6.4 LINEAR MOMENTUM OF A SYSTEM OF Eq. (6.15), this also means that when the PARTICLES total external force on the system is zero the velocity of the centre of mass remainsLet us recall that the linear momentum of a constant. (We assume throughout the particle is defined as discussion on systems of particles in this p = m v (6.12) chapter that the total mass of the system Let us also recall that Newton’s second law remains constant.) written in symbolic form for a single particle is Note that on account of the internal forces, dp i.e. the forces exerted by the particles on one F = (6.13) another, the individual particles may have dt Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 101 complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle. The vector Eq. (6.18a) is equivalent to three scalar equations, Px = c1, Py = c2 and Pz = c3 (6.18 b) (a) (b) Here Px, Py and Pz are the components of the total linear momentum vector P along the x–, y– Fig. 6.14 (a) Trajectories of two stars, S1 (dotted line) and z–axes respectively; c1, c2 and c3 are and S2 (solid line) forming a binary constants. system with their centre of mass C in uniform motion. (b) The same binary system, with the centre of mass C at rest. move back to back with their centre of mass remaining at rest as shown in Fig.6.13 (b). In many problems on the system of particles, as in the above radioactive decay problem, it is convenient to work in the centre of mass frame rather than in the laboratory frame of reference. (a) (b) In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star Fig. 6.13 (a) A heavy nucleus radium (Ra) splits into moves like a free particle, as shown in Fig.6.14 a lighter nucleus radon (Rn) and an alpha (a). The trajectories of the two stars of equal particle (nucleus of helium atom). The CM mass are also shown in the figure; they look of the system is in uniform motion. complicated. If we go to the centre of mass (b) The same spliting of the heavy nucleus radium (Ra) with the centre of mass at frame, then we find that there the two stars rest. The two product particles fly back are moving in a circle, about the centre of to back. mass, which is at rest. Note that the position of the stars have to be diametrically opposite As an example, let us consider the to each other [Fig. 6.14(b)]. Thus in our frame radioactive decay of a moving unstable particle, of reference, the trajectories of the stars are a combination of (i) uniform motion in a straightlike the nucleus of radium. A radium nucleus line of the centre of mass and (ii) circulardisintegrates into a nucleus of radon and an orbits of the stars about the centre of mass.alpha particle. The forces leading to the decay As can be seen from the two examples,are internal to the system and the external separating the motion of different parts of aforces on the system are negligible. So the total system into motion of the centre of mass andlinear momentum of the system is the same motion about the centre of mass is a verybefore and after decay. The two particles useful technique that helps in understanding produced in the decay, the radon nucleus and the motion of the system. the alpha particle, move in different directions in such a way that their centre of mass moves 6.5 VECTOR PRODUCT OF TWO VECTORS along the same path along which the original decaying radium nucleus was moving We are already familiar with vectors and their [Fig. 6.13(a)]. use in physics. In chapter 5 (Work, Energy, Power) If we observe the decay from the frame of we defined the scalar product of two vectors. An reference in which the centre of mass is at rest, important physical quantity, work, is defined as the motion of the particles involved in the decay a scalar product of two vector quantities, force looks particularly simple; the product particles and displacement. Reprint 2025-26 102 PHYSICS We shall now define another product of two A simpler version of the right hand rule is vectors. This product is a vector. Two important the following : Open up your right hand palm quantities in the study of rotational motion, and curl the fingers pointing from a to b. Your namely, moment of a force and angular stretched thumb points in the direction of c. momentum, are defined as vector products. It should be remembered that there are two angles between any two vectors a and b . In Definition of Vector Product Fig. 6.15 (a) or (b) they correspond to θ(as shown) A vector product of two vectors a and b is a and (3600– θ). While applying either of the above vector c such that rules, the rotation should be taken through the (i) magnitude of c = c = ab sinθ where a and b smaller angle (<1800) between a and b. It is θ are magnitudes of a and b and θ is the here. angle between the two vectors. Because of the cross (×) used to denote the (ii) c is perpendicular to the plane containing vector product, it is also referred to as cross product. a and b. • Note that scalar product of two vectors is (iii) if we take a right handed screw with its head commutative as said earlier, a.b = b.a lying in the plane of a and b and the screw The vector product, however, is not perpendicular to this plane, and if we turn commutative, i.e. a × b ≠ b × a the head in the direction from a to b, then The magnitude of both a × b and b × a is the the tip of the screw advances in the direction same ( ab sin θ ); also, both of them are of c. This right handed screw rule is perpendicular to the plane of a and b. But the illustrated in Fig. 6.15a. rotation of the right-handed screw in case of Alternately, if one curls up the fingers of a × b is from a to b, whereas in case of b × a it right hand around a line perpendicular to the is from b to a. This means the two vectors are plane of the vectors a and b and if the fingers in opposite directions. We have are curled up in the direction from a to b, then a × b = − b × a the stretched thumb points in the direction of • Another interesting property of a vector c, as shown in Fig. 6.15b. product is its behaviour under reflection. Under reflection (i.e. on taking the plane mirror image) we have x →− x , y →−y and z →− z . As a result all the components of a vector change sign and thus a →−a , b →−b . What happens to a × b under reflection? a × b →−( a ) × ( − b ) = a × b Thus, a × b does not change sign under reflection. • Both scalar and vector products are distributive with respect to vector addition. Thus, a.( b + c ) = a.b + a.c a × ( b + c ) = a × b + a × c (a) (b) • We may write c = a × b in the component form. For this we first need to obtain some elementary cross products: Fig. 6.15 (a) Rule of the right handed screw for (i) a × a = 0 (0 is a null vector, i.e. a vector defining the direction of the vector with zero magnitude) product of two vectors. This follows since magnitude of a × a is (b) Rule of the right hand for defining the direction of the vector product. a 2 sin0° = 0 . Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 103 From this follow the results ˆ ˆ ˆ i j k (i) ˆi × ˆi = 0, ˆj × ˆj = 0, kˆ × kˆ = 0 a × b = 3 − 4 5 = 7 ˆi − ˆj − 5 kˆ (ii) ˆi × ˆj = kˆ − 2 1 − 3 Note that the magnitude of ˆi × ˆj is sin900 Note b × a = −7ˆi + ˆj + 5 kˆ ⊳ or 1, since ˆi and ˆj both have unit magnitude and the angle between them is 900. 6.6 ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY Thus, ˆi × ˆj is a unit vector. A unit vector In this section we shall study what is angular perpendicular to the plane of ˆi and ˆj and velocity and its role in rotational motion. We related to them by the right hand screw rule is have seen that every particle of a rotating body moves in a circle. The linear velocity of the ˆk . Hence, the above result. You may verify particle is related to the angular velocity. The similarly, relation between these two quantities involves ˆ j × kˆ = ˆi and kˆ × ˆi = ˆj a vector product which we learnt about in the last section. From the rule for commutation of the cross Let us go back to Fig. 6.4. As said above, inproduct, it follows: rotational motion of a rigid body about a fixed ˆ j × ˆi = − kˆ , kˆ × ˆj = − ˆi, ˆi × kˆ = − ˆj axis, every particle of the body moves in a circle, Note if ˆi, ˆj, kˆ occur cyclically in the above vector product relation, the vector product is positive. If ˆi, ˆj, kˆ do not occur in cyclic order, the vector product is negative. Now, a × b = (a x ˆi + a y ˆj + a z kˆ ) × (b x ˆi + b y ˆj + b z kˆ ) = a x b y kˆ − a x b z ˆj − a y b x kˆ + a y b z ˆi + a z b x ˆj − a z b y ˆi = (a y b z − a z b y )i + (a z b x − a x b z ) j + (a x b y − a y b x )k We have used the elementary cross products in obtaining the