other axis when you calculate the
metacentric height it will definitely be greater.
This is known as metacentric height. And really
when you calculate different metacentric height
based on different views these represent different
types of angular motion like rolling, pitching
these are different technical names based
on with respect to what type of axis it is
tilting.
So, but just for simple design you can consider
the smaller one and you divide it by the volume,
not the total volume but the volume of the
submerged part. So, that you can easily calculate
based on what part is submerged? Let us say
that d is the submerged depth. It is easy
to calculate this because you can use the
equilibrium that for equilibrium the buoyancy
force must be equal to the weight.
So, from that if you know the density of fluid
and density of the solid body you can come
up with what should be the equilibrium d by
very simple equating of the two forces. So,
you can calculate what is the volume which
is immersed in the fluid, center of buoyancy
location you can get from the centroid of
the displaced volume. Center of gravity also
has a fix position with respect to the body.
So, you can clearly substitute these dimensions
and find out what is the metacentric height.
So this effectively requires only the calculation
of the immersed volume, the calculation of
the second moment of area with respect to
these axis and specification of the locations
of the center of buoyancy and center of gravity
with respect to the axis fixed on the body.
So, that is a sufficient information to calculate
the metacentric height.
So, what we can clearly see is that if M is
above G that makes it stable. That mens lower
the location of the center of gravity it is
having a greater chance that it will be stable.
Because then it’s a greater chance that
M is above G, lower the location of G. So,
location of G if it goes higher and higher
it might make a previously stable system or
convert it into an unstable . Let us look
into an animated example to consider this
case.
So, in this animated example what we will
see. We will see that how the stability may
be disturbed because of the shifting of the
distribution of the weight of the body. So,
just look into it carefully. So, there is
a body which has given a slight displacement
and you see that for a small displacement
it oscillates like a pendulum. We may easily
derive what is the time period of it. Now,
you see that the distribution of the weight
is being altered.
So the center of gravity is being shifted
higher and higher by putting the load more
and more towards the top and you see that
it topples. So, obviously this is clear illustration
of this concept that we have learnt in this
example that if you have metacentre above
the center of gravity it makes it more stable
and otherwise it is not so. So, now for small
oscillations or for small deflection you could
see that it oscillates like a pendulum.
So, when it oscillates like a pendulum it
has a time period and that may be calculated
by calculating the moment of the resultant
force with respect to the axis. So, if you
have a tilted axis like this and if you have
a force FB which is the buoyancy force. This
force has a movement with respect to the axis
of the body. So what is the movement of this
force with respect to the axis of the body
you can just find out the perpendicular distance
of this force from the axis of the body.
What will be the perpendicular distance of
this buoyancy force, line of action of the
buoyancy force? I mean you can see that it
eventually passes through the axis of the
body, right. So, this force individually which
is important it is the couple moment that
you are having the so called the restoring
couple. So, you have also the W and it is
basically the couple moment of these forces
that you need to consider. So, the perpendicular
distance between these two.
In terms of the metacentric height would you
express this so if you have this as the metacentre
M. So, we are interested about these distance
so it is possible to express it in terms of
the metacentric height. So, this angle being
theta, what is this perpendicular distance?
GM sin theta. So, the moment of this force
is W GM sin theta and the resultant moment
of all the forces is nothing but if it is
like rotation of a rigid body with respect
to a fixed axis.
It is a restoring moment so you should have
a – sign associated with one. That means
whatever is the chosen positive direction
of theta, theta. theta ..this has a direction
opposite to that bring it to its original
configuration. So, you can write this as 
so I is what now? I is the mass moment of
inertia. It is not the same I that we were
talking about it is the real mass moment of
inertia with respect to the axis of the body
relative to which it is tilted.
So, this plus for small theta again this is
approximately equal to theta that is equal
to 0. So, it is just like equation of a spring
mass system mx ..+ kx = 0. So, what is the
natural frequency of oscillation of this system?
Root over equivalent stiffness/ equivalent
mass. So, W* GM/I that is the natural frequency
of oscillation of the system. It is angular
oscillation not a linear one so you can clearly
see that greater the metacentric height, greater
will be the frequency of oscillation.
