IIT Physics help Topics:

General Physics | Mechanics | Wave Motion | Thermal Physics | Electric Current |Electrostatics | Magnetism | Electromagnetic induction | Wave Optics | Ray Optics |Modern Physics  

General Physics

1. Dimensions

2. Applications of Dimensions

3. Scalars and Vectors

4. Addition and Subtraction of Vectors

5. Multiplication of Vectors

6. Vector Components

 Dimensions

By international agreement a small number of physical quantities such as length, time etc. are chosen and assigned standards. These quantities are called ‘base quantities’ and their units as ‘base units’. All other physical quantities are expressed in terms of these ‘base quantities’. The units of these dependent quantities are called ‘derived units’. 

The standard for a unit should have the following characteristics.

        (a) It should be well defined.         (b) It should be invariable (should not change with time)         (c) It should be convenient to use         (d) It should be easily accessible 

The 14th general conference on weights and measures (in France) picked seven quantities as base quantities, thereby forming the International System of Unitsabbreviated as SI (System de International) system. Base quantities and their units

                                           The seven base quantities and their units are 

        Base quantity Unit SymbolLength Metre MMass Kilogram KgTime Second Sec

Electric current Ampere ATemperature Kelvin K

Luminous intensity Candela CdAmount of substance Mole Mole

Derived units 

We can define all the derived units in terms of base units. For example, speed is defined to be the ratio of distance to time. 

                        Unit of Speed = (unit of distance (length))/(unit of time) 

                               = m/s = ms-1 (Read as metre per sec.) 

                          SOME DERIVED SI UNITS AND THEIR SYMBOLS 

Quantity Unit Symbol Express in base units

Force newton N Kg-m/sec2

Work joules J Kg-m2/sec2

Power watt W Kg-m2/sec3

Pressure pascal Pa Kg m-1/S2

 

Important:

The following conventions are adopted while writing a unit.

(1) Even if a unit is named after a person the unit is not written capital letters. i.e. we write joules not Joules.

(2) For a unit named after a person the symbol is a capital letter e.g. for joules we write ‘J’ and the rest of them are in lowercase letters e.g. seconds is written as ‘s’. 

(3) The symbols of units do not have plural form i.e. 70 m not 70 ms or 10 N not 10Ns. 

(4) Not more than one solid’s is used i.e. all units of numerator written together before the ‘/’ sign and all in the denominator written after that. i.e. It is 1 ms-2 or 1 m/s-2 not 1m/s/s. 

(5) Punctuation marks are not written after the unit e.g. 1 litre = 1000 cc not 1000 c.c.

It has to be borne in mind that SI system of units is not the only system of units that is followed all over the world. There are some countries (though they are very few in number) which use different system of units. For example: the FPS (Foot Pound Second) system or the CGS (Centimeter Gram Second) system.

Dimensions 

The unit of any derived quantity depends upon one or more fundamental units. This dependence can be expressed with the help of dimensions of that derived quantity. In other words, the dimensions of a physical quantity show how its unit is related to the fundamental units.

To express dimensions, each fundamental unit is represented by a capital letter. Thus the unit of length is denoted by L, unit of mass by M. Unit of time by T, unit of electric current by I, unit of temperature by K and unit of luminous intensity by C. 

Remember that speed will always remain distance covered per unit of time, whatever is the system of units, so the complex quantity speed can be expressed in terms of length L and time T. Now,we say that dimensional formula of speed is LT-1. We can relate the physical quantities to each other (usually we express complex quantities in terms of base quantities) by a system of dimensions.

Dimension of a physical quantity are the powers to which the fundamental quantities must be raised to represent the given physical quantity. 

Example

Density of a substance is defined to be the mass contained in unit volume of the substance.

                Hence, [density] = ([mass])/([volume]) = M/L3 = ML-3 

    So, the dimensions of density are 1 in mass, -3 in length and 0 in time. 

                Hence the dimensional formula of density is written as

                                           [ρ]= ML-3T0

It is to be noted that constants such as ½ π, or trigonometric functions such as “sin wt” have no units or dimensions because they are numbers, ratios which are also numbers.

Units and Dimensions are important from IIT JEE perspective. Objective questions are framed on this section. AIEEE definitely has 1-2 questions every year directly on these topics. Sometimes both IIT JEE and AIEEE do not ask questions on units and dimensions directly but they change units and involve indirect application. So it’s very important to master these concepts at early stage as this forms the basis of your preparation for IIT JEE and AIEEE Physics.

At askIITians we provide you free study material on units and dimensions so that you get all the professional help needed to get through IIT JEE and AIEEE easily. AskIITians also provides live online IIT JEE preparation and coaching where you can attend our live online classes from your home! 

Applications of Dimensions

Broadly speaking, dimension is the nature of a Physical quantity. Understanding of this nature helps us in many ways.

Following are some of the applications of the theory of dimensional analysis in Physics: 

(i) To find the unit of a given physical quantity in a given system of units:

By expressing a physical quantity in terms of basic quantity we find its dimensions. In the dimensional formula replacing M, L, T by the fundamental units of the required system, we get the unit of physical quantity. However, sometimes we assign a specific name to this unit. 

Illustration:

              Force is numerically equal to the product of mass and acceleration 

                                 i.e. Force = mass x acceleration 

          or [F] = mass x velocity/time= mass x displacement/(time)2 ) = mass x                                                                     length/(time)2)

                               = [M] x [LT-2] = [MLT-2]

Its unit in SI system will be Kgms<sup<>-2 which is given a specific name “newton (N)”. Similarly, its unit in CGS system will be gmcms-2 which is called “dyne”.

(ii) To find dimensions of physical constants or coefficients:

The dimension of a physical quantity is unique because it is the nature of the physical quantity and the nature does not change. If we write any formula or equation incorporating the given physical constant, we can find the dimensions of the required constant or co-efficient.

Illustration:

      From Newton’s law of Gravitation, the exerted by one mass upon another is

                          F=G (m1 m2)/r2 or G=(Fr2)/(m1 m2 )

                         or [G] = ([MLT]2][L-2]) / ([M][M]) = [M-1 L3 T-2 ]

                       We can find its SI unit which is m3/Kgs2.

(iii) To convert a physical quantity from one system of units to another:

This is based on the fact that for a given physical quantity, magnitude x unit = constant So, when unit changes, magnitude will also change. 

Illustration:

                          Convert one Newton into dyne </sup<>

Solution: 

                        Dimensional formula for Newton = [MLT-2] 

               Or 1 N = 1 Kg m/s2 ; But 1 kg = 103 g and 1 m = 102 cm

                 Therefore 1 N = ((103 g)(102 cm))/s2 = 105 g cm/s2 = 105 dyne

(iv) To check the dimensional correctness of a given physical relation:

This is based on the principle that the dimensions of the terms on both sides on an equation must be same. This is known as the ‘principle of homogeneity’. If the dimensions of the terms on both sides are same, the equation is dimensionally correct, otherwise not. 

Caution: It is not necessary that a dimensionally correct equation is also physically correct but a physically correct equation has to be dimensionally correct. 

Illustration: 

(i) Consider the formula, T=2∏√(l/g)

Where T is the time period of oscillation of a simple pendulum in a simple harmonic motion, l and g are the length of the pendulum and gravitational constants respectively. Check this formula, whether it is correct or not, using the concept of dimension. 

As we know [g] = [LT2] 

                     Therefore  [T] = √(([L])/([LT-2])) = [T] s

Thus the above equation is dimensionally correct (homogenous) and later you will come to know that it is physically also correct.

(ii) Consider the formula s=ut -1/3 at2. Check this formula whether it is correct or not, using the concept of dimension.

Dimensionally 

                            [L] = [LT-1] [L] – [LT-2] [T2] 

                                   => [L] = [L] – [L] 

In this case also the formula is dimensionally correct but, you know that it is physically incorrect as the correct formula is given by                                    S = ut + 1/3at2 

(v) As a research tool to derive new relations: 

One of the aims of scientific research is to discover new laws relating different physical quantities. The theory of dimensions (in the light of principal of homogeneity) provides us with a powerful tool of research in the preliminary stages of investigation [It must be again emphasized that mere dimensional correctness of an equation does not ensure its physical correctness]

Limitations of the theory of dimensions 

The limitations are as follows:

(i) If dimensions are given, physical quantity may not be unique as many physical quantities have the same dimension. For example, if the dimensional formula of a physical quantity is [ML2T-2] it may be work or energy or even moment of force. 

(ii) Numerical constants, having no dimensions, cannot be deduced by using the concepts of dimensions. 

(iii) The method of dimensions cannot be used to derive relations other than product of power functions. Again, expressions containing trigonometric or logarithmic functions also cannot be derived using dimensional analysis, e.g. 

              s = ut + 1/3at2    or   y = a sincot    or    P= P0e (–Mgh)/RT 

    cannot be derived. However, their dimensional correctness can be verified. 

(iv) If a physical quantity depends on more than three physical quantities, method of dimensions cannot be used to derive its formula. For such equations, only the dimensional correctness can be checked. For example, the time period of a physical pendulum of moment of inertia I, mass m and length l is given by the following equation.

T = 2∏√(I/mgl) (I is known as the moment of Inertia with dimensions of [ML2] through dimensional analysis), though we can still check the dimensional correctness of the equation

(Try to check it as an exercise).

(v) Even if a physical quantity depends on three Physical quantities, out of which two have the same dimensions, the formula cannot be derived by theory of dimensions, and only its correctness can be checked e.g. we cannot derive the equation. 

Scalars and Vectors

Scalars 

Physical quantities which can be completely specified by a number and unit, and therefore have the magnitude only, are scalars. Some physical quantities which are scalar are mass, length, time, energy etc. These examples obey the algebraic law of addition. 

Vectors 

Vectors are physical quantities, which besides having both magnitude and direction also obey the law of geometrical addition. (The law of geometrical addition, i.e. the law of triangular addition and law of parallelogram are discussed later in this chapter). Some physical quantities, which are vectors, are displacement, velocity, force etc. 

Representation of a Vector 

Since vectors have directions, any representation of them has to include the direction. 

To represent a vector we use a line with an arrow head. The length of the line represents the magnitude of vector and direction of the arrow represents the direction of the vector. Let us start with a vector quantity called displacement. In the enclosed figure the change of position from point P1 to P2 is represented graphically by the directed line segment with an arrowhead to represent direction of motion. 

                                            

Vector is a Physical quantity and all physical quantities have units. Hence, the vectors also have units, they are called unit vectors. 

Unit Vectors: A unit vector is a vector having a magnitude of unity. Its only purpose is to describe a direction in space. On x-y co-ordinate system ĩ denote unit vector in positive x direction and ĵ denotes unit vector in positive y direction.

                                   

Any vector in x – y plane can be represented in terms of these unit vectors ĩ & ĵ. 

Similarly any vector in a 3 dimensional x y z space can be represented in terms of unit vectors ĩ, ĵ and k where, k is the unit vector in the positive z direction, as shown in figure above. 

Parallel Vectors: Two or more vectors are said to be parallel when they are parallel to the same line. In the figure below, the vectors A B and C are all parallel. 

                                            

Equal Vectors: Two or more, vectors are equal if they have the same magnitude (length) and direction, whatever their initial points. In the figure above, the vectors A and B are equal. 

Negative Vectors: Two vectors which have same magnitude (length) but their direction is opposite to each, other called the negative vectors of each other. In figure above vectors A and C or B and C are negative vectors. 

Null Vectors: A vector having zero magnitude is called zero vector or ‘null vector’and is written as = O vector. The initial point and the end point of such a vector coincide so that its direction is indeterminate. 

The concept of null vector is hypothetical but we introduce it only to explain some mathematical results. 

Invariance of the vector: Any vector is invariant so it can be taken anywhere in the space keeping its magnitude and direction same. In other words, the vectors remain invariant under translation. 

IIT JEE study material is available online free of cost at askIITians.com. Study Vectors, Scalars, Units & Dimensions – General Physics at askIITians website and be a winner. We offer numerous live online courses as well for live online IIT JEE preparation – you do not need to travel anywhere any longer – just sit at your home and study for IIT JEE live online

with askIITians.com 

Addition and Subtraction of Vectors

Geometrical method 

                           

To find a  + b , shift  vector b  such that its initial point coincides with the terminal point of vector a. Now, the vector whose initial point coincides with the initial point of vector a , and terminal point coincides with the terminal point of vector b  represents (a +b ) as shown in the above figure. 

To find (b +a ), shift a  such that its initial point coincides with the terminal point b . A vector whose initial point coincides with the initial point of b  and terminal point coincides with the terminal point of a  represents (b +a ).

                         

Law of Parallelogram of Vectors 

The addition of two vectors may also be understood by the law of parallelogram. 

According to this law if two vectors P and Q are represented by two adjacent sides of a parallelogram both pointing outwards as shown in the figure below , then the diagonal drawn through the intersection of the two vectors represents the resultant (i.e. vector sum of P and Q). If Q is displacement from position AD to BC by displacing it parallel to itself, this method becomes equivalent to the triangle method. 

