In the last module, we have discussed the
first step of the kinematics analysis under Meme it

the title displacement analysis. Today, we
shall start the second step of kinematics Meme it
analysis which you call velocity analysis. Meme it
By velocity analysis, we mean that for a given
mechanism at a particular configuration, if Meme it
the input velocity characteristic is specified,
we should be able to determine the velocity Meme it
characteristics of all other links of the
same mechanism. Let me repeat, by velocity Meme it
analysis we determine the velocity characteristics
which may be linear velocity of a point or Meme it

angular velocity of a link of all the links
of a mechanism at a given configuration, when Meme it
the input velocity is specified. So, by velocity
analysis basically we mean the velocity analysis Meme it
of interconnected rigid bodies in plane motion.
Because we are discussing on the planar linkage Meme it
but before getting in to the details of this
velocity analysis of interconnected rigid Meme it
body, let me recapitulate some basic points
of kinematics of one rigid body which is under Meme it
plane motion. Meme it

First let me recapitulate some basic points
of the plane motion of a rigid body. We should Meme it
remember that in plane motion the angular
motion of a rigid body is the same as that Meme it
of a line on that rigid body in the plane
of motion. This I will explain a little later. Meme it
The second point is that the angular velocity
and angular acceleration vectors of that rigid Meme it
body are always perpendicular to the plane
of motion. Because the rigid bodies in plane Meme it
motion, all angular quantities will be represented
by the vectors which are perpendicular to Meme it

this plane of motion. We should also remember
that a general plane motion consists of translation Meme it
of the rigid body plus the rotation of that
rigid body. In a mechanism, a particular link Meme it
may be undergoing pure rotary motion and another
link may be undergoing pure translatory motion, Meme it
but there may be some links which undergo
both translation and rotation simultaneously. Meme it
What do you mean by translation is that, if
a body is under pure translation, then all Meme it
points on that body have identical motion.
The difference between the motion of two points Meme it

on the rigid body is entirely due to its rotation.
So translation means identical motion of all Meme it
points and if there is rotation then there
will be difference in the motion between various Meme it
points of the same rigid body. Meme it
Let me now explain these points with reference
to this figure. Here we have a rigid body Meme it
in plane motion and the plane of motion is
the plane of the screen. Let at the time instant Meme it
t, the rigid body is in this configuration
point. After the small interval of time, say Meme it

delta t, that is at time t plus delta t, the
rigid body comes to this configuration which Meme it
is given by 2. So during time delta t the
rigid body moves from configuration 1 to configuration Meme it
2. You consider two points on the rigid body
namely B and C. We say B1 C1 at the locations Meme it
of B and C at configuration 1 and B2 and C2
at the locations of same points B and C in Meme it
configuration 2. Now, this is a general plane
motion of this rigid body. As I said, the Meme it
plane of motion is the plane of the screen.
Let us say X and Y. XY is the plane of motion. Meme it

So all angular quantities like angular velocity
or angular acceleration of this rigid body Meme it
will be perpendicular to the plane of motion
that is, perpendicular to the screen, which Meme it
we will denote by the Z axis with unit vector
K. As I said earlier, for this plane motion, Meme it
the angular motion of the rigid body will
be same as the angular motion of a line on Meme it
this rigid body for example, the line B1 C1.
So what do you see? That the line B1 C1 has Meme it
come to the position B2 C2. Meme it

The line B1 C1 has moved and also rotated
and occupied the position B2 C2. So if I draw Meme it
B2 C1 prime which is parallel to B1 C1, then
we can see rotation of this line B1 C1 during Meme it
this time delta t is given by delta theta.
So let us consider this general plane motion Meme it
of this rigid body during the time interval
delta t when it moved from configuration one Meme it
to configuration two. This can be decomposed
under two headings namely, a translation which Meme it
takes from configuration one to configuration
one prime, when all the points moved by the Meme it

identical amount and in identical direction.
That means, B1 goes to B2, C1 goes to C1 prime Meme it
by the same amount. All points of the rigid
body move identically and the rigid body goes Meme it
from one to one prime that is what we call
translation. Then from one prime to two, the Meme it
point B2 does not move, it undergoes rotation
by an amount delta theta. Meme it
So, the angular velocity vector of this rigid
body can be defined as: omega is limit delta Meme it
t tending to 0, delta theta by delta t. The
direction of this vector is along Z which Meme it

