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

6.7 Relationship between to such a small degree that only a very Equilibrium Constant K, minute quantity of product is formed. Reaction Quotient Q and Gibbs Energy G Problem 6.10 The value of Kc for a reaction does not depend The value of ∆G  for the phosphorylation of on the rate of the reaction. However, as you glucose in glycolysis is 13.8 kJ/mol. Find have studied in Unit 5, it is directly related the value of Kc at 298 K. to the thermodynamics of the reaction and Solutionin particular, to the change in Gibbs energy, ∆G. If, ∆G  = 13.8 kJ/mol = 13.8 × 103J/mol • ∆G is negative, then the reaction is Also, ∆G  = – RT lnKc spontaneous and proceeds in the forward Hence, ln Kc = –13.8 × 103J/mol direction. (8.314 J mol–1K–1 × 298 K) • ∆G is positive, then reaction is considered ln Kc = – 5.569 non-spontaneous. Instead, as reverse reaction would have a negative ∆G, the Kc = e–5.569 products of the forward reaction shall be Kc = 3.81 × 10–3 converted to the reactants. Problem 6.11• ∆G is 0, reaction has achieved equilibrium; Hydrolysis of sucrose gives, at this point, there is no longer any free energy left to drive the reaction. Sucrose + H2O Glucose + Fructose A mathematical expression of this Equilibrium constant Kc for the reaction is thermodynamic view of equilibrium can be 2 ×1013 at 300K. Calculate ∆G  at 300K. described by the following equation: Solution ∆G = ∆G + RT lnQ (6.21) ∆G  = – RT lnKcwhere, G is standard Gibbs energy. ∆G  = – 8.314J mol–1K–1× At equilibrium, when ∆G = 0 and Q = Kc, 300K × ln(2×1013) the equation (6.21) becomes, ∆G  = – 7.64 ×104 J mol–1 ∆G = ∆G + RT ln K = 0 6.8 FACTORS AFFECTING EQUILIBRIA ∆G = – RT lnK (6.22) One of the principal goals of chemical lnK = – ∆G / RT synthesis is to maximise the conversion of the Reprint 2025-26 EQUILIBRIUM 185 reactants to products while minimising the “When the concentration of any of the expenditure of energy. This implies maximum reactants or products in a reaction at yield of products at mild temperature and equilibrium is changed, the composition pressure conditions. If it does not happen, of the equilibrium mixture changes so as then the experimental conditions need to be to minimize the effect of concentration adjusted. For example, in the Haber process changes”. for the synthesis of ammonia from N2 and Let us take the reaction, H2, the choice of experimental conditions is of real economic importance. Annual world H2(g) + I2(g) 2HI(g) production of ammonia is about hundred If H2 is added to the reaction mixture million tones, primarily for use as fertilisers. at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, Equilibrium constant, Kc is independent the reaction proceeds in a direction whereinof initial concentrations. But if a system at equilibrium is subjected to a change in the H2 is consumed, i.e., more of H2 and I2 react to form HI and finally the equilibrium shiftsconcentration of one or more of the reacting in right (forward) direction (Fig.6.8). This is insubstances, then the system is no longer at accordance with the Le Chatelier’s principleequilibrium; and net reaction takes place in which implies that in case of addition of asome direction until the system returns to reactant/product, a new equilibrium willequilibrium once again. Similarly, a change be set up in which the concentration of thein temperature or pressure of the system may reactant/product should be less than what italso alter the equilibrium. In order to decide was after the addition but more than what itwhat course the reaction adopts and make was in the original mixture.a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to all physical and chemical equilibria. We shall now be discussing factors which can influence the equilibrium. 6.8.1 Effect of Concentration Change In general, when equilibrium is disturbed by the addition/removal of any reactant/ products, Le Chatelier’s principle predicts that: • The concentration stress of an added reactant/product is relieved by net Fig. 6.8 Effect of addition of H2 on change reaction in the direction that consumes of concentration for the reactants the added substance. and products in the reaction, • The concentration stress of a removed H2(g) + I2 (g) 2HI(g) reactant/product is relieved by net reaction in the direction that replenishes The same point can be explained in terms the removed substance. of the reaction quotient, Qc, or in other words, Qc = [HI]2/ [H2][I2] Reprint 2025-26 186 chemistry Addition of hydrogen at equilibrium concentration of [Fe(SCN)]2+ decreases, the results in value of Qc being less than Kc . Thus, intensity of red colour decreases. in order to attain equilibrium again reaction Addition of aq. HgCl2 also decreases redmoves in the forward direction. Similarly, colour because Hg2+ reacts with