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Today, I am going to discuss about the kinematics of a particle. Now, what is a kinematics?
Kinematics is the area of mechanics concerned with the study of motion of particles and
rigid bodies without consideration of what has caused the motion. When, when we take
into consideration the factors causing the motion, the area of the study is called dynamics.
Now, just see this slider crank mechanism. Here, when I am moving the crank, then this
slider is moving. If I know that what is the speed of the slider I need to have, then for
that how much rpm crank should rotate? If this type of question is posed to us then
we must know the kinematics. See here, if I am moving the crank the slider is moving.
If I am rotating it by, let us say, 3 rpm; then, what is the slider speed? Or conversely,
if I am moving the slider at certain speed what will be the crank speed? Such types of
questions are answered by kinematics. In kinematics, we are not concerned with the forces; we are
concerned with the displacement, velocity and acceleration. All of you are familiar
with these terms.
Now, quite often, we only deal with the kinematics. For example, if I want to design this mechanism
and let us say, this is just a toy mechanism, then probably we need not carry out extensive
force analysis. But nevertheless, we have to do kinematic analysis in order to find
out the relation between the motion of various things. Similarly, suppose you are going from
Pan Bazaar to Paltan Bazaar and you are going in a cycle rickshaw, you are concerned at
what speed the rickshaw moves or if you are going by train from Delhi to Bombay, you must
know that what is the speed of the train. In these types of questions you take interest.
Generally, when we design certain machines then only we pay attention into these forces.
Therefore, an engineer is usually concerned with the forces whereas kinematics has got
much wider application. Another point is that when we take into account the forces and we
then we must know the motion also. Therefore, dynamics encompasses kinematics also. Dynamics
encompasses kinematics because Newton's law says force is equal to mass into acceleration.
If you want to calculate the force, you must know the acceleration also. So therefore,
the study of kinematics is very important.
Now, why are we talking about kinematics of a particle? Now, in this is you see, that
kinematics this one, kinematics is the, in kinematics of a particle, what is a particle?
Particle is the concept which is having a mass but not having any size. A rigid body
can be considered as a combination of a particle. Therefore, the concepts learned for a particle
can be helpful in understanding the kinematics of a rigid body also. Moreover, sometimes,
what happens is that bodies can be treated as particles. For example, I play this animation
for you. Let me just play this car. So, this is here, in this ...
Another thing I observed, one minute, that
you know you cannot continue kinematics in this area. One slide you missed that particle I said,
know, that you have to take the velocity and acceleration. There was one slide in which
the definition of a particle was given but that slide he missed. But anyway, I have spoken
about it. But one slide we missed. I said that you know that in one, kinematics is the
area of this one. After that second point was there. That slide he missed. When this
type of thing happens, I become nervous and this is the thing . This is why, I am not
able to do. See, he has missed one and since, here I was not able to move to this one, see.
This one, now I am moing. See, it is very difficult. Now I am able to do. I am continuing
from here. Let me play the animation here. This is the animation and this is a car. Why
this pen, sometimes it comes sometimes it does not come. See, sometimes it is coming,
sometimes it is not coming. No, animation; some it comes, sometimes it does not comes.
Ake thum ne hata be dhiya oos slide dheko particle vala. Particle and ... this thing.
Aye thumnae hataya tha, tho us kae baadh may bola ki esko laga, per vo particle ko hatabi
dhiya thumnae. Mistake, thumnae blank slide laakhae es may daal dhi. Yae jo hae ho gaya.
Kinematics of a particle tho particle ki definition thi. Hai na, hai na. Tho this is the thing.
Aab is may kya hotha ki animation nahi hotha. Thi isliyae esmay .... Aab aaya, kabi kabi
aaa jaatha hai and kabhi kabhi nahi aatha hai. Dhekiyae, abhi Abhi hai, abhi aajayaega.
Kesi hai? Soft hai? What soft hai? Ake minute dhekiyae.
I am starting now. So, now, I am playing the animation. See, this car is moving on a long
highway. So therefore, compared to the distance it travels, the size of the car can be treated
as a small particle. Therefore, we can apply kinematics of particles to this problem.
Next, I will discuss about how we can find out velocity and acceleration of a particle.
If path of motion of a particle is known and one knows the position of the particle as
a function of time, then the velocity of the particle can be easily calculated.
How do we calculate this? This was done during Newton's time. Now, let us say that we have
got a article P. This particle goes into P dash in time delta t. I will play this animation.
Here, this is, you think this is, this animation, ... See, this is creating problem, hai na?
See, I am just, see either mouse should be given to me hai na? See, you are also does
not able to. See, ometimes it comes, sometimes it does not comes. Kya, you are all trained
with technique, I think there is problem with ... mera mathlab, paesa kam rakhne hai? Nahi,
but ... may koye dhoosra technique hai. I think that here. see, yae jaisae kiya, yaha,
ake minute, acha, acha, acha, okay okay okay fine. You go there.
Now, I am playing one animation again. See, this is a particle, this one and this has
gone to this one from in delta t time, it has gone to P dash time. It has gone very
fast, animation should have been somewhat slow. Now here, this distance PP dash, is
PP dash, this line will be basically delta r. It is a vector and it is the difference
of 2 vectors. The vector, one vector is OP dash. This is O; this one point is origin
O and this is P dash and this is OP. So, the difference between the two vectors r t plus
delta t minus r t will be PP dash. P P dash is a vector. It is denoted by delta r. I will
always be denoting the vectors by boldfaced letters. In your notebook, probably, you can
put an arrow on top of the alphabet to indicate that it is a vector.
Now, arc PP dash is the total distance travelled by the particle and that is denoted by delta
s.
One has to understand the difference between a vector and this distance delta s. Delta
s is not a vector whereas delta r is a vector. Now, so, velocity is the rate of change of
the position of the particle. Calculus was invented in order to define velocity properly
by Newton. In the xyz reference system, if at time t the particle is at P with position
vector r(t) and at time t plus delta t, the particle is at P dash with position vector
r(t plus delta t) then the velocity vector V will be defined as V will be limit delta
t tends to 0, that means the time interval is very very small. This is r (t plus delta
t) minus r(t) divided by delta t. Now, r (t plus delta t) minus r(t) is nothing but the
vector PP dash delta r. Therefore, this is a vector and if I divide it by a scalar quantity
delta t entire thing is a vector. Therefore, the velocity is a vector only.
Now, this is indicated in the language of calculus, this is called dr by dt, where dr
does not mean the product of d and r. It simply indicates a differential operator. So, you
cannot cancel d and d here and say that it is equal to r by t. Just like sin theta does
not mean sin into theta, it is basically symbol for the operator. Similarly, here your d is
basically one operator. This can also be written as dr by ds into ds by dt, where I have multiplied
by a small change in the arc length divided by the same quantity. This is basically a
type of chain of derivatives.
Next in this figure, delta r is the displacement given by the chord PP dash and delta s is
the distance given by arc PP dash. In the limit, delta r divided by delta s becomes
a unit vector tangent to the trajectory because in the limit, if you know that time is very
small and distance between P and P dash is very small, in that case, this delta length
chord will be equal to arc. Therefore, the magnitude of delta r is same as delta s. Nevertheless,
delta r is a vector and its direction is tangent to the direction at the point P. If you find
out the tangent at P then this delta r is in the direction of the tangent. That is how
we define the tangent also.
So therefore, therefore, we can write velocity equal to dr by dt. Same thing can be written
as ds by dt into dr by ds. ds by dt can be called as speed. It is a scalar quantity;
rate of change of distance with respect to time multiplied by unit tangent vector. Here
cap et is a unit tangent vector. Therefore, velocity of a particle is a vector having
a magnitude equal to the speed of a particle and direction tangent to the trajectory. That
means, path of the motion of the particle.
Now, let me discuss the acceleration. The acceleration vector of a particle is basically
the rate of change of the velocity vector. a is equal to dv by dt which can also be written
as d square r by dt square where d square, it is not indicating any square; but rather,
it is indicating the double derivative. These symbols have been used - one must be very
clear about that. d square r by dt square means the double derivative of position vector
with respect to time. We can obtain a also in the following manner. We can write a is
equal to dv by dt. Again, dv by ds into ds by dt; ds by dt is nothing but the magnitude
of the velocity which is the speed. So, this is the magnitude of the velocity into dv by
ds.
So this is Now, since the velocity is a vector, therefore it can be expressed in component
forms also very easily. V(t) is equal to dr(t) by dt, rate of change of the position vector,
r is basically xi plus yj plus zk, where unit vector i is along x-axis, unit vector j is
along y-axis and unit vector k is along z-axis. So therefore, dr(t) by dt is dx by dt i plus
dy by dt j plus dz by dt k which can be written as x dot (t)i y dot (t)j and z dot (t)k. x
dot t indicates that this component is a function of time. Similarly, the other component is
also a function of time. Dot indicates the derivative with respect to time; single dot
will indicate first derivative and two dots on the alphabet will indicate double derivative.
Acceleration can be written as d square rt by dt square is equal to a. This is equal
to x double dot (t)i plus y double dot (t)j plus z double dot (t)k where double dot indicates
the double differentiation with respect to time.
Now, if the acceleration is given as a function of time, then velocity and displacements can
be found by integration. Integration is the reverse process of differentiation. For example,
suppose a particle moves in a straight line from rest with an acceleration of t, with
an acceleration of t where t is the time; in 5 seconds, where will the particle be from
the starting point?
If this is the problem, then we can solve it like this. Here, the motion is one-dimensional.
So we can do away with the vectors. In one dimension only algebraic quantities are good
enough. x can be positive or negative. If it does not have, we need not give any directions.
So, given a is equal to d square x by dt square, which is basically the x component of the
vector or if the motion is one dimensional, then the acceleration itself that has been
given as a time t. If we integrate it twice, we get x is equal to t cube by 6 plus At plus
B, where A and B are the constants of integration. The constants of integration can be found
by putting the initial conditions. Since the particle starts from rest at the origin, therefore
at time t is equal to 0, x is equal to 0 and x dot is equal to 0.
Now, this gives A is equal to B is equal to 0. Thus, x is equal to t cube by 6. So, we
have found the position of the particle with respect to time by this formula. With respect
to the origin at what distance is the particle? We can substitute the value of t, we can find
out. For example, at t is equal to 5, x is equal to 125 by 6 units. In a psi system,
it will be meter.
Now, let us discuss the example of a two dimensional motion. We will study the motion of projectiles.
Now, freely projected particles which move under the combined effect of a vertical and
horizontal motion are called projectiles. It has only downward vertical acceleration
due to gravity. On earth's surface, the value of this gravitational acceleration is 9.81m
per second square.
The motion of the projectile on the earth's surface is of great interest. This finds application
in the defense and in the space applications. Following are some examples of a projectile
motion. Suppose, we have got a missile, missile starting from one point and see this is the
thing, I am starting the animation here. I am starting, missile is going and now it is
reaching at the top, then it is falling down and it has gone there. I will play it again.
This is starting; this is the thing, it is going, reaching the top maximum height, coming
down again. If we know our target, then we must properly know at what velocity and what
angle I should project the missile so that it properly falls on the target. Here, a missile
fired from one point reaches at the target at some distance away.
Another example that if we see this sloping surface, on this I am putting a ball and if
the ball starts rolling down, if there was no gravity then the motion of the ball would
have been in straight path. This is gravity free path in which ball moves from this point
and it goes up to this point. However, because there is gravity is present, it will deviate
from that straight path and ball will start moving downward. Therefore, its path will
not be straight, rather it will be forwarding in x direction but at the same time it will
come down in the y direction. Let us see this in animation. This is So, this is the path
of this one and it has fallen here. We can play the animation again. This is this is
the thing. See it is falling down.
Similarly, this one; a basketball player is throwing a ball in the basket. He must know,
with what velocity and what angle he should throw the ball so that it falls into the basket.
Of course, he does not do any calculation. He does not study the kinematics but he does
that mental calculation and by that he is able to throw the ball. Let me see this animation,
it is going like this. We can play this animation again, yes this has fallen there. These are
the examples of the projectile motion.
Now, we will discuss some terminology of the projectile motion. Here, this shows the trajectory
of a projectile. One particle has started from point A and it has reached another point.
Now, velocity of projection is the velocity with which the particle is projected. This
is in meter per second.
Suppose in this case, u is the velocity of the projection. Then here, the range of the
particle is also shown. Range is the horizontal distance it travels, so horizontal range.
Let me just Angle of projection is the angle between the direction of projection and horizontal
direction is called as angle of projection alpha. The trajectory is the path traced by
the projectile. Here, it is the trajectory and generally, it has parabolic shape.
Then, horizontal range - the horizontal distance through which the projectile travels in its
flight is called the horizontal range. In this case, the range is this. Then, time of
flight; the time interval during which the projectile is in motion is called time of
flight.
Next, we will now study the case when the projectile flies over an inclined plane, because
moving on the horizontal plane is a special case of moving on inclined plane.
Horizontal plane is also an inclined plane with angle of inclination from horizontal
equal to 0. So, a horizontal surface is a special case of an inclined plane. Let AB
be a plane inclined at an angle beta to the horizontal as shown. Now, a projectile is
fired up the plane from point A with initial velocity u meter per second and angle alpha.
This is actually angle alpha. This will be angle alpha; actually this is not A, this
is alpha. The range on an inclined plane AB and the time of flight are to be determined
if you want to find out the range. Suppose, I know the initial velocity u that is velocity
of projection, then I know the projection angle alpha, the angle which it makes from
the horizontal surface.
I also know the inclination angle beta. My task is to find out the time of flight and
also the range AB. For that purpose, for the given projectile, we have to find out, if
we take the x-axis along the inclined plane and y-axis perpendicular to that then we can
study the motion in x y plane. The acceleration is downward and that is half about minus g
but it has the components along the plane and normal to the plane. Normal to the plane,
the component of the acceleration ay that is indicated by y double dot is equal to minus
g cos beta; y double dot means d square y by dt square is the y component of the acceleration.
Similarly, in the horizontal direction ax is equal to x double dot is equal to minus
g sin beta. If you square them, a square plus ax square will be equal to g square. Therefore,
the magnitude of the acceleration is g only.
Now, this one, so we can integrate these both equations with respect to time and we get
the equation y dot which is basically the velocity component in the vertical direction
that is equal to minus g cos beta into t plus A1 where capital A1 is a constant. Similarly,
x dot is equal to minus g sin beta t plus A2 where A2 is another constant. At time t
is equal to 0, vy that is the vertical velocity y dot is equal to u sin alpha minus beta because
this is the component which is normal to the plane and the horizontal component along the
plane will be vx, which is equal to x dot which is equal to u cos alpha minus beta.
Therefore, if we put these values here then we can get the values of the constants. We
get A1 is equal to u sin alpha minus beta and A2 is equal to u cos alpha minus beta.
Hence, x dot is equal to u cos alpha minus beta minus g sin beta into t and y dot is
equal to u sin alpha minus beta minus g cos beta into t. These are the two equations in
the form of the velocity components. Therefore, the velocity as a function of time has been
attained up to this point. If we integrate both the above equations with respect to time,
then we get x is equal to u cos alpha minus beta into t minus half g sin beta into t square
plus C1. Similarly, y is equal to u sin alpha minus beta into t minus half g cos beta t
square plus C2 where constant C1 and C2 are determined from the initial conditions. At
time t is equal to 0, x is equal to y is equal to 0 because we are taking that as the origin.
Hence, C1 and C2 both are zeroes.
Therefore, the equation of the trajectory in parametric form is given as x is equal
to u cos alpha minus beta t minus half g sin beta t square and y is equal to u sin alpha
minus beta t minus half g cos beta t square. Here, x and y both are functions of t. So,
t is the parameter knowing which I can find out x and y. Therefore, this form is called
parametric form.
After we have found out the equations of this motion, let me discuss the range on inclined
plane. In this figure, AB is the range of the particle on inclined plane. It starts
from A, reaches some distance here, again it falls back to B. We have to find distance
AB with our coordinate system. We have to find x-coordinate of point B when the y-coordinate
is 0 because at A and B, at both the places y-coordinate is 0.
Therefore, if we put y is equal to 0 is equal to u sin alpha minus beta t minus half g cos
beta t square which gives two solutions; t is equal to 0 or t is equal to 2u sin alpha
minus beta divided by g cos beta. That means y component is 0 at the beginning, at time
t is equal to 0 and then later on at time 2u sin alpha minus beta divided by g cos beta.
Therefore, this must be the time of flight. Time of flight is given by 2u sin alpha minus
beta by g cos beta. In this, beta is the inclination of the plane. If it is horizontal surface
then beta will be 0 and time of flight will be 2u sin alpha divided by g. So, for horizontal
plane, the time of flight is 2u sin alpha divided by g which you have studied in your
twelfth class.
Now next, so, once we have found time t, we can put this expression in the expression
for x and then we get R is equal to u cos alpha minus beta into 2u sin alpha minus beta
divided by g cos beta minus half g sin beta 4 u square sin square alpha minus beta divided
by g square cos square beta. We get this lengthy expression.
Let us see if we can simplify this. This can be written as u square by g cos beta into
sin 2 alpha minus beta. We know the formula where sin 2a is equal to 2 sin a into cos
a. Then similarly, the other part can be written as minus u square g cos square beta 2 sin
square alpha minus beta into sin beta. Here, I can take sin alpha minus beta common, u
square by g cos square beta as common. Then, this will become cos alpha minus beta into
cos beta minus sin beta into sin alpha minus beta. Using some trigonometric formula, we
try to simplify this and this is equal to 2 u square divided by g cos square beta into
sin alpha minus beta into cos alpha. This is the expression for the range.
So therefore, if you repeat it again, time of flight is basically when y is equal to
0. In the beginning y is equal to 0; when the particle has fallen down, then again the
time of flight is 0. We have already found that t is equal to 0 or t is equal to this
thing. Hence, the time of flight is 2u sin alpha minus beta divided by g cos beta.
We are going to discuss the angle of projection which gives the maximum range. We have the
expression for the range R which is given by R is equal to 2u square g cos beta into
sin alpha minus beta into cos alpha. For a fixed beta, R will be maximum. If beta is
fixed then that means the plane is fixed. Therefore, sin alpha minus beta into cos alpha
should be maximum. When sin alpha minus beta into cos alpha is maximum, then you get the
maximum range. Sin alpha minus beta into cos alpha is basically half sin 2 alpha minus
beta minus sin beta. Now sin beta, beta is fixed. This portion will be maximum when 2
alpha minus beta is equal to 90 degree or pi by 2 radian because sin is maximum at an
angle of pi by 2. So, 2 alpha minus beta is equal to pi by 2 or alpha is equal to pi by
4 plus beta by 2. Therefore, this is the angle of projection for getting the maximum range.
If it is a horizontal surface then beta will be 0 and therefore, the alpha will come out
to be pi by 4. Therefore, on a horizontal surface the angle of projection is kept 45
degrees for getting the maximum range.
If you fire something at an angle of 45 degrees, then the range will be maximum. This is about
kinematics of a particle. Basically, in essence, in kinematics of a particle you need to carryout
vector calculus. If you are given the displacement as vector then you can differentiate it and
apply vector calculus, you can find out the velocity and acceleration of the particle.
Also, if somebody has given you acceleration as a function of time, then you can integrate
and you can find out the velocity and you can find out the displacement also.
Now, Let me, number of problems you can solve. For example, suppose somebody gives you acceleration
of a particle which is a vector a is equal to 3 t i; i is a unit vector minus 2 t square
j plus 5 t cube k meter per second square. If we say particle starts with 0 velocity
at the origin, we write the condition that x dot at 0 is equal to 0, x is 0 at 0. Therefore,
if somebody wants to find out particles position, displacement, distance travelled after 5 seconds,
then we can integrate this expression and find out the particle's position. We can find
out the particle's position displacement and distance. Suppose, somebody wants to find
out the distance travelled, then of course, we know that small distance ds will be equal
to dx square plus dy square plus dt square.
Therefore, ... therefore, what happens, if you want so between two time limits, now dx
can be expressed in terms of dt. You will be getting some expression for x which will
come in the form of dt. Then, you have to integrate between 0 to t; this quantity dx
square plus dy square plus dt square which will basically come dz square. This is dz
square which will come in the form of dt. Then, you will be able to find out the distance
travelled also. You can find out velocity, you can find out speed and acceleration. So,
this is what that this one.
Another problem can be of this type. Here, if you have a slider crank mechanism, this
is a crank then this is a connecting rod and this is a slider which is moving. Now, you
want to find out the velocity and acceleration of a slider crank mechanism as a function
of theta. That can be easily done, provided we can express, we will be moving the crank
radius r and we will be moving this length as connecting rod length l. Then, we can find
out the distance of this particle, which is in the slider, from the origin as a function
of r and as a function of theta. Finally, we can differentiate this expression to find
out the velocity also as a function of theta. Velocity will contain the term of d theta
by dt which is nothing but the angular velocity of the link. Similarly, there you can solve
the problems of the projectiles. Now here, this is For example, this is the thing, projectile. If this is at 30 degrees,
it is fired at say 45 degrees from the horizontal. When you know how it falls and what is the
range, these types of problems you can solve. Similarly, there may be problems of this type.
If you look, there are two columns; this is this one and this is column B, this is B and
this is D; A B C D. Now, suppose you fire some shot from here, now at what angle should
it be projected so that it hits here? What should be the angle of projection? Naturally,
if you hit along this line, it will not touch D. So, therefore we have to decide the angle
of projection. So, the similar types of problems can be solved.
Now, in my next lecture, I will be talking about kinematics of a particle moving on a
curve. That you know the particle which is constrained to move on a particular curve,
then how we study the kinematics?
For example, in this case the slider is constrained to move in this guide way. See this slider,
it is not free to move anywhere; it is just constrained to move. This is a straight surface
but sometimes the slider maybe constrained to move on a curved surface also. So, these
types of things also we will study in the next lecture.