I write as the unit vector k. Angular velocity
vector comes out as theta dot that is the Meme it
derivative of theta with respective to time
in the direction k. Similarly, angular acceleration Meme it
of this can be written as alpha which is second
derivative of theta with respect to time that Meme it
is theta double dot and in the direction k.
So this is what we mean by the general plane Meme it
motion of the rigid body in the XY plane with
the angular velocity vector omega and the Meme it
angular acceleration vector alpha. Meme it

Now as we see during the translation, there
is no difference between the motion of various Meme it
points of the rigid body like B and C. They
are under identical motion. The difference Meme it
in the motion between C and D is entirely
due to rotation that is due to this angular Meme it
velocity vector omega. Positive k means counter
clockwise direction. Now, the difference of Meme it
velocity between the points C and D, let me
write it as difference of velocity of point Meme it
C minus velocity of point B. This is the difference
of velocity we use the symbol, VCB. As we Meme it

have noticed, this is entirely due to the
rotation of the rigid body that means the Meme it
point C with respect to B moves in a circle
and the angular velocity of line BC is omega. Meme it
So, from our basic knowledge of dynamics or
kinematics of rigid bodies, we know that for Meme it
the circular motion, this difference of velocity
is given by omega cross the vector BC. So, Meme it
this is one very basic result which we will
be using very frequently in our subsequent Meme it
velocity analysis, that if we consider two
different points B and C on the same rigid Meme it

body whose angular velocity is omega, then
the difference of velocity VC minus VB is Meme it
given by omega cross BC. Meme it
Now we consider the difference of acceleration
between these two points namely C and B. As Meme it
you have already noticed with respect to B,
the C is going in a circle and the angular Meme it
velocity because BC never changes, so C goes
on the circle with B as center and the angular Meme it
velocity of the line BC is omega and the angular
acceleration of line BC is alpha, which is Meme it

nothing but the angular velocity and angular
acceleration of the rigid body. Meme it
So now we can write the difference of acceleration
between these two points namely aC minus aB Meme it
that as before I used the symbol, aCB which
is the different between the acceleration Meme it
of the two points namely C and B on the same
rigid body. There is the centripetal acceleration Meme it
due to the circular motion which is given
by omega cross omega cross BC. There is a Meme it
tangential component which along the vector
t1 that is given by alpha cross BC. Because Meme it

omega vector is along the k direction and
the BC vector is in the XY plane, it can be Meme it
easily shown that this will always turn out
to be, the first term will be omega square Meme it
CB that is towards B from C, that is the vector
CB and the magnitude will be omega square Meme it
into BC. Plus the tangential component will
be alpha cross BC. That means, it is be perpendicular Meme it
to the vector BC is the sense of alpha, if
alpha is counter clockwise then it will be Meme it
this direction, if alpha is clockwise then
it will be in the opposite direction. So we Meme it

get the relationship between the difference
of two points of the velocity like B and C Meme it
given by omega cross BC. Meme it
At this stage, we should know that if we know
the velocity of any point say namely VB which Meme it
is a two dimensional vector that is the two
components of VB and the vector omega that Meme it
is the angular velocity of the rigid body.
Then I can determine velocity of any other Meme it
point namely C using this relation that is
velocity of any other point C, VC can be obtain Meme it

from VB plus omega cross BC. You should know
that there can be only three independent angles Meme it
namely, two components of VB and omega. Then
velocity of all other points on that same Meme it
rigid body is completely determined by this
equation. Meme it
Similarly, if we are given acceleration of
a particular point aB which has two components Meme it
that is two unknowns and the angular velocity
vector omega and the angular acceleration Meme it
vector alpha then acceleration of any point
C can be determined using this equation, that Meme it

aC will be aB plus these two components. So
there can be four unknowns, namely two components Meme it
of aB and omega and alpha. Then acceleration
of all other points on that same rigid body Meme it
is completely known. This information as we
shall see later will be very useful for velocity Meme it
analysis of mechanisms. Meme it
Now that it is clear, how to correlate the
velocity and acceleration of two points on Meme it
the same rigid body, let us develop another
very important concept for velocity analysis Meme it

that is called instantaneous center of velocity
or in short IC. Meme it
For a rigid body in plane motion, we will
be able to show that always there exists a Meme it
point on the rigid body of course in plane
of motion whose instantaneous velocity is Meme it
0 and this point is called the instantaneous
center of velocity or IC. It must be mentioned Meme it
that this IC may lie outside the physical
boundary of the rigid body but it lies in Meme it
the plane of motion of that rigid body. If
a rigid body is under pure translation, then Meme it

obviously no point has 0 velocity, in that
case we say that the IC lies at infinity in Meme it
the direction perpendicular to the direction
of translation. Just like a prismatic pair, Meme it
we consider a revolute pair in a direction
perpendicular to the direction of sliding. Meme it
Let me now show that there does exist such
a point that which we call the instantaneous Meme it
center of velocity. To prove the existence
of instantaneous center of velocity for a Meme it
rigid body under plane motion, let us consider
this figure. Meme it

We consider two points namely, A and B whose
velocities are given by VB and VA. We have Meme it
already seen that if VA is given and omega
is given, then VB is completely determined. Meme it
That means both VB and VA having four components
cannot be independent. Now, let me draw perpendicular Meme it
to velocity vector. At B I draw the line BI
which is at 90 degree to VB. At A, I draw Meme it
a BI line AI which is at 90 degree to the
velocity vector VA. These two perpendicular Meme it
lines drawn at A and B meet at I. We will
show that the velocity of this point I of Meme it

the rigid body at this configuration must
be zero and that is what is called the instantaneous Meme it
center of velocity for this configuration.
To show that VI is 0, let us consider the Meme it
component of the velocity VA in the direction
of AB. That will be VA cos theta. VA cos theta Meme it
is the velocity A in the direction AB. The
component of VB in the direction of AB is Meme it
VB cos phi where phi is the angle between
the direction VB and the direction AB. Now Meme it
the distance AB never changes that means,
these two velocity components must be same. Meme it

That is the component of VA along AB must
be same as the component of VB along AB because Meme it
the distance AB never changes. So I can put
VA cos theta is same as VB cos phi. Meme it
Let us consider the triangle IAB, because
IA is perpendicular to VA and IB is perpendicular Meme it
to VB, we can easily see that this angle is
pi by 2 minus phi and this angle that is between Meme it
IA and AB is phi by 2 minus theta. Considering
this triangle IAB if we use the sin law, I Meme it
can write, IA divided by sin of IBA that is
sin of pi by 2 minus phi that is cos phi is Meme it

same as IB divided by sin of the angle IAB
that is sin of phi by 2 minus theta which Meme it
is cos theta. So we get from this equation
VA divided by cos phi is same as VB divided Meme it
by cos theta. From the geometry IA by cos
phi is same as IB by cos theta. So I can easily Meme it
write, VA by IA is equal to VB by IB. This
ratio I call omega that is the angular velocity Meme it
of the rigid body. So VA is IA into omega
and VB is IB into omega and the velocity vectors Meme it
VB and VA are by construction perpendicular
to the direction of IA and IB, which means Meme it

I is the center of rotation. At this instant
the body is rotating about the point I with Meme it
angular velocity omega.
In fact, you can also prove that velocity Meme it
of I is zero by considering the velocity of
point I in two different directions namely, Meme it
IA and IB. To find the velocity of point I
along IA, I come from the point A. Velocity Meme it
of A is along VA, so the component along IA
is zero and I and A are two points from the Meme it
same rigid body. So velocity of I along IA
is also 0, because the relative velocity of Meme it

