Untitled Flashcards Set

In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided 3.1 Introduction 3.2 Scalars and vectors 3.3 Multiplication of vectors by real numbers 3.4 Addition and subtraction of vectors — graphical method 3.5 Resolution of vectors 3.6 Vector addition — analytical method 3.7 Motion in a plane 3.8 Motion in a plane with constant acceleration 3.9 Projectile motion 3.10 Uniform circular motion Summary Points to ponder Exercises 2024-25 28 PHYSICS just as the ordinary numbers*. For example, if the length and breadth of a rectangle are 1.0 m and 0.5 m respectively, then its perimeter is the sum of the lengths of the four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = 3.0 m. The length of each side is a scalar and the perimeter is also a scalar. Take another example: the maximum and minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of 2.7 kg, then its volume is 10–3 m3 (a scalar) and its density is 2.7×103 kg m–3 (a scalar). A vector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the parallelogram law of addition. So, a vector is specified by giving its magnitude by a number and its direction. Some physical quantities that are represented by vectors are displacement, velocity, acceleration and force. To represent a vector, we use a bold face type in this book. Thus, a velocity vector can be represented by a symbol v. Since bold face is difficult to produce, when written by hand, a vector is often represented by an arrow placed over a letter, say r v . Thus, both v and r v represent the velocity vector. The magnitude of a vector is often called its absolute value, indicated by |v| = v. Thus, a vector is represented by a bold face, e.g. by A, a, p, q, r, ... x, y, with respective magnitudes denoted by light face A, a, p, q, r, ... x, y. 3.2.1 Position and Displacement Vectors To describe the position of an object moving in a plane, we need to choose a convenient point, say O as origin. Let P and P′ be the positions of the object at time t and t′, respectively [Fig. 3.1(a)]. We join O and P by a straight line. Then, OP is the position vector of the object at time t. An arrow is marked at the head of this line. It is represented by a symbol r, i.e. OP = r. Point P′ is represented by another position vector, OP′ denoted by r′. The length of the vector r represents the magnitude of the vector and its direction is the direction in which P lies as seen from O. If the object moves from P to P′, the vector PP′ (with tail at P and tip at P′) is called the displacement vector corresponding to motion from point P (at time t) to point P′ (at time t′). Fig. 3.1 (a) Position and displacement vectors. (b) Displacement vector PQ and different courses of motion. It is important to note that displacement vector is the straight line joining the initial and final positions and does not depend on the actual path undertaken by the object between the two positions. For example, in Fig. 3.1(b), given the initial and final positions as P and Q, the displacement vector is the same PQ for different paths of journey, say PABCQ, PDQ, and PBEFQ. Therefore, the magnitude of displacement is either less or equal to the path length of an object between two points. This fact was emphasised in the previous chapter also while discussing motion along a straight line. 3.2.2 Equality of Vectors Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction.** Figure 3.2(a) shows two equal vectors A and B. We can easily check their equality. Shift B parallel to itself until its tail Q coincides with that of A, i.e. Q coincides with O. Then, since their tips S and P also coincide, the two vectors are said to be equal. In general, equality is indicated Addition and subtraction of scalars make sense only for quantities with same units. However, you can multiply and divide scalars of different units. * In our study, vectors do not have fixed locations. So displacing a vector parallel to itself leaves the vector unchanged. Such vectors are called free vectors. However, in some physical applications, location or line of application of a vector is important. Such vectors are called localised vectors. 2024-25 MOTION IN A PLANE 29 as A = B. Note that in Fig. 3.2(b), vectors A′ and B′ have the same magnitude but they are not equal because they have different directions. Even if we shift B′ parallel to itself so that its tail Q′ coincides with the tail O′ of A′, the tip S′ of B′ does not coincide with the tip P′ of A′. 3.3 MULTIPLICATION OF VECTORS BY REAL NUMBERS Multiplying a vector A with a positive number λ gives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A : λ A = λ A if λ > 0. For example, if A is multiplied by 2, the resultant vector 2A is in the same direction as A and has a magnitude twice of |A| as shown in Fig. 3.3(a). Multiplying a vector A by a negative number −λ gives another vector whose direction is opposite to the direction of A and whose magnitude is λ times |A|. Multiplying a given vector A by negative numbers, say –1 and –1.5, gives vectors as shown in Fig 3.3(b). The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A. For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector. 