So, we shall move onto the next module of
the course on dynamics. I will start with
Particle Dynamics. It is mostly like a review
because you know quite a bit of particle dynamics.
And you know, in this course, I have always
tried to give an importance to modeling to
the extent possible.
I have a situation like this; the car is travelling
on a road. I have another situation; it is
very unfortunate that this met with an accident.
And I would like you to observe what happened
to the car, ok.
Now, the question is, I have two situations;
how do I model it for the purpose of analysis
of situation one? What is the minimum requirement
to model situation two? You can always complicate
the problem. Bringing in complication is very
very simple. Engineers have to solve the problem
with suitable idealizations. And obviously,
this is a huge body; in a car, you know, four
or five people can sit. The situation was;
it was moving in a straight line; how can
I model for the purpose of dynamic analysis?
Can I model it as a particle? You can model
it as a particle. That is how you have solved
the problems. I can model this as a particle.
Again, the same car; that has met with an
accident, I wanted you to see what happened
in the accident; what is it that you have
noticed? There is a particular thing that
happened to the car. I brought it out, very
deliberately. The car was rotating. Turning
turtle is very common; if we apply brakes
very heavily, the car can turn turtle. So,
what you see there is; it has a rotation.
The moment the object has a rotation, you
can no longer analyze it as a particle. You
will have to bring in a minimum of idealizing
that as a rigid body.
In an actual situation, you know you will
also have to bring in the deformability; you
must say that this is elastic. You may want
to find out for a car moving on a speed breaker
and we have funny speed breakers here. And
you know, if you have to find out the design
of suspension system and all other aspects
of the car, you may have to consider that
as deformable, make your life complicated
and have complicated mathematics.
And one of the simplest things that you come
across in 
particle dynamics is projectile motion. A
particle that moves in vertical plane with
some initial velocity but its acceleration
is always the free-fall acceleration g is
a projectile. You have learnt it thoroughly,
solved thousands of problems in your earlier
exposure. Nevertheless, we will have a quick
overview of it.
And in the case of statics, I have said that
when I apply a load, try to visualize; how
the load could be applied? That brings in
certain amount of engineering visualization.
In dynamics, you must visualize; when I say
it has some initial velocity, how the initial
velocity has come about? Think about it. That
will make you understand, what is the real
problem situation? Or when you have a real
problem situation, you can use that information
to model it appropriately. It will help you
to understand the problem better.
And you have a fun fact here, you have many
games. Although, I have not played this game,
my student was so enthusiastic; he had put
it in this. And probably, the game developers
would have used projectile motion to simulate
all this movement of the bird because the
bird here does not have its own energy. You
are only making it move and you have to recognize
that this is a free motion.
Let us understand this little more. And before
we get into it, let us see the commonly used
equations. Please write this down even though
you know these equations. There are many symbolisms
that are used; I can have the velocities represented
as v and u; then I have the velocity v as
u plus a t. I can also represent it as v and
v-naught. And I will also bring in the initial
distance, initial velocity and so on.
So, I have a relation relating v to u plus
a t, v squared equal to u squared plus 2 a
s and the distance travelled is s equal to
u t plus half at squared. And in all these
cases, you should note that the acceleration
is constant. You have a variety of problems
in engineering where you have acceleration
to be constant, fine. So, here large class
of problems can be handled. And you can also
have other symbolisms like u x, u y, v x,
v y; so many symbolisms, people have adopted.
And you should remember that these equations
are valid only if the acceleration is constant.
Let us understand the vertical motion; I have
a particle that is dropped from this height
h. Now, I do it differently in the blue diagram.
I will drop two balls; one ball horizontally
and one ball vertically; that means, I give
a horizontal velocity. Now, you have to visualize;
from where this horizontal velocity can come?
We will also solve one problem where I will
tell you how this could be visualized, fine.
What do you think will happen? When I drop
the ball vertically down; when I put it like
this. What happens? You look at the animation.
What is the striking observation? Both of
them reached the ground down at the same time.
