3.6 VECTOR ADDITION – ANALYTICAL then, a vector T = a + b – c has components : METHOD T x = a x + b x − c x Although the graphical method of adding vectors Ty = a y + b y − c y (3.23b) helps us in visualising the vectors and the T z = a z + b z − c z .resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors ⊳ Example 3.2 Find the magnitude andby combining their respective components. direction of the resultant of two vectors AConsider two vectors A and B in x-y plane with and B in terms of their magnitudes and components Ax, Ay and Bx, By : angle θ between them. ɵ ɵ A = A x i + A y j (3.18) * Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. Reprint 2025-26 34 PHYSICS ⊳ Example 3.3 A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat. Answer The vector vb representing the velocity of the motorboat and the vector vc representing Fig. 3.10 the water current are shown in Fig. 3.11 in Answer Let OP and OQ represent the two vectors directions specified by the problem. Using the A and B making an angle θ (Fig. 3.10). Then, parallelogram method of addition, the resultant using the parallelogram method of vector R is obtained in the direction shown in the addition, OS represents the resultant vector R : figure. R = A + B SN is normal to OP and PM is normal to OS. From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + B cos θ SN = B sin θ OS2 = (A + B cos θ)2 + (B sin θ)2 or, R2 = A2 + B2 + 2AB cos θ R = A 2 + B 2 + 2AB cosθ (3.24a) In ∆ OSN, SN = OS sinα = R sinα, and in ∆ PSN, SN = PS sin θ = B sin θ Therefore, R sin α = B sin θ R B or, = (3.24b) sin θ sin α Fig. 3.11 Similarly, PM = A sin α = B sin β We can obtain the magnitude of R using the Law A B or, = (3.24c) of cosine : sin β sin α Combining Eqs. (3.24b) and (3.24c), we get R = v 2b + v c2 + 2v bv c cos120 o R A B = = (3.24d) = 25 2 + 10 2 + 2 × 25 × 10 ( -1/2 ) ≅ 22 km/h sin θ sin β sin α To obtain the direction, we apply the Law of sines Using Eq. (3.24d), we get: R vc v c B = sin θ or, sin φ = sin α = sin θ (3.24e) sin θ sin φ R R where R is given by Eq. (3.24a). 10 × sin120 10 3 = = ≅ 0.397 SN B sin θ 21.8 2 × 21.8or, tan α= = (3.24f) OP + PN A + B cos θ φ ≅ 23.4 ⊳ Equation (3.24a) gives the magnitude of the 3.7 MOTION IN A PLANEresultant and Eqs. (3.24e) and (3.24f) its direction. Equation (3.24a) is known as the Law of cosines In this section we shall see how to describe and Eq. (3.24d) as the Law of sines. ⊳ motion in two dimensions using vectors. Reprint 2025-26 MOTION IN A PLANE 35 3.7.1 Position Vector and Displacement Suppose a particle moves along the curve shown The position vector r of a particle P located in a by the thick line and is at P at time t and P′ at plane with reference to the origin of an x-y time t′ [Fig. 3.12(b)]. Then, the displacement is : reference frame (Fig. 3.12) is given by ∆r = r′ – r (3.25) ɵ ɵ and is directed from P to P′. r = x i + y j We can write Eq. (3.25) in a component form: where x and y are components of r along x-, and y- axes or simply they are the coordinates of ɵ ɵ ɵ ɵ ∆r = x' i + y' j − x i + y j ( ) ( )the object. ɵ ɵ = i ∆ x + j ∆ y where ∆x = x ′ – x, ∆y = y′ – y (3.26) Velocity v The average velocity ( ) of an object is the ratio of the displacement and the corresponding time interval : ɵ ɵ ∆ r ∆ x i + ∆y j ɵ ∆ x ɵ ∆ y v = = = i + j (3.27) ∆ t ∆ t ∆ t ∆ t Or, v = v x ˆi + v y j (a) ∆ r Since v = , the direction of the average velocity ∆t is the same as that of ∆r (Fig. 3.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : ∆ r d r v = lim = (3.28) ∆ t → 0 ∆ t d t The meaning of the limiting process can be easily understood with the help of Fig 3.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P1, P2 and (b) P3 represent the positions of the object after Fig. 3.12 (a) Position vector r. (b) Displacement ∆r and times ∆t1,∆t2, and ∆t3. ∆r1, ∆r2, and ∆r3 are the average velocity v of a particle. displacements of the object in times ∆t1, ∆t2, and Fig. 3.13 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. Reprint 2025-26 36 PHYSICS ∆t3, respectively. The direction of the average velocity v is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t1,∆t2, and ∆t3, (∆t1 > ∆t2 > ∆t3). As ∆t → 0, ∆r → 0 and is along the tangent to the path [Fig. 3.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion. We can express v in a component form : Fig. 3.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that dr vx = v cos θ, vy = v sin θ. v = dt The acceleration (instantaneous acceleration)  ∆x ɵ ∆y ɵ  is the limiting value of the average acceleration = lim  i + j  (3.29) ∆t → 0 ∆t ∆t  as the time interval approaches zero : ∆ v ∆x ∆y ɵ ɵ a = lim (3. 32a) + = i lim j lim ∆t → 0 ∆t ∆t → 0 ∆t ∆ t → 0 ∆ t ɵ ɵ ɵ dx ɵ dy ɵ ɵ Since ∆ v = ∆ v x i + ∆ v y j, we have + jOr, v = i = v x i + v y j . dt dt ɵ ∆ v x ɵ ∆ v y dx dy a = i lim + j lim ∆ t → 0 ∆ t ∆ t → 0 ∆ twhere v x = dt , v y = dt (3.30a) ɵ ɵ So, if the expressions for the coordinates x and Or, a = a x i + a y j (3.32b) y are known as functions of time, we can use d v y d v xthese equations to find vx and vy. (3.32c)* where, a x = , a y = The magnitude of v is then d t d t 2 2 As in the case of velocity, we can understand v = v x + v y (3.30b) graphically the limiting process used in defining and the direction of v is given by the angle θ : acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 3.15(a) to  v y  (d). P represents the position of the object at v y −1 tanθ = , θ = tan     (3.30c) time t and P1, P2, P3 positions after time ∆t1, ∆t2,  v x  v x ∆t3, respectively (∆t 1> ∆t2>∆t3). The velocity vectors vx, vy and angle θ are shown in Fig. 3.14 for a at points P, P1, P2, P3 are also shown in Figs. 3.15 velocity vector v at point