SCN– ions to we can say that removal of a product also form stable complex ion [Hg(SCN)4]2–. Removalboosts the forward reaction and increases of free SCN– (aq) shifts the equilibrium the concentration of the products and this in equation (6.24) from right to left to has great commercial application in cases replenish SCN– ions. Addition of potassium of reactions, where the product is a gas or a thiocyanate on the other hand increases the volatile substance. In case of manufacture of colour intensity of the solution as it shift the ammonia, ammonia is liquified and removed equilibrium to right. from the reaction mixture so that reaction keeps moving in forward direction. Similarly, 6.8.2 Effect of Pressure Change in the large scale production of CaO (used A pressure change obtained by changing the as important building material) from CaCO3, volume can affect the yield of products in constant removal of CO2 from the kiln drives case of a gaseous reaction where the total the reaction to completion. It should be number of moles of gaseous reactants and remembered that continuous removal of a total number of moles of gaseous products are product maintains Qc at a value less than Kc different. In applying Le Chatelier’s principle and reaction continues to move in the forward to a heterogeneous equilibrium the effect direction. of pressure changes on solids and liquids can be ignored because the volume (and Effect of Concentration – An experiment concentration) of a solution/liquid is nearly This can be demonstrated by the following independent of pressure. reaction: Consider the reaction, Fe3+(aq)+ SCN–(aq) [Fe(SCN)]2+(aq) (6.24) CO(g) + 3H2(g) CH4(g) + H2O(g)yellow colourless deep red Here, 4 mol of gaseous reactants (CO + 3H2) become 2 mol of gaseous products (CH4 + H2O). Suppose equilibrium mixture (for above (6.25) reaction) kept in a cylinder fitted with a piston at constant temperature is compressed to A reddish colour appears on adding two one half of its original volume. Then, totaldrops of 0.002 M potassium thiocynate solution pressure will be doubled (according to to 1 mL of 0.2 M iron(III) nitrate solution due pV = constant). The partial pressure and to the formation of [Fe(SCN)]2+. The intensity therefore, concentration of reactants and of the red colour becomes constant on products have changed and the mixture is no attaining equilibrium. This equilibrium can be longer at equilibrium. The direction in which shifted in either forward or reverse directions the reaction goes to re-establish equilibrium depending on our choice of adding a reactant can be predicted by applying the Le Chatelier’s or a product. The equilibrium can be shifted principle. Since pressure has doubled, in the opposite direction by adding reagents the equilibrium now shifts in the forward that remove Fe3+ or SCN– ions. For example, direction, a direction in which the number oxalic acid (H2C2O4), reacts with Fe3+ ions of moles of the gas or pressure decreases (we to form the stable complex ion [Fe(C2O4)3]3–, know pressure is proportional to moles of the thus decreasing the concentration of free gas). This can also be understood by using Fe3+(aq). In accordance with the Le Chatelier’s reaction quotient, Qc. Let [CO], [H2], [CH4] principle, the concentration stress of removed and [H2O] be the molar concentrations at Fe3+ is relieved by dissociation of [Fe(SCN)]2+ equilibrium for methanation reaction. When to replenish the Fe3+ ions. Because the volume of the reaction mixture is halved, the Reprint 2025-26 EQUILIBRIUM 187 partial pressure and the concentration are Production of ammonia according to the doubled. We obtain the reaction quotient by reaction, replacing each equilibrium concentration by N2(g) + 3H2(g) 2NH3(g);double its value. ∆H= – 92.38 kJ mol–1  CH 4 ( g )  H 2 O ( g ) is an exothermic process. According to Qc = 3  CO ( g )  H 2 ( g ) Le Chatelier’s principle, raising the temperature shifts the equilibrium to left As Qc < Kc , the reaction proceeds in the and decreases the equilibrium concentration forward direction. of ammonia. In other words, low temperature is favourable for high yield of ammonia, but In reaction C(s) + CO2(g) 2CO(g), when practically very low temperatures slow downpressure is increased, the reaction goes in the the reaction and thus a catalyst is used.reverse direction because the number of moles of gas increases in the forward direction. Effect of Temperature – An experiment Effect of temperature on equilibrium can6.8.3 Effect of Inert Gas Addition be demonstrated by taking NO2 gas (brown If the volume is kept constant and an inert gas in colour) which dimerises into N2O4 gas such as argon is added which does not take (colourless). part in the reaction, the equilibrium remains 2NO2(g) N2O4(g); ∆H = –57.2 kJ mol–1undisturbed. It is because the addition of an inert