You can easily explain it from your projectile
motion, fine. But for a commoner, it may be
very difficult for them visualize. They will
only say, if I drop the ball straight, it
will reach faster; if I throw the ball horizontally,
it will take more time to reach. That is how
your normal thinking will happen; your so-called
intuition cannot help. And if you look at
the analysis, you should recognize that the
horizontal motion and vertical motion are
independent of each other.
Though the horizontal motion of this ball
is different, the vertical motion is identical
to the red ball or the blue ball that is dropped
straight. You exploit the equation that you
have in both the cases u y is zero and then
I have y t equal to zero into t plus half
g t squared. So, I have this as half g t squared
and get t equal to root of 2 h by g. So, the
time taken to reach the ground is the same.
It is a very important learning when you learn
particle dynamics, to start with.
And let us move on to projectile motion. And
here again, recognize the horizontal and vertical
motions are independent. I use different symbols
for different problems so that you will get
accustomed to different symbolisms; people
have used different symbols. So, I have a
ball which is thrown at a velocity u oriented
at an angle theta and this goes in a parabola.
Even before you know the problem, you understand
all these quantities.
So, I have what is called the range of this
projectile; what is the maximum height it
can reach? And what is the condition when
it reaches the maximum height? And we can
also derive what the equation of this curve
is? It is all very very simple. We are only
recapitulating what you have learnt already.
So, let me write for horizontal motion separately
and vertical motion separately. There is no
acceleration in the horizontal motion. I have
acceleration in the vertical motion; that
is minus g. I have given the reference axis
this way.
And I can find out the x-component of the
velocity as u cos theta and y-component of
the velocity is u sine theta minus g t. And
x is given as u cos theta into t and y is
given as u sine theta into t minus half g
t squared; all of this you know. You have
to treat these motions independent. And then,
you can also go about getting the expression
for the range, for the height, for the time
taken, etc.
So, if I substitute t equal to x by u cos
theta, I am in a position to get an expression
of this curve. When I simplify this, I get
the expression for y as x tan theta minus
g x squared divided by 2 u squared cos squared
theta and this is nothing but a second degree
curve. So, I have y equal to a x plus b x
squared. This is nothing but an equation of
a parabola. That is the reason why you put
the curve, the projectile motion as a parabola.
And what is the other understanding that we
have?
Let us say, T is the time taken to reach the
ground; that means, from here to this, it
takes time T. And mind you, in all these analyses,
we have not considered the effect of air around
it; there is no friction acting on the projectile,
fine. And it will reach the peak point at
T by 2. So, I have an expression relating
u y to u sine theta minus g t. From this,
I can easily get an expression for what is
the time taken t equal to 2 u sine theta g.
Then I can also find out range from the expression
for x, u cos theta into t. I get the range
as u squared sine 2 theta divided by g. Now,
I have to find out, what is the height H?
So, if you consider the x-motion and y-motion
independently, you can derive all these quantities
comfortably; there is no great difficulty
in doing that.
To summarize, I have u y zero is u sine theta,
u y t by 2 equal to zero, a y is minus g and
I have this expression; so, from this, I am
in a position to calculate the height; maximum
height possible as u squared sine squared
theta divide by 2 g. And we have already got
the expression for T and also the expression
for R. What is the range? What is the time
taken? What is the maximum height reached?
And this is one of the first things you do
when you learn particle dynamics and this
has many many applications in day-to-day life;
right from cricket onwards, you have many
many applications. That is the reason why
you have learnt all of this.
And this is a very nice animation. It visually
summarizes what happens in a projectile. See,
I start with a velocity and look here, the
vertical component goes to zero; only horizontal
component remains and another point is; for
the entire motion, the horizontal component
remains same.
Visually you understand. See, visual understanding
and appreciation will stay much longer in
your system than just mathematical expressions.
And when it comes and hits the ground, it
is not coming and sitting. It is coming and
hitting with a velocity. It is not coming
and simply hitting; coming and landing on
this with zero velocity, ok. So, this gives
a nice appreciation of what happens to a projectile.
And now, we will solve some interesting problems.