p. (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. ByAcceleration definition, the direction of average acceleration The average acceleration a of an object for a is the same as that of ∆v. We see that as ∆t time interval ∆t moving in x-y plane is the change decreases, the direction of ∆v changes and in velocity divided by the time interval : consequently, the direction of the acceleration ɵ ɵ changes. Finally, in the limit ∆t g0 [Fig. 3.15(d)], j v x i + v y ∆ v ∆ ( ∆ v y ɵ ) ∆ v x ɵ a = = = i + j (3.31a) the average acceleration becomes the ∆ t ∆ t ∆ t ∆ t instantaneous acceleration and has the direction ɵ ɵ as shown. Or, a = a x i + a y j . (3.31b) * In terms of x and y, ax and ay can be expressed as Reprint 2025-26 MOTION IN A PLANE 37 x (m) Fig. 3.15 The average acceleration for three time intervals (a) ∆t1, (b) ∆t2, and (c) ∆t3, (∆t1> ∆t2> ∆t3). (d) In the limit ∆t g0, the average acceleration becomes the acceleration. Note that in one dimension, the velocity and the acceleration of an object are always along -1  v y  − 1  4  ° θ = tan   = tan   ≅ 53 with x-axis. the same straight line (either in the same  v x   3  direction or in the opposite direction). ⊳ However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° 3.8 MOTION IN A PLANE WITH CONSTANT between them. ACCELERATION ⊳ Suppose that an object is moving in x-y plane Example 3.4 The position of a particle is and its acceleration a is constant. Over an given by interval of time, the average acceleration will r = 3.0t ˆi + 2 .0t 2 ˆj + 5 .0 kˆ equal this constant value. Now, let the velocity where t is in seconds and the of the object be v0 at time t = 0 and v at time t. coefficients have the proper units for r to Then, by definition be in metres. (a) Find v(t) and a(t) of the v − v 0 v − v 0 particle. (b) Find the magnitude and a = = t − 0 t direction of v(t) at t = 1.0 s. Or, v = v 0 + a t (3.33a) Answer In terms of components : v x = v ox + a x t dr d ɵ ɵ ɵ v ( t ) = = 3.0 t i + 2.0t2 j + 5.0 k ( ) v y = v oy + a y t (3.33b) d t dt ɵ ɵ = 3.0 i + 4.0t j Let us now find how the position r changes with d v time. We follow the method used in the one- a ( t ) = = +4.0ɵj dt dimensional case. Let ro and r be the position a = 4.0 m s–2 along y- direction vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then,At t = 1.0 s, v = 3.0ˆi + 4.0 ˆj over this time interval t, the average velocity is 2 2 -1 (vo + v)/2. The displacement is the average It’s magnitude is v = 3 + 4 = 5.0 m s velocity multiplied by the time interval : and direction is Reprint 2025-26 38 PHYSICS ˆ  v + v 0   ( v 0 + at ) + v 0  = 5.0 t + 1.5 t 2 i + 1.0 t 2 ˆj ( ) t t = r − r0 = 2  2    2 Therefore, x (t ) = 5.0 t + 1.5 t 1 2 y (t ) = +1.0 t 2 at = v 0 t + 2 Given x (t) = 84 m, t = ? 1 2 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s Or, r = r0 + v 0t + at (3.34a) At t = 6 s, y = 1.0 (6)2 = 36.0 m 2 d r 2.0 t ˆjIt can be easily verified that the derivative of Now, the velocity v = = ( 5.0 + 3.0 t ) ˆi + d t d r ɵ ɵEq. (3.34a), i.e. gives Eq.(3.33a) and it also At t = 6 s, v = 23.0 i + 12.0 j d t satisfies the condition that at t=0, r = ro. speed = v = 232 + 122 ≅ 26 m s−1 . ⊳Equation (3.34a) can be written in component form as 3.9 PROJECTILE MOTION 1 2 x = x 0 + v ox t + a x t As an application of the ideas developed in the 2 previous sections, we consider the motion of a 1 2 projectile. An object that is in flight after being y = y 0 + v oy t + a y t (3.34b) thrown or projected is called a projectile. Such 2 a projectile might be a football, a cricket ball, a One immediate interpretation of Eq.(3.34b) is that baseball or any other object. The motion of a the motions in x- and y-directions can be treated projectile may be thought of as the result of two independently of each other. That is, motion in separate, simultaneously occurring components a plane (two-dimensions) can be treated as two of motions. One component is along a horizontal separate simultaneous one-dimensional direction without any acceleration and the other motions with constant acceleration along two along the vertical direction with constant perpendicular directions. This is an important acceleration due to the force of gravity. It was result and is useful in analysing motion of objects Galileo who first stated this independency of the in two dimensions. A similar result holds for three horizontal and the vertical components ofdimensions. The choice of perpendicular projectile motion in his Dialogue on the greatdirections is convenient in many physical world systems (1632).situations, as we shall see in section 3.9 for In our discussion, we shall assume that theprojectile motion. air resistance has negligible effect on the motion ⊳ Example 3.5 A particle starts from origin of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which θo with the x-axis as shown in Fig. 3.16. produces a constant acceleration of (3.0iɵ+2.0jɵ) m/s 2. (a) What is the After the object has been projected, the acceleration acting on it is that due to gravity y-coordinate of the particle at the instant which is directed vertically downward: its x-coordinate is 84 m ? (b) What is the ɵ speed of the particle at this time ? a = −g j Or, ax = 0, ay = – g (3.35) Answer From Eq. (3.34a) for r0 = 0, the position The components of initial velocity vo are : of the particle is given by 1 at 2 vox = vo cos θo r (t ) = v 0 t + 2 2 voy= vo sin θo (3.36) t 3.0ˆi + 2.0 ˆj = 5.0 ˆi t + (1/2 )( ) Reprint 2025-26 MOTION IN A PLANE 39 Now, since g, θo and vo are constants, Eq. (3.39) is of the form y = a x + b x2, in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 3.17). Fig 3.16 Motion of an object projected with velocity vo at angle θ0. If we take the initial position to be the origin of the reference frame as shown in Fig. 3.16, we have : xo = 0, yo = 0 Then, Eq.