gas at constant volume does not NO2 gas prepared by addition of Cu change the partial pressures or the molar turnings to conc. HNO3 is collected in two 5 mL test tubes (ensuring same intensityconcentrations of the substance involved in of colour of gas in each tube) and stopperthe reaction. The reaction quotient changes sealed with araldite. Three 250 mL beakersonly if the added gas is a reactant or product 1, 2 and 3 containing freesing mixture, waterinvolved in the reaction. at room temperature and hot water (363K), 6.8.4 Effect of Temperature Change respectively, are taken (Fig. 6.9). Both the test tubes are placed in beaker 2 for 8-10 minutes.Whenever an equilibrium is disturbed by After this one is placed in beaker 1 and thea change in the concentration, pressure or other in beaker 3. The effect of temperaturevolume, the composition of the equilibrium on direction of reaction is depicted very wellmixture changes because the reaction in this experiment. At low temperatures inquotient, Qc no longer equals the equilibrium beaker 1, the forward reaction of formation ofconstant, Kc. However, when a change in temperature occurs, the value of equilibrium N2O4 is preferred, as reaction is exothermic, and thus, intensity of brown colour dueconstant, Kc is changed. to NO2 decreases. While in beaker 3, high In general, the temperature dependence temperature favours the reverse reaction of of the equilibrium constant depends on the sign of ∆H for the reaction. • The equilibrium constant for an exothermic reaction (negative ∆H) decreases as the temperature increases. • The equilibrium constant for an endothermic reaction (positive ∆H) increases as the temperature increases. Temperature changes affect the Fig. 6.9 Effect of temperature on equilibrium for the reaction, 2NO2 (g) N2O4 (g)equilibrium constant and rates of reactions. Reprint 2025-26 188 chemistry formation of NO2 and thus, the brown colour Similarly, in manufacture of sulphuric intensifies. acid by contact process, Effect of temperature can also be seen in 2SO2(g) + O2(g) 2SO3(g); Kc = 1.7 × 1026 an endothermic reaction, though the value of K is suggestive of reaction [Co(H2O)6]3+(aq) + 4Cl–(aq) [CoCl4]2–(aq) + going to completion, but practically the 6H2O(l) oxidation of SO2 to SO3 is very slow. Thus, pink colourless blue platinum or divanadium penta-oxide (V2O5) is used as catalyst to increase the rate of the At room temperature, the equilibrium reaction.mixture is blue due to [CoCl4]2–. When cooled Note: If a reaction has an exceedingly smallin a freesing mixture, the colour of the mixture K, a catalyst would be of little help.turns pink due to [Co(H2O)6]3+. 6.9 IONIC EQUILIBRIUM IN SOLUTION6.8.5 Effect of a Catalyst Under the effect of change of concentrationA catalyst increases the rate of the chemical on the direction of equilibrium, you havereaction by making available a new low energy pathway for the conversion of reactants to incidently come across with the following products. It increases the rate of forward equilibrium which involves ions: and reverse reactions that pass through the Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq) same transition state and does not affect There are numerous equilibria that involve equilibrium. Catalyst lowers the activation ions only. In the following sections we will energy for the forward and reverse reactions study the equilibria involving ions. It is well by exactly the same amount. Catalyst does known that the aqueous solution of sugar not affect the equilibrium composition of does not conduct electricity. However, when a reaction mixture. It does not appear in common salt (sodium chloride) is added the balanced chemical equation or in the to water it conducts electricity. Also, the equilibrium constant expression. conductance of electricity increases with an Let us consider the formation of NH3 increase in concentration of common salt. from dinitrogen and dihydrogen which is Michael Faraday classified the substances highly exothermic reaction and proceeds into two categories based on their ability with decrease in total number of moles to conduct electricity. One category of formed as compared to the reactants. substances conduct electricity in their Equilibrium constant decreases with increase aqueous solutions and are called electrolytes in temperature. At low temperature rate while the other do not and are thus, referred to decreases and it takes long time to reach at as non-electrolytes. Faraday further classified equilibrium, whereas high temperatures give electrolytes into strong and weak electrolytes. satisfactory rates but poor yields. Strong electrolytes on dissolution in water German chemist, Fritz Haber discovered are ionized almost completely, while the weak that a catalyst consisting of iron catalyse electrolytes are only partially dissociated. the