So, if it is a ship it will be more uncomfortable
to the passenger. So, these are two conflicting
designs. See greater the metacentric height
you expect it to be safe in terms of stability
but it will have more oscillation within that
stability regime. That means for a passenger
it may be quite uncomfortable at the same
time if it is used for a particular say critical
purpose like warfare and so on.
So, there stability of ship more than the
comfort and they are obviously it is the stability
that should be the driving factor for designs.
So, when a ship is designed you have two conflicting
things. One is the comfort another is the
stability and comfort factor comes from this
high frequency of oscillation and the stability
comes from the metacentric height or location
of metacentre relative to the center of that.
Now, that we have seen the stability criteria
and so on we will come to our final topic
fluid statics which ironically is not fluid
statics but fluid with rigid body motion.
And as we discuss earlier that we are going
to address this issue within the purview of
fluid statics for a very simple reason that
when you have fluid under rigid body motion
it is still fluid element without any shear.
So, when it is without any shear it is just
the normal force which is acting on the surface
so the distribution of force in terms of pressure
on the surface remains unaltered. No matter
whether it is at rest or under rigid body
motion when the fluid elements are deforming
then only you have the shear. So, let us take
a simple example for fluid under rigid body
motion.
Let us say that we have a tank 
partly filled up with water and the tank is
accelerating towards the right with an acceleration
of ax along x, fixed acceleration and neglect
the deformation of the water that is there
in the tank. In reality that deformation is
there so whatever analysis we are presenting
here is not perfectly correct. Because in
reality there will be shear and deformation
and so on.
But just as an idealization just like many
we discussed about frictionless surfaces not
that they are there but through this idealization
we learnt certain concepts. So, let us say
that there is no deformation. So, the water
which is there within the tank just get deflected
in its configuration in terms of the free
surface like a rigid body. So, how we calculate
that what should be the new location of the
free surface because of this acceleration
that is what we want to see.
Say initially the height of water in the tank
was h0, may be let us specify the width of
the tank as b and may be the width perpendicular
to the plain of the board as W. Let us say
that this is rectangular plain. Now, recall
that we derive the sudden expressions with
regard to distribution of pressure in presence
of body forces. So, far as I remember this
is the expression that we derive some time
back when we were actually starting with our
discussion on fluid statics.
So, we will try to use this expression here.
So, along x we have minus because there is
no body force acting along x. Let us say that
the y direction so we have x and y as our
chosen directions. Let us say that the y direction
is the direction opposite to the acceleration
due to gravity. So, for y direction what you
can write? So, what is ay, ay is 0 it is accelerating
along x. So, ay is 0. What is by-g?
So, you can get –rho g. Now when you have
a free surface, the free surface is characterized
by what? The free surface has the same pressure
throughout because it is exposed to the atmosphere
which has same pressure throughout. So, there
is a pressure equilibrium between that and
the atmosphere. So for the free surface you
must have dp =0 that should be the governing
parameter for locating the free
So, when you have dp =0 remember that now
p is a function of both x and y. So, you can
write dp as this one. So, you can substitute
in place of the partial derivative with respect
to x as –rho ax dx and this one is –rho
gdy =0 that means what is dy/dx that is = -ax/g.
What does this dydx represent? It represent
the slop of the free surface. So, can you
tell now weather the slop of this surface
will be like this or like this. 1 or 2?
1 because you can clearly see that this will
represent a kind of tilt like this so this
will become the new free surface and the angle
that you are considering for the slop is basically
this one. Because this is negative ax is positive
and g is positive so this is negative. So,
it must be an . So, in the direction in which
the tank is accelerating the liquid will be
more down and in the other direction it will
be more up.
And if you specify this angle as theta then
that theta is nothing but 180 degree – this
slope angle. So, you can say that tan theta
is nothing but equal to ax/g. Because theta
is nothing but 180 degree minus this. So,
you can find out that what is the extent to
which the water level will rise on one side
and may be fall on the other side. Let us
say that this rise is delta h on one side
because of symmetry it will be a fall of equivalent
delta h on the other side.
So, it will be as if with respect to the center.
So, we can calculate what is delta h. So,
if you calculate delta h what will be that?
It is nothing but delta h/p/2 is tan theta
that is ax/ G. So, from here you can calculate
what is delta h? ax/g. Now, let us come to
a critical condition if we are happy with
this sometimes we are deceived. How? Let us
say that the total height of the tank is H
if it so happens that let us say from calculation
we get delta h + h 0 > H practically that
is not possible.
Because the liquid cannot occupy a height
which is greater than that is provided by
the tank. So, what this will mean? This will
means some water has spilled out. So, this
is a condition from which you can say that
it has actually spilled out. When it has spilled
out it is more disconfigeration. So, when
it has spilled out what will happen? What
type of configuration you expect?
So before spilling out it tried it best to
climb up to the top most level and then it
spilled out. So, this will be one end of the
surface and may be the other end is like this.
So, with spill out may be this will spill
out. Irrespective of whether it has spilled
out or not you still have this expression
applicable. So, now this will be the angle
theta you can find out that what should be
this length let us say d1, what should be
this length d1.
Because you can say H/b1 =tan theta which
is ax/g. H being the height of the tank know
so from there you can find out what is b1
and therefor you can calculate what is the
volume which is there now within the tank.
And the difference between the original volume
and that volume will give you what is the
volume that has got spilled. So, you see that
it is not just like your solution should not
be driven by a magic formula.
But based on the numerical data given you
have to come to a decision whether the water
is there inside or it has got spilled or so
on. Let us take a variant of this example.
So, the variant of the example is that now
we have the tank located on a inclined plain.
With an angle of inclination alpha tank is
there on the plain and it is accelerating
say downwards with an acceleration of a, which
is a uniform acceleration may be because of
a resultant force which act along that direction.
So, in this case it may be more convenient
it you fix up your coordinate axis relating
to the inclined plain say x and y. So, the
similar equation will be applied and let us
just do it very quickly.
Because it is very straight forward. So, you
will have minus partial derivative of p with
respect to x plus what will be px now? bx
is g sin theta + rho g sin theta =rho here
it is not theta we have given a name alpha.
So rho g sin alpha = rho a, ax is a. What
about y? So, -this so y will be –g cos theta
so –rho g cos alpha =0. So, from here you
can find out when you have dp=0 that means
you can substitute the expression so it will
be.
Let us write it at the top the expression
now becomes in place of partial derivative
with respect of x you can substitute rho g
sin alpha – rho a dx + -rho g cos alpha
dy. So, dy/dx =g sin alpha – a/ g cos alpha
and this is now = tan theta where theta is
the angle relative to the original location
of the free surface or like assumed x direction.
So, you can see now that there is depending
on the magnitude of a this may be positive
or negative.
So, you may have a case when sin alpha is
> a or g sin alpha < a. So, here you cannot
say that whether it is 1 or 2, case 1 or case
2. Again you see that it is not a magic formula
that should drive your decision it depends
on what is the physical situation that is
prevailing. Let us consider a 3 example. “Professor
- student conversation starts” Surface should
always be normal on the resultant.
See here what we are doing here we are implicitly
applying the same condition. You see that
we are having one very important assumption.
The assumption is that you do not have an
oscillation in the surface. So sometime because
of this displacement the surface oscillates
it becomes like a wavy situation and that
is known as sloshing of tanks. So, we are
not going into that details so we are assuming
that the surface remains flat.
And when the surface remains flat and under
these conditions when you have dp=0 that is
exactly the same condition. See vector analysis
is not, it is also mathematics right when
you say that when you are dealing with mathematical
analysis I feel this is too mathematical I
do not see any difference. Like if you have
vector analysis this is also vector analysis
just dealing with scalar component.
So, it is better to be habituated with these
because again I am telling you there are situations
when it would not be as straight forward as
this. So, we have to use this fundamental
equation. So, whenever you are solving a problem
try to adhere to the fundamental situations.
I am getting your point why you are telling
this because you have been habituated in solving
problems in that way through your entrance
exam.