(3.34b) becomes : x = vox t = (vo cos θo ) t and y = (vo sin θo ) t – ( ½ )g t2 (3.37) Fig. 3.17 The path of a projectile is a parabola. The components of velocity at time t can be Time of maximum height obtained using Eq.(3.33b) : How much time does the projectile take to reach the vx = vox = vo cos θo maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Eq. (3.38): vy = vo sin θo – g t (3.38) vy = vo sinθo – g tm = 0 Equation (3.37) gives the x-, and y-coordinates Or, tm = vo sinθo /g (3.40a) of the position of a projectile at time t in terms of The total time Tf during which the projectile is two parameters — initial speed vo and projection in flight can be obtained by putting y = 0 in angle θo. Notice that the choice of mutually Eq. (3.37). We get : perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in Tf = 2 (vo sin θo )/g (3.40b) a simplification. One of the components of Tf is known as the time of flight of the projectile. velocity, i.e. x-component remains constant We note that Tf = 2 tm , which is expected throughout the motion and only the because of the symmetry of the parabolic path. y- component changes, like an object in free fall Maximum height of a projectile in vertical direction. This is shown graphically The maximum height hm reached by theat few instants in Fig. 3.17. Note that at the point projectile can be calculated by substituting of maximum height, vy= 0 and therefore, t = tm in Eq. (3.37) : -1 v y 2 = oθ = tan    g  v 0 sinθ0 v 0 sinθ0 v x y = h m = ( v 0 sinθ0 )  −   Equation of path of a projectile  g  2  g  What is the shape of the path followed by the ( v 0 sinθ0 ) 2 projectile? This can be seen by eliminating the Or, h m = (3.41) 2gtime between the expressions for x and y as given in Eq. (3.37). We obtain: Horizontal range of a projectile g 2 The horizontal distance travelled by a projectile from x y = ( tan θo ) x − (3.39) its initial position (x = y = 0) to the position where it 2 (v o cosθo )2 passes y = 0 during its fall is called the horizontal Reprint 2025-26 40 PHYSICS range, R. It is the distance travelled during the time y (t) = yo + voy t +(1/2) ay t2 of flight Tf . Therefore, the range R is Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2, R = (vo cos θo) (Tf ) vox = 15 m s-1. The stone hits the ground when y(t) = – 490 m. =(vo cos θo) (2 vo sin θo)/g – 490 m = –(1/2)(9.8) t2. 2 sin 2θ0 This gives t =10 s. v 0 Or, R = (3.42a) The velocity components are vx = vox and g vy = voy – g t Equation (3.42a) shows that for a given so that when the stone hits the ground : projection velocity vo , R is maximum when sin vox = 15 m s–1 2θ0 is maximum, i.e., when θ0 = 450. voy = 0 – 9.8 × 10 = – 98 m s–1 The maximum horizontal range is, therefore, Therefore, the speed of the stone is 2 v 0 v 2x + v y2 = 15 2 + 98 2 = 99 m s −1 ⊳ R m = (3.42b) g ⊳ ⊳ Example 3.6 Galileo, in his book Two new Example 3.8 A cricket ball is thrown at a sciences, stated that “for elevations which speed of 28 m s–1 in a direction 30° above exceed or fall short of 45° by equal amounts, the horizontal. Calculate (a) the maximum the ranges are equal”. Prove this statement. height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point Answer For a projectile launched with velocity where the ball returns to the same level. vo at an angle θo , the range is given by v 02 sin2θ0 R = Answer (a) The maximum height is given by g 2 2 sin sin 30 ( 28 ( v 0 θo ) ° ) m = m =Now, for angles, (45° + α) and ( 45° – α), 2θo is h 2 2 g ( 9.8 )(90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are 14 × 14 = = 10.0 mthe same, equal to that of cos 2α. Therefore, 2 × 9.8 ranges are equal for elevations which exceed or fall short of 45° by equal amounts α. ⊳ (b) The time taken to return to the same level is Tf = (2 vo sin θo )/g = (2× 28 × sin 30° )/9.8 ⊳ = 28/9.8 s = 2.9 s Example 3.7 A hiker stands on the edge (c) The distance from the thrower to the point of a cliff 490 m above the ground and where the ball returns to the same level is throws a stone horizontally with an initial speed 2 of 15 m s-1. Neglecting air resistance, v o sin2θo 28 × 28 × sin60o ( ) = = 69 m ⊳ find the time taken by the stone to reach R = g 9.8 the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). 3.10 UNIFORM CIRCULAR MOTION Answer We choose the origin of the x-,and y- When an object follows a circular path at a axis at the edge of the cliff and t = 0 s at the constant speed, the motion of the object is called instant the stone is thrown. Choose the positive uniform circular motion. The word “uniform” direction of x-axis to be along the initial velocity refers to the speed, which is uniform (constant) and the positive direction of y-axis to be the throughout the motion. Suppose an object is vertically upward direction. The x-, and y- moving with uniform speed v in a circle of radius components of the motion can be treated R as shown in Fig. 3.18. Since the velocity of the independently. The equations of motion are : object is changing continuously in direction, the x (t) = xo + vox t object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Reprint 2025-26 MOTION IN A PLANE 41 Fig. 3.18 Velocity and acceleration of an object in uniform circular motion. The time interval ∆t decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle. Let r and r′ be the position vectors and v and r′ be ∆θ. Since the velocity vectors v and v′ are v′ the velocities of the object when it is at point P always perpendicular to the position vectors, the and P′ as shown in Fig. 3.18(a). By definition, angle between them is also ∆θ . Therefore, the velocity at a point is along the tangent at that triangle CPP′ formed by the position vectors and point in the direction of motion. The velocity the triangle GHI formed by the velocity vectors vectors v and v′ are as shown in Fig. 3.18(a1). v, v′ and ∆v are similar (Fig. 3.18a). Therefore, ∆v is obtained in Fig. 3.18 (a2) using the triangle the ratio of the base-length to side-length for law of vector addition. Since the path is circular, one of the triangles is equal to that of the other v is perpendicular to r and so is v′ to r′. triangle. That is : Therefore, ∆v is perpendicular to ∆r. Since  ∆v  ∆ v ∆ r average acceleration is along ∆v  a =  , the =  ∆t  v R average acceleration a is perpendicular to ∆r. If we place ∆v on the line that bisects the angle ∆ r Or, ∆ v = v between r and r′, we see that it is directed towards R the centre of the circle. Figure 3.18(b) shows the Therefore, same quantities for smaller time interval. ∆v and ∆ v v ∆ r v ∆r hence a is again directed towards the centre. a = lim = lim = lim t ∆ t → 0 R ∆ t R ∆ t → 0 ∆ tIn Fig. 3.18(c), ∆tŽ 0 and the average ∆ t → 0 ∆ acceleration becomes the instantaneous If ∆t is small, ∆θ will also be small and then arc acceleration. It is directed towards the centre*. PP′ can be approximately taken to be|∆r|: Thus, we find that the acceleration of an object ∆ r ≅v∆ t in uniform circular motion is always directed ∆r towards the centre of the circle. Let us now find ≅ v the magnitude of the acceleration. ∆ t The magnitude of a is, by definition, given by ∆ r lim = v ∆v Or, ∆ t → 0 ∆ t a = lim ∆ t → 0 ∆t Let the angle between position vectors r and Therefore, the centripetal acceleration ac is : * In the limit ∆tŽ0, ∆r becomes perpendicular to r. In this limit ∆v→ 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path. Reprint 2025-26 42 PHYSICS 2 2 2  v  v R ω 2  R ac = = = ω a c =  R v = v2/R (3.43) R R Thus, the acceleration of an object moving with 2 a c = ω R (3.46) speed v in a circle of radius R has a magnitude v 2/R and is always directed towards the centre. The time taken by an object to make one revolution This is why this acceleration is called centripetal is known as its time period T and the number ofacceleration (a term proposed by Newton). A revolution made in one second is called itsthorough analysis of centripetal acceleration was frequency ν (=1/T). However, during this timefirst published in 1673 by the Dutch scientist the distance moved by the object is s = 2πR.Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. Therefore, v = 2πR/T =2πRν (3.47) “Centripetal” comes from a Greek term which means In terms of frequency ν, we have ‘centre-seeking’. Since v and R are constant, the ω = 2πν magnitude of the centripetal acceleration is also v = 2πRν constant. However, the direction changes — ac = 4π2 ν2R (3.48)pointing always towards the centre. Therefore, a ⊳centripetal acceleration is not a constant vector. Example 3.9 An insect trapped in a We have another way of describing the circular groove of radius 12 cm moves along velocity and the acceleration of an object in the groove steadily and completes 7 uniform circular motion. As the object moves revolutions in 100 s. (a) What is the from P to P′ in time ∆t (= t′ – t), the line CP angular speed, and the linear speed of the (Fig. 3.18) turns through an angle ∆θ as shown motion? (b) Is the acceleration vector a in the figure. ∆θ is called angular distance. We constant vector ? What is its magnitude ? define the angular speed ω (Greek letter omega) as the time rate of change of angular Answer This is an example of uniform circular displacement : motion. Here R = 12 cm. The angular speed ω is ∆θ given by ω = ∆t (3.44) ω = 2π/T = 2π × 7/100 = 0.44 rad/s The linear speed v is :Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then : v =ω R = 0.44 s-1 × 12 cm = 5.3 cm s-1 ∆ s The direction of velocity v is along the tangent v = ∆t to the circle at every point. The acceleration is directed towards the centre of the circle. Sincebut ∆s = R ∆θ. Therefore : this direction changes continuously, ∆θ v = R = R ω acceleration here is not a constant vector. ∆ t However, the magnitude of acceleration is v = R ω (3.45) constant: a = ω2 R = (0.44 s–1)2 (12 cm)We can express centripetal acceleration ac in terms of angular speed : = 2.3 cm s-2 ⊳ Reprint 2025-26 MOTION IN A PLANE 43 SUMMARY 1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature. 2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra. 3. A vector A multiplied by a real number λ is also a vector, whose magnitude is λ times the magnitude of the vector A and whose direction is the same or opposite depending upon whether λ is positive or negative. 4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method. 5. Vector addition is commutative : A + B = B + A It also obeys the associative law : (A + B) + C = A + (B + C) 6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties : A + 0 = A λ0 = 0 0 A = 0 7. The subtraction of vector B from A is defined as the sum of A and –B : A – B = A+ (–B) 8. A vector A can be resolved into component along two given vectors a and b lying in the same plane : A = λ a + µ b where λ and µ are real numbers. 9. A unit vector associated with a vector A has magnitude 1 and is along the vector A: A ˆn = A ɵ ɵ ɵ The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system. 