reaction to occur at a satisfactory rate For example, an aqueous solution of at temperatures, where the equilibrium sodium chloride is comprised entirely of concentration of NH3 is reasonably favourable. sodium ions and chloride ions, while that Since the number of moles formed in the of acetic acid mainly contains unionized reaction is less than those of reactants, the acetic acid molecules and only some acetate yield of NH3 can be improved by increasing ions and hydronium ions. This is because the pressure. there is almost 100% ionization in case Optimum conditions of temperature of sodium chloride as compared to less and pressure for the synthesis of NH3 using than 5% ionization of acetic acid which is catalyst are around 500°C and 200 atm. a weak electrolyte. It should be noted Reprint 2025-26 EQUILIBRIUM 189 that in weak electrolytes, equilibrium is exists in solid state as a cluster of positively established between ions and the unionized charged sodium ions and negatively charged molecules. This type of equilibrium involving chloride ions which are held together due to ions in aqueous solution is called ionic electrostatic interactions between oppositely equilibrium. Acids, bases and salts come charged species (Fig.6.10). The electrostatic under the category of electrolytes and may act forces between two charges are inversely as either strong or weak electrolytes. proportional to dielectric constant of the medium. Water, a universal solvent, possesses

9.20 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. f1 = 30 cm and u1 = – , give v1 = + 30 cm. This image becomes a virtual object for the second lens. f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives, v2 = – 220 cm. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm. This image becomes a real object for the second lens: f2 = + 30 cm, u2 = – (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f2, and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system. (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); 112 × 20 f2 = – 20 cm which gives v2 = − cm 92 Magnitude of magnification due to the second (concave) 347 Reprint 2025-26 Physics lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm 9.21 If the refracted ray in the prism is incident on the second face at the critical angle ic, the angle of refraction r at the first face is (60°–ic). Now, ic = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° 1 1 1 9.22 (a) + = v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. Each square in the virtual image has an area 10 × 10 × 1 mm2 = 100 mm2 = 1 cm2 (b) Magnifying power = 25/9 = 2.8 (c) No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is |(v/u)| and magnifying power is (25/ |u|). Only when the image is located at the near point |v| = 25 cm, are the two quantities equal. 9.23 (a) Maximum magnifying power is obtained when the image is at the near point (25 cm) u = – 7.14 cm. (b) Magnitude of magnification = (25/ |u|) = 3.5. (c) Magnifying power = 3.5 Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. 9.24 Magnification = ( 6.25 / 1) = 2.5 v = +2.5u 1 1 1    2.5u u 10 i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly. 9.25 (a) Even though the absolute image size is bigger than the object size, the angular size of the image is equal to the angular size of the object. The magnifier helps in the following way: without it object would be placed no closer than 25 cm; with it the object can be placed much closer. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved. (b) Yes, it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The Reprint 2025-26 Answers effect is negligible if the image is at a very large distance away. [Note: When the eye is separated from the lens, the angles subtended at the eye by the first object and its image are not equal.] (c) First, grinding lens of very small focal length is not easy. More important, if you decrease focal length, aberrations (both spherical and chromatic) become more pronounced. So, in practice, you cannot get a magnifying power of more than 3 or so with a simple convex lens. However, using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which increases if fe is smaller. Further, magnification of the objective v O 1 = is given by | u O | (| u O |/ f O ) − 1 which is large when |u O | is slightly greater than fO. The micro- scope is used for viewing very close object. So |u O | is small, and so is fO. (e) The image of the objective in the eye-piece is known as ‘eye-ring’. All the rays from the object refracted by objective go through the eye-ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. If we position our eyes on the eye-ring and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece. When you view through a microscope by placing your eyes on one end,the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument.