Kinematics ... 2
Introduction to Motion in One Dimension ... 2
Motion under Gravity ... 11
Motion in Two Dimensions ... 22
Circular Motion ... 35
Relative Motion ... 41
Solved Examples on Mechanics:- ... 49
Newton’s Laws of Motion:- ... 63
Classification of Forces:- ... 64
Newton’s First Law of Motion:- ... 67
Newton’s Second Law of Motion:- ... 69
Newton’s Third Law of Motion:- ... 77
Friction:- ... 90
Free body diagram ... 97
Centripetal Force ... 108
Conservation of Linear Momentum:- ... 130
Work, Energy and Power:- ... 138
Work:- ... 139
Energy (E):- ... 147
Power:- ... 174
Gravitation ... 191
Gravitational Field and Intensity ... 194
Satellites and Planetary Motion ... 206
Fluid Mechanics ... 213
Kinematics
Introduction to Motion in One Dimension
A body:-
A certain amount of matter limited in all directions and consequently having a finite size, shape and occupying some definite space is called a body.
Particle:-
A particle is defined as a portion of matter infinitesimally small in size so that for the purpose of investigation, the distance between its different parts may be neglected. Thus, a particle has only a definite position, but no dimension. In the problems we are going to discuss, we will consider a body to be a particle for the sake of simplicity.
Motion:-
The position of object can change on a straight line (like on x-axis with respect to origin) or on a plane with respect to some fixed point on frame. So we can define motion as follows:-
An object or a body is said to be in motion if its position continuously changes with time with reference to a fixed point (or fixed frame of reference).
But note that, the moving object is either a particle, a point object (such as an electron) or an object that moves like a particle. A body is said to be moving like if every portion of it moves in the same direction and at the same rate.
Motion in one dimension:-
When the position of object changes on a straight line i.e. motion of object along straight line is called motion in one dimension. To understand the essential concepts of one dimensional motion we have to go through some basic definitions.
Frame of reference:-
One can see the platform from a running train, and it seems that all the objects placed on platform are continuously changing their position. But one, who is on platform, concludes that the objects on the platform are at rest. It means if we will take the trains are reference frame the objects are not stationary and taking reference frame as platform the objects are stationary. So the study of motion is a combined property of the object under study and the observer. Hence there is a need to define a frame of reference under which we have to study the motion of an object.
Definition
A frame of reference is a set of coordinate axes which is fixed with respect to a space point (a body or an object can also be treated as a point mass therefore it can become a site for fixing a reference frame), which we have arbitrarily chosen as per our observer's requirement. The essential requirement for a frame of reference is that, it should be rigid.
Position of an object
The position of an object is defined with respect to some frame of reference. As a convention, we define position of a point
representing a 3-dimensional space and each point in this space can be uniquely defined with the help of a set of X, Y and Z coordinate, all three axes being mutually perpendicular to each other. The line drawn from origin to the point represents the position vector of that point.
Position vector
It describes the instantaneous position of a particle with respect to the chosen frame of reference. It is a vector joining the origin to the particle. If at any time, (x, y, z) be the Cartesian coordinates of the particle then its position vector is given by vector = xi + yj + zk.
In one-dimensional motion: vector = xi, y = z = 0 (along x-axis) In two-dimensional motion: vector = xi + yj (in x-y plane z = 0)
In the figure above, the position of a point P is specified and vector OP is called the position vector.
Displacement
Consider a case in which the position of an object changes with time. Suppose at certain instant 't' the position of an object is x1 along the x axis and some other instant 'T' the position is x1 then the displacement Δx is defined as,
Δx = x2 - x1
It can be seen in the figure above where x1 and x2 are instantaneous position of the object at that time.
Now consider the motion of a point A with respect to a reference point O. The motion of point A makes its radius vector vary in the general case both in magnitude and in direction as shown in figure above. Suppose the point A travels from point 1 to point 2 in the time interval Δt. It is seen from the figure that the displacement vector Δ of the point A represents the increment of vector in time Δt:
Difference between distance and displacement
To understand the difference between distance and displacement, we study the motion of vertical throw of a ball with respect to point O, as shown in the left figure, to height h.
After some time it will come again to the same point O. The displacement of ball is zero but there is some distance traversed by the ball. It's because distance is a scalar quantity but displacement is a vector quantity.
Uniform and Non Uniform Motion
Speed is the rate of change of distance without regard to directions. Velocity is the rate at which the position vector of a particle changes with time. Velocity is a vector quantity whereas speed is scalar quantity but both are measured in the same unit m/sec. The motion of an object may be uniform or non-uniform depending upon its speed. In case of uniform motion the speed is constant, whereas in the non-uniform motion, the speed is variable.
In uniform motion in one dimension the velocity (v) is mathematically defined as v = (x2 - x1)/(T-t) ... (1)
Where x1 and x2 are instantaneous displacement as shown in figure above at time 't' and 'T' respectively.
Graphical representation of the uniform motion
Form the equation (1) we have the following equation
x2 = x1 + v(T - t)
where v is constant. Take t = 0, the equation becomes x2 = x1 + vT, from this equation it follows that the graph of position of object
'x2' against 'T' is a straight line, cutting off x1 on the position axis where x1 is the distance of the particle from the origin at time t = 0.
v = slope of the graph which is constant
If the graph is not a straight line, it will represent non-uniform motion.
Motion of a body cannot be correctly identified unless one knows the position of body as specified by a fixed frame of reference.Velocity Vector in Non Uniform Motion
In any non-uniform motion, we can define an average velocity over a time interval. Average velocity is the ratio of the displacement Δx (that occurs during a particle time interval Δt) to that interval of time i.e.
Now refer to the example, related to figure 2.3, the ratio of Δ /Δt is called the average velocity < > during the time interval Δt. The direction of the vector < > coincides with that of Δ . Average velocity is also a vector quantity.
Note: The ratio of total distance traveled and time taken during the motion is called average speed. Average speed is a scalar
quantity.
If at any time t1 position vector of the particle is 1 and at time 2 position vector is 2 then for this
interval
Instantaneous velocity
Instantaneous velocity is defined as the rate of change of displacement.
Velocity
The velocity at any instant is obtained from the average velocity shrinking the time interval closer to zero. As Δt tends to zero, the average velocity approaches a limiting value, which is the velocity at that instant, called instantaneous velocity, which is a vector quantity, mathematically we can define it as
The magnitude v of the instantaneous velocity is called the speed and is simple the absolute value of In the example related with figure given below, the instantaneous velocity is
Hence instantaneous velocity is the rate at which a particle's position is changing with respect to time at a given instant. The velocity of a particle at any instant is the slope (tangent) of its position curve at the point representing that instant of time, as shown in figure above.
Speed
Speed is defined as rate of change of distance with time. In any interval of time, average speed is defined as
<speed> = (total distance)/(total time taken) = Δs/Δt. As Δs > |Δ |, hence <speed> > <velocity>
Think : (i)Can a body have a constant speed and still have a varying velocity?
The dispalcement remains unaffected due to shifting of origin from one point to the other.
The displacement can have positive, negative or zero value.
The dispalcement is never greater than the actual distance travelled.
The displacement has unit of length.
Velocity can be considered to be a combination of speed and direction.
A change in either speed or direction of motion results in a change in velocity.
It is not possible for a particle to possess zero speed with a non-zero velocity.
A particle which completes one revolution, along a circular path, with uniform speed is said to possesss zero velocity and non-zero speed.
In case a body moves with uniform velocity, along a straight line, its average speed is equal to its instantaneous speed.Acceleration
Acceleration is the rate of change of velocity with time. The concept of acceleration is understood in non-uniform motion. It is a vector quantity.
Average acceleration is the change in velocity per unit time over an interval of time.
Instantaneous acceleration is defined as
Acceleration vector in non uniform motion
Suppose that at the instant t1 a particle as in figure above, has velocity and at t2, velocity is . The average
Variable Acceleration
The acceleration at any instant is obtained from the average acceleration by shrinking the time interval closer zero. As Δt tends to zero average acceleration approaching a limiting value, which is the acceleration at that instant called instantaneous acceleration which is vector quantity.
i.e. the instantaneous acceleration is the derivative of velocity.
Hence instantaneous acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Instantaneous acceleration at any point is the slope of the curve v (t) at that point as shown in figure above.
Equations of motion
The relationship among different parameter like displacement velocity, acceleration can be derived using the concept of average acceleration and concept of average acceleration and instantaneous acceleration.
When acceleration is constant, a distinction between average acceleration and instantaneous acceleration loses its meaning, so we can write
where is the velocity at t = 0 and is the velocity at some time t. So,
... (2) This is the first useful equation of motion. Similarly for displacement
... (3)
in which is the position of the particle at t0 and is the average velocity between t0 and later time t. If at t0 and t the velocity
This is the second important equation of motion.
Now from equation (2), square both side of this equation we get
This is another important equation of motion. Note:
The equation of motion derived above are possible only in uniformly accelerated motion i.e. the motion in which the acceleration is constant.Graphical Representation and Equations of Motion
While studying motion of bodies we have to keep two things in mind. We have to study various characteristics associated with moving bodies while we are to explore and study in detail, about the cause of producing motion. Kinematics is the branch of physics which deals only with the description of motion of bodies. Mainly there are two type of graphs are belonging to the kinematics. One is position-time graph and the other is velocity-time graph.
These graphs will have validity only if motion under study is along a straight line. Then, displacement, velocity and
acceleration vectors are collinear and can be treated as algebraic quantities. Let x-axis be the path of motion. Then, x-coordinate
represents the magnitude of position vector.
Position - Time Graph:-
If we plot time t along the x-axis and the corresponding position (say x) from the origin O on the y-axis, we get a graph which is called the position-time graph. This graph is very convenient to analyse different aspects of motion of a particle. Let us consider the following cases.
(i) In this case, position (x) remains constant but time changes. This indicates that the particle is stationary in the given reference frame. Hence, the straight line nature of position-time graph parallel to the time axis represents the state of rest. Note that its slope (tan θ) is zero.
.
(ii) When the x-t graph is a straight line inclined at some angle (θ≠) with the time axis, the particle traverses equal displacement Δx in equal intervals of time Δt. The motion of the particle is said to be uniform rectilinear motion. The slope of the line measured by Δx/Δt = tθ represents the uniform velocity of the particle.
(iii) When the x-t graph is a curve, motion is not uniform. It either speeds up or slows down depending upon whether the slope (tan θ successively increases or decreases with time. As shown in the figure the motion speeds up from t = 0 to t=t1 (since the slope tan
θ increases). From t=t1 to t=t2, AB represents a straight line indicating uniform motion. From t=t2 to t=t3, the motion slows down and
for t>t3 the particle remains at rest in the reference frame.
The Velocity - Time Graph:-
The velocity-time graph gives three types of information.
(i) The instantaneous velocity.
(ii) The slope of the tangent to the curve at any point gives instantaneous acceleration.
a = dv/dt = tan θ
(iii) The area under the curve gives total displacement of the particle.
Now, let us consider the uniform acceleration. The velocity-time graph will be a straight line.
The acceleration of the object is the slope of the line CD. a = tan θ = BC/BD = (v-u)/t
v = u +at …... (1)
The total displacement of the object is area OABCD
s = Area OABCD = Area of rectangle OABD + Area of triangle BCD s = ut + (1/2)at2 …... (2)
Again, s = Area OABCD
= 1/2(AC + OD) x OA = 1/2(v + u)xt
The acceleration-time graph:-
Acceleration time curves give information about the variation of acceleration with time. Area under the acceleration time curve gives the change in velocity of the particle in the given time interval.
Refer this simulation that shows a car traveling along a graph:-
The purpose is to show that, the car can traverse the graph, then it's first derivative is a continuous function.
The displacement can have positive, negative or zero value.
The displacement is never greater than the actual distance travelled.
A change in either speed or direction of motion results in a change in velocity.
It is not possible for a particle to possess zero speed with a non-zero velocity.Motion under Gravity
Whenever a body is released from a height, it travels vertically downward towards the surface of earth. This is due to the force of gravitational attraction exerted on body by the earth. The acceleration produced by this force is called acceleration due to
gravity and is denoted by „g‟. Value of „g‟ on the surface of earth is taken to the 9.8 m/s2 and it is same for all the bodies. It means all bodies (whether an
iron ball or a piece of paper), when dropped (u=0) from same height should fall with same rapidity and should take same time to reach the earth. Our daily observation is contrary to this concept. We find that iron ball falls more rapidly than piece of paper. This is due to the presence of air which offers different resistance to them. In the absence of air both would have taken same time to reach the surface of earth.
When a body is dropped from some height (earth's radius = 6400 km), it falls freely under gravity with constant acceleration g (= 9.8 m/s2) provided the air resistance is negligible small. The same set of three equations of kinematics (where the acceleration remains constant) are used in solving such motion. Here, we replace by and choose the direction of y-axis conveniently. When the y-axis is chosen positive along vertically downward direction, we take as positive and use the equation as
v = u + gt, v2 = u2 + 2gh, and h = ut + 1/2gt2
where h is the displacement of the body and u is initial velocity of projection in the vertically downward direction. However, if an object is projected vertically upward with initial velocity u, we can take y-axis positive in the vertically upward direction and the set of equations reduces to
v = u - gt, v2 = u2 - 2gh, and h = ut - 1/2gt2
In order to avoid confusion in selecting as positive or negative, it is advisable to take the y-axis as positive along vertically upward direction and point of projection as the origin. We can now write the set of three equations in the vector form:
,
and
where h is the displacement of the body.
Illustration:
The motion of a particle is described by the equation u = at. The distance travelled by the particle in the first 4 second.
Solution:
Because for the motion u = at. So acceleration is uniform which is equal to a. Therefore, Distance traveled = 1/2[(a)(4)2] = 8a
Illustration:
A body moving with a constant retardation in straight line travels 5.7 m and 3.9 m in the 6th and 9th second, respectively. When will the body come momentarily to rest?
Solution:
A body moving with initial velocity u and acceleration a, traverses distance Sn in nthsecond of its motion.
Sn = u + (1/2)(2n - 1)a => 5.7 = u + (1/2)(2 x 6 - 1) a
or 5.7 = u + (11/2) a
and 3.9 = u + (1/2)(2 x 9 - 1)a or, 3.9 = u + (17/2) a Solving eqns. (1) and (2) we get, u = 9 m/s and a = -0.6 m/s2.
If the body stops moving after t seconds, then from the relation v=u+at Thus, 0 = 9 + (-0.6)t or, t = (9/0.6)s = 15s
Illustration (JEE Advanced):
A stone, thrown up is caught by the thrower after 6s. How high did it go and where was it 4 s after start? g = 9.8 m/s2
Solution:
Time to go up and come back = 6s
Thus, time to reach the highest point = (6/2) s = 3s
From point of projection to the highest point we have
u = ?, v = 0, a = -9.8 m/s2, t = 3s
Using the relation, v = u + at 0= u – 9.8 3
Thus, u = 29.4 m/s2
Maximum height, H = u2/2g = [(29.4)2/2(9.8)] m =44.1 m
Using the relation, S = ut + ½ at2
So, h = [(29.4)(4)-1/2 (9.8) (4)2]m
= [117.6-78.4] m = 39.2 m
From the above observation we conclude that, the height would be 39.2 m.
Acceleration of all these bodies is constant.
Acceleration is always directed downward.
All bodies, when dropped (u=0) from same height should fall with same rapidity and should take same time to reach the earth. Motion in a Straight Line with AccelerationVelocity of a body is defined as the time rate of displacement, where as acceleration is defined as the time rate of change of velocity. Acceleration is a vector quantity. The motion may be uniformly accelerated motion or it may be non-uniformly
accelerated, depending on how the velocity changes with time.
Uniform acceleration
The acceleration of a body is said to be uniform if its velocity changes by equal amounts in equal intervals.
Non-uniform acceleration
The acceleration of a body is said to be non-uniform if its velocity changes by unequal amounts in equal intervals of time.
Average velocity
Average acceleration
Illustration:
A particle moves with a velocity v(t) = (1/2)kt2 along a straight line. Find the average speed of the particle in time T.
Solution:
Illustration:
A particle having initial velocity is moving with a constant acceleration 'a' for a time t. (a)Find the displacement of the particle in the last 1 second.
(b)Evaluate it for u = 2 m/s, a = 1 m/s2 and t = 5 sec.
Solution:
(a) The displacement of a particle at time t is given s = ut + 1/2at2 At time (t - 1), the displacement of a particle is given by
S' = u (t-1) + 1/2a(t-1)2
So, Displacement in the last 1 second is, St = S - S'
= ut + 1/2 at2 – [u(t-1)+1/2 a(t-1)2 ] = ut + 1/2at2 - ut + u - 1/2a(t - 1)2
= 1/2at2 + u - 1/2 a (t+1-2t) = 1/2at2 + u - 1/2at2 - a/2 + at S = u + a/2(2t - 1)
(b) Putting the values of u = 2 m/s, a = 1 m/s2 and t = 5 sec, we get S = 2 + 1/2(2 x 5 - 1) = 2 + 1/2 x 9
= 2 + 4.5 = 6.5 m
Illustration:
Position of a particle moving along x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. (a)Find the position of the particle at t = 2 s.
(b)Find the displacement of the particle in the time interval from t = 0 to t = 4 s. (c)Find the average velocity of the particle in the time interval from t = 2s to t=4s. (d)Find the velocity of the particle at t = 2 s.
Solution: (a) x(t) = 3t - 4t2 + t3 => x(2) = 3 x 2 - 4 x (2) 2 + (2)3 = 6 - 4 x 4 + 8 = -2m. (b) x(o) = 0 X(4) = 3 x 4 - 4 x (4)2 + (4)3 = 12 m. Displacement = x(4) - x(0) = 12 m. (c) < v > = X(4)X(2)/(4-2) = (12-(-2))/2 m/s = 7 m/s (d) dx/dt = 3 - 8t + 3t2 v(2) (dx/dt)2 = 3 - 8 x 2 + 3 x (2)2 = -1m/s Illustration:
Two trains take 3 sec to pass one another when going in the opposite direction but only 2.5 sec if the speed of the one is increased by 50%. The time one would take to pass the other when going in the same direction at their original speed is
(a) 10 sec (b) 12 sec
(c) 15 sec (d) 18 sec
Solution:
t = d/vr We have, 3 = d/v1+v2 2.5 = d/1.5v1+v2 Solving we get, v1 = 2d/15 and v2 = d/5
When they are going in same direction,
vr = v2 – v1 = d/15
Thus, t = d/vr = d/(d/15) = 15 s
From the above observation we conclude that, option (c) is correct.
Analysis of Uniformly Accelerated Motion
Case-I:
For uniformly accelerated motion with initial velocity u and initial position x0.
Velocity time graph
In every case tanθ = a0
Position time graph
Initial position x of the body in every case is x0 (> 0)
Case II:
For uniformly retarded motion with initial velocity u and initial position x0.
Velocity time graph
In every case tanθ = -a0
Position time graph
Initial position x of the body in every case is x0 (> 0)
A particle is moving rectilinearly with a time varying acceleration a = 4 - 2t, where a is in m/s2 and t is in sec. If the particle is starting
its motion with a velocity of -3 m/s from x = 0. Draw a-t, v-t and x-t curve for the particle.
Solution:
a = 4-2t
v = 4t-t2-3
x = 2t2 – t3/3– 3t
Acceleration
Acceleration is the rate of change of velocity with time. The concept of acceleration is understood in non-uniform motion. It is a vector quantity.
Average acceleration is the change in velocity per unit time over an interval of time.
Instantaneous acceleration is defined as
Acceleration vector in non uniform motion
Suppose that at the instant t1 a particle as in figure above, has velocity and at t2, velocity is . The average
Variable Acceleration
The acceleration at any instant is obtained from the average acceleration by shrinking the time interval closer zero. As Δt tends to zero average acceleration approaching a limiting value, which is the acceleration at that instant called instantaneous acceleration which is vector quantity.
i.e. the instantaneous acceleration is the derivative of velocity.
Hence instantaneous acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Instantaneous acceleration at any point is the slope of the curve v (t) at that point as shown in figure above.
Equations of motion
The relationship among different parameter like displacement velocity, acceleration can be derived using the concept of average acceleration and concept of average acceleration and instantaneous acceleration.
When acceleration is constant, a distinction between average acceleration and instantaneous acceleration loses its meaning, so we can write
where is the velocity at t = 0 and is the velocity at some time t Now,
Hence,
…... (2) This is the first useful equation of motion. Similarly for displacement
…... (3)
in which is the position of the particle at t0 and is the average velocity between t0 and later time t. If at t0 and t the
…... (4)
From equation (3) and (4), we get,
…... (5)
This is the second important equation of motion.
Now from equation (2), square both side of this equation we get
[Using equation (4)]
Using equation (3), we get,
…... (6)
This is another important equation of motion.
Caution:The equation of motion derived above are possible only in uniformly accelerated motion i.e. the motion in which the
acceleration is constant.
Refer this simulation for motion in a straight line:- Illustration:
The nucleus of helium atom (alpha-particle) travels inside a straight hollow tube of length 2.0 meters long which forms part of a particle accelerator. (a) If one assumes uniform acceleration, how long is the particle in the tube if it enters at a speed of 1000 meter/sec and leaves at 9000 meter/sec? (b) What is its acceleration during this interval?
Solution:
(a) We choose x-axis parallel to the tube, its positive direction being that in which the particle is moving and its origin at the tube entrance. We are given x and vx and we seek t. The acceleration ax is not involved. Hence we use equation 3, x = x0 + <v> t.
We get
x = v0 + ½ (vx0) + vx) t, with x0 = 0 or
t = 2x/(vx0+vx),
t = ((2)(2.0 meters))/((1000+9000)meters/sec) = 4.0/10-4 sec Ans. (b) The acceleration follows from equation 2, vx = vx0 + axt
=> ax = (v0-vx0)/t = ((9000-1000)meters/sec)/(4.0×10(-4) sec)
= 2.0 × 107 meter/sec2 Ans.
Or,
Let at t = 0,
then,
Or,
Further we know that,
or
Integrating,
Or,
At, t = 0, x = x0 then c' = x0
Hence,
Thus, we have derived the same equation of motion using calculus.
To understand the use of calculus in solving the kinematics problems we can look into the following illustrations.
Illustration:
The displacement x of a particle moving in one dimension, under the action of a constant force is related to the time t by the equation t = √x + 3 where x is in meter and t is in seconds. Find the displacement of the particle when its velocity is zero.
Solution:
Here t = √x + 3 => √x = t - 3
Squaring both sides, we get x = t2 - 6t + 9,
As we know velocity, v = dx/dt Hence we get v = dx/dt = 2t - 6 Put v = 0, we get, 2t - 6 = 0 So, t = 3s
When t = 3s, x = t2 - 6t + 9 = 9 - 6(3) + 9 = 0
Hence the displacement of the particle is zero when its velocity is zero.
Illustration:
A particle starts from a point whose initial velocity is v1 and it reaches with final velocity v2, at point B which is at a distance 'd' from
point A. The path is straight line. If acceleration is proportional to velocity, find the time taken by particle from A to B.
Solution:
Here acceleration a is proportional to velocity v. Hence a α v => a = kv, where k is constant => dv/dt = kv ... (1) => (dv/ds)(ds/dt) = kv => (dv/ds) v = kv From equation (1), dv/v = kdt or, Or, ln (v2/v1) = kt Or, t = ln (v2/v1) /k = [d ln (v2/v1)/(v2-v1)]
The dispalcement remains unaffected due to shifting of origin from one point to the other.
The displacement can have positive, negative or zero value.
The dispalcement is never greater than the actual distance travelled.
The displacement has unit of length.
Velocity can be considered to be a combination of speed and direction.
A change in either speed or direction of motion results in a change in velocity.
It is not possible for a particle to possess zero speed with a non-zero velocity.
A particle which completes one revolution, along a circular path, with uniform speed is said to possesss zero velocity and non-zero speed.Motion in Two Dimensions
In this part, we discuss motions in two dimensions like the motion of a particle moving on a circular path or on a parabolic path. Consider a particle moving on x-y plane along a curved pat at time t, as shown in figure given below. Its displacement from origin is measured by vector its velocity is indicated by vector (tangent to the path of the particle) and acceleration as shown in the figure.
The vectors , and are inter related and can be expressed in terms of their components, using unit vector notation as,
From the above relations we can say that two-dimensional motion is nothing but superimpose of two one dimension motions. This is for the reason that in any two-dimensional motion, all the parameters can be resolved in two mutually perpendicular directions.
Frame of reference:-
A frame of reference is a set of coordinate axes which is fixed with respect to a space point (a body or n object can also be treated as a point mass and therefore it can become a site for fixing a reference frame), which we have arbitrarily chosen as per our observer requirement. It is constituted of two components:
(i) Body on which observer is apparently sitting to observe the motion, i.e. the origin.
(ii) A coordinate system fixed on this body so that whatever is being observed can be measured or mathematically determined with the help of the coordinates of space points, which are defined by a unique set of (x, y, z). To fix a frame of
reference one has to fix the origin to a chosen space point and then fix the co-ordinate system on it. Once the origin and coordinate system are fixed, then we have done the exercise of fixing the frame of reference and we can proceed to solve problems and study various parameters i.e. position, velocity, accelerations as functions o time.
Here, an important thing to notice is that earth becomes a natural choice for the frame of reference as in most of the cases it is very easy to visualize the motion with respect to it. As it is a routine in our daily life we innocently (even in early childhood) refer and perceive all motion phenomenon with respect to earth only.
While analyzing any motion we will take the help of the coordinate system. What we shall do is that we shall keep ourselves free to take it there. This is called a frame of reference because we shall refer to all positions in this coordinate system fixed on that body.
Here, remembers that we fix all three axis of coordinate system i.e.
X-axis. Y-axis and Z-axis as well as the origin on the body and the coordinates of origin are (0, 0, 0). It is the origin where we mostly assume that the observer is sitting to observe the motion in the concerned reference frame.
Choice of a frame of reference:-
Let us come back to the concept of motion. Do you believe that all what you see moving is in motion and what you see not moving is at rest! Like vehicles on road! You should be warned that motion or observation of motion is really not that simple!
Let us take an example: The moon in the night sky! Some time at night you see
that moon is travelling across the clouds towards east or west. And in some other nights it doesn't move at all! We shall all agree that it cannot be. It cannot sometimes move faster and sometime slower and sometimes become stationary. Then, what is the phenomenon there?
Motion in Two Dimensions:-
We start with the first point that whenever we speak of any motion it should be with respect to some fixed frame of reference. So, when we see the moon moving. It is not with certainly one can point out that in the night when it looked stationary there were no clouds. So it is an effect of the clouds. The fact is that if we have clouds in some nights moving towards west or east then we see moon travelling
towards east or west in the sky respectively. How to explain it with the help of reference frames? It is like this. When we see two objects in the sky one moon and other a cloud, then due to relatively bigger size of clouds (as seen by us) one gets his eyes fixed on clouds and therefore frame of reference gets fixed on the clouds (Figure shown above). It is equivalent to say that we are sitting on the clouds itself, which looks stationary to us and the feeling comes, as if moon is travelling across this stationary cloud in the opposite direction to the clods motion (see figure given above). Similarly, we see things going backwards when we travel on a road.
Now let us deliberately fix our eyes on the moon and see. What we are doing now is that we are fixing frame of reference on the moon. It is equivalent to say that we are sitting on the moon to observe the cloud's motion and now we can see clouds drifting away (which is the fact). See Figure shown above.
An important observation to remember is that once we choose an object as a frame of reference, we can't see it moving with respect to itself. Naturally, in most of the cases of observations the Earth becomes a natural choice for the reference frame. Also, based on the choice of reference frames one can have many apparent motion of an object under consideration.
Here one important point to be noted is that in all previous discussion we have assumed that the reference frames are not accelerating, which means it's acceleration is zero. These are called inertial or un accelerated frames of reference. Now obviously, we are left with one more choice of class of reference frames that is the accelerated ones. These are called
non-inertial or accelerated frame of reference. But in all the forthcoming exercises in this chapter we shall deal with and look only
for inertial frames of reference.
Relative motion:-
Consider a observer S is fixed to the earth. The other observer S' is moving on the earth, say a passenger sitting on a bus. Hence other frame of reference S' is fixed with bus. Each observer is studying the motion of a cyclist in his or her own frame of reference. Each observer will record a displacement, velocity and acceleration for the cyclist measured relative to his reference frame. Now the question is "How will we compare their results"?. To do this we need the concept of relative motion. Let us do it.
In figure shown above at the top, the reference frame S, represented by the x and y-axes, can be thought of as fixed to the earth. The other reference frame S', represented by x'- and y'-axes, which moves along the x-axis with a constant velocity u, measured in the s-system, fixed with the bus.
Initially, a particle is at a point called A in the S-frame and called A' in the S'-frame. At a time t later, the reference frame S' has moved a distance ut to the right and particle has moved to B. The displacement of the particle from its position in the S-frame is the vector from A to B. The displacement of the particle from its initial position in the S'-frame is the vector r' from A' to B. These are different vectors because the reference point A' of the moving frame has been displaced a distance ut along the x-axis during the motion. From the second figure shown above we see that is the vector sum of and :
= +
Differentiating equation leads to
But,
is the instantaneous velocity of the particle measured in the s-frame.
and
is the instantaneous velocity of the same particle measured in the S' frame, so that = ' + ... [A]
Hence the velocity of the particle relative to the S-frame, is the vector sum of the velocity of the particle to the S' frame, , and the velocity of the S' frame relative to the S-frame.
Once again differentiating equation [A] We get,
Hence the acceleration of particle is the same in all reference frame moving relative to one another with constant velocity
The relative velocity between S and S' must be a constant.
At any instant, the velocity the velocity of a particle as measured by S is equal to the velocity of the particle as measured by S'.
Law of transformation of velocity permits us to transform a measurement of velocity made by an observer in one frame reference to another frame of reference.
The reference frames which are not accelerating are called inertial frame of reference.Problem 1:-
A particle starts with initial velocity (3i + 4j) m/s and with constant acceleration (8i + 6j) m/s2. Find the final velocity and final
displacement of the particle after time t = 4 sec.
Solution:-
As the velocity and acceleration vectors are not parallel, the velocity and acceleration cannot be treated as a scalar. Now divide the whole phenomenon in two one-dimension motions.
In x-direction,
Initial velocity = vx0 = 3m/s, ax = 8 m/s2
Hence velocity after 4 second, vx = 3 + 8 × 4 = 35 m/s
Displacement after 4 second sx = x = 3 × 4 + ½8 (4)2 = 76 m
In y-direction,
Initial velocity vy0 = 4m/s, ay = 6 m/s2 Velocity after 4 second vy = 4 + 6 × 4 = 28 m/s Displacement after 4 second sy = 4 × 4 + ½ 6 (4)2 = 64 m
It means final velocity v = vx i + vy j = (35 i + 28 j) m/s
Problem 2:-
There are two boats B1 and B2. They are continuously moving towards a bank. Water flow velocity is 10m/sec away from the bank.
When at one instant of time, observation was taken we found that B1 and b2 are at a position (figure is gen below) which is 100 m
from the bank and after 1 sec, b1 is 50 away and B2 is 25m away from the bank. Find out all possible (motion) velocities observed
from different reference frames choices.
Solution:-
Here we have four (at least) immediate choices of reference frames with respect to which we can observe velocities. (i) Bank (earth)
(ii) B1
(iii) B2
(iv) Water
(i) Bank: (See figure shown above) First we shall fix the frame of reference on the bank.
Since B1 travels 50 m towards the bank in 1 sec., its velocity towards the Bank (that is in +y direction) is 50m/sec.
Similarly B2 travels 75 m/s towards the bank
Thus, its velocity with respect to bank or as observed by an observer sitting at rest on the bank is 75m/sec (+y direction)
Since water moves away from bank at speed of 10m/sec. Therefore, water's velocity w.r.t. the bank is 10m/sec (-y direction).
(ii) Boat B1: Now, let us fix frame of reference on Boat B1.
(a) Since B1 cannot move w.r.t. itself (only in case of deformation, a part of this boat can be seen moving w.r.t. other parts)
Thus, VB1 = 0 (relative velocity w.r.t. to itself)
(b) at t = 0, B1 and B2 were in the same line and had no
at t = 1 B2 leads B1 by 25 m
that is B2 has moved 25m in 1 sec w.r.t. B1
Thus, VB2 = 25 m/sec (+ direction)
(c) Bank, at t = 0 was away by 100 m.
and at t = 1 it has come closer by 50 m towards B1
Therefore, bank is approaching B1 by a speed of
Vbank = 50m/sec (-y direction)
(d) To visualize the motion of water w.r.t. B1 we can imagine a wooden piece of negligible mass, floating on the water surface. Its
velocity will be equal to that of water. At t = 0 the piece was in the same line joining B1 and B2 at a distance of 100 m from the bank.
At t = 1 sec it will move away by 10m from the bank that is its distance from b1 will be 60m (see figure shown above)
Following a similar approach find out various relative velocities as seen from frames of reference, which are fixed, on boat B2 (figure
Problem 3:-
A windblown rain is filling, 20 degrees from the vertical, at 5 m/s. Passengers in a car see the rain falling vertically. What is the magnitude and direction of car's velocity?
Solution:-
Here we have two frames S and S'. Frame S is attached with ground while S' attached with car. To determine the magnitude and directions of car's velocity, we use the concept of relative motion discussed earlier.
Let vRG = velocity of rain with respect to ground
vRC= velocity of rain with respect to car
vc= velocity of car with respect to ground
From the equation for relative velocity in two frames, we get
From the vector diagram we see that car should be travelling along the inclined direction of rain with velocity vC = 5 sin 20
o
m/s
Motion of Projectile
Now we discuss some example of curved motion or two dimensional motion of constant acceleration such as the motion of constant acceleration such as the motion of a particle projected at certain angle with the horizontal in vertical x-y plane (this type of motion is called projectile motion). Air resistance to the motion of the body is to be assumed absent in this type of motion.
To analyze the projectile motion we use the following concept "Resolution of two dimensional motion into two one dimension motion" as discussed earlier. Hence it is easier to analyze the motion of projectile as composed of two simultaneous rectilinear motions which are independent of each other:
(a) Along the vertical y-axis with a uniform downward acceleration 'g' and (b) Along the horizontal x-axis with a uniform velocity forward.
Consider a particle projected with an initial velocity u at an angle θ with the horizontal x-axis as shown in figure shown below. Velocity and accelerations can be resolved into two components:
Velocity along x-axis = ux = u cos θ
Acceleration along x-axis ax = 0
Velocity along y-axis = uy = u sin θ
Acceleration along y-axis ay = -g
Here we use different equation of motions of one dimension derived earlier to get the different parameters. …... (a)
…... (b) v2 = v0
2 – 2g (y-y
0) …... (c)
Total Time of flight:-
When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation b. Therefore, 0 = (u sin θ) t - (½)gt2
,
Horizontal Range:-
Horizontal Range (OA=X) = Horizontal velocity × Time of flight = u cos θ × 2 u sin θ/g
So horizontal range,
Maximum Height:-
At the highest point of the trajectory, vertical component of velocity is zero. Therefore, 0 = (u sin θ)2
- 2g Hmax
So, maximum height would be,
Equation of Trajectory:-
Assuming the point of projection as the origin of co-ordinates and horizontal direction as the x-axis and vertical direction as the y-axis. Let P (x, y) be the position of the particle at instant after t second.
Then x = u cos θ.t and y = u sin θ.t - 1/2 gt2
Eliminating 't' form the above equations, we get, y = x tan θ - (gx2
/2u2cos2θ)
This is the equation of trajectory which is a parabola (y = ax + bx2).
Problem 1 (JEE
Advanced):-The speed of a projectile when it is at its greatest height is √2/5 times its speed at half the maximum height. Find out the angle of projection.
Solution
H = u2sin2θ/2g So, gH = u2sin2θ/2 Now, vH = u cosθ Or, vH2 = u2 cos2θ …... (1) vH/22 = u2-2g(H/2) = u2-gH vH/2 2 = u2-(u2sin2θ/2) …...(2) Now it is given that,
vH = [√2/5] vH/2
Or, vH2 = (2/5) vH/22
Substituting the values from equations (1) and (2), we get, (u2 cos2θ) = 2/5 [u2
-(u2sin2θ/2)] Or, 5cos2θ = 2[1-(sin2θ/2)] Or, 5(1-sin2θ) = 2-sin2θ
Or, sin2θ = ¾
Or, sinθ = (√3)/2, Or, θ = 60o
Thus from the above observation, we conclude that, the angle of projection would be 60o.
Problem
2:-A gun moving at a speed of 30m/sec fires at an angle 30o with a velocity 150m/s relative to the gun. Find the distance between the
gun and the projectile when projectile hits the ground. (g = 10 m/sec)
Solution
:-Vertical component of velocity = 150 sin 30o = 75 m/sec
Horizontal component of velocity relative to gun = 150 cos 30o = 75√3 m/sec
Horizontal component of velocity relative to ground = 75√3 + 30 ≈ 160m/sec Time of flight = (2 75)/g = 15 s
Range of projectile = 160 × 15 = 2400 m
Distance moved by the gun and projectile = 2400 - 450 = 1950 m.
Horizontal projection:-
Assuming the point of projection O as the origin of coordinates and horizontal direction as the X-axis and vertical direction as Y-axis. Let P (x, y) be the position of the particle after t seconds.
So, x = horizontal distance covered in time t = ut. ... (1) y = vertical distance covered in time t = ½gt2 ... (2) Eliminate t from equations (1) and (2) then we get,
y = (1/2)(g/u2) (x2)
This is the equation of parabola passing through the origin, with its vertex at the origin O. Hence the trajectory is a parabola.
Problem
3:-A stone is thrown at a speed of 19.6 m/sec at an angle 30o above the horizontal from a tower of height 490 meter. Find the time during which the stone will be in air. Also find the distance from the foot of the tower to the point where stone hits the ground?
Solution
:-Let us consider the motion of stone in the horizontal and vertical directions separately. (i) Vertical motion (downward direction negative) :
Initial vertical velocity y = 19.6 sin 30o
Acceleration a = g = -9.8 m/s2
Vertical distance covered = h = 490 m Using, h = ut + 1/2gt2
We have, 490 = - 9.8t + (1/2) 9.8t2
100 = - 2t + t2 or t2 - 2t - 100 = 0
t = 11.05 sec (ii) Horizontal motion:
Initial horizontal velocity y = 19.6 sin 30o = 9.8 m/s
Hence distance from the foot of tower to the point where stone hits the ground = Horizontal component × time of flight
= 19.6 cos 30o × 11.05 = 188 m
Projectile Motion on an inclined plane:-
Let the particle strike the plane at A so that OA is the range of the projectile on inclined plane. This initial velocity can be resolved into two components:
(i) u cos (α - β) along the plane (ii) u sin (α - β) perpendicular to the plane.
The acceleration due to gravity g can be resolved into two components: (i) g sin β parallel to the plane
(ii) g cos β perpendicular to the plane.
Time of Flight:-
Let t be the time taken by the particle to go from A to B. In this time the displacement of the projectile to the plane is zero. Hence, 0 = u sin (α-β) t - ½g β t2
=> t = 2u sin(α-β)/gcosβ
Range:-
During time of flight, the horizontal velocity u cos α remains constant. Hence, horizontal distance,
OB = (ucosα) t = 2u2sin(α-β)cosα/gcosβ
Now, OA = OB/cosβ = 2u2sin(α-β)cosα/gcosβ
The greatest distance of the projectile from the inclined plane is u2sin2 (α-β)/2gcosβ .
Problem
4:-A Particle is projected with a velocity 39.2 m/sec at an angle of 30o to an inclined plane (inclined at an angle of 45o to the horizontal).
Find the range on the incline (a) when it is projected upward (b) when it is projected downward.
Solution
:-Time of flight will be same in both cases because the acceleration perpendicular to the plane is same. Therefore, 0 = 39.2 sin 30o t - (½) g cos 45o t2
(a) Range upward = 39.2 cos 30o t - (½) g sin 30o t2 = 39.2 × √3/2 × 4√2- (1/2) × 9.8 × (1/2) × (4√2)2 = 113.7m (b) Range downward = 39.2 cos 30o × t + (½) g sin 30o t2 = 39.2 × √3/2 × 4√2 + (1/2) × 9.8 × (1/2) × (4√2)2 = 270.5m Ans.
Motion down the plane:-
Let the particle be thrown at a velocity v0 at angle „α‟ with the horizontal as shown in figure. v0 sin (α+β)T- 1/2 gcosβT2=0 [for y'=0]
=> T = (2v0 sin(α+β))/gcosβ
R = v0 cos(α+ β)T+ 1/2 g sin βT2= (v02)/g [(sin(2α+β)+sinβ)/(1-sin2β)]
Since α is the variable and maximum value of sin function is 1, therefore for R to be maximum, sin (2α+β)=1 and Rmax (v02)/g [(1+sinβ)/(1-sin2β)]= (v02)/(g(1-sinβ)) down the plane.
Every projectile experiences one single force and that is due to gravity only.
Horizontal velocity of a projectile remains the same throughout its flight (it may be zero also).
No projectile ever experiences any acceleration in the horizontal direction.
Vertical acceleration of every projectile is „-g‟.
The path of projectile is parabolic except for those projected along vertical direction. In that case it is a straight line.Circular Motion
Now we shall discuss another example of two-dimensional motion that is motion of a particle on a circular path. This type of motion is called circular motion. The motion of a body is said to be
circular if it moves in such a way that its distance from a certain fixed point always remains the same.
Direction of motion of body at any instant:-
If the string breaks suddenly, the stone shall fly tangentially to the path of motion. So, instantaneous direction of motion of the body is always along the tangent to the curve at that point. Consider a particle P is moving on circle of radius r on X-Y plane with origin O as centre.
The position of the particle at a given instant may be described by angle θ, called angular position of the particle, measured in radian. As the particle moves on the path, its angular position θ changes. The rate of change of angular position is called angular velocity, ω, measured in radian per second.
= dθ/dt = ds/rdt = v/r
The rate of change of angular velocity is called angular acceleration, measured in rad/s2. Thus, the angular acceleration is
α = dω/dt = d2θ/dt2
Relation between These Parameters:-
It is easy to derive the equations of rotational kinematics for the case of constant angular acceleration with fixed axis of rotation. These equations are of the same form as those for on-dimensional transitional motion.
ω = ω0 + αt ... (a) ϕ = ϕ0 + ω0t + αt2
/2 ... (b) ω2 = ω02 + 2α (ϕ - ϕ0) ... (c) ϕ = ϕ0 + (ω0 + ω)/(2t) ... (d)
Here, ϕ0 is the initial angle and ω0 is the initial angular speed.
(a) What is the angular velocity of the minute and hour hands of a clock?
(b) Suppose the clock starts malfunctioning at 7 AM which decelerates the minute hand at the rate of 4Π radians/day. How much time would the clock loose by 7 AM next day?
Solution:-
(a) Angular speed of,
minute hand : ωmh = 2π rad/hr = 48π rad/day = (Π/1800) rad/sec
hour hand : ωhh = (π/6) rad/hr = 4π rad/day = (Π/21600) rad/sec
(b) Assume at t = 0, ϕ0 = 0, when the clock begins to malfunction.
Use equation (ii) to get the angle covered by the minute hand in one day. So, ϕ = ω0(1 day) 1/2α(1 day)
2 = 46π rad
Hence the minute hand complete 23 revolutions, so the clock losses 1 hour.
Problem 2:
A particle is rotating in a circular path having initial angular velocity 5 rad/sec and the angular acceleration α = 0.5 ω, where ω is angular velocity at that instant. Find the angular velocity, after it moved an angle π?
Solution:-
Here angular acceleration is α = 0.5 ω => dω/dt = 0.5ω => (dω/dθ) (dθ/dt) = 0.5ω => ω dω/dθ = 0.5ω Or, => ω - 5 = 0.5 × π => ω = 5 + 0.5 ×π = 6.57 rad/sec.
Motion of a particle in a circular path:-
It is a special kind of two-dimensional motion in which the particle's position vector always lies on the circumference of a circle. In order to calculate the acceleration parameter it is helpful to first consider circular motion with constant speed, called uniform circular motion. Let there be a particle moving along a circle of radius r with a velocity , as shown in figure given below, such that = v = constant. For this particle, it is our aim to calculate the magnitude and direction of its acceleration. We know that,
Now, we have to find an expression for in terms of known quantities. For this, consider the particle velocity vector at two points A and B. Displacing , parallel to itself and placing it back to back with , as shown in figure given below. We have
= –
Consider ΔAOB, angle between OA and OB is same as angle between and because is perpendicular to vector OA and is also perpendicular to vector OB.
OB = OA = r and
Thus,From geometry we have, Δv/v = AB/r Now AB is approximately equal to vΔt.
In the limit Δt → 0 the above relation becomes exact, we have
This is the magnitude of the acceleration. The direction is instantaneously along a radius inward towards the centre of the circle, because of this is called radial or centripetal acceleration. The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Problem 3:
The moon revolved about the earth making a complete revolution in 2.36 mega second. Assume that the orbit is circular and has a radius of 385 mega meter. What is the magnitude of the acceleration of the moon towards the earth?
Solution:-
Here first of all we calculate speed v of the moon which is given by v = 2πR/T where R = 385 mega meter = 385 × 106 m
and T = 2.36 mega second = 2.36 × 106 sec. Hence v = 1020 m/sec.
The magnitude of centripetal acceleration is a = v2/R = 0.00273 m/sec2.
From the above observation we conclude that, the magnitude of the acceleration of the moon towards the earth would be 0.00273 m/sec2.
In the previous enquiry we have discussed the uniform circular motion in which the particle has constant speed. If the particle's speed varies with time then the motion will be no more uniform but a non-uniform circular motion. Let us discuss about this motion using the concept of vectors.
Simulation for Car and Curves:-
This animation is used to explain why a passenger slides to the "outside" of a curve while riding inside a car is NOT an example of centrifugal forces. Instead is is a combination of centripetal force and inertia. It emphasize that when an object moves to the outside of a circle it is because of a lack of enough centripetal force and inertia keeps it moving in a straight line.
Non uniform circular motion:-
Let us use the vector method to discuss non-uniform circular motion.
In the side figure, and are unit vectors along radius and tangent vector respectively. In terms of er and eθ the motion of a particle moving counter clockwise in a circle about the origin in figure 2.30 can be described be the vector equation.
In this case, not only but v also varies with time. We can obtain instantaneous acceleration as,
=(d )/dt = θ dθ/dt + v(d θ)/dt Again,
Here, aT = dv/dt and aR = v2/r
The first term, is the vector component of that is tangential to the path of the particle and arises from a change in the magnitude of the velocity in circular motion, called tangential acceleration whereas aR centripetal acceleration.
The magnitude of is
Point A travels along an arc of a circle of radius r as shown in figure given below. Its velocity depends on the arc coordinates l as v = A √l where A is a constant. Let us calculate the angle α between the vectors of the total acceleration and of the velocity of the point as a function of the coordinate l.
Solution:-
It is seen from figure shown above that the angle α can be found by means of the formula tan α = aR/aT. Let us find aR and aT.
aR = v2/r=(A2 l)/r; aT= dv/dt = dv/dl = A/(2√l)×A√l = A2/2
Hence tan α = 2l/r.
Problem 5 (JEE Main):
Position vector of a particle moving in x-y palne at time t is .
What will be the path of the particle?
Solution:-
x = a (1-cos ωt) y = a sinωt
From first equation, we get acos ωt = a-X …... (1) From second equations, we get, a sin ωt = y
Squaring and adding Eqs. (1) and (2), we get, (X-a)2 + y2 = a2
This is an equation of a circle of radius a and ceter at (a,0).
Centripetal and centrifugal forces are equal in magnitude and opposite in direction.
Centripetal and centrifugal forces cannot be termed as action and reaction since action and reaction never act on same body.
Basic requirement for a body to complete motion in a vertical circle, under limiting conditions, is that the tension in the string must not vanish before it reaches the highest point. If it vanishes earlier, it will be devoid of the necessary centripetal force required to keep the body moving in a circle.
Relative Motion
Relative Velocity
In spite of our best efforts we could not find a fixed point with respect to which we should study absolute rest and absolute motion. So, the study of relative rest and relative motion assumes importance.
Consider two cars A and B running parallel to each other on same road with same velocities. No doubt the speedo meter of each indicates motion but the two drivers will always find themselves facing each other.
Relative velocity of a body A with respect to another body B, when both are in motion, is the velocity with which A appears to move to B.
The position, velocity and acceleration of a particle depend on the reference frame chosen.
A particle P is moving and is observed from two frames S and S'. The frame S is stationary and the frame S' is in motion. Let at any time position vector of the particle P with respect to S is,
From vector triangle OO'P, we get,
Physical Significance of Relative Velocity
Let two cars move unidirectional. Two persons A and B are sitting in the vehicles as shown in figure. Assume, VA = 10 m/s & VB = 4
m/s. The person A notices person B to be moving towards him with a speed of (10-4) m/s = 6 m/sec. That is the velocity of B with respect to (or relative to) A. That means is directed from B to A.
Similarly A seems to move towards B with a speed 6 m/sec. That means the velocity of A relative to B ( ) has the magnitude 6 m/sec & directed from A to B as shown in the figure.
Therefore, In general, So, vAB = √ vA2 + vB2 - 2vAvB cosθ and θ = tan-1{(v Bsinθ)/(vA – vBcosθ)}
Relative Motion between Rain and Man
We know that, vr = vrg = velocity of rain w.r.t. ground, vm ≡ vmg.
Velocity of man w.r.t. ground and
velocity of rain w.r.t. man. So,
That means the vector addition of the velocity of rain with respect of man ( ) and the velocity of man (vehicle) ( ) yield the actual velocity of rain . The magnitude and direction of can be given as,
vr = √((vrm)2+(vm)2+ 2vrm vm cosθ)
ϕ = tan-1
((vrm sinθ)/(vrm cosθ+ vm )) with horizontal
Illustration (JEE Main):
Four particles A, B, C and D are in motion. The velocities of one with respect to other are given as is 20 m/s towards north, is 20 m/s towards east and is 20 m/s towards south. Then is
(a) 20 m/s towards north
(b) 20 m/s towards south
(c) 20 m/s towards east
(d) 20 m/s towards west
Solution:
…... (1)
…... (2)
…... (3)
Equation (1) – (2) + (3) gives:
or
That is, is 20 m/s towards west.
Thus from the above observation we conclude that, option (d) is correct.
Illustration:
A stationary person observes that rain is falling vertically down at 30 km/hr. A cyclist is moving on the level road, at 10 km/hr. In which direction should the cyclist hold his umbrella to project himself from rain?
Solution:
Relative to stationary frame, velocity of rain is 30 km/hr downward. Take horizontal axis as x-axis and vertical axis as y-axis and i,j are the unit vectors along x and y axes respectively.
= 0-30ĵ, = 10î
= -30j - 10i = -10i - 30j
If angle between horizontal and is θ, then, tan θ = -30/-10 = 3
=> θ = tan-1 3 => θ=72°
Therefore, to protect himself from rain the cyclist should hold the umbrella at an angle of 72° from horizontal.
Illustration:
A man walking eastward at 5 m/s observes that wind is blowing from the north. On doubling his speed eastward, he observes that wind is blowing from north-east. Find the velocity of the wind.
Solution:
Let velocity of the wind is, vw=(v1i+v2j)m/s
vm=5i
So, vwm = vw- vm=(v1-5)i + v2j
In first case,
v1- 5 = 0 => v1= 5 m/s.
In the second case, tan 45o = v 2/(v1- 10)
=> v2= v1 - 10 = -5 m/s.
=> vw= (5i - 5j) m/s.
Illustration:
From a lift moving upward with a uniform acceleration 'a', a man throws a ball vertically upwards with a velocity v relative to the lift. The ball comes back to the man after a time t.
Show that a + g = 2 v/t
Solution:
Let us consider all the motion from lift frame. Then the acceleration, displacement and velocity everything will be considered from the lift frame itself. As the ball comes back to the man, therefore displacement from the lift frame is zero. Again, the velocity with respect to the lift frame is v.
g - (-a) = a + g (↓) downwards
Now, s = ut + 1/2at2 => 0 = vt - 1/2 (a+g)t2
or a + g = 2(v/t) .
Relative Motion of a Swimmer in Flowing Water
Take = velocity of man = velocity of flow of river,
= velocity of swimmer w.r.t. river
can be found by the velocity addition of and .
Crossing of the River with Minimum Drift
Case 1: vmw > w
A man intends to reach the opposite bank at the point directly opposite to the stationary point. He has to swim at angle θ with a given speed w.r.t. water, such that his actual velocity will direct along AB, that is perpendicular to the bank (or velocity of water ).
=> For minimum drift, ⊥
You can realize the situation by a simple example. If you want to reach the directly opposite point or cross the river perpendicularly, a man, that is to say, Hari, must report you that, you are moving perpendicular to the shore. What does this report signify? Since Hari observes your actual velocity ( ) to be perpendicular to the bank is perpendicular to .
Observing the vector-triangle vw = vmw sinθ & vm = vmw cosθ
=> θ = sin-1
(vw/vmw ) & vm = √((vmw )2- (vw)2)
=> The time of crossing, t = d/vm
=> t = d/√(vmw ) 2 - (vw) 2 ) Case 2 : vw > vmw
Let the man swim at an angle θ‟ with normal to the bank for minimum drift. Suppose the drift is equal to zero. For zero drift, the velocity of the man along the bank must be zero.
=> vm= vw- vmw sin θ' = 0
This gives, sinθ' = vw / vmw, since vw > vmw, sinθ' > 1 which is impossible. Therefore, the drift cannot be zero.
Now, let the man swim at an angle θ with the normal to the bank to experience minimum drift. Suppose that the drifting of the man
during time t when the reaches the opposite bank is,
BC = x
x = (vm)x (t) … (1)
where t = AB/((vm )y cosθ) = d/(vmw cosθ) … (2)
and (vm)x = vw – vmw sin θ … (3)
Using (1), (2) & (3), we obtain x = (vw- vmw sinθ d/(vmw cosθ))
= (vw/vmw sec θ-tanθ)d
x = (vw/vmw sec θ-tanθ)d … (4)
For x to be minimum,
dx/dθ = (vw/vmw secθ - tanθ - sec2θ)d = 0
vw/vmw tanθ = (sec θ) => sinθ = vmw/vw
θ = sin-1(vmw/vw)
Substituting the value of θ in (4), we obtain, x = [√(vw2 – vmw2)/vmw] d
Crossing of the River in Minimum Time Case 1: To reach the opposite bank for a given vmw
Let the man swim at an angle θ with AB. We know that the component of the velocity of man along shore is not responsible for its crossing the river. Only the component of velocity of man (vm) along AB is responsible for its crossing along AB.
The time of crossing = t = AB/(vmw cosθ)
Time is minimum when cos θ is maximum.
The maximum value of cos θ is 1 for θ = 0.
That means the man should swim perpendicular to the shore. => mw ⊥ w
=> Then tmin = d/(vmw cosθ)|(θ=0) = d/vmw => tmin= d/vmw
Case 2:
To reach directly opposite point on the other bank for a given vmw & velocity v of walking along the shore.
To attain the direct opposite point B in the minimum time, let the man swim at an angle θ with the direction AB. The total time of journey t = the time taken from A to C and the time taken from C to B.
where tAC = AB/vmvcosθ & tCB = BC/v where v = walking speed of the man from C to B.
=> t = AB/vmvcosθ + BC/v
Again BC = (vm)xt
=> BC = (vw - vmwsinθ) (AB/vmwcosθ)
Using (1) & (2) we obtain,
t = AB/vmwcosθ + ((vw - vmwsinθ)/v(vmvcosθ))
=> t = AB[(1+vw/v)secθ/vmv - tanθ/v]
=> t = d/vmv[(1+vw/v)secθ/vmv - tanθ/v]
Putting dt/dθ = 0, For minimum t we get, dt/dθ = d/dθ[d/vmv (1+vw/v) secθ/vmv - tanθ/v]
= [secθ/vmv - tanθ/v (1+vw/v) (sec2θ)/v] = 0
=>tanθ/vmv (1+vw/v) secθ/v
=> sin θ = (vmw/v+vw)
=> θ = sin-1
(vmw/v+vw)
This expression is obviously true when vmw < v + vw.
Velocity of Separation/Approach, Relative Angular Velocity
Let thane be two particles A and B with velocity and at any instant as visualized from ground frame. If we visualize the motion of B from frame of A the velocity of particle B would be .
If α, β be the angle made with line AB,
then, VB cos β - VA cos α is relatively velocity of B w.r.t. A along line AB.
If VB cos β - VA cos α > 0; it is called as velocity of separation.
If VB cos β - VA cos α < 0; it is called as velocity of approach.VB sin β - VA sin α is relative velocity of B w.r.t. A along direction perpendicular to AB. If length of AB is,