10. A vector A can be expressed as ɵ ɵ A = A x i + Ay j where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle θ A y 2 2 with the x-axis, then Ax = A cos θ, Ay=A sin θ and A = A = A x + A y , tanθ = . A x 11. Vectors can be conveniently added using analytical method. If sum of two vectors A and B, that lie in x-y plane, is R, then : ɵ R = R x i + Ry ɵ,j where, Rx = Ax + Bx, and Ry = Ay + By ɵ ɵ 12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r’ is given by ∆r = r′− r ɵ ɵ = ( x ′ − x ) i + (y ′ − y ) j ɵ ɵ = ∆x i + ∆y j 13. If an object undergoes a displacement ∆r in time ∆t, its average velocity is given by ∆ r v = . The velocity of an object at time t is the limiting value of the average velocity ∆t as ∆t tends to zero : Reprint 2025-26 44 PHYSICS ∆ r d r v = lim = . It can be written in unit vector notation as : ∆ t →0 ∆ t dt ɵ ɵ ɵ dx dy dz = y v = v x i + v y j + v z k where v x = dt , v dt , v z = dt When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object. 14. If the velocity of an object changes from v to v′in time ∆t, then its average acceleration v − v' ∆ v is given by: a = = ∆ t ∆t The acceleration a at any time t is the limiting value of a as ∆t Ž0 : ∆ v dv a = lim = ∆ t → 0 ∆ t dt ɵ ɵ ɵ In component form, we have : a = a x i + a y j + a z k dvy dv x dv z where, a x = , a y = , a z = dt dt dt 15. If an object is moving in a plane with constant acceleration a = a = a x2 + a y2 and its position vector at time t = 0 is ro, then at any other time t, it will be at a point given by: 1 2 r = ro + v o t + at 2 and its velocity is given by : v = vo + a t where vo is the velocity at time t = 0 In component form : 1 2 x = x o + v ox t + a x t 2 1 2 y = yo + v oy t + a y t 2 v x = v ox + a x t v y = v oy + a y t Motion in a plane can be treated as superposition of two separate simultaneous one- dimensional motions along two perpendicular directions 16. An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity vo making an angle θo with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by : x = (vo cos θo) t y = (vo sin θo) t − (1/2) g t2 vx = vox = vo cos θo vy = vo sin θo − g t The path of a projectile is parabolic and is given by : 2 gx y = ( tanθ0 ) x – 2 2 (v o cosθo ) The maximum height that a projectile attains is : Reprint 2025-26 MOTION IN A PLANE 45 (v o sinq o )2 h m = 2g The time taken to reach this height is : v o sinθo t m = g The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is : v o2 R = sin2θo g 17. When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is ac = v2 /R. The direction of ac is always towards the centre of the circle. The angular speed ω, is the rate of change of angular distance. It is related to velocity v by v = ω R. The acceleration is ac = ω 2R. If T is the time period of revolution of the object in circular motion and ν is its frequency, we have ω = 2π ν, v = 2πνR, ac = 4π2ν2R Reprint 2025-26 46 PHYSICS POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The vector equations (3.33a) and (3.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes. 4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing. 5. An object subjected to two velocities v1 and v2 has a resultant velocity v = v1 + v2. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v12= v1 − v2. Here v1 and v2 are velocities with reference to some common reference frame. 6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. 7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion ( initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions. EXERCISES 3.1 State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. 3.2 Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. 3.3 Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. 3.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. 3.5 Read each statement below carefully and state with reasons, if it is true or false : (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector. 3.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| Reprint 2025-26 MOTION IN A PLANE 47 (c) |a−b| < |a| + |b| (d) |a−b| > ||a| − |b|| When does the equality sign above apply? 3.7 Given a + b + c + d = 0, which of the following statements are correct : Q (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of ( b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ? 3.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For Fig. 3.19 which girl is this equal to the actual length of path skate ? 3.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 3.20 3.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case. 3.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? 3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ? 3.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? Reprint 2025-26 48 PHYSICS 3.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ? 3.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. 3.16 Read each statement below carefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector 3.17 The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 kˆ m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ? 3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in ɵ ɵ the x-y plane with a constant acceleration of 8.0 i + 2.0 j m s-2. (a) At what time is ( ) the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ? 3.19 ɵi and ɵj are unit vectors along x- and y- axis respectively. What is the magnitude ɵ ɵ ɵ ɵ and direction of the vectors i + j , and i − j ? What are the components of a vector ɵ ɵ ɵ ɵ ɵ ɵ A= 2 i + 3 j along the directions of i + j and i −?j [You may use graphical method] 3.20 For any arbitrary motion in space, which of the following relations are true : (a) vaverage = (1/2) (v (t1) + v (t2)) (b) v average = [r(t2) - r(t1) ] /(t2 – t1) (c) v (t) = v (0) + a t (d) r (t) = r (0) + v (0) t + (1/2) a t2 (e) a average =[ v (t2) - v (t1 )] /( t2 – t1) (The ‘average’ stands for average of the quantity over the time interval t1 to t2) 3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes. 3.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, wat is the speed of the aircraft ? Reprint 2025-26 CHAPTER FOUR LAWS OF MOTION 4.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that 4.1 Introduction uniform motion needs the concept of velocity alone whereas 4.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in 4.3 The law of inertia addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this4.4 Newton’s first law of motion basic question.4.5 Newton’s second law of Let us first guess the answer based on our common motion experience. To move a football at rest, someone must kick it.4.6 Newton’s third law of motion To throw a stone upwards, one has to give it an upward