above relation. The expression for a × b can be put in a determinant form which is easy to remember. ˆ ˆ ˆ i j k a × b = a x a y a z b x b y b z u Example 6.4 Find the scalar and vector Fig. 6.16 Rotation about a fixed axis. (A particle (P) of the rigid body rotating about the fixed products of two vectors. a = (z-) axis moves in a circle with centre (C) and b = on the axis.) Answer which lies in a plane perpendicular to the axis a i b = (3ˆi − 4 ˆj + 5 kˆ )i( − 2ˆi + ˆj − 3 kˆ ) and has its centre on the axis. In Fig. 6.16 we redraw Fig. 6.4, showing a typical particle (at a = −6 − 4 − 15 point P) of the rigid body rotating about a fixed = −25 axis (taken as the z-axis). The particle describes Reprint 2025-26 104 PHYSICS a circle with a centre C on the axis. The radius and points out in the direction in which a right of the circle is r, the perpendicular distance of handed screw would advance, if the head of the the point P from the axis. We also show the screw is rotated with the body. (See Fig. 6.17a). linear velocity vector v of the particle at P. It is The magnitude of this vector is ω = d θ dt along the tangent at P to the circle. referred as above. Let P′ be the position of the particle after an interval of time ∆t (Fig. 6.16). The angle PCP′ describes the angular displacement ∆θ of the particle in time ∆t. The average angular velocity of the particle over the interval ∆t is ∆θ/∆t. As ∆t tends to zero (i.e. takes smaller and smaller values), the ratio ∆θ/∆t approaches a limit which is the instantaneous angular velocity dθ/dt of the particle at the position P. We denote the instantaneous angular velocity by ω (the Greek letter omega). We know from our study Fig. 6.17 (a) If the head of a right handed screw of circular motion that the magnitude of linear rotates with the body, the screw velocity v of a particle moving in a circle is advances in the direction of the angular related to the angular velocity of the particle ω velocity ω. If the sense (clockwise or by the simple relation υ = ωr , where r is the anticlockwise) of rotation of the body changes, so does the direction of ω.radius of the circle. We observe that at any given instant the relation v = ωr applies to all particles of the rigid body. Thus for a particle at a perpendicular distance ri from the fixed axis, the linear velocity at a given instant vi is given by v i = ωri (6.19) The index i runs from 1 to n, where n is the total number of particles of the body. For particles on the axis, r = 0 , and hence v = ω r = 0. Thus, particles on the axis are stationary. This verifies that the axis is fixed. Note that we use the same angular velocity ω for all the particles. We therefore, refer to ω as the angular velocity of the whole body. We have characterised pure translation of a body by all parts of the body having the same Fig. 6.17 (b) The angular velocity vector ω is directed velocity at any instant of time. Similarly, we along the fixed axis as shown. The linear may characterise pure rotation by all parts of velocity of the particle at P is v = ω × r. the body having the same angular velocity at It is perpendicular to both ωωωωω and r and any instant of time. Note that this is directed along the tangent to the circle described by the particle. characterisation of the rotation of a rigid body about a fixed axis is just another way of saying We shall now look at what the vector as in Sec. 6.1 that each particle of the body moves product ω × r corresponds to. Refer to Fig. in a circle, which lies in a plane perpendicular 6.17(b) which is a part of Fig. 6.16 reproduced to the axis and has the centre on the axis. to show the path of the particle P. The figure In our discussion so far the angular velocity shows the vector ω directed along the fixed (z–) appears to be a scalar. In fact, it is a vector. We axis and also the position vector r = OP of the shall not justify this fact, but we shall accept particle at P of the rigid body with respect to it. For rotation about a fixed axis, the angular the origin O. Note that the origin is chosen to velocity vector lies along the axis of rotation, be on the axis of rotation. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 105 Now ω × r = ω × OP = ω × (OC + CP) If the axis of rotation is fixed, the direction But ω × OC = 00000 as ωωωωω is along OC of ωωωωω and hence, that of α is fixed. In this case Hence ω × r = ω × CP the vector equation reduces to a scalar equation dω α = (6.22) The vector ω × CP is perpendicular to ω, i.e. dt to the z-axis and also to CP, the radius of the circle described by the particle at P. It is 6.7 TORQUE AND ANGULAR MOMENTUM therefore, along the tangent to the circle at P. In this section, we shall acquaint ourselves with Also, the magnitude of ω × CP is ω (CP) since two physical quantities (torque and angular ω and CP are perpendicular to each other. We momentum) which are defined as vector products shall denote CP by ⊥r and not by r, as we did of two vectors. These as we shall see, are earlier. especially important in the discussion of motion Thus, ω × r is a vector of magnitude ωr⊥ of systems of particles, particularly rigid bodies. and is along the tangent to the circle described by the particle at P. The linear velocity vector v 6.7.1 Moment of force (Torque) at P has the same magnitude and direction. We have learnt that the motion of a rigid body, Thus, in general, is a combination of rotation and v = ωωωωω × r (6.20) translation. If the body is fixed at a point or along In fact, the relation, Eq. (6.20), holds good a line, it has only rotational motion. We know even for rotation of a rigid body with one point that force is needed to change the translationalfixed, such as the rotation of the top [Fig. 6.6(a)]. In this case r represents the position vector of state of a body, i.e. to produce linear the particle with respect to the fixed point taken acceleration. We may then ask, what is the as the origin. analogue of force in the case of rotational We note that for rotation about a fixed motion? To look into the question in a concrete axis, the direction of the vector ω does not situation let us take the example of opening or change with time. Its magnitude may, closing of a door. A door is a rigid body which however, change from instant to instant. For can rotate about a fixed vertical axis passing the more general rotation, both the magnitude and the direction of ωωωωω may change through the hinges. What makes the door from instant to instant. rotate? It is clear that unless a force is applied the door does not rotate. But any force does not 6.6.1 Angular acceleration do the job. A force applied to the hinge line You may have noticed that we are developing cannot produce any rotation at all, whereas a the study of rotational motion along the lines force of given magnitude applied at right angles of the study of translational motion with which to the door at its outer edge is most effective in we are already familiar. Analogous to the kinetic producing rotation. It is not the force alone, but variables of linear displacement (s) and velocity how and where the force is applied is important (v) in translational motion, we have angular in rotational motion. displacement (θ) and angular velocity (ω) in The rotational analogue of force in linear rotational motion. It is then natural to define motion is moment of force. It is also referred to in rotational motion the concept of angular as torque or couple. (We shall use the words acceleration in analogy with linear acceleration moment of force and torque interchangeably.) defined as the time rate of change of velocity in We shall first define the moment of force for the translational motion. We define angular special case of a single particle. Later on we acceleration α as the time rate of change of shall extend the concept to systems of particles angular velocity. Thus, including rigid bodies. We shall also relate it to d ω a change in the state of rotational motion, i.e. is α = (6.21) dt angular acceleration of a rigid body. Reprint 2025-26 106 PHYSICS of the line of action of F from the origin and F⊥=( F sin θ) is the component of F in the direction perpendicular to r. Note that τ = 0 if r = 0, F = 0 or θ = 00 or 1800 . Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin. One may note that since r × F is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same. 6.7.2 Angular momentum of a particle Just as the moment of a force is the rotational analogue of force in linear motion, the quantity angular momentum is the rotational analogue Fig. 6.18 τττττ ===== r × F, τττττ is perpendicular to the plane of linear momentum. We shall first define containing r and F, and its direction is angular momentum for the special case of a given by the right handed screw rule. single particle and look at its usefulness in the context of single particle motion. We shall then If a force acts on a single particle at a point extend the definition of angular momentum to P whose position with respect to the origin O is systems of particles including rigid bodies. given by the position vector r (Fig. 6.18), the Like moment of a force, angular momentum moment of the force acting on the particle with is also a vector product. It could also be referred respect to the origin O is defined as the vector to as moment of (linear) momentum. From this product term one could guess how angular momentum τ = r × F (6.23) is defined. The moment of force (or torque) is a vector Consider a particle of mass m and linear quantity. The symbol τττττ stands for the Greek momentum p at a position r relative to the origin letter tau. The magnitude of τττττ is O. The angular momentum l of the particle with τ = r F sinθ (6.24a) respect to the origin O is defined to be l = r × p (6.25a)where r is the magnitude of the position vector r, i.e. the length OP, F is the magnitude of force The magnitude of the angular momentum F and θ is the angle between r and F as vector is shown. l = r p sinθ (6.26a) Moment of force has dimensions M L2 T -2. where p is the magnitude of p and θ is the angle Its dimensions are the same as those of work between r and p. We may write or energy. It is, however, a very different physical l = r p⊥ or r ⊥ p (6.26b)quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of where r⊥ (= r sinθ) is the perpendicular distance moment of force is newton metre (N m). The of the directional line of p from the origin and magnitude of the moment of force may be p ⊥=( p sin θ) is the component of p in a directionwritten perpendicular to r. We expect the angular τ = (r sin θ)F = r⊥ F (6.24b) momentum to be zero (l = 0), if the linear or τ = r F sin θ = rF ⊥ (6.24c) momentum vanishes (p = 0), if the particle is at the origin (r = 0), or if the directional line of p where r⊥ = r sinθ is the perpendicular distance passes through the origin θ = 00 or 1800. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 107 The physical quantities, moment of a force and angular momentum, have an important An experiment with the bicycle rim relation between them. It is the rotational Take a analogue of the relation between force and linear bicycle rim momentum. For deriving the relation in the and extend context of a single particle, we differentiate its axle on l = r × p with respect to time, both sides. Tie two d l d = ( r × p ) s t r i n g s d t d t at both ends Applying the product rule for differentiation A and B, to the right hand side, as shown in the d d r d p ( r × p ) = × p + r × a d j o i n i n g d t d t d t figure. Hold Now, the velocity of the particle is v = dr/dt both the and p = m v Initially After s t r i n g s together in dr one hand such that the rim is vertical. If you Because of this × p = v × m v = 0, dt leave one string, the rim will tilt. Now keeping the rim in vertical position with both the stringsas the vector product of two parallel vectors in one hand, put the wheel in fast rotation vanishes. Further, since dp / dt = F, around the axle with the other hand. Then leave d p one string, say B, from your hand, and observe r × = r × F = t dt what happens. The rim keeps rotating in a vertical plane d and the plane of rotation turns around the string Hence ( r × p ) = τ A which you are holding. We say that the axis dt of rotation of the rim or equivalently or (6.27) its angular momentum precesses about the string A. Thus, the time rate of change of the angular The rotating rim gives rise to an angular momentum of a particle is equal to the torque momentum. Determine the direction of this acting on it. This is the rotational analogue of angular momentum. When you are holding the the equation F = dp/dt, which expresses rotating rim with string A, a torque is generated. (We leave it to you to find out how the torque isNewton’s second law for the translational motion generated and what its direction is.) The effect of a single particle. of the torque on the angular momentum is to make it precess around an axis perpendicular Torque and angular momentum for a system to both the angular momentum and the torque. of particles Verify all these statements. To get the total angular momentum of a system of particles about a given point we need to add vectorially the angular momenta of individual particles. Thus, for a system of n particles, particle has mass mi and velocity vi) We may write the total angular momentum of a system of particles as (6.25b) The angular momentum of the ith particle is given by li = ri × pi This is a generalisation of the definition of angular momentum (Eq. 6.25a) for a singlewhere ri is the position vector of the ith particle particle to a system of particles.with respect to a given origin and p = (mivi) is Using Eqs. (6.23) and (6.25b), we getthe linear momentum of the particle. (The Reprint 2025-26 108 PHYSICS d L d d l Note that like Eq.(6.17), Eq.(6.28b) holds = ( l ) = ∑ = ∑ τ (6.28a) good for any system of particles, whether it is a d t d t i d t i rigid body or its individual particles have all where τi is the torque acting on the ith particle; kinds of internal motion. τi = ri × Fi Conservation of angular momentum The force Fi on the ith particle is the vector ext If τext = 0, Eq. (6.28b) reduces to Fi sum of external forces acting on the particle d L = 0 and the internal forces iFint exerted on it by the dt other particles of the system. We may therefore or L = constant. (6.29a) separate the contribution of the external and Thus, if the total external torque on a system the internal forces to the total torque of particles is zero, then the total angular momentum of the system is conserved, i.e. τ = ∑ τ i = ∑ ri × Fi as remains constant. Eq. (6.29a) is equivalent to i i three scalar equations, τ = τext + τ int , Lx = K1, Ly = K2 and Lz = K3 (6.29 b) Here K1, K2 and K3 are constants; Lx, Ly and τ ext = ∑ri × Fi ext Lz are the components of the total angular where i momentum vector L along the x,y and z axes respectively. The statement that the total i × Fiint τ int = ∑r and angular momentum is conserved means that i each of these three components is conserved. We shall assume not only Newton’s third law Eq. (6.29a) is the rotational analogue of of motion, i.e. the forces between any two particles Eq. (6.18a), i.e. the conservation law of the total of the system are equal and opposite, but also that linear momentum for a system of particles. these forces are directed along the line joining the Like Eq. (6.18a), it has applications in many two particles. In this case the contribution of the practical situations. We shall look at a few of internal forces to the total torque on the system is the interesting applications later on in zero, since the torque resulting from each action- this chapter. reaction pair of forces is zero. We thus have, τint = 0 and therefore τ = τττext.ττ u Example 6.5 Find the torque of a force Since τ = ∑ τ i , it follows from Eq. (6.28a) + – about the origin. The force acts on a particle whose position vector is .that d L = τ ext (6.28 b) Answer Here r = ˆi − ˆj + kˆ d t and F = 7 ˆi + 3 ˆj − 5 kˆ . Thus, the time rate of the total angular We shall use the determinant rule to find themomentum of a system of particles about a τ = r × Fpoint (taken as the origin of our frame of torque reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point. Eq. (6.28 b) is the generalisation of the single particle case of Eq. (6.23) to a system of particles. Note that when we have only one or ⊳ particle, there are no internal forces or torques. Eq.(6.28 b) is the rotational analogue of Example 6.6 Show that the angular u momentum about any point of a single d P = Fext (6.17) particle moving with constant velocity d t remains constant throughout the motion. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 109 Answer Let the particle with velocity v be at acceleration nor angular acceleration. This means point P at some instant t. We want to calculate (1) the total force, i.e. the vector sum of the the angular momentum of the particle about an forces, on the rigid body is zero; arbitrary point O. n F1 + F2 + ... + Fn = =∑i 1 Fi = 0 (6.30a) If the total force on the body is zero, then the total linear momentum of the body does not change with time. Eq. (6.30a) gives the condition for the translational equilibrium of the body. (2) The total torque, i.e. the vector sum of the torques on the rigid body is zero, n τ + τ i = 0 (6.30b) 1 2 + ... + τ n = =∑i 1 τ Fig 6.19 If the total torque on the rigid body is zero, The angular momentum is l = r × mv. Its the total angular momentum of the body does magnitude is mvr sinθ, where θ is the angle not change with time. Eq. (6.30 b) gives the between r and v as shown in Fig. 6.19. Although condition for the rotational equilibrium of the the particle changes position with time, the line body. of direction of v remains the same and hence One may raise a question, whether theOM = r sin θ. is a constant. rotational equilibrium condition [Eq. 6.30(b)] Further, the direction of l is perpendicular remains valid, if the origin with respect to whichto the plane of r and v. It is into the page of the the torques are taken is shifted. One can showfigure.This direction does not change with time. Thus, l remains the same in magnitude and that if the translational equilibrium condition direction and is therefore conserved. Is there [Eq. 6.30(a)] holds for a rigid body, then such a any external torque on the particle? ⊳ shift of origin does not matter, i.e. the rotational equilibrium condition is independent of the 6.8 EQUILIBRIUM OF A RIGID BODY location of the origin about which the torques are taken. Example 6.7 gives a proof of this result We are now going to concentrate on the motion in a special case of a couple, i.e. two forcesof rigid bodies rather than on the motion of acting on a rigid body in translationalgeneral systems of particles. equilibrium. The generalisation of this result to We shall recapitulate what effect the external forces have on a rigid body. (Henceforth n forces is left as an exercise. we shall omit the adjective ‘external’ because Eq. (6.30a) and Eq. (6.30b), both, are vector unless stated otherwise, we shall deal with only equations. They are equivalent to three scalar external forces and torques.) The forces change equations each. Eq. (6.30a) corresponds to the translational state of the motion of the rigid n n n body, i.e. they change its total linear =∑i 1 Fix = 0 , =∑i 1 Fiy = 0 and =∑i 1 Fiz = 0 (6.31a)momentum in accordance with Eq. (6.17). But this is not the only effect the forces have. The where Fix, Fiy and Fiz are respectively the x, y and total torque on the body may not vanish. Such z components of the forces Fi. Similarly, Eq. a torque changes the rotational state of motion (6.30b) is equivalent to three scalar equations of the rigid body, i.e. it changes the total angular n n 0momentum of the body in accordance with τix = 0 , τiy = and (6.31b) =∑i 1 =∑i 1 Eq. (6.28 b). where τix, τiy and τiz are respectively the x, y and A rigid body is said to be in mechanical z components of the torque τi .equilibrium, if both its linear momentum and Eq. (6.31a) and (6.31b) give six independentangular momentum are not changing with time, conditions to be satisfied for mechanicalor equivalently, the body has neither linear Reprint 2025-26 110 PHYSICS equilibrium of a rigid body. In a number of problems all the forces acting on the body are coplanar. Then we need only three conditions to be satisfied for mechanical equilibrium. Two of these conditions correspond to translational equilibrium; the sum of the components of the forces along any two perpendicular axes in the plane must be zero. The third condition corresponds to rotational equilibrium. The sum of the components of the torques along any axis Fig. 6.20 (b) perpendicular to the plane of the forces must be zero. The force at B in Fig. 6.20(a) is reversed in Fig. 6.20(b). Thus, we have the same rod with The conditions of equilibrium of a rigid body two forces of equal magnitude but acting inmay be compared with those for a particle, which opposite diretions applied perpendicular to the we considered in earlier chapters. Since rod, one at end A and the other at end B. Here consideration of rotational motion does not apply the moments of both the forces are equal, but to a particle, only the conditions for translational they are not opposite; they act in the same sense equilibrium (Eq. 6.30 a) apply to a particle. Thus, and cause anticlockwise rotation of the rod. The for equilibrium of a particle the vector sum of total force on the body is zero; so the body is in all the forces on it must be zero. Since all these translational equilibrium; but it is not in forces act on the single particle, they must be rotational equilibrium. Although the rod is not fixed in any way, it undergoes pure rotation (i.e.concurrent. Equilibrium under concurrent rotation without translation).forces was discussed in the earlier chapters. A pair of forces of equal magnitude but acting A body may be in partial equilibrium, i.e., it in opposite directions with different lines of may be in translational equilibrium and not in action is known as a couple or torque. A couple rotational equilibrium, or it may be in rotational produces rotation without translation. equilibrium and not in translational When we open the lid of a bottle by turning equilibrium. it, our fingers are applying a couple to the lid Consider a light (i.e. of negligible mass) rod [Fig. 6.21(a)]. Another known example is a compass needle in the earth’s magnetic field as(AB) as shown in Fig. 6.20(a). At the two ends (A shown in the Fig. 6.21(b). The earth’s magneticand B) of which two parallel forces, both equal field exerts equal forces on the north and southin magnitude and acting along same direction poles. The force on the North Pole is towards are applied perpendicular to the rod. the north, and the force on the South Pole is toward the south. Except when the needle points in the north-south direction; the two forces do not have the same line of action. Thus there is a couple acting on the needle due to the earth’s magnetic field. Fig. 6.20 (a) Let C be the midpoint of AB, CA = CB = a. the moment of the forces at A and B will both be equal in magnitude (aF ), but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium, but it will not be in translational fingers apply a couple to turnequilibrium; F ≠ 0 Fig. 6.21(a) Our ∑ the lid. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 111 length. This point is called the fulcrum. A see- saw on the children’s playground is a typical example of a lever. Two forces F1 and F2, parallel to each other and usually perpendicular to the lever, as shown here, act on the lever at distances d1 and d2 respectively from the fulcrum as shown in Fig. 6.23. Fig. 6.21(b) The Earth’s magnetic field exerts equal and opposite forces on the poles of a Fig. 6.23 compass needle. These two forces form a couple. The lever is a system in mechanical equilibrium. Let R be the reaction of the supportu Example 6.7 Show that moment of a at the fulcrum; R is directed opposite to the couple does not depend on the point about forces F1 and F2. For translational equilibrium, which you take the moments. R – F1 – F2 = 0 (i) Answer For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero, d1F1 – d2F2 = 0 (ii) Normally the anticlockwise (clockwise) moments are taken to be positive (negative). Note R acts at the fulcrum itself and has zero moment about the fulcrum. Fig. 6.22 In the case of the lever force F1 is usually Consider a couple as shown in Fig. 6.22 some weight to be lifted. It is called the load and acting on a rigid body. The forces F and -F act its distance from the fulcrum d1 is called the respectively at points B and A. These points have load arm. Force F2 is the effort applied to lift the position vectors r1 and r2 with respect to origin load; distance d2 of the effort from the fulcrum O. Let us take the moments of the forces about is the effort arm. the origin. Eq. (ii) can be written as The moment of the couple = sum of the d1F1 = d2 F2 (6.32a) moments of the two forces making the couple or load arm × load = effort arm × effort = r1 × (–F) + r2 × F The above equation expresses the principle = r2 × F – r1 × F of moments for a lever. Incidentally the ratio = (r2–r1) × F F1/F2 is called the Mechanical Advantage (M.A.); But r1 + AB = r2, and hence AB = r2 – r1. F1 d 2 The moment of the couple, therefore, is M.A. = = (6.32b) F2 d1AB × F. Clearly this is independent of the origin, the If the effort arm d2 is larger than the load point about which we took the moments of the arm, the mechanical advantage is greater than forces. ⊳ one. Mechanical advantage greater than one means that a small effort can be used to lift a 6.8.1 Principle of moments large load. There are several