                                

In case of addition of two vectors by parallelogram method as shown in figure, the magnitude of resultant will be given by, 

                                       (AC)2 = (AE)2 + (EC)2 

                               or R2 = (P + Q cos θ)2 (Q sin θ)2 

                                           or R = √(P2+ Q2 )+ 2PQcos θ 

                    And the direction of resultant from vector P will be given by                               tanǾ = CE/AE = Qsinθ/(P+Qcosθ) 

                                   Ǿ=tan-1 [Qsinθ/(P+Qcosθ)]  

Magnitude and direction of the resultant 

 The magnitude of resultant will be 

                          R = √(P2+ P2+2P2 cos"α" ) = 2Pcosα/2 Ans.

The direction of the resultant is 

                    Ǿ = tan-1 [Psinθ/(P+Pcosθ)] = tan-1tan(θ/2) = Ǿ = θ/2 

Vector subtraction 

Suppose there are two vectors A and B, shown in figure A and we have to subtract B and A. It is just the same thing as adding vectors – B to A. The resultant is shown in figure B. 

                                                                                     Figure (A) 

                                                                                        Figure (B) 

             

Properties of Vector Addition 

    1. Vector addition is commutative

        i.e        

                   

                                          

     2. Vector addition is associative 

      i.e  

               

                                      

Magnitude and direction of a+b

Let angle between vector a and b be θ

In the figure vector (OA) = vector a , vector (AB) = vector b 

From     △ADB

                                 AD = b cos θ

                                  BD = b sin θ

                                

In right angled △ODB

                              OD = a + b cosθ

                  BD = b sin θ        Therefore  OB = √(OD2+BD2 )

                      => |a +b |=√(a2+b2+2ab cos θ)

                   |a +b |max   =  a+b   when θ = 2n∏

                   |a +b |min   =  |a – b|   when θ = (2n + 1)∏                                       (where n = 0, 1, 2, …..)

If a + b  is inclined at an angle α with vector a , then

                        tan α = ((b sin θ)/(a+b cosθ))△AskIITians provides free study material for IIT JEE, AIEEE and other engineering entrance examinations. Vectors, their addition and subtraction rules, law of parallelogram addition of vectors are very important from engineering entrance exam point of view as there are many situations in which these concepts and rules are applied. Many a times questions are directly asked from these topics.AskIITians provides live online IIT JEE and AIEEE coaching. You can sit at your home and attend our live online coaching classes. 

Multiplication of Vectors

1. Multiplication of vector by a scalar 

Let vector a is multiplied by a scalar m. If m is a positive quantity, only magnitude of the vector will change by a factor ‘m’ and its direction will remain same. If m is a negative quantity the direction of the vector will be reversed. 

2. Multiplication of a vector by a vector      (i) Dot product or scalar product    (ii) Cross product or vector product 

Dot product or scalar product 

   The dot product of two vectors a and b is defined as                               a-> . b->  = ab cosθ 

where a and b are the magnitudes of the respective vectors and θ is the angle between them. The final product is a scalar quantity. If two vectors are mutually perpendicular then θ = 900 and cos 90 = 0, Hence, their dot product is zero. 

   Some examples of dot product: work = F-> . s->  = Fs cosθ

                                         

Here, 

              

The dot product obeys commutative law 

      i.e.     a ->. b-> =b-> .a->  

Hence,   a-> . b->  =  axbx + ayby + azbz 

Illustration : 

Find the angle between the vectors A and B where 

                                                    Solution : 

We know 

              A->.B->  = |A||B| cosθ   where |A| = √(22 +32 + 32) = √22, |B| =√(12 + 22+ 32) = √14 

Hence cosθ = (A-> .B-> )/(|A||B|)=((2i +3j +3k  )(i +2j -3k ))/(√22×√14)                                                 =(2+6-9)/(2√77)=(-1)/(2√77) 

                             => θ = cos-1((-1)/(2√77)) 

Cross product or vector product 

          The cross product of the two vectors a and b is defined as 

                                 a->  × b->  = c->  

     Here, |c-> |=|a->|×|b->| sinθ, where θ is the angle between the vectors. 

Vector product is defined as a vector quantity with a direction perpendicular to the plane containing vectors A and B then C = AB sin θ n, where n is a unit vector perpendicular to the plane of vector A and vector B. To specify the sense of the vector C, refer to the figure given below.

                                  

Imagine rotating a right hand screw whose axis is perpendicular to the plane formed by vectors A and B so as to turn it from vectors A to B trough the angle θ between them. Then the direction of advancement of the screw gives the direction of the vector product vectors A-> × B->. 

Illustration: 

Obtain a unit vector perpendicular to the two vectors A-> = 2i + 3j + 3k, B-> = i – 2j + 3k 

Solution: 

We know that A-> × B-> = AB sin θ n  

                                     n  =  (A->×B->)/ABsinθ 

We have A-> × B->  = 17i-2j-7k  

                                      A = √29 B = √14 

and θ = cos-1 8/(√14 √29) (Use concept of dot product to find θ). 

From the above values we can find n 

Solving we get, 

               n = (17i-2j-7k)/(√29 √14 sinθ)     where θ cos-1 8/(√14 √29)Cross Product of Parallel vectors 

If two vectors are parallel or antiparallel, then θ is either 00 or 1800. Since sin 00and sin 1800 both equals zero. Hence magnitude of their cross product is zero. 

The vector product does not follow commutative law.

                                           

Product of unit vectors

                                                

           

Multiplication of vectors is a very important topic from IIT JEE, AIEEE and other engineering exams perspective. Dot product or scalar product and cross product or vector product find their use both in physics and mathematics while preparing for IIT JEE, AIEEE and other engineering entrance exams. 

askIITians.com provides free of cost study material for exam preparation - you can use these resources as much as you can. askIITians also provides IIT JEE, AIEEE and other engineering entrance exams preparation live online. It’s a live online coaching where you can study from your home – get all the benefits of classroom coaching sitting at your home from askIITians.

Vector Components

Components of a Vector: From the figure given below we can write 

                                           It means B ,C  and D vectors are the components of vector A 

                                      . 

Note that we can also write A =X +Y. So vectors X and Y are also components of vector A. It implies that we can draw any number of set of components in any desired direction. 

For resolving a vector A along two directions making angles α and with it as shown in figure given below, we use the following:- 

                                      

                               Aα = (A sinβ)/(sin(α + β))

Perpendicular Components 

Representation of any vector lying in the x – y plane, as shown in figure given below, as the sum of two vectors, one parallel to the x-axis and the other parallel to y-axis, is extremely useful in physical analysis because both have mutually independent effects. These two vectors are labeled A and A. These are called theperpendicular or rectangular component of vector A and are expressed as: 

                                

If magnitude and direction of vector A are known then Ax = A cosθ and Ay = A sinθ. Hence we can write 

                              Ax2 + Ay

2 = A2 and tanθ = Ay/Ax

Now, it should be clear that if we have to add 30 vectors, we will resolve each of these 30 vectors in rectangular components in any x and y direction. Then simply add all the components in the x direction and all the components in the y direction. Adding these two resultant perpendicular components will give us the final resultant.

Illustration:

A vector quantity of magnitude L acts on a point A along the direction making an angle of 450 with the vertical, as shown in the figure given below. Find the component of this vector in the vertical direction?

                                         

Solution:

The component of the vector in the vertical direction will be

                                       L*(cos ∏/4) = L/√2.

Learning components of a vector is very important from IIT JEE, AIEEE and other engineering exams perspective. Most of the Physics is based on usage of vectors and components of vectors. Vectors and their components are also useful in trigonometry and mathematics as a whole. This is a very useful concept and should be learned thoroughly at the beginner’s level so that it paves a way for great understanding of statistics, physics and mathematics at higher levels.

AskIITians.com provides free study material for students aspiring for IIT JEE AIEEE and other engineering entrance exams. We also provide live online IIT JEE and AIEEE preparation over internet. You can study in our live online batches and do your exam preparation with us from your home. Utilize askIITians free resources like:- FREE Study Material, FREE A Question A Day, FREE Test Series while preparing for IIT JEE, AIEEE and other engineering entrance exams. 

************** ------- Mechninics ------ **************************

1. Kinematics2. Newton’s Law3. Conservation of Momentum4. Work Power And Energy5. System of particles6. Gravitation7. Rotational Motion8. Simple Harmonic Motion9. Hooke’s Law; Young’s modulus10. Fluid mechanics

Kinematics

1. Introduction to Motion in One Dimension

2. Graphical Representation and Equations of Motion

3. Motion under Gravity

4. Motion in a Straight Line with Acceleration

5. Motion in Two Dimensions

6. Motion of Projectile

7. Circular Motion

8. Relative Motion

9. Solved Examples

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 IN ONE DIMENSION

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).

Caution :  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 (essentially we treat body as a point mass) with the help of three co-ordinates X, Y and Z. Hence X, Y, Z is a set of coordinate axes 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.

Caution:    Motion of a body cannot be correctly identified unless one knows the position of body as specified by a fixed frame of reference.

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:

Δ  =  2 -  t

 

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 figure below, 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

Note:  If the graph is not a straight line, it will represent non-uniform motion.

 

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.

                                                

Illustration:

This question contains statement-1 (Assertion) and Statement-2 (Reason). Question has 4 choices (A), (B), (C) and (D) out of which only one is correct.

Statement-1

A bus moving due north take a turn and starts moving towards east with same speed. There will be no change in the velocity of the bus.

Statement-2

Velocity is a vector quantity.

(A)  Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-1.

(B)  Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for statement-1.

(C)  Statement-1 is true, Statement-2 is false.

(D)  Statement-1 is false, Statement-2 is true.

Solution (D)

     This is so because bus is changing its direction of motion.

Illustration:

      A man started running form origin and went up to (2, 0) and returned back to (-1, 0) as shown in figure 2.7. In this process total time taken by man is 2 seconds. Find the average velocity and average speed.

 

Solution:

        The man is displaced form origin to (-1, 0)

        Hence displacement, s = 

        So average velocity  = Displacement/total time =  /2 = (-1/2) m/sec.

        where as, since the total distance traveled by man

                                = (0, 0) to (2, 0)+(2, 0) to (0, 0)+(0, 0) to (-1, 0)

                                = 2 + 2 + 1 = 5 m

        Hence average speed

                                = (Total distance)/(Total time)=5/2 m/sec.

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?                (ii) Can a body have a constant velocity and still have a varying speed?

 

Illustration:

A cyclist moves 12 km due to north and then 5 km due east in 3 hr. Find (a) his average speed, (b) average velocity, in m/s.

Solution:

 In the figure, A shows the initial position and C the final position of the cyclist. The total distance covered by the cyclist AB+BC= (12+5)km = 17 km.

 .·.   Its average speed = 17/3 km/hr = 1.57 m/s

Its displacement is AC and the magnitude is given by

AC = √(AB2 + BC2) = √(122 + 52) km = 13 km

.·.    Its average velocity = 13/3 km/hr

 = 1.2 m/s along AC, i.e at tan-1(5/12) or 22.6o East of North.

 

Illustration:

        A train is moving with a constant speed of 5 m/s and there are two persons A and B standing at a separation of 10 m inside the train. Another person C is standing on the ground. Then, find

(a)    displacement covered by A, if he moves towards B and back to its position in 10 seconds in frame of reference of train and in frame of reference of C.

(b)    distance covered by A in frame of reference of train and in frame of reference of C.

Solution:

(a)    In the frame of train, displacement covered by A is zero and in frame of reference of C, displacement covered by A=0 + 5x10 = 50 m.

(b)    Distance covered by A in frame of reference of train is 20m and distance covered by A in frame of reference of C is (20 + 50) = 70 m.

 

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  1and at t2,

velocity is  2. The average acceleration <   > during the motion is defined as

 

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  0 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  0 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 of particle is

                    

        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.

 

Caution:   The equation of motion derived above are possible only in uniformly accelerated motion i.e. the motion in which the acceleration is constant.

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.

Pause :     The above equations of motion are, however, universal and can be derived by using differential calculus as given below:

               

        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         .·. 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

        =>  dv = k   ds

        => k = (v2-v1)/d

        From equation (1)

        => (dv/v) k.dt. dv/v = K dt = ln v2/v1 = kt

        => t = (d/(v2-v1))ln v2/v1                         Ans.

 

<< Back |

 

IIT JEE study material is available online free of cost at askIITians.com. Study Physics, Chemistry and Mathematics at askIITians website and be a winner. We offer numerous live online courses as well for live online IIT JEE and AIEEE preparation - you do not need to travel anywhere any longer - just sit at your home and study for IIT JEE and AIEEE live online with askIITians.com

************* Graphical Representation and Equations of Motion **********

 

(i)          Position - Time Graph

(ii)         Velocity - Time Graph

Note : 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.

 

Illustration:

The adjacent figure shows the displacement-time graph of a particle moving on the x-axis. Choose the correct option given below.

 

(A)      The particle is continuously going in positive x direction.

(B)      The particle is at rest.

(C)      The particle moves at a constant velocity upto a time t0, and then stops.

(D)      The particle moves at a constant acceleration upto a time t0,and then stops.

 

Solution (C)

Up to time t0, particle is said to have uniform rectilinear motion and after that comes to rest as the slope is zero.