A and I along AI is 0. Velocity of I with
respective to A along AI is also 0, so velocity Meme it
of I along this direction AI is 0. Coming
from the point V, I see that the velocity Meme it
of the point V is perpendicular to IB. So
component of VB along BI is 0.Velocity of Meme it
I along VI is same as the velocity B along
BI. So velocity of I along BI is also zero. Meme it
Thus the velocity of the point I is zero along
two directions, along AI and it is also zero Meme it
along BI. So if VI is a two dimensional vector
and if it is zero in two different directions, Meme it

then the velocity of I must be zero. We get,
velocity of the point I which I write VI is Meme it
zero at this configuration and this is what
is called the instantaneous center of velocity. Meme it
Now that we have explained the concept of
instantaneous center of velocity, let us extend Meme it
this concept between two moving bodies. Previously
we said one rigid body which is moving and Meme it
there exists an instantaneous center of velocity.
Now let us consider two moving bodies say Meme it
number two and three. Meme it

Then we can define a relative instantaneous
center say, I23 as a point on body three having Meme it
zero relative velocity, that is same absolute
velocity with respect to a coincident point Meme it
on body two. Let me repeat, we are talking
unto two moving rigid bodies say, number two Meme it
and three. We define a relative instantaneous
center between these two bodies two and three, Meme it
we call it I23. What is I23? This is a point
on body three having zero relative velocity Meme it
with respect to a coincident point on body
2. So from this definition it is very clear Meme it

that I23 is same as I32. In light of this
relatives instantaneous center, we can redefine Meme it
the absolute instantaneous center of velocity
which we discussed earlier as the relative Meme it
IC with respective fixed link one. Meme it
As you know in mechanism, there will be a
number of rigid bodies, interconnected rigid Meme it
bodies in plane motion and there is a fixed
length which we always number one. So out Meme it
of this any N-link mechanism, we will have
number of relative instantaneous centers and Meme it

the relative instantaneous centers with respective
to the fixed link one are nothing but the Meme it
absolute instantaneous center of velocity
that is at that instant that particular body Meme it
is rotating about that instantaneous center.
So if you have the N-link mechanism then according Meme it
to this kind of definition of relative IC's,
we will have NC2 that is 2 links taken at Meme it
a time. How many different combination we
can make? That is Nc2 or N into (N minus one) Meme it
by two relative IC's. So for an N-link mechanism,
we will have N into N minus one by two so Meme it

many number of relative instantaneous centers.
At this stage, let us discuss how by looking Meme it
at the interconnections between various rigid
bodies in a mechanism, we can determine the Meme it
relative instantaneous centers. As you know,
in a planer mechanism, rigid bodies will be Meme it
connected either by revolute pair or by prismatic
pair or by some kind of higher pair. Meme it
If the two bodies are connected by a revolute
pair, then the relative instantaneous centers Meme it
between those two bodies is obviously at the
center of the R-pair because as the mechanism Meme it

moves, the center of the R-pair between these
two bodies always move together. So there Meme it
is no relative velocity between those two
coincident points of the two rigid bodies Meme it
connected by the R-pair. The center of the
R-pair would be the relative IC. Meme it
If the two bodies are connected by a prismatic
pair, then the relative instantaneous center Meme it
is at infinity in a direction perpendicular
to the axis of the P-pair. Because if the Meme it
two bodies are the connected by a prismatic
pair, then as we know the relative motion Meme it

is a pure translation and in case of translation,
the instantaneous center lies at infinity Meme it
in a direction perpendicular to the direction
of sliding. Meme it
We can recall that a P-pair is nothing but
an R-pair at the infinity in a direction perpendicular Meme it
to the direction of relative sliding. So these
two sentences imply that for a P-pair the Meme it
relative IC is at infinity in a direction
perpendicular to the axis of the P-pair. Meme it
Suppose we have a higher pair with a kinematic
constant that, one body rolls over the other Meme it

without slipping. That means it is not a normal
higher pair, it is the higher pair with a Meme it
kinematic constant that there is no slip at
the point of contact. So from no slip condition, Meme it
I know that velocity of the two points of
contact belonging to two different bodies Meme it
must be the same. That is the condition for
no slip. Consequently, the point of contact Meme it
itself will be the relative IC between those
two bodies which are connected by a higher Meme it
pair without slip. But for general higher
pair where slipping is allowed, that means Meme it

if one body simultaneously rolls and slides
over the other when it constitutes a general Meme it
higher pair, then the relative IC lies somewhere
on the common normal at the point of contact. Meme it
We cannot determine it exactly where on this
line, common normal will lie that required Meme it
some other information, but we know that it
must lie on the common normal at the point Meme it
of contact. At this stage, let me say that
there is a very important theorem called Aronhold-Kennedy Meme it
Theorem of three centers, which will be very
useful to determine all the relative IC's Meme it

of a mechanism along with the information
that we have just now discussed. That is, Meme it
at the R-pair the relative IC is at the center
of the R-pair. At the P-pair, the relative Meme it
IC is at infinity in a direction perpendicular
to the direction of relative sliding. If it Meme it
is a higher pair without any slip, then it
is at the point of contact and if it is a Meme it
general higher pair with slip, then it is
somewhere along the common normal at the point Meme it
of contact. With this information plus the
application of the Aronhold-Kennedy Theorem, Meme it

we should be able to determine all the relative
IC's of planar mechanism having up to six Meme it
links which is sufficient for all purposes. Meme it
So next, let me state and prove the Aronhold-Kennedy
Theorem. Aronhold-Kennedy Theorem of three Meme it
centers is the following, three rigid bodies
are in relative motion. We are talking now Meme it
three rigid bodies which are in relative motion
but they may not be interconnected, but with Meme it
three bodies we have three relative IC's because
3C2, that is C into C minus one divided by Meme it

two gives you three. So three rigid bodies
in relative motion, there are the three relativities Meme it
the IC's. The Aronhold-Kennedy Theorem says
that these three relatives IC's must lie on Meme it
a line. That is three relative IC's are collinear.
We shall prove the Aronhold-Kennedy Theorem Meme it
by contradiction. That means, we will assume
them to be non-collinear and show that is Meme it
impossible, which means they have to be collinear. Meme it
To prove Aronhold-Kennedy Theorem, let as
consider three rigid bodies namely, one, two Meme it

and three which are in relative planar motion.
Because we are considering relative motion, Meme it
we do not lose any generality, if I assume
one of the bodies to be fixed. So let me assume Meme it
that body one is fixed as indicated by this
hatched lines. So to consider the relative Meme it
planar motion between three rigid bodies namely,
one, two, and three, I consider the body one Meme it
to be fixed. Let us say that, I12 is the relative
instantaneous center for body number one and Meme it
two. Similarly, I13 is the relative instantaneous
center between body number one and three. Meme it

That means, at this instance, body two is
under pure rotation about the point I12 with Meme it
respect to body one. Similarly, body three
is under pure rotation about the point I13 Meme it
with respect to body one. Meme it
Now let us say that Aronhold-Kennedy Theorem
is wrong. That means, the instantaneous center Meme it
I23 need not lie on the line joining I12 and
I13. Let us assume it is at this location Meme it
I23 prime. Now if I23 prime is the relative
instantaneous center between body number two Meme it

and three, then it must have the same velocity
whether I consider it to be a point on body Meme it
two or I consider it to be a point on body
three. That is the definition of relative Meme it
IC I23. If I consider it to be a point on
body two, it must have the same velocity if Meme it
I consider it to be a point on body three.
That is the definition of I23. Meme it
Suppose it lies here, then this vector velocity
I23 if I consider it to be a point on body Meme it
two must be in this direction. That is, perpendicular
to the line joining I12 and I23. This is the Meme it

line I12, I23, then velocity of this point
as a point on body two which is rotating about Meme it
this point must be in this direction perpendicular
to this line. But if I consider this point Meme it
to be a point on body three which is rotating
at this instant about I13, then the velocity Meme it
must be perpendicular to this line joining
I13 and I23 that is in the direction. These Meme it
two directions are different so this equation
can never hold good. This equation can never Meme it
hold good because these two vectors V I23
as a point on body three and V I23 as a point Meme it

on body two are in different directions. So
we immediately get a contradiction. These Meme it
two vectors can act same direction only if
I23 lies somewhere on this line, say here Meme it
or there or there or there. Then only the
rotation about I12 if I consider to be a point Meme it
on body two which will be perpendicular to
this line and if I consider to be a point Meme it
an body three which is rotating about I13
also is perpendicular to this line. So these Meme it
two velocity vectors can have the same direction
which is necessary condition for it to be Meme it

