4.7 Conservation of momentum push. A breeze causes the branches of a tree to swing; a 4.8 Equilibrium of a particle strong wind can even move heavy objects. A boat moves in a 4.9 Common forces in mechanics flowing river without anyone rowing it. Clearly, some external 4.10 Circular motion agency is needed to provide force to move a body from rest. 4.11 Solving problems in Likewise, an external force is needed also to retard or stop mechanics motion. You can stop a ball rolling down an inclined plane by applying a force against the direction of its motion. Summary In these examples, the external agency of force (hands, Points to ponder wind, stream, etc) is in contact with the object. This is not Exercises always necessary. A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance. This shows that external agencies (e.g. gravitational and magnetic forces ) can exert force on a body even from a distance. In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body. So far so good. But what if a body is moving uniformly (e.g. a skater moving straight with constant speed on a horizontal ice slab) ? Is an external force required to keep a body in uniform motion? Reprint 2025-26 50 PHYSICS 4.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has to imagine a world in which uniform motion isThe question posed above appears to be simple. possible with no frictional forces opposing. ThisHowever, it took ages to answer it. Indeed, the is what Galileo did.correct answer to this question given by Galileo in the seventeenth century was the foundation 4.3 THE LAW OF INERTIA of Newtonian mechanics, which signalled the Galileo studied motion of objects on an inclined birth of modern science. plane. Objects (i) moving down an inclined plane The Greek thinker, Aristotle (384 B.C– 322 accelerate, while those (ii) moving up retard. B.C.), held the view that if a body is moving, (iii) Motion on a horizontal plane is an interme- something external is required to keep it moving. diate situation. Galileo concluded that an object According to this view, for example, an arrow moving on a frictionless horizontal plane must shot from a bow keeps flying since the air behind neither have acceleration nor retardation, i.e. it the arrow keeps pushing it. The view was part of should move with constant velocity (Fig. 4.1(a)).an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe. Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus: An external force (i) (ii) (iii)is required to keep a body in motion. Fig. 4.1(a) Aristotelian law of motion is flawed, as we shall Another experiment by Galileo leading to thesee. However, it is a natural view that anyone same conclusion involves a double inclined plane.would hold from common experience. Even a A ball released from rest on one of the planes rollssmall child playing with a simple (non-electric) down and climbs up the other. If the planes are toy-car on a floor knows intuitively that it needs smooth, the final height of the ball is nearly the to constantly drag the string attached to the toy- same as the initial height (a little less but never car with some force to keep it going. If it releases greater). In the ideal situation, when friction is the string, it comes to rest. This experience is absent, the final height of the ball is the same common to most terrestrial motion. External as its initial height. forces seem to be needed to keep bodies in If the slope of the second plane is decreased motion. Left to themselves, all bodies eventually and the experiment repeated, the ball will still come to rest. reach the same height, but in doing so, it will What is the flaw in Aristotle’s argument? The travel a longer distance. In the limiting case, when answer is: a moving toy car comes to rest because the slope of the second plane is zero (i.e. is a the external force of friction on the car by the floor horizontal) the ball travels an infinite distance. opposes its motion. To counter this force, the child In other words, its motion never ceases. This is, has to apply an external force on the car in the of course, an idealised situation (Fig. 4.1(b)). direction of motion. When the car is in uniform motion, there is no net external force acting on it: the force by the child cancels the force ( friction) by the floor. The corollary is: if there were no friction, the child would not be required to apply any force to keep the toy car in uniform motion. The opposing forces such as friction (solids) and viscous forces (for fluids) are always present in the natural world. This explains why forces by external agencies are necessary to overcome the frictional forces to keep bodies in uniform motion. Now we understand where Aristotle Fig. 4.1(b) The law of inertia was inferred by Galileo went wrong. He coded this practical experience from observations of motion of a ball on a in the form of a basic argument. To get at the double inclined plane. Reprint 2025-26 LAWS OF MOTION 51 In practice, the ball does come to a stop after accomplished almost single-handedly by Isaac moving a finite distance on the horizontal plane, Newton, one of the greatest scientists of all times. because of the opposing force of friction which Newton built on Galileo’s ideas and laid the can never be totally eliminated. However, if there foundation of mechanics in terms of three laws were no friction, the ball would continue to move of motion that go by his name. Galileo’s law of with a constant velocity on the horizontal plane. inertia was his starting point which he formu- Galileo thus, arrived at a new insight on lated as the first law of motion: motion that had eluded Aristotle and those who Every body continues to be in its state