These are all fictitious problems. To bring
in some life, you attach it to some kind of
idealistic situation. And you know, the monkeys
in IIT are very, very intelligent, fine. So,
if there is a gunshot, this fellow will go
and hide. That is a natural instinct; when
there is some danger, you would like to find
out the least possible distance where you
can go and hide.
But, in the problem, it states: startled by
the sound, the monkey drops at the same time
the bullet leaves the barrel of the gun. So,
it is fictitious, fine. That means, it has
a free-fall. The question is, if the monkey
has a free-fall, will the bullet hit the monkey?
That is what we have to investigate, fine.
So, the question is, if it falls, will the
bullet come and hit the monkey? A very interesting
problem and you can analyze it comfortably.
First thing what you learn is, when you aim
at the monkey and fire your gunshot, it is
not going to go travel straight. Look, the
people who are not trained in mechanics will
not be able to visualize this; they will only
think that it will go straight. Because of
acceleration due to gravity, it has a projectile
motion, fine. Now, we have fictitious situation.
The moment it heard its sound, it starts falling
down freely. So, this blue; this green ball
falls down straight and this red ball follows
the projectile motion.
So, let us look at what happens to the monkey.
I have a coordinate system like this for the
monkey and I write the motion s equal to v
t plus half g t squared and it starts from
rest so v is zero. So, I get s equal to half
g t squared and I can also write this s with
respect to the ground; that is the height
y which is nothing but h minus s. So, I have
an expression; at any point in time t, the
height of the monkey from the ground is h
minus half g t squared.
Now, what we will have to see is; what is
the height of the gunshot which follows a
projectile motion from the ground? If these
two heights match, then there will be a hit
on the monkey. It is a fictitious problem
but very interesting to learn concepts related
to projectile motion. So, for the bullet,
I can write the motion as u sine theta into
t minus half g t squared; mainly because it
gets released from the gun with an initial
velocity, whereas, the monkey drops down with
zero initial velocity.
And what are the other things that I know?
x equal to u cos theta into t and from the
geometry, I can get tan theta equal to h by
x equal to h by u cos theta into t. So, when
I go and replace this here, I get this as;
I get from this simplification, I can get
h equal to u sine theta into t. I will replace
this here and I find that the height of the
bullet from the ground is h minus half g t
squared. At any point in time, no matter what
the distance is between the gun and the monkey;
whatever distance; from the strength of our
mathematics; provided there is a free-fall,
provided that the monkey has a good hearing
and there is no friction, there could be a
possible hit. That is what your mathematics
says. So, it is a very interesting but fictitious
problem.
And now, you have another interesting problem
which relates your circular motion and projectile
motion. You might have played this in your
school days. A boy whirls a stone in horizontal
circle of radius 1.5 meter and at a height
of 2 meters above the ground. The string breaks
and the stone flies off horizontally and strikes
the ground after travelling a horizontal distance
of 10 meters; somebody has done the measurement
and given it to you.
So, this gives you an idea. When I say something
moves with an initial velocity, you get an
idea I have the swirling of the stone like
this. And suddenly, the string snaps and it
comes with an initial velocity. When I have
a gun, you have a gun powder which provides;
which bursts and provides the initial velocity
to the bullet. And here, the initial velocity
comes from circular motion.
You know, in many of the problems that we
have solved, I have consciously taken effort
to bring in the physical reality to the extent
possible to aid your visualization, fine.
How do I go and solve the problem? I have
a circular motion initially and I have a projectile
motion to follow. And you know, if you look
at a problem in dynamics, there are many,
many ways of solving it. There is no one unique
method to solve the problem.
The method which is comfortable to you, you
should adopt and then solve. And we will solve
it from first principles; from the simplest
way of handling it. And I can look at the
horizontal and vertical motion as independent
and investigate; what is the height from which
the ball has to fall on the ground? That is
h equal to u t plus half g t squared and I
get t equal to root of 2 h by g.
So, for the time taken to reach the ground,
when I substitute the quantities, this turns
out to be 0.64 seconds. So, I have looked
at the vertical motion initially and I have
calculated the time. And let me look at the
horizontal motion next. See, before I go into
this, I would like to caution; the reason
why I said, think from where the initial velocity
comes. When you want to coin a problem, you
simply say there is an initial velocity but
you should also visualize; in which of the
applications you can get that initial velocity?
See, one of the commonest things that I find
is; you know, young mothers; when they bring
their children to the school in IIT, they
come at a terrific speed and the children
are not clinging onto the vehicle or the mother
firmly, when they come. Imagine, if there
is some obstruction and the boy or the girl
travelling in the scooter falls off; they
do not realize that it can be a cause for
danger, ok. If you are travelling at slow
speed of 20 kilometers per hour; when you
dislodge yourself, you have 20 kilometers
per hour.
When you are firmly fixed to the body, nothing
happens to you. That is the reason why they
want you to wear seatbelt in a car or when
you are sitting in a scooter, you should put
the child properly. See, elders will know
how to manage but children will not know how
to cling onto it. And if you really observe
the scooters, children are not kept in a safe
position. And if you go abroad, there are
regulations on: what should be the child seat?
How they should be secured? They will never
be allowed to seat in the front seat of a
car; a car is much safer than a two-wheeler,
fine.
So, as educated students, when you have the
role of parenting, please apply your mechanics
understanding and handle them safely. It is
very unsafe; what you see on the common ground
is very unsafe. And you should also realize,
there are thumb rules which say; if you are
travelling at a speed of 20 kilometers and
have a fall, it is equivalent to falling from
a two-storey building.
You realize that when you go to a storey building.
You will not jump off; you know that it is
very dangerous. On the other hand, you think
20 kilometers per hour is not a very great
speed. And 40 kilometers per hour is like
falling from fourth floor. So, do not make
your life very uncomfortable. You learn mechanics;
bring in that understanding in your day-to-day
dealings, fine. It is not for just learning
and getting the grades, it should percolate
into your lifestyle, where it is needed.
So, considering the horizontal motion, I get
v t equal to 10 and v is 10 by t I get 15.63
meters per second. And now, I have this as
a circular motion, I can easily find out what
is the centripetal acceleration a equal to
v squared by r; I get it as 163 meters per
second squared. So, it is very simple but
it is also very interesting. So, always visualize;
when I say there is an initial velocity; what
is the source of that initial velocity? That
will make you a better engineer because you
start visualizing and then relate it to physical
situations, fine.
I can also solve this by another method with
just one equation, provided I know the equation
which I would not recommend because you do
not have to remember anything. But here, I
should know the projectile path. If I know
the projectile path and I substitute the value,
I get the velocity straightforward. I get
the same answer; I cannot get any different
answer. There are multiple ways that you can
solve a given problem. Choose a method that
is comfortable to you. I would always prefer
a method where I start from first principles
where I do not have to remember anything,
fine. So, that is my recommendation to you.
So far, we have never looked at the air-resistance.
What happens when there is air-resistance?
I am not going to solve it mathematically
but I would give you an idea as to what would
happen to the range. Obviously, the range
will get shortened and that needs to be accommodated.
When you are designing a missile; you want
to hit your enemy, you should calculate all
of this. For intercontinental ballistic missiles,
you need to even bring in relativity theory
and then calculate.
If they travel for 500 kilometers, it is one
thing; if they travel for 20,000 kilometers,
the story is different. You cannot apply the
same mechanics for short distance and long
distance. So, there, it is all the more important
for you to learn mechanics.
And let us look at another situation. We have
seen what happens when I drop a ball? Because
of acceleration due to gravity, it acquires
speed. Now, I do not drop the ball from a
short distance; I go to a very tall building
and drop the ball. What happens? So, you have
the concept of terminal velocity. You should
thank God. God has created all of this. It
is not that anything you drop will acquire
infinite velocity, ok. The velocity cannot
reach beyond a particular point.
When an object starts falling, it starts with
a zero velocity. It gains velocity as it falls
due to gravitation but you should also look
at air resistance. See, in engineering, we
bring in features when they are significant.
When you have learnt projectile motion, there
was air around you; you have not done that
experiment in vacuum. When it is not significant,
you do not bring it into your modeling. When
it is significant, you bring it into your
modeling and then understand what happens.
Drag resistance develops. And you have a drag
force Fd and that is given as one half of
rho v squared A into Cd where, Cd is the drag
coefficient depending on which fluid you are
moving in, A is the projected area, v is the
velocity and this keeps increasing. So, it
reaches a situation that it balances the weight
of the object. At one point of time, the drag
force equals the gravitational force and the
body reaches equilibrium.
That is what you see here, F d equal to mg
and I get this final expression. And I can
also find out, what is the limiting velocity?
It will move at a constant velocity downwards
and this is equal to square root of 2 mg divided
by rho A into Cd.
So, this is the velocity of the body at this
point and you have called this as terminal
velocity. People have also calculated for
different situations, what is the terminal
velocity possible? And if you take a shot-put,
it can have terminal velocity of 145 meters
per second. It takes 2500 meters for it to
achieve this; for it to reach 95 percent of
the velocity. And for a skydiver, he will
have a highest velocity of 60 meters per second.
You need to be trained to be a skydiver, everybody
cannot do skydiving. Because now, you have
all these sports have come. In abroad, you
also have skydiving possible; people go for
skydiving and come back. And if your health
is not alright then do not try out all of
this.
There is a difference between a skydiver and
a parachutist; parachutist has a velocity
of only 5 meters per second, whereas a skydiver;
he will have a terminal velocity of 60 meters
per second and your body should be strong
enough to withstand. It is constitutional.
Strong does not mean age. You can have our
Honorable past President Dr. Abdul Kalam travelling
in a jet airplane and come back safely at
the age close to 75 or 76 while even a young
boy can have a problem, ok. So, do not try
out skydiving without knowing this.
And another interesting thing is the raindrop.
See, all of these are falling at some known
height. Raindrop, it comes from the clouds.
Imagine it acquires velocity; then you will
have injections. You will not go out in the
rain; you cannot have a nice film song in
rain fine, ok. Thankfully, it has a terminal
velocity of only 7 meters per second and it
acquires this within a very short distance
of 6 meters, ok.
See, in dynamics, what you do is; in problems
wherever you can see absolute motion, you
try to solve based on absolute motion. That
is what we have seen earlier but you can handle
a variety of practical problems if you understand
the concept of relative motion, fine. And
you know, you had recent floods in Kerala.
So, our air force was dropping food to all
of these people. The question here is; for
an observer on the Earth; he would see the
packet following a parabolic path because
it follows a projectile motion. What would
be the motion perceived by the person sitting
in the aircraft? That is what we have to find
out, ok.
Let us understand the concept of relative
motion. So, I have given the axis like this.
The particle is labeled as A and this moving
object is labeled as B. And let us analyze
the motion. And it is given in the problem,
the plane moves at an acceleration a. You
are given the acceleration a in the x-direction.
It is not moving with constant velocity. It
is picking up speed because it has to go from
one place to another place fast enough, fine.
So, the velocity of plane is v equal to u
plus a t.
And velocity of body in x-direction is this;
because at this point when it dropped, it
has the same velocity as the plane. So, this
is what I said; from where does the initial
velocity come? If it is dislodged, from the
moving object, it has a same velocity as that,
ok.
So, the velocity of body in x-direction is
u x equal to u and the velocity of body in
x direction with respect to plane is u x with
respect to p is u minus v p. And substituting
this relevant expression, I get this as minus
at and for position of body in x-direction
with respect to plane; because I want to find
out what happens to this with respect to plane?
the x-position is given as minus half a t
squared and I can also find out that velocity
of body in y direction u y equal to g t and
vertical velocity of body with respect to
plane is same; there is no change because
it is moving in the same height.
Position of body in y-direction with respect
to plane is y half g t squared. Now, I combine
these two expressions, I get an equation as
y plus g by a x equal to zero. So, this is
nothing but a straight line.
So, if a person is sitting in the airplane,
he would see the object falling as a straight
line; this is what he would see. So, this
is just to give you an idea on relative motion.
Then we move onto see; how to apply Newton’s
law? You will also have to appreciate an inertial
frame of reference. And lifts are very, very
common these days. And here, you have a few
storeys but if you go to New York, you have
75 storeys, 100 storeys. So, you really travel
for a very, very long distance and they have
very efficient lifts which will cover 60 floors
in 10 seconds or 15 seconds and so on, fine.
So, there, you will definitely feel that there
is an acceleration, steady motion and deceleration;
all that, you will have an opportunity to
feel. In a short distance of two floors or
four floors, you may not feel that. Now, what
I do is; I have a spring scale in the lift
and the man is standing on it. And let us
analyze, what happens to this. As again, we
just model this as a particle and I have this
weight acting downwards. And I have the reaction
from the floor; the floor here is the spring
scale. The idea is that we can also note down
what is the reading of the spring scale. Is
the diagram complete?
This is where you have to note the difference
between statics and problems dealing with
dynamics. The particle is having an acceleration
a; you have to indicate that very clearly.
Only then, the diagram is complete, fine.
And I am going to write the Newton’s law,
standing on the ground. I am not going to
stand inside the lift and then write it. Here,
I can say F equal to m a, fine. I can write
the motion is in the y-direction, so I can
put F y equal to m a y and from this, I get
R minus m g equal to m a and what is the surprising
answer you are getting? I get R equal to m
(g plus a). What would the spring scale show
as the reading on it? It will only show the
reaction, fine.
So, you will find the weight of the man is
more than what he actually is because the
spring scale shows only the force that it
exerts on the man. So, the apparent weight
shown by the scale is higher. So, do not get
annoyed when you are on weight reduction because
that is a common thing that is happening;
governments are putting levies on people who
are overweight. And you are on a regimen of
dieting and then you find that you are overweight;
as fun, ok. Do not be annoyed. You can also
be underweight when you are travelling in
a lift, ok. So, now, you comment and think
about it; what will spring scale show when
the lift is coming down and decelerating,
ok.
Let us move onto another situation; it has
come to this level and then it is moving at
a constant velocity. Imagine that I am going
for 100 floors in Rockfeller tower. And let
us write the free body diagram. I take this
as a particle, I have reaction R and I have
m g. Now, I recognize that this is moving
at a constant velocity; there is no acceleration.
We have learnt from Galileo’s principle
of relativity; moving at a constant velocity
or stationary is one and the same. So, in
this case, I will have Fy equal to m a y and
I have R minus m g equal to zero. And I get
the reaction equal to the actual weight.
Now, let me think of a third situation. It
is coming down accelerating; let us see what
happens; I put the free body diagram. I have
this. The diagram is not complete; I should
also put in the acceleration. Until then,
this diagram is not complete. We are dealing
with problems in dynamics. Then, simply apply
these expressions Fy equal to m a y R minus
m g equal to minus m a because the acceleration
is in the opposite direction of the coordinate
axis; so I get this as r equal to m g minus
a.
So, if you were on a weight reduction regimen,
do not be happy that your weight is low when
you are coming down in the lift. You have
to look at the actual weight and then decide
about it. Apparent weight shown by the scale
is lower. And now, comment on what will the
spring scale show when the lift is going up
and decelerating.
Let me move onto another unfortunate situation
that the lift is snapped. So, it is falling
at what acceleration? And you know now, all
these space agencies, they want to generate
more resources; they want to send people and
feel zero gravity so that millionaires can
fund some of their space programs. Earlier,
only astronauts were doing it. You get something
similar to this in this case; when I put the
equation Fy equal to m a y. And when I put
the free fall acceleration, I get an answer
the reaction is zero, ok. So, you feel weightless;
apparent weight shown by the scale is zero.
Now, let us go back and analyze from a different
perspective. Now, I have an observer inside
the lift; noting down what happens in the
spring scale. It is going up and accelerating,
fine. We have already seen this diagram earlier.
The same diagram and I will also have to put
the acceleration. And F equal to m a y, I
can apply only from an inertial frame of reference.
When I am standing on the ground, it is an
initial frame of reference. It is not having
any acceleration. I can write this expression
R minus m g equal to m a and the spring scale
shows only the force that it exerts on the
man.
However, for an observer standing outside;
for an observer standing inside the lift,
the expression for net force is again mathematically
R minus m g. We are not qualifying what the
value R is; what is the value of this one.
This is same for the observer inside the lift
also. Can the person inside the lift use this
and write the Newton’s law? Because, this
is where people make a mistake. See, you will
have to understand when to apply Newton’s
law? You can apply it only in an inertial
frame of reference. For an observer going
in a lift with an acceleration, does he perceive
any acceleration? He will perceive no acceleration.
He cannot write the Newton’s law there.
One will get only absurd answers as the lift
is accelerating and not an inertial frame
of reference. Very, very important for you
to note, ok. So, for an observer in the lift
acceleration is zero. So, apply the Newton’s
law very carefully. See, this what we are
going to learn, how do we analyze relative
motion from various perspectives for a rigid
body? Then move onto kinetics where we apply
the Newton’s law properly, fine.
And we would also see one more interesting
application; because in India, the roads have
become very good now and also have very powerful
cars. And often, we hear accidents happen
by young drivers, fine. You learn mechanics;
please apply your mechanics when you drive.
And you have to recognize that when the car
has to come to a stop, you need a finite distance.
Though, in advertisement, they say that we
have a good braking system; it is going to
be instant. Do not believe all of that. People
are definitely developing braking systems;
you have ABS, so many parameters, people are
finding out so that it does not become a turtle.
All that, they are doing; let them do all
of it.
And if you look at this, I can analyze this
as a particle. So, I get the distance required
for it to stop as a function of the velocity.
And I look at this, I have normal reaction;
I have the weight of the car. And you have
a frictional force generated because of your
braking action between the tire and the road.
And this has an acceleration like this. It
is actually deceleration but we do it like
this and mathematics will help me to get the
correct sign. So, this happens when f is f
max.
So, f max is mu times N; I have mu s m g and
I am in a position to get the expression which
relates the distance as a function of the
quantities finally. The minimum distance required
is a function of the velocity as well as the
coefficient of friction.
See now, we are having rains in Chennai. You
have to be very careful when you are driving;
because of water, the friction decreases,
fine. And what you find is that if the friction
decreases by half, the d minimum increases
by two times.
So, you have to understand; there is a minimum
stopping distance that is required. So, do
not drive the vehicle on a rainy day, carelessly.
You also have a table which shows this. I
do not think any driver in India maintains;
when they travel at 40 kilometers per hour,
a distance of 7.87 meters. So, accidents are
waiting to happen. In many of the industries,
they will say accidents do not happen, they
are caused, ok; so, this one way of bringing
in safety.
So, you have to recognize; you have to maintain
a respectful distance on the road; in case
of any eventuality and you need to brake your
car. You often hear in the Yamuna expressway,
you have a pile of cars. When somebody applies
a brake, you have 100 cars come and hit which
used to happen in the west, occasionally.
Now, it is happening in India. That is why
you need all these cars to have crash worthiness.
Earlier, we were travelling at very slow speed;
we were not travelling at very high speed.
So, once you are travelling at a high speed,
you have to bring in airbags for your protection,
seatbelts for your protection. And as engineers,
when you drive the car, recognize that there
is a minimum stopping distance, which is the
function of the velocity as well as the friction;
coefficient of friction. And the friction
drops when you have water on the ground because
of rain. So, you educate yourself. And educate
your friends and relatives and have a safe
drive.
So, in this class, we have looked at an overview
of particle dynamics. And we find that a variety
of situations can be modeled as a particle.
And we have also looked at what is known as
terminal velocity and we have also looked
at how to apply the Newton’s law when the
lift is accelerating or decelerating. And
finally, we have also looked at what is the
minimum distance for a car to stop which is
very, very essential; particularly India,
nowadays because we have very good roads and
very highly powerful cars. Thank you.