3.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is 2 2 2 2 −1 −1 R = v r + v w = 35 + 12 m s = 37 m s The direction θ that R makes with the vertical is given by v w 12 tan θ = = = 0.343 Fig. 3.8 (a) Two non-colinear vectors a and b. v r 35 (b) Resolving a vector A in terms of vectors Or, θ = tan-1 ( 0.343 ) = 19° a and b. Therefore, the boy should hold his umbrella We say that A has been resolved into two in the vertical plane at an angle of about 19o component vectors λ a and µ b along a and b with the vertical towards the east. ⊳ Reprint 2025-26 32 PHYSICS respectively. Using this method one can resolve and A2 is parallel to ɵj, we have :a given vector into two component vectors along a set of two vectors – all the three lie in the same A1= Ax ɵi , A2 = Ay ɵj (3.11) plane. It is convenient to resolve a general vector where Ax and Ay are real numbers.along the axes of a rectangular coordinate system using vectors of unit magnitude. These Thus, A = Ax ɵi + Ay ɵj (3.12) are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and This is represented in Fig. 3.9(c). The quantities points in a particular direction. It has no Ax and Ay are called x-, and y- components of the dimension and unit. It is used to specify a vector A. Note that Ax is itself not a vector, but direction only. Unit vectors along the x-, y- and ɵi is a vector, and so is Ay ɵj. Using simplez-axes of a rectangular coordinate system are Ax trigonometry, we can express Ax and Ay in terms denoted by ɵi , ɵj and ˆk , respectively, as shown of the magnitude of A and the angle θ it makes in Fig. 3.9(a). with the x-axis : Since these are unit vectors, we have Ax = A cos θ Ay = A sin θ (3.13) ˆi  = ˆj  = ˆk =1 (3.9) As is clear from Eq. (3.13), a component of a These unit vectors are perpendicular to each vector can be positive, negative or zero other. In this text, they are printed in bold face depending on the value of θ. with a cap (^) to distinguish them from other Now, we have two ways to specify a vector A vectors. Since we are dealing with motion in two in a plane. It can be specified by : dimensions in this chapter, we require use of (i) its magnitude A and the direction θ it makes only two unit vectors. If we multiply a unit vector, with the x-axis; or say ˆn by a scalar, the result is a vector (ii) its components Ax and Ay λ = λ ˆn. In general, a vector A can be written as If A and θ are given, Ax and Ay can be obtained using Eq. (3.13). If Ax and Ay are given, A and θ A = |A| ˆn (3.10) can be obtained as follows : where ˆn is a unit vector along A. 2 2 2 2 2 2 A x + A y = A cos θ + A sin θ We can now resolve a vector A in terms = A2 of component vectors that lie along unit vectors iˆ and ɵj. Consider a vector A that lies in x-y Or, A = A 2x + Ay2 (3.14) plane as shown in Fig. 3.9(b). We draw lines from the head of A perpendicular to the coordinate Ay A y tan θ = , θ = tan− 1 axes as in Fig. 3.9(b), and get vectors A1 and A2 And A x A x (3.15) such that A1 + A2 = A. Since A1 is parallel to ɵi Fig. 3.9 (a) Unit vectors ɵi , ɵj and ɵk lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components Ax and Ay along x-, and y- axes. (c) A1 and A2 expressed in terms of ɵi and ɵj . Reprint 2025-26 MOTION IN A PLANE 33 So far we have considered a vector lying in ɵ ɵ B = B x i + B y jan x-y plane. The same procedure can be used Let R be their sum. We haveto resolve a general vector A into three components along x-, y-, and z-axes in three R = A + B dimensions. If α, β, and γ are the angles* ɵ ɵ ɵ ɵ = + (3.19a) ( A x i + A y j ) ( B x i + B y j )between A and the x-, y-, and z-axes, respectively [Fig. 3.9(d)], we have Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (3.19a) as convenient to us : ɵ ɵ j (3.19b) R = ( A x + B x ) i + ( A y + B y ) ɵ ɵ SinceR = R x i + R y j (3.20) we have, R x = A x + B x , R y = A y + B y (3.21) Thus, each component of the resultant vector R is the sum of the corresponding components of A and B. In three dimensions, we have ɵ ɵ ɵ A = A x i + Ay j + A z k ɵ ɵ ɵ B = B x i + B y j + B z k (d) ɵ ɵ ɵ R = A + B = R x i + R y j + R z kFig. 3.9 (d) A vector A resolved into components along x-, y-, and z-axes with R x = A x + B x y = A y + B yA x = A cos α, A y = A cos β, A z = A cos γ (3.16a) R In general, we have R z = A z + B z (3.22) A = Ax ˆi + Ay ˆj + Az kˆ (3.16b) This method can be extended to addition and The magnitude of vector A is subtraction of any number of vectors. For A = A x2 + Ay2 + A z2 (3.16c) example, if vectors a, b and c are given as ɵ ɵ ɵ A position vector r can be expressed as a = a x i + a y j + a z k ɵ ɵ ɵ r = x i + y j + z k (3.17) ɵ ɵ ɵ b = b x i + b y j + b z k where x, y, and z are the components of r along ɵ ɵ ɵ x-, y-, z-axes, respectively. c = c x i + c y j + c z k (3.23a)

6.12 ANGULAR MOMENTUM IN CASE OF For computing the total angular momentum ROTATION ABOUT A FIXED AXIS of the whole rigid body, we add up the contribution of each particle of the body. We have studied in section 6.7, the angular momentum of a system of particles. We already Thus know from there that the time rate of total We denote by L ⊥ and L z the components of angular momentum of a system of particles L respectively perpendicular to the z-axis andabout a point is equal to the total external torque along the z-axis;on the system taken about the same point. When OC i × m i v i (6.42a)the total external torque is zero, the total angular L ⊥= ∑ momentum of the system is conserved. where mi and vi are respectively the mass and We now wish to study the angular momentum the velocity of the ith particle and Ci is the centrein the special case of rotation about a fixed axis. of the circle described by the particle; The general expression for the total angular momentum of the system of n particles is N and ˆ (6.42b) L = =∑i 1 ri × p i (6.25b) or L z = Iωk We first consider the angular momentum of The last step follows since the perpendicular a typical particle of the rotating rigid body. We distance of the ith particle from the axis is ri; andthen sum up the contributions of individual by definition the moment of inertia of the body particles to get L of the whole body. m i ri2 . For a typical particle l = r × p. As seen in the about the axis of rotation is I =∑ last section r = OP = OC + CP [Fig. 6.17(b)]. With Note L = L z + L ⊥ (6.42c)p = m v , l = ( OC × m v ) + ( CP × m v ) The rigid bodies which we have mainly considered in this chapter are symmetric about The magnitude of the linear velocity v of the the axis of rotation, i.e. the axis of rotation is particle at P is given by v = ωr⊥ where r⊥ is the one of their symmetry axes. For such bodies, for length of CP or the perpendicular distance of P a given OCi, for every particle which has a from the axis of rotation. Further, v is tangential velocity vi , there is another particle of velocity at P to the circle which the particle describes. –vi located diametrically opposite on the circle Using the right-hand rule one can check that with centre Ci described by the particle. TogetherCP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) such pairs will contribute zero to L ⊥ and as a is ˆk . Hence result for symmetric bodies L ⊥ is zero, and CP × m v = r⊥ (mv ) kˆ hence z = Iωkˆ (6.42d) = mr⊥2ω kˆ (since υ = ωr⊥ ) L = L Similarly, we can check that OC × v is For bodies, which are not symmetric about perpendicular to the fixed axis. Let us denote the axis of rotation, L is not equal to Lz and hence the part of l along the fixed axis (i.e. the z-axis) L does not lie along the axis of rotation. by lz, then Referring to Table 6.1, can you tell in which l z = CP × m v = mr⊥2ωkˆ cases L = Lz will not apply? Let us differentiate Eq. (6.42b). Since ˆk is a and l = l z + OC × m v fixed (constant) vector, we get We note that lz is parallel to the fixed axis, ˆbut l is not. In general, for a particle, the angular I ω) k d ( L z ) =  d (  d t momentum l is not along the axis of rotation, d t i.e. for a particle, l and ω are not necessarily Now, Eq. (6.28b) states parallel. Compare this with the corresponding dL fact in translation. For a particle, p and v are = τ dtalways parallel to each other. Reprint 2025-26 122 PHYSICS As we have seen in the last section, only We have already derived this equation using those components of the external torques which the work - kinetic energy route. are along the axis of rotation, need to be taken into account, when we discuss rotation about a 6.12.1 Conservation of angular momentum fixed axis. This means we can take τ = τkˆ . We are now in a position to revisit the principle of conservation of angular momentum in the Since L = L z + L ⊥ and the direction of Lz (vector context of rotation about a fixed axis. From Eq. ˆk ) is fixed, it follows that for rotation about a (6.43c), if the external torque is zero, fixed axis, Lz = Iω = constant (6.44) For symmetric bodies, from Eq. (6.42d), Lz d L z = τkˆ (6.43a) may be replaced by L .(L and Lz are respectively d t the magnitudes of L and Lz.) This then is the required form, for fixed axis d L rotation, of Eq. (6.29a), which expresses theand ⊥= 0 (6.43b) dt general law of conservation of angular momentum Thus, for rotation about a fixed axis, the of a system of particles. Eq. (6.44) applies to many component of angular momentum perpendicular situations that we come across in daily life. You may do this experiment with your friend. Sit on a to the fixed axis is constant. As L z = Iωkˆ , we swivel chair (a chair with a seat, free to rotate get from Eq. (6.43a), about a pivot) with your arms folded and feet not resting on, i.e., away from, the ground. Ask your d ( Iω) = τ (6.43c) friend to rotate the chair rapidly. While the chair d t is rotating with considerable angular speed If the moment of inertia I does not change with stretch your arms horizontally. What happens? time, Your angular speed is reduced. If you bring back d dω your arms closer to your body, the angular speed ( Iω) = I = Iα increases again. This is a situation where thed t d t principle of conservation of angular momentumand we get from Eq. (6.43c), is applicable. If friction in the rotational τ = I α (6.41) Fig 6.32 (a) A demonstration of conservation of Fig 6.32 (b) An acrobat employing the principle of angular momentum. A girl sits on a conservation of angular momentum in swivel chair and stretches her arms/ her performance. brings her arms closer to the body. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 123 mechanism is neglected, there is no external A circus acrobat and a diver take advantage torque about the axis of rotation of the chair and of this principle. Also, skaters and classical, hence Iω is constant. Stretching the arms Indian or western, dancers performing a increases I about the axis of rotation, resulting in pirouette (a spinning about a tip–top) on the toes decreasing the angular speed ω. Bringing the of one foot display ‘mastery’ over this principle. arms closer to the body has the opposite effect. Can you explain? SUMMARY 1. Ideally, a rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them. 2. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions. 3. In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time. 4. In pure translation, every particle of the body moves with the same velocity at any instant of time. 5. Angular velocity is a vector. Its magnitude is ω = dθ/dt and it is directed along the axis of rotation. For rotation about a fixed axis, this vector ω has a fixed direction. 6. The vector or cross product of two vector a and b is a vector written as a×b. The magnitude of this vector is absinθ and its direction is given by the right handed screw or the right hand rule. 7. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by v = ω × r, where r is the position vector of the particle with respect to an origin along the fixed axis. The relation applies even to more general rotation of a rigid body with one point fixed. In that case r is the position vector of the particle with respect to the fixed point taken as the origin. 8. The centre of mass of a system of n particles is defined as the point whose position vector is ri ∑m i R = M 9. Velocity of the centre of mass of a system of particles is given by V = P/M, where P is the linear momentum of the system. The centre of mass moves as if all the mass of the system is concentrated at this point and all the external forces act at it. If the total external force on the system is zero, then the total linear momentum of the system is constant. 10. The angular momentum of a system of n particles about the origin is n L = ri × pi i =∑1 The torque or moment of force on a system of n particles about the origin is τ = ∑ri × Fi 1 The force Fi acting on the ith particle includes the external as well as internal forces. Assuming Newton’s third law of motion and that forces between any two particles act along the line joining the particles, we can show τint = 0 and Reprint 2025-26 124 PHYSICS dL = τ ext dt 11. A rigid body is in mechanical equilibrium if (1) it is in translational equilibrium, i.e., the total external force on it is zero : Fi = 0 , ∑ and (2) it is in rotational equilibrium, i.e. the total external torque on it is zero : Fi = 0 . ∑ τi = ∑ri × 12. The centre of gravity of an extended body is that point where the total gravitational torque on the body is zero. 13. The moment of intertia of a rigid body about an axis is defined by the formula I m i ri2 =∑ where ri is the perpendicular distance of the ith point of the body from the axis. The 1 2 kinetic energy of rotation is K = Iω . 2 POINTS TO PONDER 1. To determine the motion of the centre of mass of a system no knowledge of internal forces of the system is required. For this purpose we need to know only the external forces on the body. 2. Separating the motion of a system of particles as the motion of the centre of mass, (i.e., the translational motion of the system) and motion about (i.e. relative to) the centre of mass of the system is a useful technique in dynamics of a system of particles. One example of this technique is separating the kinetic energy of a system of particles K as the kinetic energy of the system about its centre of mass K′ and the kinetic energy of the centre of mass MV2/2, K = K′ + MV2/2 3. Newton’s Second Law for finite sized bodies (or systems of particles) is based in Newton’s Second Law and also Newton’s Third Law for particles. 4. To establish that the time rate of change of the total angular momentum of a system of particles is the total external torque in the system, we need not only Newton’s second law for particles, but also Newton’s third law with the provision that the forces between any two particles act along the line joining the particles. 5. The vanishing of the total external force and the vanishing of the total external torque are independent conditions. We can have one without the other. In a couple, total external force is zero, but total torque is non-zero. 6. The total torque on a system is independent of the origin if the total external force is zero. 7. The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 125 8. The angular momentum L and the angular velocity ω are not necessarily parallel vectors. However, for the simpler situations discussed in this chapter when rotation is about a fixed axis which is an axis of symmetry of the rigid body, the relation L = Iω holds good, where I is the moment of the inertia of the body about the rotation axis. EXERCISES 6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ? 6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. 6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ? 6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b. 6.5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c. 6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. 6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken. 6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end. Fig. 6.33 6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. Reprint 2025-26 126 PHYSICS 6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time. 6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? 6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? 6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping. 6.14 To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient. 6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body. 6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? 6.17 The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule. Reprint 2025-26 CHAPTER SEVEN GRAVITATION 7.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything