Now let’s just look at a simple example.
We are almost on the equator of the earth.
The earth is rotating. Let’s assume that
this is the polar direction, its rotating.
The earth is rotating, I am standing on this.
Will I have an acceleration? Let’s look
at it. This is the center and it is rotating
with an angular velocity. Is it angular velocity
or angular acceleration? Almost angular velocity
because it’s a constant angular rotation
that it is having and it’s not very difficult
for me to find out. It rotates about 360 degrees
in a matter of only 24 hours. I can always
find out omega. Now the question I have here
is, is this man who is standing, I am standing
on the equator, am I accelerating? Let’s
say the fixed frame of reference is with respect
to the center. The answer is yes.
Because as the earth is rotating, I am also
going all round it. What would be the acceleration
here? If this is given by, this is B, this
is A. I am going to use the same notation
so that it’s easy. So that I have rB with
respect to A. A is also the origin, so it
makes it easy rA equals 0. Therefore if I
have to find out r dot of B, it is nothing
but r dot of B with respect to A. r double
dot of B with respect to A is what I am interested
in. That will be given by, as we have already
derived alpha cross rB with respect to A,
which is the tangential component of acceleration
plus the inward radial component of acceleration
is given by this. Omega is known, omega is
known, r of B with respect to A is also known
here. I can find out and therefore this is
not zero whereas this is zero. A man standing
on the equator will see an angular acceleration
inward. When he moves from here to here and
that’s given by this. Rest of it is not
very difficult for you to find out what will
be... the direction is very simple.
The direction is, it is directed along, acceleration
is directed along the centre of the earth
or towards the axis. Is this clear? Let’s
see if we can answer this question. What is
the man doing? The man is just standing still,
so no problem. What if the man is moving on
the equator?
Now you have two things, the earth itself
is moving and there is this man who is also
moving. Probably he may be accelerating along
lets say the equatorial line. Can I find out
what is the actual acceleration, velocity
of that particular person? Let’s do that
exercise and see whether we can do that. What
is the difference right now compared to what
we did earlier? In the earlier exercise this
point B and point A had one particular relationship.
What was that? What was the particular relationship
that we knew? Let’s make it a little more
complicated so that you can think a little
differently.
Let’s say he fitted himself, like our shastra,
he fitted himself with a particular rocket
booster and he is going like this. He just
shot up like a bird and he is going like this
with an acceleration. Can I find out the actual
acceleration of this particular person? Let’s
try to answer the question. One another way
of looking at it is supposing I have a body
and let’s say there is a small marble on
this body which is moving independent of or
let’s say there is an insect. There is 
an insect that is moving around with a velocity
or acceleration.
Now remember this body also is moving and
accelerating and can I find out with respect
to fixed frame of reference, what would be
the acceleration of this particular insect?
Why am I interested in acceleration of that
particular insect or velocity of that insect
with respect to the fixed frame? What’s
the reason, why can’t I just live with what
ever acceleration that I am seeing, when I
am standing on this? Because the acceleration
that I am seeing, when I am standing on this
rigid body is a relative acceleration whereas
the Newton’s law can be applied only if
I have been able to fix a particular frame,
a non-accelerating frame.
In this particular case those are these capital
X, capital Y frame. If I can define with respect
to this frame then whatever acceleration that
I find out, can be used for f equals m a equation.
The entire exercise is to make sure that I
am able to use the equations of motion, that’s
the idea. One thing that I already know is
if I have a point 
and relative to lets say this is A, relative
to this point A we already found out what
would be the relationship of any other particle
that is fixed to this rigid body with respect
to A. Now what we want to do is we want to
find out, what would be the acceleration of
this guy, actual acceleration or acceleration
with respect to fixed frame rather than a
frame attached to this, which means first
thing I have to do is I have to attach a frame
with this body.
Let’s attach a frame. So we attach a frame
like this. What about this frame? If this
rigid body is rotating, this frame will also
be rotating, that understanding should be
clear. Let me call this as small x, small
y and this can be taken to be r, the position
of the insect with respect to this frame that
is attached to the rigid body. If this insect
were not moving, it’s easy for me to find
out what it is. Because this distance is maintained.
The only difference that I see is that this
distance is not maintained as far as the insect
is concerned. I have to consider that in my
equations. Let’s proceed to do that exercise.
Let’s take this as rA, let’s say this
is, shall we call this as B. Let’s just
call for now this particular insect to be
positioned at B and this B may be moving with
respect to A. Let’s say this is r and this
is rA and this is rB, like what we have done.
How do I find out rB? rB is given by rA which
is a location of the point at which we have
fixed the axis plus r. What is r? r tells
me the location of this particular insect.
Now I have to find out the velocity and velocity
is not very difficult. If I have to find out
velocity of this insect with respect to fixed
frame, what would that be? It would be r dot
A plus r dot. This is very simple, this is
nothing but VA because I have already defined
r A with respect to fixed axis.
There is no problem in defining VA. Now we
have an r dot, remember this has a magnitude
and a direction. I have VA plus, I am going
to now write this as r dot magnitude or rather
r magnitude times the direction, the entire
thing dot. It makes it easy for us now to
understand, which is nothing but VA and if
I differentiate by parts, it is the derivative
of this. Let me just call that as r dot times
r cap plus magnitude of r times r cap dot.
Is it confusing? It’s fine. Let me just
write it a little differently, plus r magnitude
times r cap dot.