3.4 ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD As mentioned in section 4.2, vectors, by definition, obey the triangle law or equivalently, the parallelogram law of addition. We shall now describe this law of addition using the graphical method. Let us consider two vectors A and B that lie in a plane as shown in Fig. 3.4(a). The lengths of the line segments representing these vectors are proportional to the magnitude of the vectors. To find the sum A + B, we place vector B so that its tail is at the head of the vector A, as in Fig. 3.4(b). Then, we join the tail of A to the head of B. This line OQ represents a vector R, that is, the sum of the vectors A and B. Since, in this procedure of vector addition, vectors are Fig. 3.2 (a) Two equal vectors A and B. (b) Two vectors A′ and B′ are unequal though they are of the same length. Fig. 3.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2. (b) Vector A and resultant vectors after multiplying it by a negative number –1 and –1.5. (c) (d) Fig. 3.4 (a) Vectors A and B. (b) Vectors A and B added graphically. (c) Vectors B and A added graphically. (d) Illustrating the associative law of vector addition. 2024-25 30 PHYSICS arranged head to tail, this graphical method is called the head-to-tail method. The two vectors and their resultant form three sides of a triangle, so this method is also known as triangle method of vector addition. If we find the resultant of B + A as in Fig. 3.4(c), the same vector R is obtained. Thus, vector addition is commutative: A + B = B + A (3.1) The addition of vectors also obeys the associative law as illustrated in Fig. 3.4(d). The result of adding vectors A and B first and then adding vector C is the same as the result of adding B and C first and then adding vector A : (A + B) + C = A + (B + C) (3.2) What is the result of adding two equal and opposite vectors ? Consider two vectors A and –A shown in Fig. 3.3(b). Their sum is A + (–A). Since the magnitudes of the two vectors are the same, but the directions are opposite, the resultant vector has zero magnitude and is represented by 0 called a null vector or a zero vector : A – A = 0 |0|= 0 (3.3) Since the magnitude of a null vector is zero, its direction cannot be specified. The null vector also results when we multiply a vector A by the number zero. The main properties of 0 are : A + 0 = A λ 0 = 0 0 A = 0 (3.4) Fig. 3.5 (a) Two vectors A and B, – B is also shown. (b) Subtracting vector B from vector A – the result is R2 . For comparison, addition of vectors A and B, i.e. R1 is also shown. What is the physical meaning of a zero vector? Consider the position and displacement vectors in a plane as shown in Fig. 3.1(a). Now suppose that an object which is at P at time t, moves to P′ and then comes back to P. Then, what is its displacement? Since the initial and final positions coincide, the displacement is a “null vector”. Subtraction of vectors can be defined in terms of addition of vectors. We define the difference of two vectors A and B as the sum of two vectors A and –B : A – B = A + (–B) (3.5) It is shown in Fig 3.5. The vector –B is added to vector A to get R2 = (A – B). The vector R1 = A + B is also shown in the same figure for comparison. We can also use the parallelogram method to find the sum of two vectors. Suppose we have two vectors A and B. To add these vectors, we bring their tails to a common origin O as shown in Fig. 3.6(a). Then we draw a line from the head of A parallel to B and another line from the head of B parallel to A to complete a parallelogram OQSP. Now we join the point of the intersection of these two lines to the origin O. The resultant vector R is directed from the common origin O along the diagonal (OS) of the parallelogram [Fig. 3.6(b)]. In Fig.3.6(c), the triangle law is used to obtain the resultant of A and B and we see that the two methods yield the same result. Thus, the two methods are equivalent. 2024-25 MOTION IN A PLANE 31 ⊳ Example 3.1 Rain is falling vertically with a speed of 35 m s–1. Winds starts blowing after sometime with a speed of 12 m s–1 in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella ? Fig. 3.7 Answer The velocity of the rain and the wind are represented by the vectors vr and vw in Fig. 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 R v v r 2 w 2 = + = + = − − 35 12 m s 37 m s 2 2 1 1 The direction θ that R makes with the vertical is given by 12 tan 0.343 35 w r v v θ = = = Or, θ = tan ( ) = ° - . 1 0 343 19 Therefore, the boy should hold his umbrella in the vertical plane at an angle of about 19o with the vertical towards the east. ⊳ Fig. 3.6 (a) Two vectors A and B with their tails brought to a common origin. (b) The sum A + B obtained using the parallelogram method. (c) The parallelogram method of vector addition is equivalent to the triangle method. 