examples of a lever An ideal lever is essentially a light (i.e. of around you besides the see-saw. The beam of a negligible mass) rod pivoted at a point along its balance is a lever. Try to find more such Reprint 2025-26 112 PHYSICS examples and identify the fulcrum, the effort and The CG of the cardboard is so located that effort arm, and the load and the load arm of the the total torque on it due to the forces m1g, m2g lever in each case. …. etc. is zero. You may easily show that the principle of If ri is the position vector of the ith particle moment holds even when the parallel forces F1 of an extended body with respect to its CG, then and F2 are not perpendicular, but act at some the torque about the CG, due to the force of angle, to the lever. gravity on the particle is τi = ri × mi g. The total gravitational torque about the CG is zero, i.e. 6.8.2 Centre of gravity i × m i g = 0 (6.33) τ g = ∑ τ i = ∑r Many of you may have the experience of We may therefore, define the CG of a body balancing your notebook on the tip of a finger. as that point where the total gravitational torque Figure 6.24 illustrates a similar experiment that on the body is zero. you can easily perform. Take an irregular- We notice that in Eq. (6.33), g is the same shaped cardboard having mass M and a narrow for all particles, and hence it comes out of the tipped object like a pencil. You can locate by trial summation. This gives, since g is non-zero, and error a point G on the cardboard where it ir = 0. Remember that the position vectorscan be balanced on the tip of the pencil. (The ∑mi cardboard remains horizontal in this position.) (ri) are taken with respect to the CG. Now, in This point of balance is the centre of gravity (CG) accordance with the reasoning given below of the cardboard. The tip of the pencil provides Eq. (6.4a) in Sec. 6.2, if the sum is zero, the origin a vertically upward force due to which the must be the centre of mass of the body. Thus, cardboard is in mechanical equilibrium. As the centre of gravity of the body coincides with shown in the Fig. 6.24, the reaction of the tip is the centre of mass in uniform gravity or gravity- equal and opposite to Mg and hence the cardboard is in translational equilibrium. It is also in rotational equilibrium; if it were not so, due to the unbalanced torque it would tilt and fall. There are torques on the card board due to the forces of gravity like m1g, m2g …. etc, acting on the individual particles that make up the cardboard. Fig. 6.25 Determining the centre of gravity of a body Fig. 6.24 Balancing a cardboard on the tip of a of irregular shape. The centre of gravity G pencil. The point of support, G, is the lies on the vertical AA1 through the point centre of gravity. of suspension of the body A. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 113 free space. We note that this is true because = 30 cm, PG = 5 cm, AK1= BK2 = 10 cm and K1G = the body being small, g does not K2G = 25 cm. Also, W= weight of the rod = 4.00 vary from one point of the body to the other. If kg and W 1= suspended load = 6.00 kg; the body is so extended that g varies from part R1 and R2 are the normal reactions of the to part of the body, then the centre of gravity support at the knife edges. and centre of mass will not coincide. Basically, For translational equilibrium of the rod, the two are different concepts. The centre of R1+R2 –W1 –W = 0 (i) mass has nothing to do with gravity. It depends Note W1 and W act vertically down and R1 only on the distribution of mass of the body. and R2 act vertically up. In Sec. 6.2 we found out the position of the For considering rotational equilibrium, we centre of mass of several regular, homogeneous take moments of the forces. A convenient point objects. Obviously the method used there gives to take moments about is G. The moments of us also the centre of gravity of these bodies, if R2 and W1 are anticlockwise (+ve), whereas the they are small enough. moment of R1 is clockwise (-ve). Figure 6.25 illustrates another way of For rotational equilibrium, determining the CG of an irregular shaped body –R1 (K1G) + W1 (PG) + R2 (K2G) = 0 (ii) like a cardboard. If you suspend the body from It is given that W = 4.00g N and W1 = 6.00g some point like A, the vertical line through A N, where g = acceleration due to gravity. We passes through the CG. We mark the vertical take g = 9.8 m/s2. AA1. We then suspend the body through other With numerical values inserted, from (i) points like B and C. The intersection of the R1 + R2 – 4.00g – 6.00g = 0 verticals gives the CG. Explain why the method or R1 + R2 = 10.00g N (iii) works. Since the body is small enough, the = 98.00 N method allows us to determine also its centre From (ii), – 0.25 R1 + 0.05 W1 + 0.25 R2 = 0 of mass. or R1 – R2 = 1.2g N = 11.76 N (iv) From (iii) and (iv), R1 = 54.88 N, u Example 6.8 A metal bar 70 cm long and R2 = 43.12 N 4.00 kg in mass supported on two knife- Thus the reactions of the support are about edges placed 10 cm from each end. A 6.00 55 N at K1 and 43 N at K2. ⊳ kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. u Example 6.9 A 3m long ladder weighing (Assume the bar to be of uniform cross 20 kg leans on a frictionless wall. Its feet section and homogeneous.) rest on the floor 1 m from the wall as shown in Fig.6.27. Find the reaction forces of the Answer wall and the floor. Answer Fig. 6.26 Figure 6.26 shows the rod AB, the positions of the knife edges K1 and K2 , the centre of gravity of the rod at G and the suspended load at P. Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP Fig. 6.27 Reprint 2025-26 114 PHYSICS The ladder AB is 3 m long, its foot A is at from the axis, the linear velocity is υi = ir ω. The distance AC = 1 m from the wall. From kinetic energy of motion of this particle is Pythagoras theorem, BC = 2 2 m. The forces 1 2 1 2 2 on the ladder are its weight W acting at its centre k i = m i υi = m i ri ω 2 2 of gravity D, reaction forces F1 and F2 of the wall where mi is the mass of the particle. The totaland the floor respectively. Force F1 is kinetic energy K of the body is then given byperpendicular to the wall, since the wall is the sum of the kinetic energies of individualfrictionless. Force F2 is resolved into two particles,components, the normal reaction N and the force of friction F. Note that F prevents the ladder n 1 n 2 2 from sliding away from the wall and is therefore K = ∑ k i = ∑ (m i ri ω ) i =1 2 i =1 directed toward the wall. For translational equilibrium, taking the Here n is the number of particles in the body. forces in the vertical direction, Note ωis the same for all particles. Hence, taking N – W = 0 (i) ω out of the sum, Taking the forces in the horizontal direction, n 1 2 2 i ri ) F – F1 = 0 (ii) K = 2 ω ( ∑i =1 m For rotational equilibrium, taking the We define a new parameter characterisingmoments of the forces about A, the rigid body, called the moment of inertia I , 2 2 F1 −(1/2) W = 0 (iii) given by Now W = 20 g = 20 × 9.8 N = 196.0 N n 2 I = ∑ m i ri (6.34)From (i) N = 196.0 N i =1 With this definition,From (iii) F1 = W 4 2 = 196.0/4 2 = 34.6 N 1 2 From (ii) F = F1 = 34.6 N K = Iω (6.35) 2 2 2 Note that the parameter I is independent of F2 = F + N = 199.0 N the magnitude of the angular velocity. It is a The force F2 makes an angle α with the characteristic of the rigid body and the axis horizontal, about which it rotates. −1 Compare Eq. (6.35) for the kinetic energy oftan α = N F = 4 2 , α = tan (4 2) ≈ 80 ⊳ a rotating body with the expression for the kinetic energy of a body in linear (translational)6.9 MOMENT OF INERTIA motion, We have already mentioned that we are 1 2developing the study of rotational motion parallel K = m υ 2to the study of translational motion with which Here, m is the mass of the body and v is itswe are familiar. We have yet to answer one major velocity. We have already noted the analogy question in this connection. What is the between angular velocity ω (in respect of analogue of mass in rotational motion? We shall rotational motion about a fixed axis) and linear attempt to answer this question in the present velocity v (in respect of linear motion). It is then section. To keep the discussion simple, we shall evident that the parameter, moment of inertia consider rotation about a fixed axis only. Let us I, is the desired rotational analogue of mass in try to get an expression for the kinetic energy of linear motion. In rotation (about a fixed axis), a rotating body. We know that for a body rotating the moment of inertia plays a similar role as about a fixed axis, each particle of the body moves mass does in linear motion. We now apply the definition Eq. (6.34), toin a circle with linear velocity given by Eq. (6.19). calculate the moment of inertia in two simple cases.