followed him. The state of rest and the state of of rest or of uniform motion in a straight uniform linear motion (motion with constant line unless compelled by some external velocity) are equivalent. In both cases, there is force to act otherwise. Ideas on Motion in Ancient Indian Science Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body; transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air resistance. It was correctly summarised that the different kinds of motion (translational, rotational and vibrational) of an extended body arise from only the translational motion of its constituent particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a definite (small) displacement. There was considerable focus in Indian thought on measurement of motion and units of length and time. It was known that the position of a particle in space can be indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly understood; a current is a motion of particles of water under gravity and fluidity while a wave results from the transmission of vibrations of water particles. no net force acting on the body. It is incorrect to The state of rest or uniform linear motion both assume that a net force is needed to keep a body imply zero acceleration. The first law of motion can, in uniform motion. To maintain a body in therefore, be simply expressed as: uniform motion, we need to apply an external If the net external force on a body is zero, its force to ecounter the frictional force, so that acceleration is zero. Acceleration can be non the two forces sum up to zero net external zero only if there is a net external force on force. the body. To summarise, if the net external force is zero, Two kinds of situations are encountered in thea body at rest continues to remain at rest and a application of this law in practice. In somebody in motion continues to move with a uniform examples, we know that the net external forcevelocity. This property of the body is called on the object is zero. In that case we caninertia. Inertia means ‘resistance to change’. A body does not change its state of rest or conclude that the acceleration of the object is uniform motion, unless an external force zero. For example, a spaceship out in compels it to change that state. interstellar space, far from all other objects and with all its rockets turned off, has no net 4.4 NEWTON’S FIRST LAW OF MOTION external force acting on it. Its acceleration, Galileo’s simple, but revolutionary ideas according to the first law, must be zero. If it is dethroned Aristotelian mechanics. A new in motion, it must continue to move with a mechanics had to be developed. This task was uniform velocity. Reprint 2025-26 52 PHYSICS More often, however, we do not know all the The acceleration of the car cannot be accounted forces to begin with. In that case, if we know for by any internal force. This might sound that an object is unaccelerated (i.e. it is either surprising, but it is true. The only conceivable at rest or in uniform linear motion), we can infer external force along the road is the force of from the first law that the net external force on friction. It is the frictional force that accelerates the object must be zero. Gravity is everywhere. the car as a whole. (You will learn about friction For terrestrial phenomena, in particular, every in section 4.9). When the car moves with object experiences gravitational force due to the constant velocity, there is no net external force. earth. Also objects in motion generally experience The property of inertia contained in the First friction, viscous drag, etc. If then, on earth, an law is evident in many situations. Suppose we object is at rest or in uniform linear motion, it is are standing in a stationary bus and the driver not because there are no forces acting on it, but starts the bus suddenly. We get thrown because the various external forces cancel out backward with a jerk. Why ? Our feet are in touch i.e. add up to zero net external force. with the floor. If there were no friction, we would Consider a book at rest on a horizontal surface remain where we were, while the floor of the bus Fig. (4.2(a)). It is subject to two external forces : would simply slip forward under our feet and the the force due to gravity (i.e. its weight W) acting back of the bus would hit us. However, downward and the upward force on the book by fortunately, there is some friction between the the table, the normal force R . R is a self-adjusting feet and the floor. If the start is not too sudden, force. This is an example of the kind of situation i.e. if the acceleration is moderate, the frictional mentioned above. The forces are not quite known force would be enough to accelerate our feet fully but the state of motion is known. We observe along with the bus. But our body is not strictly the book to be at rest. Therefore, we conclude a rigid body. It is deformable, i.e. it allows some from the first law that the magnitude of R equals relative displacement between different parts. that of W. A statement often encountered is : What this means is that while our feet go with “Since W = R, forces cancel and, therefore, the book the bus, the rest of the body remains where it is is at rest”. This is incorrect reasoning. The correct due to inertia. Relative to the bus, therefore, we statement is : “Since the book is observed to be at are thrown backward. As soon as that happens, rest, the net external force on it must be zero, however, the muscular forces on the rest of the according to the first law. This implies that the body (by the feet) come into play to move the body normal force R must be equal and opposite to the along with the bus. A similar