3.5 RESOLUTION OF VECTORS Let a and b be any two non-zero vectors in a plane with different directions and let A be another vector in the same plane (Fig. 3.8). A can be expressed as a sum of two vectors — one obtained by multiplying a by a real number and the other obtained by multiplying b by another real number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have A = OP = OQ + QP (3.6) But since OQ is parallel to a, and QP is parallel to b, we can write : OQ = λ a, and QP = µ b (3.7) where λ and µ are real numbers. Therefore, A = λ a + µ b (3.8) Fig. 3.8 (a) Two non-colinear vectors a and b. (b) Resolving a vector A in terms of vectors a and b. We say that A has been resolved into two component vectors λ a and µ b along a and b 2024-25 32 PHYSICS Fig. 3.9 (a) Unit vectors ɵ i , ɵ j 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 ɵ i and ɵ j. respectively. Using this method one can resolve a given vector into two component vectors along a set of two vectors – all the three lie in the same plane. It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude. These are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only. Unit vectors along the x-, y- and z-axes of a rectangular coordinate system are denoted by ɵ i, ɵ j and kˆ , respectively, as shown in Fig. 3.9(a). Since these are unit vectors, we have  ˆi  = ˆj  = kˆ =1 (3.9) These unit vectors are perpendicular to each other. In this text, they are printed in bold face with a cap (^) to distinguish them from other vectors. Since we are dealing with motion in two dimensions in this chapter, we require use of only two unit vectors. If we multiply a unit vector, say nˆ by a scalar, the result is a vector λ = λnˆ . In general, a vector A can be written as A = |A|nˆ (3.10) where nˆ is a unit vector along A. We can now resolve a vector A in terms of component vectors that lie along unit vectors i ˆ and ɵ j. Consider a vector A that lies in x-y plane as shown in Fig. 3.9(b). We draw lines from the head of A perpendicular to the coordinate axes as in Fig. 3.9(b), and get vectors A1 and A2 such that A1 + A2 = A. Since A1 is parallel to ɵ i and A2 is parallel to ɵ j, we have : A1 = Ax ɵ i, A2 = Ay ɵ j (3.11) where Ax and Ay are real numbers. Thus, A = Ax ɵ i + Ay ɵ j (3.12) This is represented in Fig. 3.9(c). The quantities Ax and Ay are called x-, and y- components of the vector A. Note that Ax is itself not a vector, but Ax ɵ i is a vector, and so is Ay ɵ j. Using simple trigonometry, we can express Ax and Ay in terms of the magnitude of A and the angle θ it makes with the x-axis : Ax = A cos θ Ay = A sin θ (3.13) As is clear from Eq. (3.13), a component of a vector can be positive, negative or zero depending on the value of θ. Now, we have two ways to specify a vector A in a plane. It can be specified by : (i) its magnitude A and the direction θ it makes with the x-axis; or (ii) its components Ax and Ay 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 can be obtained as follows : A A A A x 2 y 2 2 2 2 2 + = + cos sin θ θ = A2 Or, A A A x 2 y 2 = + (3.14) And tan , tan θ θ = = A − A A A y x y x 1 (3.15) 2024-25 MOTION IN A PLANE 33 ⊳ B i j = + B B x y ɵ ɵ Let R be their sum. We have R = A + B = + + + ( A A B B x y x y ) ( ) ɵ ɵ ɵ ɵ i j i j (3.19a) Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (3.19a) as convenient to us : R = + + + ( A B A B )i ( )j x x y y ɵ ɵ (3.19b) SinceR i j = + R R x y ɵ ɵ (3.20) we have, R A B , R A B x x x y y 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 A i j k = + + A A A x y z ɵ ɵ ɵ B i j k = + + B B B x y z ɵ ɵ ɵ R A B i j k = + = + + R R R x y z ɵ ɵ ɵ with R A B x x x = + R A B y y y = + R A B z z z = + (3.22) This method can be extended to addition and subtraction of any number of vectors. For example, if vectors a, b and c are given as a i j k = + + a a a x y z ɵ ɵ ɵ b i j k = + + b b b x y z ɵ ɵ ɵ c i j k = + + c c c x y z ɵ ɵ ɵ (3.23a) then, a vector T = a + b – c has components : T a b c x x x x = + − T a b c y y y y = + − (3.23b) T a b c z z z z = + − . Example 3.2 Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle θ between them. Fig. 3.9 (d) A vector A resolved into components along x-, y-, and z-axes Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. So far we have considered a vector lying in an x-y plane. The same procedure can be used to resolve a general vector A into three components along x-, y-, and z-axes in three dimensions. If α, β, and γ are the angles between A and the x-, y-, and z-axes, respectively [Fig. 3.9(d)], we have (d) A A cos , A A cos , A A cos x α y β z = = = γ (3.16a) In general, we have ˆ ˆ ˆ A i j k = + + A A A x y z (3.16b) The magnitude of vector A is 2 2 2 A A A A = + + x y z (3.16c) A position vector r can be expressed as r i j k = + + x y z ɵ ɵ ɵ (3.17) where x, y, and z are the components of r along x-, y-, z-axes, respectively. 3.6 VECTOR ADDITION – ANALYTICAL METHOD Although the graphical method of adding vectors helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors by combining their respective components. Consider two vectors A and B in x-y plane with components Ax , Ay and Bx , By : A i j = + A A x y ɵ ɵ (3.18) 2024-25 34 PHYSICS ⊳ Fig. 3.10 Answer Let OP and OQ represent the two vectors A and B making an angle θ (Fig. 3.10). Then, using the parallelogram method of vector addition, OS represents the resultant vector R : 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 B 2AB 2 2 = + + cos θ (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 or, R B sin sin θ α = (3.24b) Similarly, PM = A sin α = B sin β or, A B sin sin β α = (3.24c) Combining Eqs. (3.24b) and (3.24c), we get R A sin sin sin θ β α = = B (3.24d) Using Eq. (3.24d), we get: sin sin α θ = B R (3.24e) where R is given by Eq. (3.24a). or, sin tan cos SN B OP PN A B θ α θ = = + + (3.24f) Equation (3.24a) gives the magnitude of the resultant and Eqs. (3.24e) and (3.24f) its direction. Equation (3.24a) is known as the Law of cosines and Eq. (3.24d) as the Law of sines. ⊳ 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 the water current are shown in Fig. 3.11 in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure. Fig. 3.11 We can obtain the magnitude of R using the Law of cosine : R v v v v = b 2 c 2 + + 2 cos120 b c o = 25 10 2 25 10 -1/2 22 km/h 2 2 + + × × ( ) ≅ To obtain the direction, we apply the Law of sines R vc sin sin θ φ = or, sin φ θ = v R c sin = 10 sin120 21.8 10 3 2 21.8 0.397 × = × ≅ φ ≅ 23.4 ⊳ 3.7 MOTION IN A PLANE In this section we shall see how to describe motion in two dimensions using vectors. 