But we will be encountering more challenging
problems then what you have solved earlier
through that type of magic situations, we
try to avoid that. So, our basic intention
will be that we have this basic equation this
should solely guide us for solving whatever
complicated problems that we are having of
these type. “Professor – student conversation
ends”. Let us consider another one.
Say example. Now let us say our close tank,
close tank partially filled with water this
is closed. Again you are doing the same thing
accelerating it along x. What will happen
to the free surface? Now there cannot be any
spilling. So, when there cannot be any spilling
there is even a chance that the free surface
is like this. So, the symmetry with respect
to the central line which was there for the
previous case without spilling is now broken.
That symmetry is not guaranteed because it
may try to escape and since it is not finding
an escape route it will break the symmetry
and get distributed in what way? In such a
way that the volume of the liquid now is conserved
because it cannot go out of the tank. So,
if you say that this dimension is y1 and this
is x1 then you have y1/x1 as tan theta. And
that is =ax/g and you can calculate y1 and
x1.
By considering that the volume which was there
originally is the same as the volume of water
after it is tilting of the intervals. It is
the new interface. So, this will give you
another relationship involving x1 and y1 and
you can solve for x1 and y1. Now, let us say
that we are interested to find out what is
the total force acting on the top surface
or the container. How will you find it out?
So, let us say this point is c.
So, you have to keep in mind that up to AC
it is in contact with a fluid. And you can
use this one. So, you can find out the pressure
distribution as a function of x from a to
A to C. Take a small element at a distance
x from A of width dx so the force acting on
that is b*dx* the third dimension integrate
it from A to C and AC you can find out from
these geometrical consideration. So, that
will give you the total force.
Let us consider may be a 4 variant which I
will not solve but just tell you that such
a variant is also possible. Say you have tank
now it is completely filled with water and
it is closed. Which p will be p 0 you assume
any one of the point say A as a reference
pressure say B, some reference B because always
when you have pressure it is relative to some
point and so you can express the pressure
distribution in terms of the pressure at the
point A and then integrate it.
So, next example is you are this tank accelerating
towards the right. But completely filled with
water but closed so water cannot escape. So,
if I tell you find out the total force on
the AB how will you do it? I will not do it
myself leave it on you as an exercise. And
I can only tell you that this is the simplest
of all the cases that we have considered.
But you have to keep in mind that same consideration
as this also should work.
That will give you a natural pressure distribution
from A. And similar to this you can just integrate
from A to B to find out here you do not have
the botheration of finding out what is the
portion that is exposed with the liquid because
if it is completely filled it should be completely
exposed. Now when we consider the rigid body
motion it is not always just translation it
may also be rotation. So, let us consider
a rigid body rotation example.
These types of examples you have seen earlier
that you have a tank filled up with water
to some height h0. It is a cylindrical tank
of radius R. So, are using a RG coordinate
system, axis symmetric coordinate system so
this is r coordinate and along this there
is z coordinate. So, this tank is rotated
with respect to its axis within angular velocity
momentum. So, when it is rotated you already
know it that it will come to its free surface.
Will come to a deform or deflected shape like
this which is a paraboloid of revolution.
Let us quickly see that how we can derive
that it should be a paraboloid of revolution
from this fundamental principles not from
any magic formula. So, let us see you start
with here so there are two directions r and
g. So, you have - dp/dr + what is the body
force along r – omega square r. See these
we are writing in the original form of the
Newton second low of motion.
Now, with respect to an accelerating difference.
So, no question of centrifugal. Now, you tell
when it is no question of centrifugal whether
the body force is there or not. Yes or no.
Say you are standing on a platform and you
are looking into it from the inertial reference
frame this is not an external force that is
applied that you can see only when you have
a rotating reference frame that is attached
to the platform that is having angular promotion.
So, obviously there will be no br that you
have to keep in mind. This is the form of
the Newton second law in an inertial reference.
With a non-inertial reference frame you have
to use a pseudo force for the inertia force
but then you do not write equation of dynamic
but equivalent of equation of static equilibrium.
You convert that in an equivalent static equation
through D'Alembert's principles.
But here we are not taking about that we are
talking about the proper acceleration. So,
you have no body force but you have acceleration.
What is the acceleration? It is nothing but
the centripetal acceleration. So, -rho omega
square r. See, if you had considered within
a rotating reference frame the right hand
side would be zero because you are writing
static equilibrium this would be represented
by the pseudo force.
So, the final equation would eventually be
the same. It is just a matter of the reference
frame with respect to which you are writing.
But this is read to with respect to the inertial
reference. Then when you come to the z direction
– dp/dz then let us say this is the acceleration
due to gravity direction –rho g = there
is no acceleration it is not vertically moving
again like whenever it is vertically moving
and so you can substitute.
So, start with this basic equation depending
on whatever information is given in the problem
you try to use it. So, for the free surface
you have dp=0. So, when you have dp =0 you
have 
this equal to 0. So, this is rho omega square
r. This is – rho g so you have dg/dr =omega
square r/g. You can now integrate this with
respect to r. So, on integration what follows
let us just complete that.
Integrating z = omega square r square/2g +
constant of integration Z0. You can set of
the constant of integration z0by choosing
a reference such that when r=0, z=0 so if
we choose this as our origin of the coordinate
so this is r and this is z. So, if you have
at r =0, z=0 then z0=0, z will be omega square
r square. This formula you have encountered
earlier. So, we can clearly see that it is
an equation like a of a parabola.
So, it’s a parabolic of evolution. It is
a three dimensional situation. And you can
calculate the other things just similar to
what we did in the previous case with an understanding
of what that again if it does not spill whatever
was the volume that should be conserved. So,
how can you calculate the volume? So, initial
volume you can calculate, initial volume is
pi R square *h0. What is the final volume?
Final volume is whatever is the depression
say delta h + the volume of the shaded parabola
and I leave it on you as an exercise if you
calculate this shaded volume you will see
that it will just be half of the volume of
the circumscribing cylinder. That means if
this height is h1 then the shaded volume will
be half of pi r square h1. By simple integration
you can find out this volume.
So, by equating the initial and the final
volume you can find out totally the deflected
configuration. Again you may have to check
that h1 + delta h if it becomes greater than
the height of the original tank then it will
spill and spilling again may have two different
cases. You may have in one case the cylinder
is rotating in such a way that you have spilling
but still it is having an interface shape
like this.
Again it may so happen that this rotating
so fast that it is spilling but only the part
of the parabolic of revolution is within the
cylinder. So then it is an imaginary parabolic
revolution even outside that you have to consider.
Find out what is the volume that is there
inside. So, it all depends on the rotational
speed the given dimension and so on. So, it
is not just like a fixed formula but you know
what is the basic principle.
We have discussed enough numbers of examples
to see that what is the basic principle and
that basic principle should guide you to find
out that what is the case. Now, if it is totally
closed cylinder as a final example say we
have cylinder that is totally closed and filled
up with liquid and rotated with respect to
this axis. What is the total force on the
top AB. Again the basic principle is the same.
You just use this dp/ dr formula to find out
how pressure varies with r. Of course with
the reference say r=0, p=p0 with the reference
because pressure you always calculate with
the reference. So, you know how p varies with
r by integrating this with respect to r. When
we integrate with respect to r, g is fixed.
So, then you can find out the total force
by taking an element. Here an element will
be 2 pi r dr.
So, integrating over that you can find out
the total force of the top surface. Total
force on the bottom surface will be that plus
the weight of the fluid which is there. So,
we will close this discussion by seeing just
one may be example where we will see that
how this type of vertex motion is generated
in practice. You see that this type of a vertex
motion that is there I means it is not exactly
a paraboloid of revolution. But it is by rotating
the fluid in a container.
Why it is not exactly a paraboloid of revolution
is because we have neglected here the viscous
effects. The shear between various fluid layers
we have assumed that the fluid rotates like
a rigid body. In reality that is not the case
and we will look into these situations more
emphatically whenever we are discussing viscous
motion. So, we close our discussion on fluid
statics with this. Thank you very much.