thing happens weight W ”. when the bus suddenly stops. Our feet stop due to the friction which does not allow relative motion between the feet and the floor of the bus. But the rest of the body continues to move forward due to inertia. We are thrown forward. The restoring muscular forces again come into play and bring the body to rest. ⊳ Example 4.1 An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a Fig. 4.2 (a) a book at rest on the table, and (b) a car constant rate of 100 m s–2. What is the moving with uniform velocity. The net force acceleration of the astronaut the instant after is zero in each case. he is outside the spaceship ? (Assume that Consider the motion of a car starting from there are no nearby stars to exert rest, picking up speed and then moving on a gravitational force on him.) smooth straight road with uniform speed (Fig. Answer Since there are no nearby stars to exert(4.2(b)). When the car is stationary, there is no gravitational force on him and the smallnet force acting on it. During pick-up, it spaceship exerts negligible gravitationalaccelerates. This must happen due to a net attraction on him, the net force acting on theexternal force. Note, it has to be an external force. Reprint 2025-26 LAWS OF MOTION 53 astronaut, once he is out of the spaceship, is act. One reason is that the cricketer allows a zero. By the first law of motion the acceleration longer time for his hands to stop the ball. As of the astronaut is zero. ⊳ you may have noticed, he draws in the hands backward in the act of catching the ball4.5 NEWTON’S SECOND LAW OF MOTION (Fig. 4.3). The novice, on the other hand, The first law refers to the simple case when the keeps his hands fixed and tries to catch the net external force on a body is zero. The second ball almost instantly. He needs to provide a law of motion refers to the general situation when much greater force to stop the ball instantly, there is a net external force acting on the body. and this hurts. The conclusion is clear: force It relates the net external force to the not only depends on the change in momentum, acceleration of the body. but also on how fast the change is brought Momentum about. The same change in momentum Momentum of a body is defined to be the product brought about in a shorter time needs a of its mass m and velocity v, and is denoted greater applied force. In short, the greater the by p: rate of change of momentum, the greater is p = m v (4.1) the force. Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. • Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. • If two stones, one light and the other heavy, are dropped from the top of a building, a Fig. 4.3 Force not only depends on the change in person on the ground will find it easier to catch momentum but also on how fast the change is brought about. A seasoned cricketer draws the light stone than the heavy stone. The in his hands during a catch, allowing greater mass of a body is thus an important time for the ball to stop and hence requires a parameter that determines the effect of force smaller force. on its motion. • Speed is another important parameter to consider. A bullet fired by a gun can easily • Observations confirm that the product of pierce human tissue before it stops, resulting mass and velocity (i.e. momentum) is basic to in casualty. The same bullet fired with the effect of force on motion. Suppose a fixed moderate speed will not cause much damage. force is applied for a certain interval of time Thus for a given mass, the greater the speed, on two bodies of different masses, initially at the greater is the opposing force needed to stop rest, the lighter body picks up a greater speed the body in a certain time. Taken together, than the heavier body. However, at the end of the product of mass and velocity, that is the time interval, observations show that each momentum, is evidently a relevant variable body acquires the same momentum. Thus of motion. The greater the change in the the same force for the same time causes momentum in a given time, the greater is the the same change in momentum for force that needs to be applied. • A seasoned cricketer catches a cricket ball different bodies. This is a crucial clue to the second law of motion. coming in with great speed far more easily • In the preceding observations, the vector than a novice, who can hurt his hands in the Reprint 2025-26 54 PHYSICS character of momentum has not been evident. ∆p ∆p In the examples so far, momentum and change F ∝ or F = k ∆t ∆ t in momentum both have the same direction. But this is not always the case. Suppose a where k is a constant of proportionality. Taking stone is rotated with uniform speed in a ∆p the limit ∆t → 0, the term becomes the horizontal plane by means of a string, the ∆t magnitude of momentum is fixed, but its derivative or differential co-efficient of p with direction changes (Fig. 4.4). A force is needed d p to cause this change in momentum vector. respect to t, denoted by . Thus dt This force is provided by our hand through the string. Experience suggests that our hand d p F = k (4.2) needs to exert a greater force if the stone is d t rotated at greater speed or in a circle of For a body of fixed mass m, smaller radius, or both. This corresponds to greater acceleration or equivalently a greater d p d d v = (m v ) = m = m a (4.3) rate of change in momentum vector. This d t d t d t suggests that the greater the rate of change i.e the Second Law can also be written as in momentum vector the greater is the force F = k m a (4.4) applied. which shows that force is proportional to the product of mass m and acceleration a. The unit of force has not been defined so far. In fact, we use Eq. (4.