2024-25 MOTION IN A PLANE 35 3.7.1 Position Vector and Displacement The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame (Fig. 3.12) is given by r i j = + x y ɵ ɵ where x and y are components of r along x-, and y- axes or simply they are the coordinates of the object. (a) (b) Fig. 3.12 (a) Position vector r. (b) Displacement ∆r and average velocity v of a particle. Suppose a particle moves along the curve shown by the thick line and is at P at time t and P′ at time t′ [Fig. 3.12(b)]. Then, the displacement is : ∆r = r′ – r (3.25) and is directed from P to P′. We can write Eq. (3.25) in a component form: ∆r = + − + ( x' y' x ) ( ) ɵ ɵ ɵ ɵ i j i j y = + ɵ ɵ i j ∆ ∆ x y where ∆x = x ′ – x, ∆y = y′ – y (3.26) Velocity The average velocity (v) of an object is the ratio of the displacement and the corresponding time interval : v r i j = = i j + = + ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ t ∆ x y t x t y t ɵ ɵ ɵ ɵ (3.27) Or, ˆ v i j = + v v x y Since v r = ∆ ∆t , the direction of the average velocity 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 : v r r = = → lim∆t t t ∆ 0 ∆ d d (3.28) 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 P3 represent the positions of the object after times ∆t 1 ,∆t 2 , and ∆t 3 . ∆r1 , ∆r2 , and ∆r3 are the displacements of the object in times ∆t 1 , ∆t 2 , 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. 2024-25 36 PHYSICS ∆t 3 , respectively. The direction of the average velocity v is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t 1 ,∆t 2 , and ∆t 3 , (∆t 1 > ∆t 2 > ∆t 3 ). 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 : v r = d dt = +       → lim x t y ∆t t ∆ ∆ ∆ 0 ∆ ɵ ɵ i j (3.29) = + → → ɵ ɵ i lim j x t lim y ∆t ∆t t ∆ ∆ ∆ 0 0 ∆ Or, v i j i j = + = + ɵ ɵ ɵ ɵ d d d d x t y t v v x y . where v x t v y t x = = y d d d d , (3.30a) So, if the expressions for the coordinates x and y are known as functions of time, we can use these equations to find vx and vy . The magnitude of v is then v v v x 2 y 2 = + (3.30b) and the direction of v is given by the angle θ : tan tan 1 θ θ = =         − v v v v y x y x , (3.30c) vx , vy and angle θ are shown in Fig. 3.14 for a velocity vector v at point p. Acceleration The average acceleration a of an object for a time interval ∆t moving in x-y plane is the change in velocity divided by the time interval : ( ) a v i j = = i j + = + ∆ ∆ ∆ ∆ ∆ ∆ ∆ t ∆ v v t v t v t x y x y ɵ ɵ ɵ ɵ (3.31a) Or, a i j = + a a x y ɵ ɵ . (3.31b) The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero : a v = → lim ∆t t ∆ 0 ∆ (3. 32a) Since ∆ ∆ ∆ v = + v v , x y ɵ ɵ i j we have a i = + j → → ɵ ɵ lim v t lim v t t x t y ∆ ∆ ∆ ∆ ∆ 0 0 ∆ Or, a i j = + a a x y ɵ ɵ (3.32b) where, a v t , a v t x x y y = = d d d d (3.32c)* As in the case of velocity, we can understand graphically the limiting process used in defining acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 3.15(a) to (d). P represents the position of the object at time t and P1 , P2 , P3 positions after time ∆t 1 , ∆t 2 , ∆t 3 , respectively (∆t 1 > ∆t 2 >∆t 3 ). The velocity vectors at points P, P1 , P2 , P3 are also shown in Figs. 3.15 (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. By definition, the direction of average acceleration is the same as that of ∆v. We see that as ∆t decreases, the direction of ∆v changes and consequently, the direction of the acceleration changes. Finally, in the limit ∆t g0 [Fig. 3.15(d)], the average acceleration becomes the instantaneous acceleration and has the direction as shown. Fig. 3.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that vx = v cos θ, vy = v sin θ. In terms of x and y, ax and ay can be expressed as 2024-25 MOTION IN A PLANE 37 ⊳ x (m) Note that in one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same 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° between them. Example 3.4 The position of a particle is given by r i j k = + + 3.0t ˆ . ˆ . 2 0 5 0 ˆ 2 t where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0 s. Answer v( ) ( ) r t i j k t t t t2 = = + + d d d d 3.0 2.0 5.0 ɵ ɵ ɵ = + 3.0 .0 ɵ ɵ i j 4 t a ( ) v t j t = d d = +4.0ɵ a = 4.0 m s–2 along y- direction At t = 1.0 s, v = i + j 3.0 4.0 ˆ ˆ It’s magnitude is 2 2 -1 v = 3 4 5.0 m s + = and direction is -1 1 4 = tan tan 53 3 y x v v θ −     °   = ≅       with x-axis. ⊳ 3.8 MOTION IN A PLANE WITH CONSTANT ACCELERATION Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity of the object be v0 at time t = 0 and v at time t. Then, by definition a v v0 v v0 = − − = − t 0 t Or, v v a 0 = + t (3.33a) In terms of components : v v a t x ox x = + v v a t y oy y = + (3.33b) Let us now find how the position r changes with time. We follow the method used in the onedimensional case. Let ro and r be the position vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then, over this time interval t, the average velocity is (vo + v)/2. The displacement is the average velocity multiplied by the time interval : 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. 