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.

           s =  v dt  

        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 =  OABD + Δ BCD

        s = ut + (1/2)at2                                                    (2)

        Again

                s = Area OABCD

                = 1/2(AC + OD) x OA = 1/2(v + u)xt

 

Illustration:

 From the velocity-time plot shown in figure, find

(a)distance travelled by the particle during the first 40 seconds.

(b)displacement travelled by the particle during the first 40  seconds.

(c) Also find the average velocity during the period.

Solution:

 (a)    Distance = area under the curve

                     = 1/2 x 20 x 5 + 1/2 x 5 x 20

                                 

                For distance measurement, the curve is plotted as in Fig. (a)

(b)    Displacement = area under the curve in Fig. (b) = 0        

(c)    Vav = Displacement/time

                        As displacement is zero,

                .·.      Vav = 0

Illustration:

 The velocity-time graph of a moving object is given in the figure. Find the maximum acceleration of the body and distance travelled by the body in the interval of time in which this acceleration exists.

                        

Solution:

 Acceleration is maximum when slope is maximum

                        amax  = (80-20)/(40-30) = 5m/s2

                        S = 20 m/s x 10 s + 1/2 x 6 m/s2 x 100s2 = 500m

 

Illustration:

 The displacement versus time curve is given:

  Column I                                    Column II

(a)    OA                             (p)    Velocity increase with time linearly

(b)    AB                             (q)    Velocity decreases with time

(c)    BC                              (r)     Velocity is independent of time

(d)    CD                             (s)    Velocity is zero

Solution:

         A → (p)      (B)  → (r)           (C)  → (q)          (D) → (s)

    This is so because slope of displacement-time curve gives instantaneous velocity.

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.

      

 

 

<< Back |

 

Register for online classroom programmes targeting IIT JEE 2010 and 2011. You can also participate in the online tests conducted by askIITians and also give answers in AQAD ( A Question  A Day) to get a number of benefits in the online live classroom courses. Visit askIITians.com to read online Study material for IIT JEE and AIEEE preparation absolutely free.

*************** Motion Under Gravity **********************

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:

        

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.

        .·. 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

.·. 0 = 9 + (-0.6)t        or,  t = (9/0.6)s = 15s

 

Illustration:

A car moving in a straight line at 30 m/s slows uniformly to  a speed of 10 m/s in 5 sec. Determine:

(a)the acceleration of the car,

(b)displacement in the third second.

Solution:

Let us take the direction of motion to be the +x direction.

(a)  For the 5 s interval, we have t = 5 s, u = 30 m/s, v = 10 m/s

      Using v = u + at, we have

      a = (v-u)/t = (10-30)/5 = -4 m/s2

(b)  s = (displacement in 3s) - (displacement in 2s)

      =

      Here u = 30 m/s, a = -4 m/s2, t2 = 2 s; t3 = 3 s

       s = 30(3-2) + 1/2(-4)(32-22)  = 20 m

Alternatively

       Sn = u + a/2(2n - 1)

Hence n = 3,

.·. s3 = 30 + ((-4))/2 [(3)- 1]= 20 m

 

Illustration:

A bullet fired into a fixed target loses half of tis velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion.

Solution:

        Let initial velocity of the bullet = u

        After penetrating 3 cm its velocity becomes = u/2

        From v2 = u2 - 2as

        (u/2)2 = u2 - 2as

.·.  a  = u2/8

Let further it will penetrate through distance x and stops at some point

        0 =  (u/2)2 - 2 (u2/8)x

.·.  x = 1 cm

 

Illustration:

An anti-aircraft shell is fired vertically upwards with a muzzle velocity of 294 m/s. Calculate (a) the maximum height reached by it, (b) time taken to reach this height, (c) the velocities at the ends of 20th and 40th second. (d) When will its height be 2450 m? Given g = 9.8 m/s2.

Solution:

(a)    Here, the initial velocity u = 294 m/s and g = 9.8 m/s2

        .·.   The maximum height reached by the shell is,

         H = us/2g = 2942/(2 x 9.8) = 4410 m = 4.41 km

(b)    The time taken to reach the height is, T = u/g = 294/9.8 = 30 s

(c)    The velocity at the end of 20th second is given by

         v = u - gt = 294 - 9.8 x 20 = 98 m/s upward,

         and the velocity at the end of 40th second is given by,

         v = 294 - 9.8 x 40 = -98 m/s

         The negative sign implies that the shell is falling downward.

(d)    From the equation

        H = ut + (1/2)gt2      or     2450 = 2941 t - 1/2 x 9.8  t2

        or t2 - 60 t + 500 = 0

        .·.  t = 10 s and 50 s.

At t = 10 s, the shell is at a height of 2450 m and is ascending, and at the end of 50 s it is at the same height, but is falling.

 

 

<< Back |

 

Motion in a Straight Line with 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

 .·.                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

 

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)

 

Illustration:

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:

            

 

**************  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.

Illustration:

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 Ans.

and displacement = s sx i + sy j = (76 i + 64 j) m Ans.

 

Frame of reference (Contd.)

       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?

                                             

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 unaccelerated 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   is the vector sum of    

and   t:

 =   +   t.

        Differentiating equation leads to d /dt = d /dt + 

        But d /dt =  , the instantaneous velocity of the particle measured in the s-frame, and

d /dt =  , 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.

 

Illustration:

        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

.·. 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.

.·. 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.

See Figure shown below.

(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)

 .·.   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

 .·.  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

 .·.   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 shown below) and when the reference frame is fixed on water. (Figure shown next after this given below)

                              

 

Illustration:

        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  RG = velocity of rain with respect to ground

              RC  = velocity of rain with respect to car

              C  = Velocity of car with respect to ground

                                                        

            From the equation for relative velocity in two frames, we get

                                             RC  =   RG -  C

        From the vector diagram we see that car should be travelling along the inclined direction of rain with velocity

                                                     C  =5 sin 20o m/s

 

  

<< Back |

 

**************** Motion of Projectile ******************

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).

Note: 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 α 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 α

        Acceleration along x-axis ax = 0

        Velocity along y-axis = uy = u sin α

        Acceleration along y-axis ay = -g

        Here we use different equation of motions of one dimension derived earlier to get the different parameters.

                         

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 α) t - (½)gt2

                        or t = 2u sin α/g                  (as t cannot equal to 0)

Horizontal Range:

Horizontal Range (OA)   = Horizontal velocity × Time of flight

                               = u cos α × 2 u sin α/g

                               = u2 sin 2 α/g

 

Maximum Height:

At the highest point of the trajectory, vertical component of velocity is zero.

Therefore,          0 = (u sin α)2 - 2g Hmax

                        Or, Hmax = u2 sin α/2g

 

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 α.t and y = u sin α.t - 1/2 gt2

        Eliminating 't' form the above equations, we get,

        y = x tan α - gx2/2u2cos2α

        This is the equation of trajectory which is a parabola (y = ax + bx2).

 

Illustration:

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 sec

                         Range of projectile = 160 × 15 = 2400 m

           Distance moved by the gun and projectile = 2400 - 450 = 1950 m.

 

Horizontal projection

        Consider a particle projected horizontally with a velocity   from a point O as shown in figure given below.

                            

        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.

.·.     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.

 

Illustration:

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 =                                 

                                        .·. t = 11.25 sec.                         Ans.

        (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 30 × 11.02 = 190 m                           Ans.

 

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

Think  :  The greatest distance of the projectile from the inclined plane is u2sin2

(α-β)/2gcosβ .

 

Illustration:

       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

        Or,    t = (2×39.2 sin 30)/(g cos 45) = 4√2  sec

        (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  Ans.

        (b)    Range downward

                 = 39.2 cos 30 × 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= ( )/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 ( )/g [(1+sinβ)/(1-sin2β)]= ( )/(g(1-sinβ)) down the plane.

**************** 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.

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.

ω = limΔt→0 Δθ/Δt = 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                           ............ (i)

                        Φ = Φ0 + ω0t + αt2/2              ............ (ii)

                        ω2 = ω02 + 2α (Φ - Φ0)            ............ (iii)

                        Φ = Φ0 + (ω0 + ω)/(2t)            ............ (iv)

         Here, Φ0 is the initial angle and ω0 is the initial angular speed.

 

Illustration:

(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.

 

Illustration:

        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

        => 

        => ω - 5 = 0.5 × Π

        => ω = 5 + 0.5 × Π = 6.57 rad/sec.                                Ans.

        Hence, when acceleration is not constant, use the method of calculus as shown in above illustration.

 

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 Δ  in terms of known quantities. For this, consider

the particle velocity vector at two points A and B. Displacing  B, parallel to itself and placing

it back to back with  A, as shown in figure given below. We have

Δ  =  B  -  A

        Consider ΔAOB, angle between OA and OB is same as angle between  A and

B because  A is perpendicular to vector OA and  B is also perpendicular to vector OB.

         OB = OA = r and | A| = | B| = v

.·.      ΔAOB is similar to the triangle formed by  A,  B  and Δ

.·.      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 

| | = limΔt → 0 Δ /Δt = v2/r

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.

Think:  The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

Illustration:

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.                                          Ans.

       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.

 

Non uniform circular motion 

Let us use the vector method to discuss non-uniform circular motion. 

Refer to figure 2.30,  r and  θ 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.

                                 

                                                  = eθv 

In this case, not only  θ but v also varies with time. We can obtain instantaneous acceleration as, 

   =(d )/dt =  θ dθ/dt + v(d θ)/dt 

Recall :    (d θ)/dt =  rv2/r

                    =  θ aT - e ̂r v2/r = aT  θ - aR  r                       Where aT= dv/dt and aR = v2/r

The first term,  θ aT, 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 

| | = √(a2T+a2

N ) 

Illustration: 

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. 

*************** Relative Motion ******************

Relative Velocity

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

                                      

    

Position vector of the origin of S' with respect to S is 

From vector triangle OO'P, we get

   

Physical Significance of Relative Velocity

Let two cars move unidirectionally. 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  BA 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 ( AB) has the magnitude 6 m/sec & directed from A to B as shown in the figure.

        

 

                                       

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  rm  =    rm -  m     velocity of rain w.r.t.man

       =>    r  =  rm +  m

That means the vector addition of the velocity of rain with respect of man ( rm) and the

velocity of man (vehicle) ( m) yield the actual velocity of rain  r. The magnitude and

direction of  r can be given as

        vr = √((vrm)2+(vm)2+ 2vrm vm cosθ)

       Ø = tan-1((vrm sinθ)/(vrm cosθ+ vm )) with horizontal  m

 

 

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.

                 R= 0-30j ̂,  c= 10i ̂

                         R,c =  R-  c

                        = -30j - 10i = -10i - 30j

 

        If angle between horizontal and  R,c 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

        And velocity of the man is

               vm=5i 

           .·. vwm = vw- vm=(v1-5)i + v2j

        In first case,

                v1- 5 = 0   =>   v1=  5 m/s.

        In the second case,  tan 45o  =  v2/(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        m  =  velocity of man

               w   =  velocity of flow of river,

               mw= velocity of swimmer w.r.t. river

               m can be found by the velocity addition of  mw and  w.

      

Crossing of the River with Minimum Drift

Case 1:    mw   >      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    mw w.r.t. water, such that his actual

velocity  m will direct along AB, that is perpendicular to the bank (or velocity of water  w).

                          

 

=> For minimum drift,  m ⊥  w 

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 ( m) to be perpendicular to the bank  m is

perpendicular to  w. 

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 =   

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

          => t = tAC + tCB

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  A and  B  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  B -  A.

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 .

Then, angular velocity B w.r.t. A is (VB sinβ - VA sinα)/l

Relative angular velocity = (VB sinβ - VA sinα)/l.      

 

Solved Examples 

Example 1:

A car A is travelling on a straight level road with a uniform speed of 60 km h-1. It is followed by another car B which is moving with a speed of 70 km h-1. When the distance between them is 2.5 km, the car b is given a deceleration of 20 km h-2. After what distance and time will B catch up with A?

Solution:

Suppose the car B catches up with the car A in t hours. The distance travelled by the car A moving with a velocity of 60 km h-1 in time hours is given by s1 = 60 t km.

The distance traveled by the car B moving with an initial velocity u = 70 km h-1 and decelerated at the rate of 20 km h-2 in time t hours is given by

s2 = ut + ½ at2 = 70t + ½ (-20)t2 = 70t - 10t2

But (s2-s1) = 2.5 km

.·. 70 t -10 t2 - 60 t = 2.5 => 10t2 - 10t + 2.5 = 0

.·. t2 - t + 0.25 = 0 => (t - 0.5)2 = 0

=> t = ½ h

Substituting t = ½ h, we get s2 = 32.5 km. Ans.

 

Example 2:

A train moves from one station to another in two hours time. Its speed during the motion is shown in the graph (Figure) Calculate:

(i)     maximum acceleration during the journey.

(ii)    distance covered during the time interval from 0.75 hour to 1 hour.

 

Solution:

(i)     Since acceleration is the rate of change of velocity, it is given by the slope of the velocity time curve. Here the slope is greatest between ¾ and 1 hour.

Change in the velocity in this interval = (60-20) km h-1

.·.     Acceleration in this interval

        = (60-20)/(1-(3/4)) km h-2 = 160 km h-2                 Ans.

(ii)    Distance travelled during this interval is given by

        s = ut + 1/2 at2 = 20 (1/4) + 1/2(160)(1/4)2 = 10 km

Note:   We can find distance covered during the time interval from 0.75 hour to one hour by assuming the concept of:"Area enclosed between the velocity time graph represents the distance travelled covered".

 

Example 3:

The speed of a train increases at a constant rate α from zero to v and then remains constant for an interval and finallydecreases to zero at a constant rate β. If l is the total distance covered, prove that total time taken is l/v + v/2 [1/α+1/β].

Solution:

 

As shown in figure, we can divide the distance covered into three stages:

                Ist stage - OA:            v = αt1,              s1 = v2/2α

                IInd stage - AB            s2 = l - (s1+s3)    = vt2

                IIIrd stage - BC           s3 = v2/2β

        Hence, total time t = t1 + t2 + t3

                                = v/α + l(-s1+s3 )/v + v/β

                                = v/α + l/v - v/2α - v/2β + v/β

                                = v/2α + v/2β + l/v

                                = l/v + v/2 [1/α+1/β]                        Hence Proved.

 

Example 4:

A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds Calculate (i) the maximum velocity reached (ii) the total distance traveled.

Solution:

The velocity time graph for the motion in question is as shown in figure. Let the car acceleration is α for time interval t1. Let v be the maximum velocity reached.

 

        Then acceleration

        α = v/t1                                    .·. t1 = v/α ............ (1)

        Retardation β = v/(t-t1)              .·. t - t1 = v/β ............ (2)

        Adding (1) and (2) we get,

        or t = v = (1/α + 1/β)           

         Hence v = t(αβ)/(α+β)                                    Ans.

To determine the total distance we can use equations of motion or area under the triangle OAB. Here it is easier to find area under the triangle OAB. Hence

Total distance traveled = area of the triangle OAB

= (1/2) × base × height = (1/2) × t × v

= 1/2 t × ((αβ)/(α+β)) t = 1/2 ((αβ)/(α+β)) t2         Ans.

 

Example 5:

A body falling freely from a given height "H" hits an inclined plane in its path at a height 'h'. As result of this impact the direction of the velocity of the body becomes horizontal. For what value of (h/H) the body will take maximum time to reach the ground.

Solution:

        In this example we use the equation of motion of the following form.

                        h = ut + 1/2 gt2

        Total time to reach the ground, t = t1 + t2

        where        t1 = time taken from O to A and

                        t2 = time taken from A to B

 

Hence

t1 = √2(H-h)/g , t2 = √2h/g

.·. t = √2(H-h)/g + √2h/g

For maximum value of t, the value of |√(H-h) + √h| is maximum. To find the maximum value, we have to use the concept of differential calculus. Hence,

                        d/dh[√(H-h) + √h] = 0

                        or (H - h)½ (-1) + h-½ = 0

                        or H - h = h

                        or h = H/2  h/H = 1/2          Ans.

  

Example 6:

A man wishes to cross a river to an exactly opposite point on the other bank. If he can pull his boat with twice the velocity of the current, find, at what inclination to the current he must keep the boat pointed.

Solution:

The man starts from A and wishes to reach the point B on the opposite bank. He has to direct his boat towards the point C, to reach the point B. If AD = 2v represents the and DE = v, the velocity of the river.

 

        Hence,

        sin α = v/2v = 1/2

        .·.    α = 30o

        Hence the inclination with the current = 30o + 90o = 120o          Ans.

 

Example 7:

Two ships are 10 km apart on a line running south to north. The one farther north is streaming west at 20 kmh-1. The other is streaming north at 20 kmh-1. What is the distance of closest approach.

Solution:

Suppose the two ships X, Y moving with velocities u, v respectively each 20 km h-1. The velocity of Y relative to X = v - u = v + (-u)

 

We therefore draw OA to represent v and add to it AB which represents -u. The relative velocity is then represented by OB.

OB = √(OA2 + AB2) = √(202 + 202)= 28.28 km h-1.

Also tanθ = AB/OA = 20/20 = 1 => θ = 45o

Thus the ship Y will be move along a direction QR relative to the ship X where QR is at 45o to PQ, the north south direction. When the relative velocity is considered, the ship X is at rest.

 

If PQ = 10 km the distance of closest approach is PN where PN is the perpendicular from P to QR.

PN = PQ sin 45o = 10 sin 45o = 7.71 km

The distance QN = 10 cos 45o = 8.071 km

Time to reach N = 7.071/28.28 = 0.25h

 

Example 8:

The speed of a motor boat with respect to still water is v = 7m/s and the speed of current in a stream is u = 3 m/s; when the boat began travelling upstream a float was dropped from it. The boat traveled 4.2 km upstream and turned about and caught up with float. How long is it before boat reaches the float?

Solution:

        Velocity of the boat while travelling upstream

        = boat velocity - stream velocity = v - u

        Time taken by the boat for travelling the distance l = 4.2 km,

        t1 = l/(v-u)

Let 't' be the time taken by the launch after dropping the float and meeting it again.

Distance traveled by the float during time t = ut

Down-stream velocity of the boat = v + u

Distance traveled down the stream by the boat = l + ut

Time taken t2 = (l+ut)/(v+u);

Total time taken by the boat t = t1 + t2

t = l/(v-u) + (l+ut)/(v+u) = l/(v-u) + l/(v+u) + ut/(v+u)

t(1-u/(v+u)) = [(v+u)+(v-u) ]l/(v-u)(v+u)

Hence, t (1-u/(v+u)) = 2vl/(v-u)(v+u)

t = 2l/(v-u) = (2×4.2×103)/((7-3)) h = (2×4.2×103)/(4×60) min = 35 min

 

Example 9:

A particle is projected vertically upwards, and 't' seconds afterwards particle is projected upwards with same initial velocity. Prove that the particles will meet after a lapse of [t/2+u/g] seconds from the instant of projection of the first particle. What are the velocities of the particles when they meet?

Solution:

Let the particles meet at a height h from the ground t0 seconds after the projection of the first particle.

For the first particle,

S = h, initial velocity is u m/s and time is t0.

Using the formula, S = ut + (1/2) at2

                          h = ut0 - (1/2) gt02                    ...... (i)

For the second particle,

S = h, initial velocity is u m/s and time is t0.

Using the same formula, we get

h = u (t0 - t) - (1/2)g(t0-t)2                                  ...... (ii)

Equation (i) and (ii)

ut0 - 1/2gt02  = u(t0 - t) - 1/2 g(t0 - t)2 = ut0 - 1/2 gp[t0

2 - 2t0t + t2]

=>     -1/2 gt2 + gt0t - ut = 0

=>     -1/2 gt + gt0 - u = 0 => t0 = (t/2+u/g) seconds       Hence proved

Velocity of the first particle = u - gt0

= u - g (t/2+u/g) = -1/2 gt                                           Ans.

 

Example 10 :

A cricket ball is thrown with a speed of 49 m/sec. Find the greatest range on the horizontal plane and the two directions in which the ball may be thrown so as to give a range 122.5 m.

Solution :

        Here initial speed u = 49 m/sec

        Horizontal range = u2sin2θ/g

        The range will be greatest when

        sin 2θ = 1 => θ = ∏/4

 

        .·. Greatest horizontal range = (49 * 49)/9.8 = 245 m                  Ans.

        Now it is given that range = 122.5 m

        .·. u2sin2θ/g = 122.5

        => (49*49)sin2θ/9.8 = 122.5

        => sin2θ = ½ => 2θ = 30o or 150o

        => θ = 15o or 75o

.·.  The required two directions which will give the range of 122.5 are 15o and 75o  Ans.

Think :  The range remains the same when the particle is projected at an angle of θ or (∏/2 - θ) with the horizontal.

Example 11:

A particle is projected form O at an elevation α and after t seconds to have an elevation β as seen from the point of projection. Prove that the initial velocity was gt cos β/[2sin(α-β)/].

 

 

Solution:

Let the velocity of projection of the particle be u. Let the particle be at P after time t such that OP = l then in time t the particle moves

Horizontal distance ON = l cos β

Vertical distance PN = l sin β

Hence considering motion of the particle in horizontal and vertical direction.

We have

l cos β = u cos αt                                                ...... (i)

l sin β= u sin α t - ½ gt2                                       ...... (ii)

Dividing (1) by (ii), we get

cosβ/sinβ = 2u cosα/(2u sinα-gt)

After solving this equation,

=> u = gt cosβ/2sin(α-β)            Hence proved.

 

Example 12:

A projectile aimed at a target, which is in a horizontal plane through the point of projection, falls a meter short of it when the angle of projection is α. It goes b meter too far when the angle of projection is β. Find the angle of projection, in terms of α, β, b and a to hit the target if the velocity of projection be the same in all cases.

 

Solution:

Let O be the point of projection of the projectile and the target is T. Here

                AT = a

                TB = b

                OT = c

Let θ be the proper angle of projection and u the velocity of projection which is the same in all cases.

When angle of projection is α, the range is

OA = c - a = u2sin2α/g                        ....... (i)

When angle of projection is β, the range is

OB = c + b = u2sin2β/g                       ....... (ii)

When angle of projection is θ, the range is

OT = c = u2sin2θ/g                            ....... (iii)

Multiplying (i) by b and (ii) by a, and adding, we have

c(b+a) = u2(b sin 2α + a sin 2β)/g        ...... (iv)

Substitute value of c from (iii) in (iv),

We get 'θ' after solving

=> θ = 1/2sin-1[(b sin 2α + a sin 2β )/(b+a)]                              Ans.

 

Example 13:

A shell bursts on contact with the ground and pieces from it fly in all directions with velocities up to 49 metres per second. Show that a man 122.5 metres away is in danger for 5√2 second.

Solution:

        Given that R = 122.5 metres and u - 49.5 m/sec.

        But R = (u2 sin 2α)/g

        .·.     122.5 = [(49)2 sin 2α]/9.8 or sin 2α = ½

As there are always two directions of projections α and (½∏-α) for a given range, hence for the range of 122.5 metres there will be two directions of projection i.e. 15o and 90o - 15o and 75o.

Let t1 and t2 be the times of flight in the two cases, then

t1 = 2u sin15°/g and t2 = 2u sin75°/g

The man is in danger for a time = t2 - t1

= 2u/g (sin 75o - sin 15o) = 2u/g (2 cos 45o sin 30o).

= 2*29/9.8*2*1/√2*1/2 = 5√2 seconds.       Hence proved

 

Example 14:

A body is projected at an angle α to the horizontal, so that it is just able to clear two walls of equal height 'a' at a distance '2a' from each other. Show that the range is equal to 2α cot α/2.

Solution:

Let us be the velocity of projection and R the required range of the particles

Then R = (2u2 sin α cos α)/g                                         ...... (i)

Also referred to horizontal and upward drawn vertical lines through the point of projection (and lying in the plane of flight) as coordinate axes, the equation of the path is

y = x tan α - gx2/2u2cos2α

As the top of the walls of height a lie on it.

.·. The distance of the walls from the point of projection of the particle be a = x tan α - gx2/2u2cos2α

or gx2 - 2u2x sin α cos α + 2au2 cos2α = 0                       ...... (ii)

Let the distances of the walls from the point of projection of the particle be x1 and x2. Then x1 and x2 are the roots of the equation (ii).

.·. x1 + x2 = (2u2 sin α cos α)/g = R,                               ...... (iii)

and x1 x2 = (2au2 cos2α)/g

Now distance between the walls = 2a = x2 - x1

Squiring 4a2 = (x2 - x1)2 = (x2 + x1)2 - 4x1x2

or 4a2 = R2 - 8αu2cos2α/g ,       from (iii) and (iv)

or 4a2 = R2 - 4a cos2 α [ R/sin α cos α], from (i)

=> R2 - 4aR cot α - 4a2 = 0

=> R = ½ [4a cot α + √{16a2 cot2a + 16a2}

(negative sign is inadmissible as R is positive)

or T = 2a cot α 2a cosec α = 2a [(cos α + 1)/sin α]

.·. R = 2a cot α/2.                                                  Hence proved

 

Example 15:

Shots are fired simultaneously from the top and bottom of a vertical cliff with elevation α and β respectively strike on object simultaneously. Show that if a be the horizontal distance of the object from the cliff, the height of the cliff is a (tan β - tan α).

 

Solution:

Let OO' is the cliff of height h. P is the object. Let u1 and u2 be the velocities of the shots fired from the top and bottom of the cliff respectively.

.·. Time taken by each shot in reaching P is the same.

Let this time be t, then as the horizontal component of velocity remains constant throughout the motion and the horizontal distance traveled si due to this component of velocity, so we have

a = u1 cos α.t = u2 cos β t                                              ...... (i)

Also referred to horizontal and vertical lines through the points of projection as coordinate axes, the equations of the paths traced out by the shots projected form O and O' are

                y = x tan α - gx2/2u12cos2α                             ...... (ii)

and           y = x tan β - gx2/2u12cos2β                             ...... (iii)

Let the height of P above O be y1, then as shown in the diagram the depth of P below O' is (h - y1),

.·. The coordinates of P referred to axes through O are (a, y1) and through O' are (a, - h + y1).

Hence from (ii) and (iii) we have

-(h - y1) = a tan α - gx2/2u12cos2α  and y1 = a tan β - gx2/2u1

2cos2β

Subtracting,

h = a(tan β - tan α) - ga2/2(1/2u12cos2β - 1/2u1

2cos2α)

= a (tan β - tan α), from (i),                                 Hence proved.

 

Example 16:

If at any instant the velocity of a particle be u, and its direction of motion θ to the horizontal, then show that it will be moving at right angles to the direction after time (u/g) cosec θ.

 

Solution:

Let at P the velocity of the particle be u, making an angle q with the horizontal. Let θ be the velocity of the particle at Q, when it is moving at right angles to its direction at P.

.·. At Q its direction of motion is inclined to the horizontal at an angle (90o - θ) as shown in the diagram.

.·. Horizontal component of velocity remains constant throughout the motion.

.·. u cos θ = v cos (90o - θ) or u cos θ = v sin θ                       ...... (i)

Also for the vertical component of velocity from "v = u + at",

 

We have v sin (90o - θ) = u sin θ - gt,

Where t is the time taken in moving from P to Q.

Or gt = u sin θ + v cos θ = u sin θ + [u cos θ/sin θ] cos θ, from (i)

= (u/sin θ) [sin2θ + cos2θ]

Or t = (u/g) cosec θ.                                    Hence proved.

 

Example 17:

A stone is thrown at an angle a with the horizontal from a point in a plane whose inclination to the horizontal is β, the trajectory lying in the vertical plane containing the line of greatest slope. Show that if γ be the elevation of that point of the path that is farthest from inclined plane, then 2 tan γ = tan α + tan β.

 

Solution:

Let B be that point of the path which is most distant from the inclined plane, then the tangent at B to the trajectory must be parallel to the inclined plane.

Let the coordinates of B be (h, k) referred to the horizontal and vertical lines through the point of projection O and lying in the plane of flight as coordinate axes.

Trajectory is y = x tan α gx2/2u2cos2α                           ....... (i)

·.· B (h, k) lies in it, so k = h tan α gh2/2u2cos2α            ....... (ii)

Also from (i) we get dy/dx = tan α - gx2/u2cos2α             ...... (iii)

Which gives the inclination of the tangent to horizontal a any point of (i). At B, tangent to (i) is inclined at an angle b to the horizontal.

.·. from (iii) at B,

tan β = tan α - -gh/2u2cos2α or - gh/2u2cos2α = tan α - tan β

Also γ is the elevation of B (h, k), so tan γ = k/h

or k = h tan γ

.·. From (ii), h tan γ = h tan α - [gh2/2u2cos2α)]

or tan γ = tan α - [tan α - tan β], from (iv)

2 tan γ = tan α + tan β.                                                Hence Proved

 

Example 18:

Two guns, situated on the top of a hill of height 10 m one shot each with the same speed 5√3 ms-1 at some interval of time. One gun fires horizontally and other fires upwards at an angle of 60o with the horizontal. The shots collide in air at a point P. Find (i) the time-interval between the firings, and (ii) the coordinates of the point P. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane.

Solution:

Let the F1 be fired upward at an angle 60o with the horizontal and the gun G2 be fired horizontally at the time interval of δ. If t1 and t2 are the respective times taken by the shots 1 and 2 to reach the common point P, we will have

                        t1 = t2 + δ                                           ...... (i)

The coordinates (x, y) of a projectile as a function of time are

                        x - x0 = (v0 cos θ) t

                y - y0 = (v0 sin θ)t - 1/2 gt2

 

For the gun G1, θ = 60o. The coordinates of the point P at which shot 1 reaches in time t1 are

x - x0 = (v0 cos 60o) t1 => x - x0 = 1/2 v0 t1

y - y0 = (v0 sin 60o) t1 - 1/2 gt12 => y = y0 + √3/2 v0 t1 - 1/2 gt1

2

For the gun G2, θ = 0o. The coordinates of the point P at which shot 2 reaches in time t2 are

x - x0 = (v0 cos 0o) t2 => x = x0 + v0t2

y - y0 = (v0 sin 0o) t2 - 1/2 gt22 => y = y0 - 1/2 gt2

2

Equating x-coordinates, we get

1/2 v0t1 = v0t2 => 1/2 t1 = t2          or     t1 = 2t2              ...... (ii)

Equating y-coordinates, we get

√3/2 v0t1 - 1/2 gt22 = - 1/2 gt2

2

or, √3/2 v0t1 + 1/2 g(t22 - t1

2) = 0

Making use of equation (ii), we get

√3/2 v0(2t0) + 1/2 g (t22 - 4t2

2) = 0

or, t2 (√3v0 - 3/2gt2) = 0

This gives t2 = 0 and t2 = (2/√3)(v0/g)

 

Substituting the values of v0 and g in the expression of t2, we get

t2 = 2/√3 (5√3ms-1)/(10ms-2) = 1s

Thus, t1 = 2t2 = 2 (1 s) = 2s          (Equation ii)

δ = t1 - t2 = 2s - 1 = 1 s              (Equation i)                Ans.

The coordinates of P at which the two shots collides are

x = x0 + v0t2 = 0 + (5√3ms-1) (1s) = 5√3 m                      Ans.

y = y0 - 1/2 gt22 = 10 m - 1/2 (10 ms-2)(1s)2 = 5 m        Ans.

We can do the above problem using another:

Taking the point of firing as the origin, the motion of projectile at an angle θ with the horizontal x-axis is

y = x tan θ 2(gx2/2  cos2α)

For the gun G2,

θ = 0o, tan θ = 0 and cos θ = 1

Thus, y = -gx2/2                                                           ...... (i)

 

For the gun G1,

θ = 60o, tan 60o = √3 , and cos 60o = 1/2

y = x√3 - gx2/2 (1/4) = x√3 - 2gx2/                                     ...... (ii)

Since the two shots meet at P (x, y), we equate the y-coordinates as given equations (i) and (ii).

-gx2/2     = x√3 - 2gx2/

Or, x√3 - 2gx2/  - gx2/2  =  3gx2/2

This gives x = 0 and x =  2 /√3g

Substituting the values of v0 and g, we get

x = (2(5√3 ms-1)2)/(√3 (10ms-1)) = 5√3 m                              Ans.

y=gx2/( ) = -(10ms-2)(5√3 m)2/2(5√3 ms-1)2 = -5m                Ans.

The coordinates of point P with respect to the bottom of the hill are (5√3 m, 5 m).

Time taken for the shot from G2 to travel a distance 5√3 m with velocity v0

t1 = x/v0 = (5√3 m)/(5√3 ms-1 ) = 1s

Time taken for the shot from G1 to travel a distance 5√3 m with velocity v0 cos 60ois t2 = x/(v0 cos 60o ) = (5√3 m)/((5√3 ms-1 )(1/2)) = 2s

Time interval between the two shots is

δ = t2 - t1 = 2 s - 1 s = 1 s                                                    Ans.

 

Example 19:

A large, heavy box is sliding without friction down to smooth plane of inclination θ. From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of the particle with respect to the box is u, and the direction of projection makes an angle a with the bottom as shown in figure.

 

(a)    Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance).

(b)    If the horizontal displacement of the particle as seen by an observer on the ground is zero, find the speed of the box with respect to the ground at the instant when the particle was projected.

Solution:

(a)    To determine the distance PQ, we consider the motion of the projectile with reference to the frame of the box, as shown in figure.

The initial velocity u of the projectile has two components

along x axis ux = u cos α                                               ... (i)

along y axis uy = u sin α                                                ... (ii)

The accelerations experienced by the projectile are

along x axis gx = g sin θ                                               ... (iii)

along y axis gy = g cos θ                                              ... (iv)

 

The displacement of the particle along the x and y axes are

x = uxt                                                                      ... (v)

y = uyt - 1/2 gyt2                                                       ... (vi)

To determine the range of the projectile along the x-axis, we set y = 0 in equation (vi) so as to determine the time at which the particle returns on the x-axis. This gives

t = 2uy/cosθ = 2u sinα/gcosθ

Substituting this in equation (v), we get

x = ux t = (u cos α) (2u sinα/gcosθ) (u sinα/gcosθ)                                  Ans.

(b)    Since the horizontal displacement of the particle as seen by an observer on the ground is zero, the box covers a distance exactly equal to the projectile range x(= u sin2α/gcosθ) in time t(= (2u sinα/gcosθ). The acceleration experienced by the box along the inclined plane

is g sin θ. If V is the velocity of the box at the instant the projectile was projected, then by using the expression s = ut + (1/2) at2, we get

 (u sin2α/gcosθ) = V (2u sinα/gcosθ) + 1/2(g sinθ)(2u sinα/gcosθ)2

After solving the above equations, we get

u cos α = V + (u sinθ sinα)/cosθ

Or, V = u (cosα cosθ - sinθ sinα)/cosθ = u(cos(α+θ)/cosθ)                         Ans.

 

Example 20:

A point moves along a circle of radius r with deceleration; at any moment the magnitudes of its tangential and normal accelerations are equal. The point was set in motion with the velocity v0. Find the velocity v and the magnitude of the total acceleration a of the point as a function of the distances covered by it.

Solution:

As we know, dv/dt = -v2/r.

We can write

dv/v = -ds/r.

The integration of this expression with regard to the initial velocity yields the following result :

V = v0e-s/r.

In this case |aT| = aN, and therefore the total acceleration

a = √2 aN = √2 v2/r, or

a = √2 v02/re2s/r.

 

Example 21:

A point moves along a plane path so that its tangential acceleration aT = a and the normal acceleration aN = bt4, where a and b are positive constants and t is time. The point started moving at the moment t = 0. Find the curvature radius r of its path and its total acceleration a as a function of the distance s covered by the point.

Solution:

The elementary velocity increment of the point dv = aT dt. Integrating this equation, we get v = at. The distance covered s = at2/2.

The curvature radius for the path can be represented as r = v2/aN = a2/bt2.

Or r = a3/2bs.

The total acceleration

a = √(a2T + a2

N) = √(1+4bs3/a2)2

 

Example 22:

A particle moves uniformly with the velocity v along a parabolic path y = ax2, where a is a positive constant. Find the acceleration w of the particle at the point x = 0.

Solution:

        Let us differentiate twice the path equation with respect to time:

        dy/dt = 2ax(dx/dt);d2y/dt2 = 2a [(dx/dt)2 + x d2x/dt2]

Since the particle moves uniformly, its acceleration at all points of the path is purely normal and at the point x = 0 is coincides with the derivative d2y/dt2 at the point. Keeping in mind that at the point x = 0 |dx/dt| = v we get

        a = 2av2.

Note that in this solution method we have avoided calculating the curvature radius of this path at the point x = 0, which is usually needed to determine the normal acceleration (aN = v2/r).

 Example 23:

A train travels due south at 30 m/s (relative to ground) in a train that is blown towards the south by the wind. The path of each raindrop makes an angle of 22owith the vertical, as measured by an observer stationary on Earth. An observer on a train, however, sees the drops fall perfectly vertical. Determine the speed of rain drops relative to Earth.

Solution:

Here we notice that a change in reference frame changes the observation and mathematics involved in it i.e. the statements made about the rain by two different observes are different. Let us draw vector diagram with the given information.

 

VTrain = 30 m/sec due south (see figure)

VDrop = VD at an /22o with vertical (observed from the earth)

VDrop = V'D vertical as seen from train

.·.  'D =  D -  T.

.·. from the Δ, VD = √(V'D2 - VT2)

VD = 80.08 m/sec

  

 

 > Physics > Mechanics > Newton’s Laws of motion  

1. Some Definitions

2. Newton's First Law of Motion

3. Newton’s Second Law of Motion

4. Newton’s Third Law of Motion

5. Inertial and Non-Inertial Reference Frames

6. Friction

7. Free Body Diagram

8. Banking of Roads

9. Solved Examples

Types of Forces

| 1 | | 2 |Force

 "Force" is an external or internal agent present to "influence" the natural state of motion of an object. So this is an influence (force) needed to change the natural state of body; that is of rest or of uniform motion.

Classification of Forces

There are different types of forces in our universe. Based on the nature of the interaction between two bodies, forces may be broadly classified as follows.

(a)  Contact Forces

        Forces that act between the bodies in contact with each other. For example, Normal Reaction, Friction etc.

(b)  Field Forces (Non-Contact Forces)

        Forces that act between bodies separated by a distance without any actual contact. For example, Tension, Spring, Weight etc.

 

Tension (T)

When a string, thread or wire is held taut, the ends of the string or thread (or wire) pull on whatever bodies are attached to them in the direction of the string. This force is known as Tension.

If the string is massless, then the tension T has the same magnitude at all points throughout the string.

The direction of tension is always from the point of attachment to the body. In the given figure, two segments of tension act at O towards A and B.  

                                 

 

Spring

Force in an extended (or compressed) spring is proportional to the magnitude of extension (or compression).

i.e. F α x, in magnitude, but opposite in direction.

So, F = -kx, where k is a positive constant, also known as the spring constant of the spring; and x is the compression or elongation from the natural length.

  

 

Normal reaction

When a body is in contact with another body, there exists a force that prevents them from penetrating each other. Each one "stops" the other by applying a force away from itself.

The forces  ,   shown in the diagram acting on bodies A and B, respectively, act away from the surface of contact, and prevent the two bodies from "occupying the same space".

If  , is the action   is reaction: they are equal in magnitude but opposite in

direction. Further,  , and,   are both perpendicular to the surfaces in contact and note that they act on two different bodies.

 

| 1 | | 2 |

Examples:

 

 

Friction

It is a force that acts between bodies in contact with each other along the surface of contact and it opposes relative motion (or tendency of relative motion) between the two bodies. The direction of friction force on A is opposite to that of force on B and magnitude is same for both.

 

  

Weight (W)

It is a field force, the force with which a body is pulled towards the centre of the earth due to its gravity. It has the magnitude mg, where m is the mass of the body and g is the acceleration due to gravity.

 

Newton’s First Law of Motion 

It states that every object persists in its natural state of motion i.e. continues to be at rest or moves in a straight line with uniform (constant) velocity! (This is what is meant by natural state of motion); in the absence of a net external force acting (impressed) on it

Mathematically, it is equivalent to say that for producing acceleration (that is for changing velocity) in a body, we need to have a net external force. (By net external force we mean

vector sum of all the external forces acting on it).

It can be easily deduced from the statement of change in the state of motion. It is directly related to a frame of reference about which we have discussed earlier. To mark the point here, we can discover that by viewing objects from different frame of references the natural state of motion as perceived by different observers will be obviously different (can only be same if the frames are truly equivalent). Therefore, the change in state will also depend on the choice of reference frame. Finally, the amount of acceleration produced in a body (or change in velocity) will depend on our choice of reference frames.

 

Inertia

The first law of motion 'describes a property of matter: No body (dead or living) can change its velocity by itself, without an external force acting on it. This property is known as inertia. When a car suddenly starts moving, the passengers in the car are thrown back due to inertia of rest. Similarly, when the car stops suddenly, the passengers stoop forward due to inertia (of motion).

 

Momentum

Momentum is the measure of motion contained in a body. It is measured by the product of mass and velocity of the body. Its direction is same as that of the velocity of the body.

          = m  , where   = momentum of the  body, m= mass of the

body,   = velocity of the body

  

Illustration:

A mass 'M' is lying (figure shown below) on a table which is at rest (w.r.t. the table on which it is kept). Explain its state with the help of Newton's First Law of motion.

 

Since 'M' is lying on a table, there is no external force acting on it (forget about gravity just for the immediate discussion). As per Newton's first law of motion it will keep on lying at rest with respect to table for infinite time.

Here, comes out a very important, intrinsic (that is inherent) property of a body which is

that it retains its state of motionlessness (as well as of motion, if it is in motion) which is termed as INERTIA of an object. This is present in all materialistic bodies in this universe.

 ewton’s Second Law of Motion

 

It states that rate of change of momentum of a body is equal to the force applied on it, in terms of the magnitude as well as in the

sense of direction. Here the momentum is defined as the product f mass and velocity i.e. m .

        Therefore we can write mathematically.

        F = d(mv)/dt, if 'm' remains constant then

        => F = m dv/dt

        but d /dt = 

Recall :      Acceleration is rate of change of velocity             Since direction of a is same as F, we can write 

                                   = m           which is mathematically Newton's second law of motion.

        Here, if F = 0 then we find a = 0. This reminds us of first law of motion. That is, if net external force is absent, then there will be no change in state of motion, that means its acceleration is zero.

        Further we can extend second law of motion, (in fact its decomposition) to three mutually perpendicular directions as per our coordinate system.

        If components in x, y and z direction are Fx, Fy & Fz respectively, the three acceleration produced when Fx, Fy & Fz act simultaneously) in the body are, Now,

                    

        If we add three forces then resultant is called net external force.

         Similarly

                                               

        is called net acceleration produced in the body.

 

Impulse:

A large force acting for a short time to produce a finite change in momentum which is called impulse of this force and the force acted is called impulsive force or force of impulse.

Mathematically it is described as the product of force and time.

       .·. Impulse (J) = F Δt (For constant force)

       .·. Impulse (J) = mv - mu and since force is variable, hence J = 

        The area under F - t curve gives the magnitude of impulse.

        Impulse is a vector quantity and its direction is same as the direction of    .

Unit of Impulse : The unit in S.I. system is kgm/sec or newton -second.

Dimension         : MLT1

 

Illustration:

In Figure shown below, let us have M = 10 kg and a new net external force in the direction as shown in figure below is 150N. Find its acceleration.

 

               = 15m/sec2

           (in the direction of force)

             .·.   = 15 m/sec2 (X-direction)

 

 

Newton’s Third Law of Motion

|  1   | |  2  | |  3  |Newton's third law of motion was discovered and formulated, during the investigation of the fact that in all experiments it appeared that "whenever a body exerts a force on a second appeared that "whenever a body exerts a force on a second body, the second body always exerts a force on the first one".

        Let us visualize and understand this phenomena with an experiment:

        Suppose, we throw a stone on a surface of good strength; and the surface is made of glass, one finds it broken (the surface). From here one concludes that a force was exerted by stone on the surface and consequently it was broken.

        Now, the question is, did that surface also exert a force on the stone. Just to know about it let us change our throwing object from stone to an egg of almost equal mass. Now, one throws this egg on the same surface of good strength with the same throwing force which he used for the stone. What happens? Obviously with your daily experience you know that the egg will be broken (And the damage to the surface will not be visible due to egg's spoiling the observation).

        This is only possible if there was a force acting on the egg at the time it hit the surface. In fact we can now conclude that there is mutual force acting on the contact point of the surface and the object thrown. The breaking of either one (or may be both) depends on their ability to absorb forces without getting damaged (that is their strength) so in precise words:-

        To every action there is always opposite and equal reaction, it is equivalent to say that mutual actions of two bodies upon each other are always equal and directed to contrary parts.

 

Note: The most important fact to notice here is that these oppositely directed equal action and reaction can never balance or cancel each other because they always act, on two different point (broadly on two different objects) For balancing any two forces the first requirement is that they should act one one and the same object. (or point, if object can be treated as a point mass, which is a common practice).

  

Illustration of Newton's Third Law:

Some of the examples of Newton's third law of motion are given below:

 1.      Book kept on a table: A book lying on a table exerts a force on the table which is equal to the weight of the book. This is the force of action. The table supports the book, by exerting an equal force on the book. This is the force of reaction, as shown in the figure. As the system is at rest, net force on it is zero. Therefore, forces of action and reaction must be equal and opposite.

 

 

|  1   | |  2  | |  3  || 1 | | 2 | | 3 |

2.     Walking: while walking a person presses the ground in the backward direction (action) by his feet. The ground pushes the person in forward direction with an equal force (reaction). The component of reaction in the horizontal direction makes the person move forward.

3.     Swimming: A swimmer pushes the water backwards (action). The water pushed the swimmer forward (reaction) with the same force. Hence the swimmer swims.

4.     Firing from a gun: When a gun is fired, the bullet moves forward (action). The gun recoils backwards (reaction).

5.     Fight of jet planes and rockets: The burnt fuel which appears in the form of hot and highly compressed gases escapes through the nozzle (action) in the backward direction. The escaping gases push the jet plane or rocket forward (reaction) with the same force, hence, the jet or rocket moves.

6.     Rebounding of a rubber ball: When a rubber ball is struck against a wall or floor it exerts a force on a wall (action). The ball rebounds with an equal force (reaction) exerted by the wall or floor on the ball.

7.     It is difficult to walk on sand or ice: This is because on pushing, sand gets displaced and reaction from sandy ground is very little. In case of ice, force of reaction is again small because friction between feet and ice is very small.

8.     Driving a nail into a wooden block without holding the block is difficult:This is because when the wooden block is not resting against a support, the block and nails both move forward on being hit with a hammer. However, when the block is held firmly against a support, and the nail is hit, an equal reaction of the support drives the nail into the block.

  

Illustration:

         Suppose in figure shown above we put one more block of 5 kg mass adjacent to 10 kg and a force of 150 N acts as shown in the figure below, then find the forces acting on the interface.

 

The combined acceleration of the two bodies when treated as one is                             a = F/((10+5))=150/15=10/sec2

        So each one moves with a = 10m/sec2 keeping their contact established.

        Here you can feel that due to 150N force the body of 5 kg feels as if it is being pushed by the 10 kg mass. There is force acting on 5kg called R1, to oppose it by third law this body exerts a force R2 on 10kg. The interface is as shown in Figure given below.

 

        Also, third law tells us that R1 = R2 in magnitude and is opposite in direction.

 

       .·. R1 = R2 = R

 

Here since 150 N force acts on the 10kg mass and only r acts on the 5kg mass. For motion in 5kg only R is responsible. We can write the initial equation as:

        F = 150 = (10 + 5) a

        150 = 10a + 5a

        Here 10a is force experienced by 10kg mass. And 5a is experienced by 5kg mass.

        .·. R = 5a   a = 10m/sec2

        .·. R = 50N

.·. Net force experienced by 10kg block is (150-R) = 10a 150-R = 1010 = 100 N

.·. R = 50

Therefore we get R = 50N for both blocks. Hence we find "action and reaction are equal and opposite". Now net force on the body of 10kg mass is 100N & Net force on the body of 5kg mass is 50N and on the interface action and reaction are both equal and also are equal to force experienced by second body.

Inertial and non-Inertial Reference Frames

| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 |In general we solve the problem of mechanics using inertial frame, which was discussed in chapter two, but as the same time it is possible to solve the same problem using a non-inertial frame. Let us discuss about the difference between these frames.

        When Newton stated his first law he made a very important distinction. He decreed the absolute equivalence between a state of rest and one f uniform motion and distinguished it specifically and absolutely from that of an accelerated motion. If the environment is completely symmetric then no direction is preferred over another and therefore if a body possesses a initial velocity (which might be zero) it will persist with that velocity. If suppose we say that the velocity will change then we will have to concede that the velocity changes in a particular direction. But why should it change in one direction and not in the other, since all direction are equally favoured. So the only way it can change is to change in all directions. But this is impossible so it will not change at all, i.e. if environment is really symmetric. Therefore if we grant a change in velocity we will also have to grant an irregularity in the environment in the same direction as the change in velocity. The acceleration, he said to be understood as an irregularity and he expressed force as that basic asymmetry in the environment which produces this irregularity. The most important aspect in all this is that force is theoretical construction to explain away the irregularities in motion and is not to be understood as a tangible entity.

                           = d(m )/dt

        Now, for constant mass system    = m  .

       If force is a tangible entity then the force in all systems on the same body should be same. Let us see if this is true.

Consider a body of mass m. We will observe its motion from three different frames.

(i)     Reference frame is at rest

The acceleration of the mass will be, say,  rest.

Therefore the force on it will be  rest

We will reason that

rest = m  rest

(ii)    Reference frame starts moving with constant velocity 

        The acceleration of frame =   = 0

        .·. Acceleration of mass m relative to frame is given by

inertial =  rest -   =  rest

        .·. Force on it will be  inertial  and we will reason that

inertial  = m  inertial  = m  rest =  rest 

(iii)    Now reference frame moves with constant acceleration

        Let the acceleration of frame be  frame

        .·. Acceleration of mass relative to frame will be  rel

rel =  inertial -  frame =  rest  - m frame

Let there be force  frame on mass we will reason, that

                           frame = m rel = m rest -  m frame

                               =  rest  + m(- frame ) =   rest +   pseudo

 

| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 |

We see that the force is not the same as that in the inertial frames.

        Therefore we postulate that under observation from an accelerated reference frame we substitute the inertial forces on the body with those same initial forces plus an additional force which numerically equal to the mass of the body under observation times the acceleration of the frame taken in the opposite direction. This force we call as pseudo force.

        Now, we can work on a problem from an accelerated reference frame by just adding a pseudo force and pretending that nothing has changed.

        Let us illustrate our point.

 

Illustration:

        We have two masses connected by a light spring of stiffness k. The mass m2is pulled to the right by a constant force F. What will be the maximum extension of the spring?

 

Solution:

(i)     Solution in inertial frame

        The spring will keep stretching for as long as v2 > v1

        Force on m2 = F2 = F - kx

        Force or m1 = F1 = kx

        where x is elongation of spring

        Now initially kx will be small therefore the acceleration of m2 will be more than m1 and have velocity of m2 will be more than that of m1 and spring will kept stretching.

        With time 'kx' will increase and so a1 will increase and a2 will decrease. After some time a2 will obviously become zero and then negative all this while a1 will increase. So after a while v1 will be equal to v2. Beyond that time v2 will become less than v1 and so spring will not stretch anywhere. Therefore at maximum elongation v1 = v2.

Force on m2

F - kx = m2 x2

Force on m1

Kx = m1 x1

Also x = x2 - x1           so x1 - x1 = x

 

 

(ii)    In the reference frame of m1, which is non-inertial frame?

        Acceleration of m1 = F1/m1 = (k/m1)x

       Also the elongation will be maximum when v1 = v1 or velocity of m2 is zero relative to m1.

        Now forces one m2 in this frame =  rest  +   pseudo.

        = F - kx - (m2/m1 )kx

       .·. F - k (1+m2/m11 )x = m2 x = m2v2dv2/dx

        Remember that in the reference frame of m1, the spring stretches by exactly the same amount as the displacement of m2.

        So 

        Since, final velocity of m2, relative to m1 is zero.

       .·. Fxmax = 1/2k ((m1+m2)/m1 ) xmax2            xmax = 2Fm1/(k(m1+m2))

        Now we introduce the concept of friction.

 

Is earth an inertial frame of reference?

Earth rotates around its axis as also revolves around the sun. In both these motion, centripetal acceleration is present. Therefore, strictly speaking earth or any frame of reference fixed on earth cannot be taken as an inertial fame. However, as we are dealing with speeds  ≈ x 108 ms-1 (speed of light) and speed of earth is only about 3 x 104 m/s, therefore when small time intervals are involved effect of rotation and revolution of earth can be ignored. Furthermore, this speed of earth can be assumed to be constant. Hence

earth or any other frame of reference set up on earth can be taken as an approximately inertial frame of reference.

On the contrary, a frame of reference which is accelerated or decelerated is a non-inertial frame.

Other examples of inertial frames of reference are:

(i)     A frame of reference remaining fixed w.r.t. stars.

(ii)    A space ship moving in outer space, without spinning and with its engine cut off.

        Inertial frames of reference are important because:

(iii)   All fundamental laws of Physics have been formulated in respect of inertial frames.

(iv)   The fundamental laws of Physics have the same mathematic form in all the inertial frames of reference.

(v)    The optical experiments performed in an inertial frame in any direction will always yield the same results. This is isotropic property of the inertial frames of reference.

  

Apparent weight of a Man in a Lift/Elevator

Suppose a person of mass m is standing on a weighing machine placed in an elevator/lift. The actual weigh of the person equal to mg. This acts on the weighting machine which offers a reaction R given by the reading of the weighing machine. This reaction exerted by the surface of contact on the person is the apparent weight of the person. We shall discuss how is related to mg in the following different situations:

 

Case I:

 When the elevator is at rest                                 

 

        Acceleration of the person = 0      .·.

        Net force on the person F = 0

        i.e., R - mg = 0 or R = mg

        i.e., apparent weights is equal to the actual weight of the person, in the figure

 

Case II:

 When the elevator is accelerating upwards                     

        Suppose uniformly upward acceleration of the person in the lift = a

        or R1 = mg + ma = m(g + a)                        ... (i)

                Thus R1 > mg

        Hence apparent weight of the person becomes more than the actual weight, when the elevator is accelerating upwards.

 

Case III:

 When the elevator is accelerating downwards

       Suppose uniform downward acceleration of the person in the lift = a

                R2 = mg - ma = m(g - a)             ... (ii)

        Thus R2 < mg

        Hence apparent weight of the person becomes less than the actual weight when the elevator is accelerating downwards.

 

Case IV:

 In free fall of a body under gravity, a = g

       .·. R2 = m(g - g) = 0

        i.e., apparent weight of the body becomes zero or the body becomes weightless.

        Note that weightlessness is felt only because the force of reaction between the person and the plane with which he is in contact vanishes.

 

Case V:

 When downward acceleration is greater than g.

       i.e., a > g, then, R2 = m (g - a), R2 becomes negative i.e., apparent weight of the person becomes negative. In that event, the person will rise from the door of the lift and stick to the ceiling of the lift.

 

Constraint Relations

The equations showing the relation of the motions of a system of bodies, in which motion of one body is constrained relations.

Applying Newton's laws alone is not sufficient in some cases where the number of equations is less than the number of unknowns.

 Note:   Relation of velocities or acceleration through constraint relation only give relation between magnitude of velocity or acceleration.

Consider the situation in the adjacent figure.                                  

 

In this case, constraint equation is very simple, a1 = a2

 

Illustration:      Find the acceleration of two blocks. Assume pulleys and strings to be massless and frictionless. Also, assume the string to be inextensible.

Solution:   Y + 2X + Z = l                                           (i)

                Y + X + l2 + X + l2 + Z - l1 = l

        Subtracting eq. (i) from eq. (ii), we get   2l2 = l1

        Differentiating twice this equation w.r.t. time we get

                A1 = 2a2

        i.e. acceleration of block m1 is twice the acceleration of block m2.

 

        For block m1:

                T1 = m1a1

        For block m2:

                M2g - T2 = m2a2

                2T1 = T2                     [·.·    Pulley is massless]

                Let a2 = a                  .·.      T = 2m1a

        .·.      a1 = 2a,                             m2g - 2T = m2a

                 T1 = T                                Solving a = (m2 g)/(m2+4m1)

        .·.      T2 = 2T                      .·.      a1 = 2a = (2 m2 g)/(m2+4m1)

 

Illustration:

 Find the accelerations of the rod A and the wedge B in the arrangement shown in the figure if the ratio of the mass of the wedge to that of the rod equals η and the friction between all surfaces is negligible.

 

Solution:   y = x tan α => (d2y)/dt2 = (d2x)/dt2 tan α

               arod = awedge.tan α (constraint equation)

                                mg - N cos α = maR and

                                N sin α = (ηm).awedge

                        =>    mg - N cos α = m aw tan α

 

                        Solving, we get

                        aWedge = g/(tanα + ηcotα) and aRod =  g/(1+ ηcot2α)

 

Illustration:

Find the acceleration of the two blocks m1 and m2. Assume that the pulleys are massless and frictionless and the strings are inextensible.

 

Solution:

Let the acceleration of blocks m1, m2 and pulley B be a1, a2 and a3, respectively.

        Constraint relationship for the string attached to block of mass m1:

                        X1 + x3 - 2C1 = constant

                Differentiating twice w.r.t. time, we get

                        (d2x1)/dt2 = (d2x3)/dt2 = 0

                => a1 = a3                                   ... (i)

 

The minus sign signifies acceleration of pulley B is opposite to that of block of mass m1.

        Constraint relationship for the string attached to block of mass m2:

                c2 - x3 + x2 - x3 = constant.

                (d2x3)/dt2 + (d2x2)/dt2 = 0

                => a2 = 2a3                        ... (ii)

                     A2 = 2a1                        ... (iii)  [from equations (i) and (iii)]

        Taking the magnitudes only and ignoring the sign,

                T1 = 2T2                             ... (iv)

                m1g - T1 = m1a1                     ... (v)

                T2 - m2g = m2a2                    ....(vi)

        Solving equations from (iii) to (vi) for a1 and a2, we get

 

                a1 = (m1- 2m2)/(m1+4m2 ) ; a2 = 2a1

                =  2(m1- 2m2)/(m1+4m2 )

 

 

Friction

| 1 | | 2 | | 3 | | 4 | | 5 | 6 |Whenever the surface of a body slides over another, each body experiences a contact force which always opposes the relative motion between the surfaces. This contact force is called frictional force.

        Intermolecular interaction arising due to elastic properties of matter is the cause of frictional force.

 

Cause of sliding friction

Old view

Earlier it was thought that roughness of the two surfaces causes friction in the figure because it can be easily seen that smoother the surfaces, lesser is the friction. Interlocking

of irregularities of the two surfaces causes hindrance to sliding. This, however, is not the current view

 

Current view

The current view is a slight deviation form the old view. Earlier we thought that interlocking of irregularities of surfaces was causing friction. Now, it is though that due to irregularities, the common surface area which is in actual contact of the two surfaces, is much less than the total overall area in contact. In one experiment, it came out to be 1/10,000th of the apparent area.

Thus, while the total interactive (action and reaction) forces between the two surfaces remain the same, the pressures at the points of contact are extremely high and cause the humps to flatten out (undergoing plastic deformation) until the increased area of contact enables the upper solid to be supported. It is thought that at the points of contact, small, cold-welded joints are formed by the strong adhesive forces between molecules which are very close together. These have to be broken away before one surface can move over the other.

 

Types of frictions

(a)    Static friction: it is the friction force that acts between surfaces at rest. It is almost independent of the area of contact (Although it is dependent on micro area of contact). It is proportional to normal force. This static friction force can vary from zero to a certain maximum value.

                Maximum static friction force is the smallest applied force necessary to start the motion of the body.

                The ratio of the magnitude of the maximum force of static friction tot eh magnitude of normal force is called the coefficient of static frictional (μs). If the magnitude of the applied force is less than the maximum frictional force, then we can say

        fS = Applied force < μs N (Maximum frictional force)

 

(b)    Kinetic Friction: The kinetic friction is present between surfaces which have relative motion. The ratio of the magnitude of maximum kinetic frictional force to the magnitude of the applied force during motion is called coefficient of kinetic friction.

 

| 1 | | 2 | | 3 | | 4 | | 5 | 6 || 1 | | 2 | | 3 | | 4 | | 5 | 6 |

Cause and direction of rolling friction

 

A wheel of radius R rolling without sliding on a flat surface will experience a resistance due to the very small local deformation that takes place, which is sometimes elastic sometimes inelastic, i.e., a kind of ridge is formed in front of the wheel as shown exaggerated in the figure. This gives rise to the force FR, whose line of action passes through the centre C of the wheel and P the horizontal force necessary to force the wheel to topple over the point M, the total clockwise torque acting on the wheel about M must be more than or marginally more than the total anticlockwise torque about M.

        F = Frictional force

.·.     P × R cos θ > mg × R sin θ

        P > mg tan θ

The value of "tan θ" is called the coefficient of rolling friction (μg). This value does not depend upon R. If the two surfaces are absolutely rigid, then no ridge will be formed and q will be zero i.e., coefficient of rolling friction will be zero.

Typical values are μR = 0.006 for steel and 0.02 - 0.04 for rubber tyres on concrete surfaces. Rolling friction is very small compared to the sliding friction. In case of pure rolling μR = 0.

 

Laws of Friction

1.      Laws of Static Friction

Static friction force is always equal and opposite to the net external force acting on the body.

2.      Laws of limiting friction

(i)     Limiting frictional force is independent of the apparent area of contact till the value of the normal reaction remains same.

(ii)    The direction of limiting frictional force is opposite to the direction in which one body is on the verge of starting its motion.

(iii)    The limiting frictional force depends upon the nature of surfaces in contact.

(iv)   Quantitatively, the magnitude of the force of limiting friction (F) on one of the two bodies in contact, is directly proportional to the normal reaction (R) on this body due to the other.

                F ∞ R

             Or F = μR

        Where m is the constant of proportionally called coefficient of friction.

 

| 1 | | 2 | | 3 | | 4 | | 5 | 6 |Co-efficient of Friction

According to the law of limiting friction,

                F µ R

          Or F = μR           ............ (3)

where μ is a constant of proportionality and is called the coefficient of limiting friction between the two surfaces in contact.

From (3),

           μ = F/R              ............ (4)

Hence coefficient of limiting friction between any two surfaces in contact is defined as the ratio of the force of limiting friction and normal reaction between them. The value of μ depends on

(i)     nature of the surfaces in contact i.e., whether dry or wet; rough or smooth; polished or not polished.

(ii)    material of the surfaces in contact.

        For example, when two polished metal surfaces are in contact, μ ≈ 0.2, when these surfaces are lubricated, μ ≈ 0.1. Between two smooth wooden surfaces, μ varies between 0.2 and 0.5. Obviously, μ has no units.

        When a body is actually moving over the surface of another body, we place F by Fx, the kinetic friction, and μ and μk.

        Therefore,

                μk = Fk/r

μk is then called the coefficient of kinetic or dynamic friction. As Fk < F, therefore, μk is always less than μ i.e. coefficient of kinetic or dynamic friction is always less than the coefficient of limiting friction.

Table gives the values of coefficient of limiting/kinetic friction between some pairs of materials:

S.No. Surface in contact

Coefficient of limiting friction

Coefficient of kinetic friction

1. Wood on wood 0.70 0.40

2. Wood on leather 0.50 0.40

3. Steel on Steel (mild)

0.74 0.57

4. Steel on Steel (hard)

0.78 0.42

5. Steel on Steel (greased)

0.10 0.05

| 1 | | 2 | | 3 | | 4 | | 5 | 6 |

Angle of friction

        It is the angle between normal Reaction and the resultant of the normal reaction and limiting frictional force. i.e. we have a body of mass m which is placed on a table and we say that the body and the surface of the table have a coefficient of friction μ between them.

 

        If we apply a small force F, the body will not move. Let us gradually increase the force until the body starts moving. At one stage the applied force will be equal to the frictional force. The coefficient of static friction μs = F/N where F is the applied force and N = mg is the normal force.

 

Angle of repose

        It is the minimum inclination of an inclined plane at which a body just starts moving under its own weight.

        Let us take a body of mass m placed on a table. The body and the surface of the table have a coefficient of friction μ between them. If we gradually increase the inclination of the table at a certain stage the body will start moving under gravity. The angle made by the table to horizontal is called the angle of repose.

 

Illustration:

A body of mass 100kg is placed on a table. The coefficients of static & kinetic frictions are 0.4 & 0.35 respectively. A force equal to the maximum static frictional force is applied and the body is slightly disturbed.

 

Analyse the motion (a) find acceleration (b) a relation between angle of repose & angle of friction.

        F = μs N and N = mg

so, for the motion of the body

        F - Fk = ma

        or a = (Fs-Fk)/m = (μsmg - μkmg)/m = (μs-μk)g

        = (0.4 - 0.5) × 10 = 0.5m/s2

 

 

| 1 | | 2 | | 3 | | 4 | | 5 | 6 |

So, only a slight disturbance will move the body with an acceleration a = 0.5 m/s. though the force which is reducing the acceleration was enough to balance the body in static friction.

        Let us now take the same body placed on the table again. Now limiting frictional force is applied without disturbing it.

 

                       R cos θ = N                                        R sin θ = μ N

               => m = tan θ                                        θ = tan-1 μ

we can say that the resultant of friction force and the normal force is making an angle θ with the normal force. θ is called limiting angle of friction :

        The same body is placed on an inclined plane and the angle of inclination is increased slowly. In the limiting case when the body starts moving. We draw the force diagram as shown in figure above.

        mg sin α = m mg cos α

        tan α = m

        where α is called angle of repose

        also since m = tan α

        α = θ         i.e. angle of repose = limiting angle of friction

 

 

 

| 1 | | 2 | | 3 | | 4 | | 5 | 6 |

(c)    Rolling Friction:

        (In pure rolling only the static friction acts which is necessary for pure rolling). This will be dealt with in detail in module 2 (with Rigid body dynamics)

        After knowing three laws of motion and friction let us see that how can we use these laws for solving problems. But just to tell you, to actually apply the Laws related to force and acceleration, we always need to know all the forces acting on the system on which we want

to apply Newton's Law's because without knowing all the forces we can't know the net external force and for that we have to convert our system in a free body diagram.

  

 

 

  

| 1 | | 2 | | 3 | | 4 | | 5 | 6 |Study Material   > Physics > Mechanics > Newton’s Laws of motion > Free body diagram

Free body diagram

| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 |It is to be noticed here that during previous examples, we were using a concept called FBD implicitly which can now be brought to you conscious attention.

        It is like this. Whenever one attempts a problem involving forces and acceleration (say of dynamics or statics) one must show all forces and acceleration (possible acceleration may be unknown also) on each part of the system treating that part separately (it is called dividing system into possible subsystems). By all forces we mean external as well as internal forces (internal forces refer to mutual reactions). Now each system is ready to get treatment of laws of motion e.g. acceleration, velocity.

        Coming to actual situation one should first identify all the component involved in the system, say mass m1, mass m2 .........., pulley 1, pulley 2, etc. Now separate them from others by cutting the string contacts (sort of imaginary separation) In effect, make them free (that's where comes the name free body) from other components and at the same time show all the forces acting on it, external as well as internal, arising due to separation from other parts. These are called mutual interaction forces (one of this type we have seen in the illustration of IIIrd law) and show all possible acceleration. The diagram thus obtained is called a free body diagram.

 

Illustration :

        Let us draw FBD for various given systems

        [Here we are assuming that all the surfaces strings and pulleys are ideal that is we are neglecting their masses & any friction present]

 

FBD (1) :  Block of mass M is resting on a frictionless rigid surface

 

        There are only two forces in the system in figure above. mg which is the weight of the block, and is acting on the surface. R2 = mg through its centre (since the body is symmetric). So, here itself from FBD we can see that net external force on the block is zero. That is why, it is stationary on the surface.

| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 |Free body diagram

|  1   | |  2  | |  3  | |  4  | |  5  | 6 | 7 | 8 | 9 | 10 | 11 |FBD (2) :  Draw the free body diagram of the block shown in figure 1.12.

 

        Where R is reaction from the surface (R = Mg)

 

FBD (3) :  Draw free body diagrams of both the blocks (figure 1.13 a), Assuming a reaction of magnitude 'R' is present at the interface.

Now a question comes Can 'R' have a direction opposite to what is shown here? Answer is, of course it can have,

[before we proceed further let me point out that once you make calculations you will find that values of R will automatically come negative, which tells us that earlier direction were

the correct ones].

 

|  1   | |  2  | |  3  | |  4  | |  5  | 6 | 7 | 8 | 9 | 10 | 11 || 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

FBD (4) :  Suppose situations is as shown in figure above that a light inextensible string pulls a block of mass M on a frictionless rigid surface.

 

Here string is acting as a force transmitting element

It will experience a tension T in it. Let us cut it (imagine) at then the situation is as shown in figure shown below.

        (We join it again then net force should become zero at that point. Since t, and T are oppositely directed their sum will come out to be zero).

So tension t is responsible for dragging mass of block.

Therefore free body representation is

 

Caution :   What happens if rope is not massless?

FBD (5) :  Draw free body in case of system shown (figure shown below).

| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 || 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

Suppose T is tension in the string, then by cutting it at 1-1' and 2-2' we can draw FBD's as shown in figure shown below.

 

R1 and R2 are reactions from plane, and from Newton's law they are equal to the weights of respective blocks.

 

FBD (6) :  Draw FBD where one of the block is resting on an inclined plane and rope goes over a frictionless massless pulley (Figure given below).

FBD becomes

Here the noticeable fact is that R2 is perpendicular to the inclined plane and mg is perpendicular to horizontal plane. Since Mg is always directed downwards. R2 is due to the component of Mg, which is acting, perpendicular to inclined plane. (Here, the rule is that reaction force is always normal to the surface, which provides the reaction).

        Here for "ease" we can resolve force Mg in two components. One parallel to inclined plane and other on perpendicular to inclined plane. As shown in figure below.

 

| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

 

        .·. free Body, of the block on the inclined plane can be represented by figure shown below.

 

 

Finally at the end of this FBD, I want to point out that T2 will always be equal to T2for a continuous homogeneous massless inextensible string passing over a massless frictionless pulley.

 

FBD (7) :  Suppose both the block are on inclined planes as shown in figure given below.

 

       For FBD cut the strings at 1-1' and 2-2' and separate the two blocks and pulleys as subsystems.

 

 

 

 

| 1 | | 2 | | 3 | | 4 | | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

FBD (8) :  If pulley is hanging from a rigid support say roof and masses are connected as shown in figure given below.

 

        FBD's become.

 

Once we are comfortable in drawing FBD then we can proceed to write equation of motion, using Newton's II Law. But prior to that, it is necessary to briefly introduce you with the concept of equilibrium of bodies.

Equilibrium:

An object is said to be in equilibrium if the vector sum of all the forces acting on the body (externally applied + forces arising due to mutual interaction;) is zero.

 

That is  1 +  2 +......+  N = 0 

        But since it is impractical to apply it, as it is, to problems, therefore we resolve all forces in the three directions X, Y & Z equilibrium and if some of the forces is zero in each direction then the body is said to be in equilibrium. (to be more specific - translator equilibrium).

        F1x + F2x + ............ + Fnx = 0

        F1y + F2y + ............ + Fny = 0

        F1z + F2z + ............ + Fnz = 0

 

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

Enquiry : How to apply equations of motion to any problem?

Writing Down Equations of Motion: Once we have made a free body diagram then we can write equations of motion for each part of the system, for which we have drawn FBD. To write down equations of motion for a sub-system for which we have already drawn the FBD the requirement is to choose two direction (if required we will need three directions) in

which we shall work to reduce the complexity say, direction X and Y   which in most of the cases is natural. In other cases, where we are dealing with inclined planes we can fix our coordinate axis x-y in any desired orientation which reduces our trouble of finding out components of the forces which are acting in various directions and by experience one knows that it is better to choose X axis parallel to the inclined plane and Y axis

perpendicular to it. For inclined plane    is the right choice of coordinate axis, tilted as per inclination. Now once can go for writing F = ma equations in two directions for any ith sub-system.

        ∑Fx = miax

        ∑Fy = miay

Let us write down equation of motion for the FBD's we have already drawn.

 

Refer to FBD (2):      Here figure is given below.

        ∑Fy = R - Mg

        It will be zero since we know that on the surface the body can't move upwards or downwards.

        .·. R - mg = 0             => R = Mg and ∑Fx = F

        And this alone should accelerate it

        ∑ F = Ma,   => a = F/m

 

Refer to FBD (3): Here Figure is given below.

        for the smaller block

        ∑Fy = R1 - mg = 0             .·. R1 = mg

        ∑Fx = F-R                        .·. F-R = m a1

        For the Large Block          ∑Fy = R2 - Mg = 0 .·. R2 = Mg

        ∑Fx = R                           .·. R = Ma2

        Since both Blocks move as on system, their accelerations will be equal

        a1 = a2                           .·. F - R = ma

        & R = Ma                    from here we can solve for R & a.

 

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

Refer to FBD (4): Here figure is given below.

 

        For block

        ∑Fy = R - Mg = 0

        .·. R = mg

        ∑Fx = T = Ma

        .·. a = T/m

 

Refer to FBD (5): Here Figure is given below.

        Here also, since system is fully connected and we assume that string is not loose, then the acceleration in different parts will be equal. That is, Blocks m and M as well as the string will move with the same acceleration 'a'.

 

        Equation for smaller Block

        ∑Fy = R1 - mg = 0

        .·. R1 = mg

        ∑Fx = T = ma ............ (1)

        For Block - 'M'            ∑Fy = R2 - Mg = 0

        .·. R2 = Mg

        ∑Fx = F - T = Ma ......... (ii)

        solving (i) & (ii) we get T & a.

 

Refer to FBD (6): Here Figure is given below.

        Since, system is fully connected by light inextensible string therefore acceleration in all the parts will be equal.

 

 

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

For Block 'm'

        R1 - mg = 0 .·. R1 = mg

        T1 = ma.

For Block 'M' (Refer to the diagram made with mg cos α & mg sin α components.

R2 - Mg cos α = 0 (since it is not moving to the plane)

        R2 = Mg cos α and is parallel to the plane

        Mg sin α - T2 = Ma                       ............ (ii)

 

It is already explained that T1 = T2 = T

.·. solving (i) & (ii) we get 'T' & 'a' for the system.

  

Refer to FBD (7) : Here Figure is given below.

T1 = T2 = T (similar to the analysis done in FBD (6))

For block m1 perpendicular to inclined plane T1 - m1g cos α = 0 R1 = m1g cos α.

Parallel to the plane.

T-m1g sin α = m1a ............. (1)

        For block m2 perpendicular to plane

R2 - m2g cos β = 0 R2 = m2g cos β and m2 g sin β - T = m2 a ...... (ii)

Solving (i) & (ii) we get 'a' & 'T.

 

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

Refer to FBD (8) : Here Figure is given below.

 

In Y-direction :

For block m1 T - m1g = m1a

For Block m2 m2g - T = m2 a

From (i) & (ii) we can solve for 'T' & 'a'.

On pulley 2T force is acting downwards. Therefore the pulley experiences a force F = 2T. Now you are well conversant with making FBD's and writing down equations of motion for the given system. Therefore, now we can write down some definite steps to follow for any given problem, which can considerably reduce your diversion or confusion or say the possibility of getting trapped into complexity of a problem. (of course for that you have to use your brain all the time with these steps).

 

(i)      Draw the fully connected clear diagram.

(ii)     Define subsystems, that is parts of the system on which you will work to get your answers.

(iii)    Draw free body diagrams for all possible subsystems.

(iv)    Resolve forces as well as accelerations in x y & z direction. (or perpendicular and || to the plane whenever require).

(v)     Write, ∑Fx, ∑Fy, ∑Fz equations with the physical constraints appearing in the problem.

(vi)    Eliminate some variables and get the required one(s).

 

Centripetal Force

If a body is moving with a constant speed in a circle, as seen from an inertial frame, it is continuously towards the centre of rotation with magnitude vr/r (known as centripetal acceleration), where v is the speed of the particle and r is the radius of the circular path.

According to Newton's second law, this body will experience net force directed towards the centre called the centripetal force.

Therefore, net force acting on the body towards the centre = mv2/r, where m is mass of body.

Centrifugal force is a pseudo force acting on the body from a rotating frame.

 

 

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |

Illustration:

A bob of mass m is suspended form a inextensible, massless describe a horizontal uniform circular motion as shown in figure 1.42. about a vertical. Analyze the dynamics of this system.

 

Solution:

        The path of the system described above is shown in figure 1.42. Let the radius of circular path of bob is r equal to l sin θ and tension in string is T. The string makes an angle θ with the vertical. Consider the bob is at A. We can draw the free body diagram of bob at a as shown in figure 1.43. The force acting on the bob is it's weight mg and tension T of the string. Tenstion T is resolved in two components T cos θ and T sin θ as shown in figure 1.43. we can write the equation of motion

                                T cos θ = mg                      T sin θ = mv2/r

 

 

|  1   | |  2 | |  3 | |  4 | |  5 | 6 | 7 | 8 | 9 | 10 | 11 |