(Refer to Fig. 6.16). For a particle at a distance Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 115 (a) Consider a thin ring of radius R and mass change in its rotational motion, it can be M, rotating in its own plane around its centre regarded as a measure of rotational inertia of with angular velocity ω. Each mass element the body; it is a measure of the way in which of the ring is at a distance R from the axis, different parts of the body are distributed at and moves with a speed Rω. The kinetic different distances from the axis. Unlike the energy is therefore, mass of a body, the moment of inertia is not a fixed quantity but depends on distribution of 1 2 1 2 2 K = M υ = MR ω mass about the axis of rotation, and the 2 2 orientation and position of the axis of rotation Comparing with Eq. (6.35) we get I = MR 2 with respect to the body as a whole. As a for the ring. measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we can define a new parameter, the radius of gyration. It is related to the moment of inertia and the total mass of the body. Notice from the Table 6.1 that in all cases, we can write I = Mk2, where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint, k 2 = L2 12, i.e. k = L 12 . Similarly, k = R/2 for the circular disc about its diameter. The length k is a geometric property of the body and axis of rotation. It is called the radius of Fig. 6.28 A light rod of length l with a pair of gyration. The radius of gyration of a body masses rotating about an axis through about an axis may be defined as the distance the centre of mass of the system and perpendicular to the rod. The total mass from the axis of a mass point whose mass is of the system is M. equal to the mass of the whole body and whose moment of inertia is equal to the moment of (b) Next, take a rigid rod of negligible mass of inertia of the body about the axis. length of length l with a pair of small masses, Thus, the moment of inertia of a rigid body rotating about an axis through the centre of depends on the mass of the body, its shape and mass perpendicular to the rod (Fig. 6.28). size; distribution of mass about the axis of Each mass M/2 is at a distance l/2 from rotation, and the position and orientation of the the axis. The moment of inertia of the masses axis of rotation. is therefore given by From the definition, Eq. (6.34), we can infer (M/2) (l/2)2 + (M/2)(l/2)2 that the dimensions of moments of inertia are Thus, for the pair of masses, rotating about ML2 and its SI units are kg m2. the axis through the centre of mass The property of this extremely important perpendicular to the rod 2 quantity I, as a measure of rotational inertia of I = Ml / 4 the body, has been put to a great practical use. Table 6.1 simply gives the moment of inertia of The machines, such as steam engine and thevarious familiar regular shaped bodies about automobile engine, etc., that produce rotationalspecific axes. (The derivations of these motion have a disc with a large moment ofexpressions are beyond the scope of this inertia, called a flywheel. Because of its largetextbook and you will study them in higher classes.) moment of inertia, the flywheel resists the As the mass of a body resists a change in its sudden increase or decrease of the speed of the state of linear motion, it is a measure of its inertia vehicle. It allows a gradual change in the speed in linear motion. Similarly, as the moment of and prevents jerky motions, thereby ensuring inertia about a given axis of rotation resists a a smooth ride for the passengers on the vehicle. Reprint 2025-26 116 PHYSICS Table 6.1 Moments of inertia of some regular shaped bodies about specific axes Z Body Axis Figure I (1) Thin circular Perpendicular to M R 2 ring, radius R plane, at centre (2) Thin circular Diameter M R2/2 ring, radius R (3) Thin rod, Perpendicular to M L2/12 length L rod, at mid point (4) Circular disc, Perpendicular to M R2/2 radius R disc at centre (5) Circular disc, Diameter M R2/4 radius R (6) Hollow cylinder, Axis of cylinder M R2 radius R (7) Solid cylinder, Axis of cylinder M R2/2 radius R (8) Solid sphere, Diameter 2 M R2/5 radius R 6.10 KINEMATICS OF ROTATIONAL MOTION translation. We wish to take this analogy further. ABOUT A FIXED AXIS In doing so we shall restrict the discussion only We have already indicated the analogy between to rotation about fixed axis. This case of motion rotational motion and translational motion. For involves only one degree of freedom, i.e., needs example, the angular velocity ω plays the same only one independent variable to describe the role in rotation as the linear velocity v in motion. This in translation corresponds to linear Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 117 motion. This section is limited only to kinematics. We shall turn to dynamics in later sections. We recall that for specifying the angular displacement of the rotating body we take any particle like P (Fig.6.29) of the body. Its angular displacement θ in the plane it moves is the angular displacement of the whole body; θ is measured from a fixed direction in the plane of motion of P, which we take to be the x′-axis, chosen parallel to the x-axis. Note, as shown, the axis of rotation is the z – axis and the plane of the motion of the particle is the x - y plane. Fig. 6.29 also shows θ0, the angular displacement at t = 0. We also recall that the angular velocity is the time rate of change of angular displacement, ω = dθ/dt. Note since the axis of rotation is fixed, there is no need to treat angular velocity as a Fig.6.29 Specifying the angular position of a rigid vector. Further, the angular acceleration, α = body. dω/dt. u Example 6.10 Obtain Eq. (6.36) from first The kinematical quantities in rotational principles. motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) Answer The angular acceleration is uniform, respectively are analogous to kinematic hence quantities in linear motion, displacement (x), dω velocity (v) and acceleration (a). We know the = α = constant (i) kinematical equations of linear motion with d t uniform (i.e. constant) acceleration: Integrating this equation, α dt + c v = v0 + at (a) ω = ∫ 1 2 x = x 0 + υ0t + at (b) = αt + c (as α is constant) 2 At t = 0, ω = ω0 (given) 2 2 υ = υ0 + 2ax (c) From (i) we get at t = 0, ω = c = ω0 Thus, ω = αt + ω0 as required. where x0 = initial displacement and v0= initial With the definition of ω = dθ/dt we may velocity. The word ‘initial’ refers to values of the integrate Eq. (6.36) to get Eq. (6.37). This quantities at t = 0 derivation and the derivation of Eq. (6.38) is left The corresponding kinematic equations for as an exercise. rotational motion with uniform angular acceleration are: u Example 6.11 The angular speed of a motor wheel is increased from 1200 rpm to ω= ω0 + αt (6.36) 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the 1 2 θ = θ0 + ω0t + αt (6.37) acceleration to be uniform? (ii) How many 2 revolutions does the engine make during and ω2 = ω0 2 + 2α(θ– θ0 ) (6.38) this time? Answer where θ0= initial angular displacement of the (i) We shall use ω = ω0 + αt rotating body, and ω0 = initial angular velocity ω0 = initial angular speed in rad/s of the body. Reprint 2025-26 118 PHYSICS = 2π × angular speed in rev/s It is, however, necessary that these correspondences are established on sound 2π × angular speed in rev/min dynamical considerations. This is what we now = 60 s/min turn to. Before we begin, we note a simplification 2π × 1200 that arises in the case of rotational motion = rad/s 60 about a fixed axis. Since the axis is fixed, only those components of torques, which are along = 40π rad/s the direction of the fixed axis need to be Similarly ω = final angular speed in rad/s considered in our discussion. Only these 2π × 3120 components can cause the body to rotate about = rad/s the axis. A component of the torque 60 perpendicular to the axis of rotation will tend to = 2π × 52 rad/s turn the axis from its position. We specifically = 104 π rad/s assume that there will arise necessary forces of constraint to cancel the effect of the ∴Angular acceleration perpendicular components of the (external) torques, so that the fixed position of the axis ω − ω will be maintained. The perpendicular α = 0 = 4 π rad/s2 t components of the torques, therefore need not be taken into account. This means that for our The angular acceleration of the engine calculation of torques on a rigid body: = 4π rad/s2 (1) We need to consider only those forces that (ii) The angular displacement in time t is lie in planes perpendicular to the axis. given by Forces which are parallel to the axis will give torques perpendicular to the axis and need 1 2 θ = ω0 t + αt not be taken into account. 2 (2) We need to consider only those components 1 2 of the position vectors which are = (40π × 16 + × 4π × 16 ) rad 2 perpendicular to the axis. Components of position vectors along the axis will result in = (640π + 512π) rad torques perpendicular to the axis and need = 1152π rad not be taken into account. 1152π = 576 ⊳ Work done by a torqueNumber of revolutions = 2π