the relative instantaneous center I23. So
we have proved Aronhold-Kennedy Theorem of Meme it
three centers by contradiction and now I will
demonstrate by use of a module. Meme it
Let me try to explain both the concept of
relative IC and the application of Aronhold-Kennedy Meme it
Theorem with the help of this model. In this
model as we see, we are talking of three rigid Meme it
bodies in relative motion namely one which
is the fixed link, two this is this red body Meme it
and three that is this blue body. Meme it

So now, 1 and 2 is connected by a revolute
pair here, so immediately I locate the relative Meme it
instantaneous center 1 2 at this point. Similarly
body 3 which is connected by a revolute pair Meme it
with 1 at this point I consider this as IC,
a relative instantaneous center 1 3. Now 3 Meme it
and 2 has a higher pair at this point of contact,
so we are talking of three rigid bodies namely, Meme it
1, 2, and 3. 1 and 2 has a revolute pair here,
1 and 3 has a revolute pair there which means Meme it
I have located two relative instantaneous
centers namely 1 2 and 1 3. Now the relative Meme it

instantaneous center 2 3, I know must lie
along the common normal at this point of contact. Meme it
This is a general higher pair. We have already
discussed that in case of general higher pair Meme it
the relative instantaneous centers in this
case between body two and body three that Meme it
is I23 must lie on this common normal. We
also know from Aronhold-Kennedy Theorem, that Meme it
1 2, 1 3 and 2 3 must be collinear. That is
how we determine the location of 2 3 at the Meme it
intersection of the common normal and this
line joining 1 2 and 1 3. 1 2, 1 3 and 2 3 Meme it

must be collinear and 2 3 must lie on this
common normal that determines this location Meme it
2 3. So this is the relative instantaneous
center 2 3 between two moving bodies two and Meme it
three. By definition at this instant, if I
consider this point to be a point on body Meme it
two, that is this red link or a point on body
three that is this blue link has the same Meme it
velocity. Meme it
So as we will notice that around this configuration,
these two points body 2 and body 3 at the Meme it

location of I23 move with same velocity. However
these two points now as we see have moved Meme it
differently. This is a point 2 3 on body 2
and this is the point 2 3 on body 3. At this Meme it
location, at this configuration, their velocities
are same. So around this configuration, they Meme it
move almost with same velocity. At this instant,
with exactly same velocity around that region, Meme it
almost with same velocity, so there is not
much of separation, however when they move Meme it
a lot as we see that point two three, which
was here as a point of body 3 has come here Meme it

and as a point on body 2 has come here. I
hope, this clears both the concept of relative Meme it
instantaneous center and the application of
Aronhold-Kennedy Theorem which we shall continue Meme it
subsequently. Meme it
Let me now summarize what we have covered
in this lecture. We started with the general Meme it
plane motion of a single rigid body and I
have shown that, if we know the velocity of Meme it
any point on that rigid body and the angular
velocity of the rigid body, then we can determine Meme it

the velocity of all other points that rigid
body. Then we have shown that if we know the Meme it
acceleration of the point on that rigid body
and the angular velocity and angular acceleration Meme it
of the rigid body, then acceleration of all
other points on that rigid body can be determined. Meme it
Next we have shown the existence of the so
called instantaneous center of velocity. That Meme it
means, for a rigid body in plane motion, there
exists a point about which at that instant Meme it
the body can be considered under pure rotation.
Then we extended the concept of instantaneous Meme it

center to relative motion of between two moving
bodies. Then we defined relative instantaneous Meme it
center exactly the same way. Meme it
Finally, we have stated and proved the Aronhold-Kennedy
Theorem of 3 centers and finally we have shown Meme it
the concept of relative instantaneous center
and the application of Aronhold-Kennedy Theorem Meme it
for determining all the relative IC's of a
particular mechanism. Meme it
In our next lecture, we will continue with
the application of this instantaneous center, Meme it

their determination and their use for continuing
the velocity analysis of planar mechanisms. Meme it