2024-25 38 PHYSICS ( ) 2 ˆ ˆ2 = + + 5.0 1.5 1.0 t t t i j Therefore, ( ) 2 x t t t = + 5.0 1.5 ( ) 2 y t t = +1.0 Given x (t) = 84 m, t = ? 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s At t = 6 s, y = 1.0 (6)2 = 36.0 m Now, the velocity ( ) d 5.0 3.0 2.0 ˆ ˆ d t t t = = + + r v i j At t = 6 s, v i j = + 23. ɵ ɵ 0 12.0 speed 2 2 1 23 12 26 m s− = = + ≅ v . ⊳ 3.9 PROJECTILE MOTION As an application of the ideas developed in the previous sections, we consider the motion of a projectile. An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independency of the horizontal and the vertical components of projectile motion in his Dialogue on the great world systems (1632). In our discussion, we shall assume that the air resistance has negligible effect on the motion of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle θo with the x-axis as shown in Fig. 3.16. After the object has been projected, the acceleration acting on it is that due to gravity which is directed vertically downward: a j = −g ɵ Or, ax = 0, ay = – g (3.35) The components of initial velocity vo are : vox = vo cos θo voy = vo sin θo (3.36) ⊳ r r v v v a v 0 0 0 0 − =  +      = ( ) + +      2 2 t t t 1 2 2 = + t t v a 0 Or, r r v a = + + 0 0t t 1 2 2 (3.34a) It can be easily verified that the derivative of Eq. (3.34a), i.e. d d r t gives Eq.(3.33a) and it also satisfies the condition that at t=0, r = ro . Equation (3.34a) can be written in component form as x x v t a t = + + 0 ox x 1 2 2 1 2 2 y y v t a t = + + 0 oy y (3.34b) One immediate interpretation of Eq.(3.34b) is that the motions in x- and y-directions can be treated independently of each other. That is, motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. This is an important result and is useful in analysing motion of objects in two dimensions. A similar result holds for three dimensions. The choice of perpendicular directions is convenient in many physical situations, as we shall see in section 3.9 for projectile motion. Example 3.5 A particle starts from origin at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0i ɵ+2.0j ɵ ) m/s 2 . (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time ? Answer From Eq. (3.34a) for r0 = 0, the position of the particle is given by ( ) 1 2 2 t t t = + 0 r v a ( )( ) ˆ ˆ ˆ 2 = + 5.0 1/2 3.0 2.0 it i j + t 2024-25 MOTION IN A PLANE 39 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) The components of velocity at time t can be obtained using Eq.(3.33b) : vx = vox = vo cos θo vy = vo sin θo – g t (3.38) Equation (3.37) gives the x-, and y-coordinates of the position of a projectile at time t in terms of two parameters — initial speed vo and projection angle θo . Notice that the choice of mutually perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in a simplification. One of the components of velocity, i.e. x-component remains constant throughout the motion and only the y- component changes, like an object in free fall in vertical direction. This is shown graphically at few instants in Fig. 3.17. Note that at the point of maximum height, vy = 0 and therefore, -1 tan o y x v v θ = = Equation of path of a projectile What is the shape of the path followed by the projectile? This can be seen by eliminating the time between the expressions for x and y as given in Eq. (3.37). We obtain: ( ) ( ) 2 o 2 o o tan 2 cos g y x x v θ θ = − (3.39) Fig 3.16 Motion of an object projected with velocity vo at angle θ0 . 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. 3.17 The path of a projectile is a parabola. Time of maximum height How much time does the projectile take to reach the maximum height ? Let this time be denoted by tm . Since at this point, vy = 0, we have from Eq. (3.38): vy = vo sinθo – g tm = 0 Or, tm = vo sinθo /g (3.40a) The total time Tf during which the projectile is in flight can be obtained by putting y = 0 in Eq. (3.37). We get : Tf = 2 (vo sin θo )/g (3.40b) Tf is known as the time of flight of the projectile. We note that Tf = 2 tm , which is expected because of the symmetry of the parabolic path. Maximum height of a projectile The maximum height hm reached by the projectile can be calculated by substituting t = tm in Eq. (3.37) : y h v( ) v g g v g m 0 0 0 = =         −         sin sin 2 sin 0 0 0 2 θ θ θ Or, ( ) h v m 0 = sin 0 θ 2 2g (3.41) Horizontal range of a projectile The horizontal distance travelled by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal 2024-25 40 PHYSICS ⊳ ⊳ ⊳ range, R. It is the distance travelled during the time of flight Tf . Therefore, the range R is R = (vo cos θo ) (Tf ) =(vo cos θo ) (2 vo sin θo )/g Or, R v g 0 2 = sin 2 0 θ (3.42a) Equation (3.42a) shows that for a given projection velocity vo , R is maximum when sin 2θ0 is maximum, i.e., when θ0 = 450 . The maximum horizontal range is, therefore, R v g m 0 2 = (3.42b) Example 3.6 Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement. Answer For a projectile launched with velocity vo at an angle θo , the range is given by sin2 0 2 v0 R g θ = Now, for angles, (45° + α) and ( 45° – α), 2θo is (90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are the same, equal to that of cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α. ⊳ Example 3.7 A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m s-1. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). Answer We choose the origin of the x-,and yaxis at the edge of the cliff and t = 0 s at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction. The x-, and ycomponents of the motion can be treated independently. The equations of motion are : x (t) = xo + vox t y (t) = yo + voy t +(1/2) a y t2 Here, xo = yo = 0, voy = 0, a y = –g = –9.8 m s-2 , vox = 15 m s-1 . The stone hits the ground when y(t) = – 490 m. – 490 m = –(1/2)(9.8) t 2 . This gives t =10 s. The velocity components are vx = vox and vy = voy – g t so that when the stone hits the ground : vox = 15 m s–1 voy = 0 – 9.8 × 10 = – 98 m s–1 Therefore, the speed of the stone is 2 2 15 98 99 m s 2 2 1 v v x y − + = + = ⊳ Example 3.8 A cricket ball is thrown at a speed of 28 m s–1 in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level. Answer (a) The maximum height is given by ( ) ( ) ( ) 2 2 o sin 28sin 30 m 2 2 9.8 0 m v h g θ ° = = = × × = 14 14 2 9.8 10.0 m (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 (c) The distance from the thrower to the point where the ball returns to the same level is R ( ) 2 o o sin2 28 28 sin60 69 m 9.8 v o g θ × × = = = ⊳ 3.10 UNIFORM CIRCULAR MOTION When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Suppose an object is moving with uniform speed v in a circle of radius R as shown in Fig. 3.18. Since the velocity of the object is changing continuously in direction, the object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. 2024-25 MOTION IN A PLANE 41 Let r and r′ be the position vectors and v and v′ the velocities of the object when it is at point P and P′ as shown in Fig. 3.18(a). By definition, velocity at a point is along the tangent at that point in the direction of motion. The velocity vectors v and v′ are as shown in Fig. 3.18(a1). ∆v is obtained in Fig. 3.18 (a2) using the triangle law of vector addition. Since the path is circular, v is perpendicular to r and so is v′ to r′. Therefore, ∆v is perpendicular to ∆r. Since average acceleration is along ∆v a v =       ∆ ∆t , the average acceleration a is perpendicular to ∆r. If we place ∆v on the line that bisects the angle between r and r′, we see that it is directed towards the centre of the circle. Figure 3.18(b) shows the same quantities for smaller time interval. ∆v and hence a is again directed towards the centre. In Fig. 3.18(c), ∆té 0 and the average acceleration becomes the instantaneous acceleration. It is directed towards the centre. Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle. Let us now find the magnitude of the acceleration. The magnitude of a is, by definition, given by a v = →∆ ∆ t 0 ∆t Let the angle between position vectors r and r′ be ∆θ. Since the velocity vectors v and v′ are always perpendicular to the position vectors, the angle between them is also ∆θ . Therefore, the triangle CPP′ formed by the position vectors and the triangle GHI formed by the velocity vectors v, v′ and ∆v are similar (Fig. 3.18a). Therefore, the ratio of the base-length to side-length for one of the triangles is equal to that of the other triangle. That is : ∆ v ∆ r v R = Or, ∆ ∆ v r = v R Therefore, a v r r = → = → = ∆ → ∆ ∆ ∆ ∆ ∆ ∆ ∆ t 0 0 t t R R 0 ∆ v t v t t If ∆t is small, ∆θ will also be small and then arc PP′ can be approximately taken to be|∆r|: ∆ ∆ r ≅ v t ∆ ∆ r t ≅ v Or, ∆ ∆ t 0 ∆t v → = r Therefore, the centripetal acceleration ac is : 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. * In the limit ∆té0, ∆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. lim lim lim lim lim 2024-25 42 PHYSICS ⊳ ac = v R       v = v 2/R (3.43) Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude v 2 /R and is always directed towards the centre. This is why this acceleration is called centripetal acceleration (a term proposed by Newton). A thorough analysis of centripetal acceleration was first published in 1673 by the Dutch scientist Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. “Centripetal” comes from a Greek term which means ‘centre-seeking’. Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes — pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector. We have another way of describing the velocity and the acceleration of an object in uniform circular motion. As the object moves from P to P′ in time ∆t (= t′ – t), the line CP (Fig. 3.18) turns through an angle ∆θ as shown in the figure. ∆θ is called angular distance. We define the angular speed ω (Greek letter omega) as the time rate of change of angular displacement : ω θ = ∆ ∆t (3.44) Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then : v s t = ∆ ∆ but ∆s = R ∆θ. Therefore : v R t = = ∆ ∆ θ R ω v = R ω (3.45) We can express centripetal acceleration ac in terms of angular speed : a v R R R c = = = R 2 2 ω 2 ω 2 a R c = ω 2 (3.46) The time taken by an object to make one revolution is known as its time period T and the number of revolution made in one second is called its frequency ν (=1/T). However, during this time the distance moved by the object is s = 2πR. Therefore, v = 2πR/T =2πRν (3.47) In terms of frequency ν, we have ω = 2πν v = 2πRν ac = 4π 2 ν 2R (3.48) Example 3.9 An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ? Answer This is an example of uniform circular motion. Here R = 12 cm. The angular speed ω is given by ω = 2π/T = 2π × 7/100 = 0.44 rad/s The linear speed v is : v =ω R = 0.44 s-1 × 12 cm = 5.3 cm s-1 The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant: a = ω2 R = (0.44 s–1) 2 (12 cm) = 2.3 cm s-21 and the tangent at P1, and α1 is the angle between F1 and the radius vector OP1 ; φ1 + α1 = 90°. The torque due to F1 about the origin is OP1 × F1 . Now OP1 = OC + OP1 . [Refer to Fig. 6.17(b).] Since OC is along the axis, the torque resulting from it is excluded from our consideration. The effective torque due to F1 is τ 1 = CP × F1 ; it is directed along the axis of rotation and has a magnitude τ 1 = r 1 F1 sinα , Therefore, dW1 = τ 1 dθ If there are more than one forces acting on the body, the work done by all of them can be added to give the total work done on the body. Denoting the magnitudes of the torques due to the different forces as τ 1 , τ 2 , … etc, d ( ...)d W 1 2 = + + τ τ θ Remember, the forces giving rise to the torques act on different particles, but the angular displacement dθ is the same for all particles. Since all the torques considered are parallel to the fixed axis, the magnitude τ of the total torque is just the algebraic sum of the magnitudes of the torques, i.e., τ = τ1 + τ2 + ..... We, therefore, have d d W = τ θ (6.39) This expression gives the work done by the total (external) torque τ which acts on the body rotating about a fixed axis. Its similarity with the corresponding expression dW= F ds for linear (translational) motion is obvious. Dividing both sides of Eq. (6.39) by dt gives d d d d W P t t θ = = = τ τω or P = τω (6.40) This is the instantaneous power. Compare this expression for power in the case of rotational motion about a fixed axis with that of power in the case of linear motion, P = Fv In a perfectly rigid body there is no internal motion. The work done by external torques is Table 6.2 Comparison of Translational and Rotational Motion Linear Motion Rotational Motion about a Fixed Axis 1 Displacement x Angular displacement θ 2 Velocity v = dx/dt Angular velocity ω = dθ/dt 3 Acceleration a = dv/dt Angular acceleration α = dω/dt 4 Mass M Moment of inertia I 5 Force F = Ma Torque τ = I α 6 Work dW = F ds Work W = τ dθ 7 Kinetic energy K = Mv2/2 Kinetic energy K = Iω2/2 8 Power P = F v Power P = τω 9 Linear momentum p = Mv Angular momentum L = Iω 2024-25 120 PHYSICS u therefore, not dissipated and goes on to increase the kinetic energy of the body. The rate at which work is done on the body is given by Eq. (6.40). This is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is d d d t d I I t ω ω ω 2 2 2 2       = ( ) We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid and also the axis does not change its position with respect to the body. Since α ω = d /d ,t we get d dt I I ω ω α 2 2       = Equating rates of work done and of increase in kinetic energy, τω ω α = I τ α = I (6.41) Eq. (6.41) is similar to Newton’s second law for linear motion expressed symbolically as F = ma Just as force produces acceleration, torque produces angular acceleration in a body. The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body. In this respect, Eq.(6.41) can be called Newton’s second law for rotational motion about a fixed axis. Example 6.12 A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 6.31. The flywheel is mounted on a horizontal axle with frictionless bearings. (a) Compute the angular acceleration of the wheel. (b) Find the work done by the pull, when 2m of the cord is unwound. (c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest. (d) Compare answers to parts (b) and (c). Answer Fig. 6.31 (a) We use I α = τ the torque τ = F R = 25 × 0.20 Nm (as R = 0.20m) = 5.0 Nm I = Moment of inertia of flywheel about its axis 2 2 MR = = 2 20.0 (0.2) 2 × = 0.4 kg m2 α = angular acceleration = 5.0 N m/0.4 kg m2 = 12.5 s–2 (b) Work done by the pull unwinding 2m of the cord = 25 N × 2m = 50 J (c) Let ω be the final angular velocity. The kinetic energy gained = 1 2 2 Iω , since the wheel starts from rest. Now, 2 2 ω ω αθ ω = + = 0 0 2 , 0 The angular displacement θ = length of unwound string / radius of wheel = 2m/0.2 m = 10 rad ω 2 2 = × × = 2 12 5 10 0 250 . . d/s)(ra ∴ (d) The answers are the same, i.e. the kinetic energy gained by the wheel = work done by the force. There is no loss of energy due to friction. ⊳ 2024-25 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 121 6.12 ANGULAR MOMENTUM IN CASE OF ROTATION ABOUT A FIXED AXIS We have studied in section 6.7, the angular momentum of a system of particles. We already know from there that the time rate of total angular momentum of a system of particles about a point is equal to the total external torque on the system taken about the same point. When the total external torque is zero, the total angular momentum of the system is conserved. We now wish to study the angular momentum in the special case of rotation about a fixed axis. The general expression for the total angular