5.10 POWER Often it is interesting to know not only the work u Example 5.10 An elevator can carry a done on an object, but also the rate at which maximum load of 1800 kg (elevator + this work is done. We say a person is physically passengers) is moving up with a constant fit if he not only climbs four floors of a building speed of 2 m s–1. The frictional force opposing but climbs them fast. Power is defined as the the motion is 4000 N. Determine the time rate at which work is done or energy is minimum power delivered by the motor to transferred. the elevator in watts as well as in horse The average power of a force is defined as the power. ratio of the work, W, to the total time t taken Answer The downward force on the elevator is W Pav = F = m g + = (1800 × 10) + 4000 = 22000 N t Ff The motor must supply enough power to balanceThe instantaneous power is defined as the this force. Hence,limiting value of the average power as time interval approaches zero, P = F. v = 22000 × 2 = 44000 W = 59 hp ⊳ d W 5.11 COLLISIONS P = (5.20) d t In physics we study motion (change in position). The work dW done by a force F for a displacement At the same time, we try to discover physical dr is dW = F.dr. The instantaneous power can quantities, which do not change in a physical also be expressed as process. The laws of momentum and energy conservation are typical examples. In this d r P = F. section we shall apply these laws to a commonly d t encountered phenomena, namely collisions. Several games such as billiards, marbles or = F.v (5.21) carrom involve collisions.We shall study the where v is the instantaneous velocity when the collision of two masses in an idealised form. force is F. Consider two masses m1 and m2. The particle Power, like work and energy, is a scalar m1 is moving with speed v1i , the subscript ‘i’ quantity. Its dimensions are [ML2T–3]. In the SI, implying initial. We can cosider m2 to be at rest. its unit is called a watt (W). The watt is 1 J s–1. No loss of generality is involved in making such The unit of power is named after James Watt, a selection. In this situation the mass m1 one of the innovators of the steam engine in the collides with the stationary mass m2 and this eighteenth century. is depicted in Fig. 5.10. There is another unit of power, namely the horse-power (hp) 1 hp = 746 W This unit is still used to describe the output of automobiles, motorbikes, etc. We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy. 100 (watt) × 10 (hour) Fig. 5.10 Collision of mass m1, with a stationary mass m2. = 1000 watt hour The masses m1 and m2 fly-off in different =1 kilowatt hour (kWh) directions. We shall see that there are = 103 (W) × 3600 (s) relationships, which connect the masses, the = 3.6 × 106 J velocities and the angles. Reprint 2025-26 84 PHYSICS 5.11.1 Elastic and Inelastic Collisions The loss in kinetic energy on collision is In all collisions the total linear momentum is 1 2 1 2 conserved; the initial momentum of the system ∆ K = m 1 v1i − (m 1 + m 2 )v f 2 2 is equal to the final momentum of the system. One can argue this as follows. When two objects 2 1 2 1 m 1 2 collide, the mutual impulsive forces acting over = m 1v1i − v1i [using Eq. (5.22)] the collision time ∆t cause a change in their 2 2 m 1 + m 2 respective momenta : 1 2  m 1  ∆p1 = F12 ∆t = 2 m 1v1i 1 − m 1 + m 2  ∆p2 = F21 ∆t where F12 is the force exerted on the first particle 1 m 1m 2 2by the second particle. F21 is likewise the force = v1i 2 m 1 + m 2exerted on the second particle by the first particle. Now from Newton’s third law, F12 = − F21. This which is a positive quantity as expected.implies ∆p1 + ∆p2 = 0 Consider next an elastic collision. Using the above nomenclature with θ1 = θ2 = 0, the The above conclusion is true even though the momentum and kinetic energy conservation forces vary in a complex fashion during the equations are collision time ∆t. Since the third law is true at every instant, the total impulse on the first object m1v1i = m1v1f + m2v2f (5.23) is equal and opposite to that on the second. 2 2 2 m 1v1i = m 1v1 f + m 2 v 2 f (5.24) On the other hand, the total kinetic energy of the system is not necessarily conserved. The From Eqs. (5.23) and (5.24) it follows that, impact and deformation during collision may generate heat and sound. Part of the initial kinetic m 1v1i (v 2 f − v1i ) = m 1v1 f (v 2 f − v1 f ) energy is transformed into other forms of energy. A useful way to visualise the deformation during or, v 2 f (v1i − v1 f ) = v12i − v12f collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains = (v1i − v1 f )(v 1i + v 1 f ) its original shape without loss in energy, then the initial kinetic energy is equal to the final Hence, ∴ v 2 f = v1i + v1 f (5.25) kinetic energy but the kinetic energy during the Substituting this in Eq. (5.23), we obtaincollision time ∆t is not constant. Such a collision is called an elastic collision. On the other hand (m 1 − m 2 ) v1 f = v1i (5.26)the deformation may not be relieved and the two m 1 + m 2 bodies could move together after the collision. A 2m 1v1icollision in which the two particles move together and v 2 f = (5.27) m 1 + m 2after the collision is called a completely inelastic collision. The intermediate case where the Thus, the ‘unknowns’ {v1f, v2f} are obtained in deformation is partly relieved and some of the terms of the ‘knowns’ {m1, m2, v1i}. Special cases initial kinetic energy is lost is more common and of our analysis are interesting. is appropriately called an inelastic collision. Case I : If the two masses are equal 5.11.2 Collisions in One Dimension v1f = 0 Consider first a completely inelastic collision v2f = v1i in one dimension. Then, in Fig. 5.10, The first mass comes to rest and pushes off the θ 1 = θ 2 = 0 second mass with its initial speed on collision. Case II : If one mass dominates, e.g. m2 > > m1 m1v1i = (m1+m2)vf (momentum conservation) v1f ~ − v1i v2f ~ 0 m 1 The heavier mass is undisturbed while the v f = v1i (5.22) m 1 + m 2 lighter mass reverses its velocity. Reprint 2025-26 WORK, ENERGY AND POWER 85 ⊳ dimensional, where the initial velocities and the Example 5.11 Slowing down of neutrons: final velocities lie in a plane. In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed 5.11.3 Collisions in Two Dimensions to 103 m s–1 so that it can have a high Fig. 5.10 also depicts the collision of a moving probability of interacting with isotope 235U92 mass m1 with the stationary mass m2. Linear momentum is conserved in such a collision. and causing it to fission. Show that a neutron can lose most of its kinetic energy Since momentum is a vector this implies three in an elastic collision with a light nuclei equations for the three directions {x, y, z}. like deuterium or carbon which has a mass Consider the plane determined by the final of only a few times the neutron mass. The velocity directions of m1 and m2 and choose it to material making up the light nuclei, usually be the x-y plane. The conservation of the heavy water (D2O) or graphite, is called a z-component of the linear momentum implies moderator. that the entire collision is in the x-y plane. The x- and y-component equations are Answer The initial kinetic energy of the neutron m1v1i = m1v1f cos θ1 + m2v2f cos θ2 (5.28)is 0 = m1v1f sin θ1 − m2v2f sin θ2 (5.29) 1 2 K 1i = m 1v1i 2 One knows {m1, m2, v1i} in most situations. There are thus four unknowns {v1f, v2f , θ1 and θ2}, andwhile its final kinetic energy from Eq. (5.26) only two equations. If θ 1 = θ 2 = 0, we regain 1 2 1  m 1 − m 2  2 2 Eq. (5.23) for one dimensional collision. K 1 f = m 1v1 f = m 1 v1i 2 2  m 1 + m 2  If, further the collision is elastic, 1 2 1 2 1 2 m1v1i = m1v1 f + m 2v 2 f (5.30) The fractional kinetic energy lost is 2 2 2 2 We obtain an additional equation. That still K 1 f  m 1 − m 2  f 1 = = leaves us one equation short. At least one of K 1i  m 1 + m 2  the four unknowns, say θ1, must be made known while the fractional kinetic energy gained by the for the problem to be solvable. For example, θ1 moderating nuclei K2f /K1i is can be determined by moving a detector in an angular fashion from the x to the y axis. Given f2 = 1 − f1 (elastic collision) {m1, m2, v1i , θ1} we can determine {v1f , v2f, θ2} 4m 1m 2 from Eqs. (5.28)-(5.30). = 2 ⊳ (m1 + m 2 ) Example 5.12 Consider the collision depicted in Fig. 5.10 to be between two One can also verify this result by substituting billiard balls with equal masses m1 = m2.from Eq. (5.27). The first ball is called the cue while the For deuterium m2 = 2m1 and we obtain second ball is called the target. The f1 = 1/9 while f2 = 8/9. Almost 90% of the billiard player wants to ‘sink’ the target neutron’s energy is transferred to deuterium. For ball in a corner pocket, which is at an carbon f1 = 71.6% and f2 = 28.4%. In practice, angle θ2 = 37°. Assume that the collision ishowever, this number is smaller since head-on elastic and that friction and rotational collisions are rare. ⊳ motion are not important. Obtain θ1. If the initial velocities and final velocities of Answer From momentum conservation, since both the bodies are along the same straight line, the masses are equal then it is called a one-dimensional collision, or head-on collision. In the case of small spherical v1i = v 1f + v 2f bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 ⋅ or v 1i 2 = ( v1 f + v 2 f ) ( v1 f + v 2 f ) which is at rest. In general, the collision is two- = v1 f 2 + v 2 f 2 + 2 v1 f .v 2 f Reprint 2025-26 86 PHYSICS 2 2 The matter simplifies greatly if we consider= (5.31) { v1 f + v 2 f + 2v1 f v 2 f cos (θ1 + 37 ° ) } spherical masses with smooth surfaces, and assume that collision takes place only when the Since the collision is elastic and m1 = m2 it follows bodies touch each other. This is what happensfrom conservation of kinetic energy that in the games of marbles, carrom and billiards. v1i 2 = v1 f 2 + v 2 f 2 (5.32) In our everyday world, collisions take place only when two bodies touch each other. But considerComparing Eqs. (5.31) and (5.32), we get a comet coming from far distances to the sun, or cos (θ1 + 37°) = 0 alpha particle coming towards a nucleus and or θ1 + 37° = 90° going away in some direction. Here we have to deal with forces involving action at a distance. Thus, θ1 = 53° Such an event is called scattering. The velocities This proves the following result : when two equal and directions in which the two particles go away masses undergo a glancing elastic collision with depend on their initial velocities as well as the one of them at rest, after the collision, they will type of interaction between them, their masses, move at right angles to each other. ⊳ shapes and sizes. SUMMARY 1. The work-energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body. Kf - Ki = Wnet 2. A force is conservative if (i) work done by it on an object is path independent and depends only on the end points {xi, xj}, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position. 3. For a conservative force in one dimension, we may define a potential energy function V(x) such that d V ( x ) F ( x ) = − d x x f F ( x ) d x or Vi − V f = ∫ x i 4. The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative. 5. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is V(x) = m g x where the variation of g with height is ignored. 5. The elastic potential energy of a spring of force constant k and extension x is 1 2 V ( x ) = k x 2 7. The scalar or dot product of two vectors A and B is written as A.B and is a scalar quantity given by :A.B = AB cos θ, where θ is the angle between A and B. It can be positive, negative or zero depending upon the value of θ. The scalar product of two vectors can be interpreted as the product of magnitude of one vector and component of the other vector along the first vector. For unit vectors : ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1 and ˆi ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ ˆi = 0 Scalar products obey the commutative and the distributive laws. Reprint 2025-26 WORK, ENERGY AND POWER 87 POINTS TO PONDER 1. The phrase ‘calculate the work done’ is incomplete. We should refer (or imply clearly by context) to the work done by a specific force or a group of forces on a given body over a certain displacement. 2. Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities. The work done by the friction or viscous force on a moving body is negative. 3. For two bodies, the sum of the mutual forces exerted between them is zero from Newton’s Third Law, F12 + F21 = 0 But the sum of the work done by the two forces need not always cancel, i.e. W12 + W21 ≠ 0 However, it may sometimes be true. 4. The work done by a force can be calculated sometimes even if the exact nature of the force is not known. This is clear from Example 5.2 where the WE theorem is used in such a situation. 5. The WE theorem is not independent of Newton’s Second Law. The WE theorem may be viewed as a scalar form of the Second Law. The principle of conservation of mechanical energy may be viewed as a consequence of the WE theorem for conservative forces. 5. The WE theorem holds in all inertial frames. It can also be extended to non- inertial frames provided we include the pseudoforces in the calculation of the net force acting on the body under consideration. 7. The potential energy of a body subjected to a conservative force is always undetermined upto a constant. For example, the point where the potential energy is zero is a matter of choice. For the gravitational potential energy mgh, the zero of the potential energy is chosen to be the ground. For the spring potential energy kx2/2 , the zero of the potential energy is the equilibrium position of the oscillating mass. 8. Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction. 9. During a collision : (a) the total linear momentum is conserved at each instant of the collision ; (b) the kinetic energy conservation (even if the collision is elastic) applies after the collision is over and does not hold at every instant of the collision. In fact the two colliding objects are deformed and may be momentarily at rest with respect to each other. Reprint 2025-26 88 PHYSICS EXERCISES 5.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 5.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results. 5.3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. Fig. 5.11 Reprint 2025-26 WORK, ENERGY AND POWER 89 5.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this Fig. 5.12 potential must ‘turn back’ when it reaches x = ± 2 m. 5.5 Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not Fig. 5.13 normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ? 5.6 Underline the correct alternative : (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. 5.7 State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. 5.8 Answer carefully, with reasons : (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ? Reprint 2025-26 90 PHYSICS (c) What are the answers to (a) and (b) for an inelastic collision ? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy). 5.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = −ˆi + 2 ˆj + 3 kˆ N where ˆ,i ˆ,j kˆ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ? 5.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J). 5.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ? 5.14 A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ? 5.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ? 5.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ? Fig. 5.14 Reprint 2025-26 WORK, ENERGY AND POWER 91 5.17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. 5.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ? 5.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a Fig. 5.15 frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ? 5.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m ? 5.21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ? 5.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? 5.23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house. Reprint 2025-26 CHAPTER SIX SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.1 INTRODUCTION In the earlier chapters we primarily considered the motion of a single particle. (A particle is ideally represented as a 6.1 Introduction point mass having no size.) We applied the results of our 6.2 Centre of mass study even to the motion of bodies of finite size, assuming 6.3 Motion of centre of mass that motion of such bodies can be described in terms of the

13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u