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is dp F = = m a (4.5) dt In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2. Let us note at this stage some important points Fig. 4.4 Force is necessary for changing the direction about the second law : of momentum, even if its magnitude is constant. We can feel this while rotating a 1. In the second law, F = 0 implies a = 0. The second stone in a horizontal circle with uniform speed law is obviously consistent with the first law. by means of a string. 2. The second law of motion is a vector law. It is These qualitative observations lead to the equivalent to three equations, one for each second law of motion expressed by Newton as component of the vectors : follows : d p x F x = = ma xThe rate of change of momentum of a body is d t directly proportional to the applied force and d p ytakes place in the direction in which the force F y = = ma y acts. d t dp zThus, if under the action of a force F for time F z = =m a z (4.6) dtinterval ∆t, the velocity of a body of mass m changes from v to v + ∆v i.e. its initial momentum This means that if a force is not parallel to the velocity of the body, but makes some anglep = m v changes by ∆ p = m ∆v . According to the with it, it changes only the component of Second Law, velocity along the direction of force. The Reprint 2025-26 LAWS OF MOTION 55 component of velocity normal to the force Answer The retardation ‘a’ of the bullet remains unchanged. For example, in the (assumed constant) is given by motion of a projectile under the vertical – 90 × 90 – u 2 gravitational force, the horizontal component m s −2 = – 6750 m s −2 a = = of velocity remains unchanged (Fig. 4.5). 2s 2 × 0.6 3. The second law of motion given by Eq. (4.5) is The retarding force, by the second law of applicable to a single point particle. The force motion, is F in the law stands for the net external force = 0.04 kg × 6750 m s-2 = 270 N on the particle and a stands for acceleration of the particle. It turns out, however, that the The actual resistive force, and therefore, law in the same form applies to a rigid body or, retardation of the bullet may not be uniform. The answer therefore, only indicates the average even more generally, to a system of particles. resistive force. ⊳ In that case, F refers to the total external force ⊳ on the system and a refers to the acceleration Example 4.3 The motion of a particle of of the system as a whole. More precisely, a is 1 2 the acceleration of the centre of mass of the mass m is described by y = ut + gt . Find 2 system about which we shall study in detail in the force acting on the particle. Chapter 6. Any internal forces in the system are not to be included in F. Answer We know 1 2 y = ut + gt 2 Now, d y v = = u + gt d t dv acceleration, a = = g d t Fig. 4.5 Acceleration at an instant is determined by Then the force is given by Eq. (4.5) the force at that instant. The moment after a F = ma = mg stone is dropped out of an accelerated train, Thus the given equation describes the motion it has no horizontal acceleration or force, if of a particle under acceleration due to gravity air resistance is neglected. The stone carries no memory of its acceleration with the train and y is the position coordinate in the direction a moment ago. of g. ⊳ 4. The second law of motion is a local relation Impulse which means that force F at a point in space We sometimes encounter examples where a large (location of the particle) at a certain instant force acts for a very short duration producing a of time is related to a at that point at that finite change in momentum of the body. For instant. Acceleration here and now is example, when a ball hits a wall and bounces determined by the force here and now, not by back, the force on the ball by the wall acts for a any history of the motion of the particle very short time when the two are in contact, yet the force is large enough to reverse the momentum (See Fig. 4.5). of the ball. Often, in these situations, the force ⊳ and the time duration are difficult to ascertain Example 4.2 A bullet of mass 0.04 kg separately. However, the product of force and time, moving with a speed of 90 m s–1 enters a which is the change in momentum of the body heavy wooden block and is stopped after a remains a measurable quantity. This product is distance of 60 cm. What is the average called impulse: resistive force exerted by the block on the bullet? Impulse = Force × time duration = Change in momentum (4.7) Reprint 2025-26 56 PHYSICS A large force acting for a short time to produce a Thus, according to Newtonian mechanics, finite change in momentum is called an impulsive force never occurs singly in nature. Force is the force. In the history of science, impulsive forces were mutual interaction between two bodies. Forces put in a conceptually different category from always occur in pairs. Further, the mutual forces ordinary forces. Newtonian mechanics has no such between two bodies are always equal and distinction. Impulsive force is like any other force – opposite. This idea was expressed by Newton in except that it is large and acts for a short time. the form of the third law of motion. ⊳ To every action, there is always an equal and Example 4.4 A batsman hits back a ball opposite reaction. straight in the direction of the bowler without changing its initial speed of 12 m s–1. Newton’s wording of the third law is so crisp and If the mass of the ball is 0.15 kg, determine beautiful that it has become a part of common the impulse imparted to the ball. (Assume language. For the same reason perhaps, linear motion of the ball) misconceptions about the third law abound. Let us note some important points about the third law, particularly in regard to the usage of theAnswer Change in momentum terms : action and reaction. = 0.15 × 12–(–0.15×12) 1. The terms action and reaction in the third law = 3.6 N s, mean nothing else but ‘force’. Using different Impulse = 3.6 N s, terms for the same physical concept in the direction from the batsman to the bowler. can sometimes be confusing. A simple and clear way of stating the third law is as This is an example where the force on the ball follows :by the batsman and the time of contact of the ball and the bat are difficult to know, but the Forces always occur in pairs. Force on a impulse is readily calculated. ⊳ body A by B is equal and opposite to the force on the body B by A.