What is this? You are taking for example if
I am doing this, it is nothing but, what is
this? This is i, j, k taken here and this
is nothing but the magnitude of r dot. It
is x dot i plus y dot j plus z dot k. Do you
understand this? This is what it means. If
I write this as this, let
me call this as Vrelative. Why? Because if
I am standing on this and having this fixed
frame of reference on the rigid body and I
am seeing the insect moving, that will be
the
derivative of x, y and z and therefore I am
going to call that as Vrelative. What I know
now is this is Vrelative. Correct? There is
one more term here, which is the magnitude
of r times the derivative of the direction
r cap. How do I find out the derivative of
the direction? If
there is a rotation occurring, so I know that
this is rotating. If it is rotating, how do
I find out this? It is nothing but omega cross
r cap. So using that I will get VB equals
V A plus
Vrelative plus omega cross r, that’s it.
As you can see if I don’t have this, it
is nothing but VB equals VA plus omega cross
r which is something that we already knew
for a rigid body. What is added to that is
Vrelative.
Now I seek to find out accelerations, acceleration
of B and I want to relate through acceleration
of A. Acceleration of B I want to know, given
acceleration of A and the rigid body movement.
How do I get it? Let me just translate this
over there because it’s easier. What I have
here is V dot B equals V dot A plus V dot
relative plus omega cross r dot, simple. Now
instead of writing V dot relative, I am now
going to use what we have already written
here, which is the r dot time’s r cap. It
is VA plus r dot times r cap plus omega dot
cross r plus omega cross r dot. Is this okay?
Remember V dot relative is a vector again
and if I take a derivative of that, I will
have to take the derivative of the magnitude
as well as the direction and that’s an important
thing we have to do, this whole thing dot.
V dot relative, V dot B equals V dot A. Here
V dot A . Let’s just rewrite V dot B equals
V dot A plus, this v dot relative again I
have to now look at the magnitude and the
direction.
Let me write it as either this way or this
way, I have V dot magnitude, let me write
it like
this V dot magnitude times, lets write it
as Vrelative magnitude times V direction.
This whole thing taken derivative of plus
omega dot is nothing but alpha, alpha cross
r that
does not change. Again I have omega cross
r dot. Now this is acceleration of B, this
is acceleration of A plus this I have to now
resolve, how do I do that? This will give
me two
terms, one term is related to derivative of
Vrelative, so I will get an arelative which
is nothing but a derivative of Vrelative magnitude
times the direction which is arelative. I
am going to write it as arelative plus Vrelative
magnitude times which will give me what? Directly
I can write this to be omega cross. Is that
okay? Plus alpha cross r plus, what is this?
What is r dot? r dot again I can take from
here, I have already written r dot here, r
dot
happens to be Vrelative plus this. I have
omega cross, I know that this is the thing
that is related to r dot. I will write it
as Vrelative plus omega cross r. Is this clear?
This is clear because if you look at this
particular expression r dot, this r dot manifests
as V relative
plus omega cross r. Therefore here that’s
how it manifests. Now we just have to group
them together. What do we get is nothing but
A plus let me write this, plus notice here
I
have two terms omega cross, omega cross r,
omega cross Vrelative. Let me write omega
cross omega cross r plus I have this term
which is arelative plus what else is left
out? Omega cross Vrelative, omega cross Vrelative
added together is two times omega cross Vrelative.
What was this? What is this that you know
of? aB equals aA plus alpha cross r plus omega
cross omega cross r, it is even that we found
out for a rigid body. If there is an insect
that is moving with an acceleration or whatever,
it will be the relative acceleration with
respect to the rigid body on which I am standing
stationary and then looking at it, plus one
more term. And this one more term is, what
is counter intuitive.
One would think, if I have a point over here
and this is a fixed point with respect to
the point that I am looking at as the pivot
and when it is rotating, this acceleration
is given by aA plus alpha cross r plus omega
cross omega cross r. Therefore if I have an
insect that is
moving with an acceleration, I should expect
an additional arelative. But what is counter
intuitive is, there is an additional term
that comes here which is omega cross Vrelative.
This counter intuitive acceleration is often
called as coriolis acceleration. Is this clear?
Otherwise it’s a very simple idea that I
can understand, but upon doing this derivation,
we find that there is an additional acceleration.
We already found out what would be the direction
of this, what would be the direction of
this. The direction of arelative is not very
difficult to understand. It is along the direction
of r and how about this? It is perpendicular
to the direction of omega which is nothing
but
the perpendicular to the board and it is also
perpendicular to Vrelative. Understood? If
I take a perpendicular direction with respect
to those two, so this and this, this is the
direction in which I will expect coriolis
acceleration to occur.
Now a simple example. Let’s say I have on
top of a tower, imagine I am the tower. On
top of it, there is a rotating restaurant.
This is a rotating restaurant, this is the
center of it. There is a person that moves
with a constant velocity. What will happen
is, there will be an acceleration that he
will experience which will be perpendicular
to his velocity that he sees and perpendicular
to the direction of the rotation. So perpendicular
to this and let’s say he is moving towards
the center, perpendicular to this and perpendicular
to this is a direction like this. He will
see himself moving, being pushed like this
and that’s what is called Coriolis force.
When you are in the relative realm, you will
seemingly see that acceleration. When you
are outside with the fixed frame of reference,
this seems an acceleration that does not give
any sense of feeling that you see when you
are inside the restaurant.