2.2 INSTANTANEOUS VELOCITY AND SPEED The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small. In other words, ∆ x v = lim (2.1a) ∆ t → 0 ∆ t Fig. 2.1 Determining velocity from position-time d x = (2.1b) graph. Velocity at t = 4 s is the slope of the dt tangent to the graph at that instant. lim where the symbol stands for the operation ∆→t 0 Now, we decrease the value of ∆t from 2 s to 1 of taking limit as ∆tg0 of the quantity on its s. Then line P1P2 becomes Q1Q2 and its slope right. In the language of calculus, the quantity gives the value of the average velocity over on the right hand side of Eq. (2.1a) is the the interval 3.5 s to 4.5 s. In the limit ∆t → 0, differential coefficient of x with respect to t and the line P1P2 becomes tangent to the position- d x time curve at the point P and the velocity at t is denoted by (see Appendix 2.1). It is the d t = 4 s is given by the slope of the tangent at rate of change of position with respect to time, that point. It is difficult to show this process graphically. But if we useat that instant. numerical method to obtain the value of We can use Eq. (2.1a) for obtaining the the velocity, the meaning of the limiting value of velocity at an instant either process becomes clear. For the graph shown graphically or numerically. Suppose that we in Fig. 2.1, x = 0.08 t3. Table 2.1 gives the want to obtain graphically the value of value of ∆x/∆t calculated for ∆t equal to 2.0 s, velocity at time t = 4 s (point P) for the motion 1.0 s, 0.5 s, 0.1 s and 0.01 s centred at t = of the car represented in Fig.2.1 calculation. 4.0 s. The second and third columns give theLet us take ∆t = 2 s centred at t = 4 s. Then, t  t    ∆ ∆by the definition of the average velocity, the t + t −     and t 2 = and the value of t1= 2 2slope of ( Fig. 2.1) gives the value of     line P1P2 average velocity over the interval 3 s to 5 s. fourth and the fifth columns give the ∆x Table 2.1 Limiting value of at t = 4 s ∆ t Reprint 2025-26 MOTION IN A STRAIGHT LINE 15 3 a + 16b – a – 4b corresponding values of x, i.e. x (t1) = 0.08 t1 = = 6.0 × b 2.0 -1 ⊳and x (t2) = 0.08 t23. The sixth column lists the = 6.0 × 2.5 =15 m s difference ∆x = x (t2) – x (t1) and the last column gives the ratio of ∆x and ∆t, i.e. the Note that for uniform motion, velocity is average velocity corresponding to the value the same as the average velocity at all of ∆t listed in the first column. instants. We see from Table 2.1 that as we decrease Instantaneous speed or simply speed is the the value of ∆t from 2.0 s to 0.010 s, the value of magnitude of velocity. For example, a velocity of the average velocity approaches the limiting + 24.0 m s–1 and a velocity of – 24.0 m s–1 — value 3.84 m s–1 which is the value of velocity at both have an associated speed of 24.0 m s-1. It should be noted that though average speed over dx t = 4.0 s, i.e. the value of at t = 4.0 s. In this a finite interval of time is greater or equal to the dt magnitude of the average velocity, manner, we can calculate velocity at each instantaneous speed at an instant is equal to instant for motion of the car. the magnitude of the instantaneous velocity at The graphical method for the determination that instant. Why so ? of the instantaneous velocity is always not a 2.3 ACCELERATIONconvenient method. For this, we must carefully plot the position–time graph and calculate the The velocity of an object, in general, changes value of average velocity as ∆t becomes smaller during its course of motion. How to describe and smaller. It is easier to calculate the value this change? Should it be described as the rate of velocity at different instants if we have data of change in velocity with distance or with of positions at different instants or exact time ? This was a problem even in Galileo’s expression for the position as a function of time. time. It was first thought that this change could Then, we calculate ∆x/∆t from the data for be described by the rate of change of velocity decreasing the value of ∆t and find the limiting with distance. But, through his studies of value as we have done in Table 2.1 or use motion of freely falling objects and motion of differential calculus for the given expression and objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is dx calculate at different instants as done in a constant of motion for all objects in free fall. dt On the other hand, the change in velocity with the following example. distance is not constant – it decreases with the ⊳ increasing distance of fall. This led to the Example 2.1 The position of an object concept of acceleration as the rate of change moving along x-axis is given by x = a + bt2 of velocity with time. where a = 8.5 m, b = 2.5 m s–2 and t is The average acceleration a over a time interval measured in seconds. What is its velocity at is defined as the change of velocity divided by t = 0 s and t = 2.0 s. What is the average the time interval : velocity between t = 2.0 s and t = 4.0 s ? v 2 – v1 ∆v (2.2)Answer In notation of differential calculus, the a = = t 2 – t1 ∆tvelocity is where v2 and v1 are the instantaneous velocities dx d 2 -1 2b t = 5.0 t m s or simply velocities at time t2 and t1 . It is thev = = ( a + bt ) = dt dt average change of velocity per unit time. The SI At t = 0 s, v = 0 m s–1 and at t = 2.0 s, unit of acceleration is m s–2 . v = 10 m s-1 . On a plot of velocity versus time, the average acceleration is the slope of the straight line x ( 4.0 ) − x ( 2.0 )Average velocity = connecting the points corresponding to (v2, t2) 4.0 − 2.0 and (v1, t1). Reprint 2025-26 16 PHYSICS Instantaneous acceleration is defined in the (c) An object is moving in negative direction same way as the instantaneous velocity : with a negative acceleration. ∆v d v (d) An object is moving in positive direction lim a = = (2.3) till time t1, and then turns back with the d t ∆→ t 0 ∆ t same negative acceleration. The acceleration at an instant is the slope of the tangent to the v–t curve at that An interesting feature of a velocity-time instant. graph for any moving object is that the area Since velocity is a quantity having both under the curve represents the magnitude and direction, a change in displacement over a given time interval. A velocity may involve either or both of these general proof of this statement requires use of factors. Acceleration, therefore, may result calculus. We can, however, see that it is true from a change in speed (magnitude), a for the simple case of an object moving with change in direction or changes in both. Like constant velocity u. Its velocity-time graph is velocity, acceleration can also be positive, as shown in Fig. 2.4. negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. 2.4 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration. Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is vo at t = 0 and v at time t, we have v − v 0 a = or, v = v 0 + a t (2.4) t − 0 Fig. 2.3 Velocity–time graph for motions with Fig. 2.2 Position-time graph for motion with constant acceleration. (a) Motion in positive (a) positive acceleration; (b) negative direction with positive acceleration, acceleration, and (c) zero acceleration. (b) Motion in positive direction with Let us see how velocity-time graph looks like negative acceleration, (c) Motion in for some simple cases. Fig. 2.3 shows velocity- negative direction with negative acceleration, (d) Motion of an object withtime graph for motion with constant acceleration negative acceleration that changesfor the following cases : direction at time t1. Between times 0 to (a) An object is moving in a positive direction t1, it moves in positive x - direction with a positive acceleration. and between t1 and t2 it moves in the (b) An object is moving in positive direction opposite direction. with a negative acceleration. Reprint 2025-26 MOTION IN A STRAIGHT LINE 17 Fig. 2.4 Area under v–t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the time axis and the area under it between t = 0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u × T = uT which Fig. 2.5 Area under v-t curve for an object with is the displacement in this time interval. How uniform acceleration. come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two coordinate axes, and you will arrive at As explained in the previous section, the area the answer. under v-t curve represents the displacement. Therefore, the displacement x of the object is : Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp 1 x = ( v – v 0 ) t + v 0 t (2.5)kinks at some points implying that the 2 functions are not differentiable at these But v − v 0 = a tpoints. In any realistic situation, the functions will be differentiable at all points 1 2 Therefore, x = a t + v 0 tand the graphs will be smooth. 2 What this means physically is that 1 2 or, x = v 0 t + at (2.6)acceleration and velocity cannot change 2 values abruptly at an instant. Changes are Equation (2.5) can also be written as always continuous. v + v 0 x = t = v t (2.7a)2.4 KINEMATIC EQUATIONS FOR 2 UNIFORMLY ACCELERATED MOTION where, For uniformly accelerated motion, we can derive some simple equations that relate displacement v + v 0 v = (constant acceleration only)(x), time taken (t), initial velocity (v0), final 2 velocity (v) and acceleration (a). Equation (2.4) (2.7b) already obtained gives a relation between final and initial velocities v and v0 of an object moving Equations (2.7a) and (2.7b) mean that the object with uniform acceleration a : has undergone displacement x with an average velocity equal to the arithmetic average of the v = v0 + at (2.4) initial and final velocities. From Eq. (2.4), t = (v – v0)/a. Substituting this in This relation is graphically represented in Fig. 2.5. Eq. (2.7a), we get The area under this curve is : Area between instants 0 and t = Area of triangle  v + v 0   v − v 0  v 2 − v 02 x = v t =ABC + Area of rectangle OACD  2   a = 2a 1 2 2 = (v – v 0 ) t + v 0 t v = v 0 + 2ax (2.8) 2 Reprint 2025-26 18 PHYSICS This equation can also be obtained by t v 0 + at ) d tsubstituting the value of t from Eq. (2.4) into Eq. = ∫ 0 ( (2.6). Thus, we have obtained three important equations : 1 2 x – x 0 = v 0 t + a t 2 v = v 0 + at 1 2 1 2 x = x 0 + v 0 t + a t x = v 0t + at 2 2 We can write v 2 = v 02 + 2ax (2.9a) d v d v d x d v a = = = v d t d x d t d x connecting five quantities v0, v, a, t and x. These or, v dv = a dxare kinematic equations of rectilinear motion for Integrating both sides,constant acceleration. The set of Eq. (2.9a) were obtained by v x v d v = a d xassuming that at t = 0, the position of the particle, ∫ v 0 ∫ x 0 x is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non- v 2 – v 02 = a ( x – x 0 ) zero, say x0. Then Eqs. (2.9a) are modified 2 (replacing x by x – x0 ) to : 2 2 v = v 0 + 2a ( x – x 0 ) v = v 0 + at The advantage of this method is that it can be used 1 2 for motion with non-uniform acceleration x = x 0 + v 0t + at (2.9b) also. 2 Now, we shall use these equations to some v 2 = v 02 + 2a ( x − x 0 ) (2.9c) important cases. ⊳ ⊳ ⊳ Example 2.3 A ball is thrown vertically Example 2.2 Obtain equations of motion upwards with a velocity of 20 m s–1 from for constant acceleration using method of the top of a multistorey building. The calculus. height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how longAnswer By definition will it be before the ball hits the ground? d v Take g = 10 m s–2. a = d t dv = a dt Answer (a) Let us take the y-axis in the Integrating both sides vertically upward direction with zero at the v t ∫ v 0 d v = ∫ 0 a d t ground, as shown in Fig. 2.6. Now vo = + 20 m s–1, t d t (a is a = – g = –10 m s–2, = a ∫ 0 v = 0 m s–1 constant) If the ball rises to height y from the point of v – v 0 = at launch, then using the equation 2 + 2 a 0 ( y – y 0 ) v = v 0 + at v 2 = v we get d x Further, v = 0 = (20)2 + 2(–10)(y – y0) d t Solving, we get, (y – y0) = 20 m. dx = v dt Integrating both sides (b) We can solve this part of the problem in two x t ways. Note carefully the methods used. ∫ x 0 dx = ∫v0 d t Reprint 2025-26 MOTION IN A STRAIGHT LINE 19 0 = 25 +20 t + (½) (-10) t2 Or, 5t2 – 20t – 25 = 0 Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration. ⊳ ⊳ Example 2.4 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is Fig. 2.6 said to be in free fall. If the height through which the object falls is small compared to the FIRST METHOD : In the first method, we split earth’s radius, g can be taken to be constant, the path in two parts : the upward motion (A to equal to 9.8 m s–2. Free fall is thus a case of B) and the downward motion (B to C) and motion with uniform acceleration. calculate the corresponding time taken t1 and We assume that the motion is in y-direction, t2. Since the velocity at B is zero, we have : more correctly in –y-direction because we v = vo + at choose upward direction as positive. Since the 0 = 20 – 10t1 acceleration due to gravity is always downward, Or, t1 = 2 s it is in the negative direction and we have This is the time in going from A to B. From B, or a = – g = – 9.8 m s–2 the point of the maximum height, the ball falls The object is released from rest at y = 0. Therefore, freely under the acceleration due to gravity. The v0 = 0 and the equations of motion become: ball is moving in negative y direction. We use v = 0 – g t = –9.8 t m s–1equation y = 0 – ½ g t2 = –4.9 t 2 m 1 2 y = y 0 + v 0t + at v2 = 0 – 2 g y = –19.6 y m2 s–2 2 These equations give the velocity and the We have, y0 = 45 m, y = 0, v0 = 0, a = – g = –10 m s–2 distance travelled as a function of time and also 0 = 45 + (½) (–10) t2 2 the variation of velocity with distance. The Solving, we get t2 = 3 s variation of acceleration, velocity, and distance, with time have been plotted in Fig. 2.7(a), (b)Therefore, the total time taken by the ball before and (c). it hits the ground = t1 + t2 = 2 s + 3 s = 5 s. SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation 1 2 y = y0 + v 0t + at 2 Now y0 = 25 m y = 0 m vo = 20 m s-1, a = –10m s–2, t = ? (a) Reprint 2025-26 20 PHYSICS traversed during successive intervals of time. Since initial velocity is zero, we have 2 y = −1 gt 2 Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ… which are given in second column of Table 2.2. If we take (–1/ 2) gτ2 as y0 — the position coordinate after first time interval τ, then third column gives (b) the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall. ⊳ ⊳ Example 2.6 Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is (c) an important factor for road safety and depends on the initial velocity (v0) and theFig. 2.7 Motion of an object under free fall. braking capacity, or deceleration, –a that (a) Variation of acceleration with time. (b) Variation of velocity with time. is caused by the braking. Derive an (c) Variation of distance with time ⊳ expression for stopping distance of a vehicle in terms of vo and a. ⊳ Example 2.5 Galileo’s law of odd Answer Let the distance travelled by the vehicle numbers : “The distances traversed, during before it stops be ds. Then, using equation of equal intervals of time, by a body falling 2 motion v2 = vo + 2 ax, and noting that v = 0, we from rest, stand to one another in the same have the stopping distance ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it. – v 02 d s = 2aAnswer Let us divide the time interval of motion of an object under free fall into many Thus, the stopping distance is proportional to equal intervals τ and find out the distances the square of the initial velocity. Doubling the Table 2.2 Reprint 2025-26 MOTION IN A STRAIGHT LINE 21 initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula. Stopping distance is an important factor considered in setting speed limits, for example, in school zones. ⊳ ⊳ Example 2.7 Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think Fig. 2.8 Measuring the reaction time. and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before Answer The ruler drops under free fall. he slams the brakes of the car is the Therefore, vo = 0, and a = – g = –9.8 m s–2. The reaction time. Reaction time depends distance travelled d and the reaction time tr are on complexity of the situation and on related by an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between Or, your thumb and forefinger (Fig. 2.8). Given d = 21.0 cm and g = 9.8 m s–2 the reaction After you catch it, find the distance d time is travelled by the ruler. In a particular case, d was found to be 21.0 cm. ⊳ Estimate reaction time. SUMMARY 1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. 2. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small : ∆ x d x v = lim v = lim = ∆→t 0 ∆→t 0 ∆t d t The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. Reprint 2025-26 22 PHYSICS 3. Average acceleration is the change in velocity divided by the time interval during which the change occurs : ∆ v a = ∆t 4. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t goes to zero : ∆v d v a = lim a = lim = ∆→t 0 ∆→t 0 ∆ t d t The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight line inclined to the time axis. 5. The area under the velocity-time curve between times t1 and t2 is equal to the displacement of the object during that interval of time. 6. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken t, initial velocity v0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion : v = v0 + at 1 2 x = v0 t + at 2 v 2 = v 02 + 2ax if the position of the object at time t = 0 is 0. If the particle starts at x = x0 , x in above equations is replaced by (x – x0). Reprint 2025-26 MOTION IN A STRAIGHT LINE 23 POINTS TO PONDER 1. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration. 2. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis. 3. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed. 4. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity. 5. In the kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs. 6. The definitions of instantaneous velocity and acceleration (Eqs. (2.1) and (2.3)) are exact and are always correct while the kinematic equations (Eq. (2.9)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion. Reprint 2025-26 24 PHYSICS EXERCISES 2.1 In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table. 2.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ; (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice). Fig. 2.9 2.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion. 2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start. 2.5 A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ? 2.6 A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance). 2.7 Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. Reprint 2025-26 MOTION IN A STRAIGHT LINE 25 2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s. 2.9 Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only]. 2.10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] Fig. 2.10 2.11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? 2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. 2.13 Figure 2.11shows the x-t plot of one- dimensional motion of a particle. Is it correct to say from the graph that the particle moves Fig. 2.11 in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph. 2.14 A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car). Reprint 2025-26 26 PHYSICS 2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12): Fig. 2.12 2.16 Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s. Fig. 2.13 2.17 Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval. Fig. 2.14 2.18 Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ? Fig. 2.15 Reprint 2025-26 CHAPTER THREE MOTION IN A PLANE 3.1 INTRODUCTION In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to 3.1 Introduction describe the motion of an object along a straight line. We 3.2 Scalars and vectors found that the directional aspect of these quantities can be taken care of by + and – signs, as in one dimension only two3.3 Multiplication of vectors by real numbers directions are possible. But in order to describe motion of an 3.4 Addition and subtraction of object in two dimensions (a plane) or three dimensions vectors — graphical method (space), we need to use vectors to describe the above- 3.5 Resolution of vectors mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors. What is a vector? How to3.6 Vector addition — analytical method add, subtract and multiply vectors ? What is the result of 3.7 Motion in a plane multiplying a vector by a real number ? We shall learn this to enable us to use vectors for defining velocity and3.8 Motion in a plane with constant acceleration acceleration in a plane. We then discuss motion of an object 3.9 Projectile motion in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail3.10 Uniform circular motion the projectile motion. Circular motion is a familiar class of Summary motion that has a special significance in daily-life situations. Points to ponder We shall discuss uniform circular motion in some detail. Exercises The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions. 3.2 SCALARS AND VECTORS In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided Reprint 2025-26 28 PHYSICS just as the ordinary numbers*. For example, represented by another position vector, OP′ if the length and breadth of a rectangle are denoted by r′. The length of the vector r 1.0 m and 0.5 m respectively, then its represents the magnitude of the vector and its perimeter is the sum of the lengths of the direction is the direction in which P lies as seen four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = from O. If the object moves from P to P′, the 3.0 m. The length of each side is a scalar vector PP′ (with tail at P and tip at P′) is called and the perimeter is also a scalar. Take the displacement vector corresponding to another example: the maximum and motion from point P (at time t) to point P′ (at time t′). minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of

6.11 Dynamics of rotational the motion of extended bodies. motion about a fixed axis A large class of problems with extended bodies can be