momentum of the system of n particles is L r p = × = ∑ i i i N 1 (6.25b) We first consider the angular momentum of a typical particle of the rotating rigid body. We then sum up the contributions of individual particles to get L of the whole body. For a typical particle l = r × p. As seen in the last section r = OP = OC + CP [Fig. 6.17(b)]. With p = m v , l = × + × (OC v CP v m m ) ( ) The magnitude of the linear velocity v of the particle at P is given by v = ωr ⊥ where r ⊥ is the length of CP or the perpendicular distance of P from the axis of rotation. Further, v is tangential at P to the circle which the particle describes. Using the right-hand rule one can check that CP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) is kˆ . Hence ( ) ˆ CP v k × = m r mv ⊥ = 2ω ˆ mr⊥ k (since υ = ωr⊥ ) Similarly, we can check that OC × v is perpendicular to the fixed axis. Let us denote the part of l along the fixed axis (i.e. the z-axis) by l z , then lz = × CP v m = 2ω ˆ mr⊥ k and l l = + × z OC v m We note that l z is parallel to the fixed axis, but l is not. In general, for a particle, the angular momentum l is not along the axis of rotation, i.e. for a particle, l and ω are not necessarily parallel. Compare this with the corresponding fact in translation. For a particle, p and v are always parallel to each other. For computing the total angular momentum of the whole rigid body, we add up the contribution of each particle of the body. Thus We denote by L ⊥ and Lz the components of L respectively perpendicular to the z-axis and along the z-axis; L OC v ⊥ = × ∑ i i i m (6.42a) where mi and vi are respectively the mass and the velocity of the i th particle and Ci is the centre of the circle described by the particle; and or ω ˆ L k z = I (6.42b) The last step follows since the perpendicular distance of the i th particle from the axis is r i ; and by definition the moment of inertia of the body about the axis of rotation is I m r =∑ i i 2 . Note L L L = + z ⊥ (6.42c) The rigid bodies which we have mainly considered in this chapter are symmetric about the axis of rotation, i.e. the axis of rotation is one of their symmetry axes. For such bodies, for a given OCi , for every particle which has a velocity vi , there is another particle of velocity –vi located diametrically opposite on the circle with centre Ci described by the particle. Together such pairs will contribute zero to L ⊥ and as a result for symmetric bodies L⊥ is zero, and hence ωˆ L L k = = z I (6.42d) For bodies, which are not symmetric about the axis of rotation, L is not equal to Lz and hence L does not lie along the axis of rotation. Referring to Table 6.1, can you tell in which cases L = Lz will not apply? Let us differentiate Eq. (6.42b). Since kˆ is a fixed (constant) vector, we get d d d d z t t ( ) L = (I ) k      ω  ˆ Now, Eq. (6.28b) states d dt = L τ 2024-25 122 PHYSICS As we have seen in the last section, only those components of the external torques which are along the axis of rotation, need to be taken into account, when we discuss rotation about a fixed axis. This means we can take ˆ τ = τk . Since L L L = + z ⊥ and the direction of Lz (vector kˆ ) is fixed, it follows that for rotation about a fixed axis, d ˆ d = τ L k z t (6.43a) and d 0 dt ⊥ = L (6.43b) Thus, for rotation about a fixed axis, the component of angular momentum perpendicular to the fixed axis is constant. As ωˆ L k z = I , we get from Eq. (6.43a), ( ) d d I t ω τ = (6.43c) If the moment of inertia I does not change with time, ( ) d d d d I I I t t ω ω α = = and we get from Eq. (6.43c), τ α = I (6.41) We have already derived this equation using the work - kinetic energy route. 6.12.1 Conservation of angular momentum We are now in a position to revisit the principle of conservation of angular momentum in the context of rotation about a fixed axis. From Eq. (6.43c), if the external torque is zero, Lz = Iω = constant (6.44) For symmetric bodies, from Eq. (6.42d), Lz may be replaced by L .(L and Lz are respectively the magnitudes of L and Lz .) This then is the required form, for fixed axis rotation, of Eq. (6.29a), which expresses the general law of conservation of angular momentum of a system of particles. Eq. (6.44) applies to many situations that we come across in daily life. You may do this experiment with your friend. Sit on a swivel chair (a chair with a seat, free to rotate about a pivot) with your arms folded and feet not resting on, i.e., away from, the ground. Ask your friend to rotate the chair rapidly. While the chair is rotating with considerable angular speed stretch your arms horizontally. What happens? Your angular speed is reduced. If you bring back your arms closer to your body, the angular speed increases again. This is a situation where the principle of conservation of angular momentum is applicable. If friction in the rotational Fig 6.32 (a) A demonstration of conservation of angular momentum. A girl sits on a swivel chair and stretches her arms/ brings her arms closer to the body. Fig 6.32 (b) An acrobat employing the principle of conservation of angular momentum in her performance. 2024-25 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 123 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 R r = ∑m M i i 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 L r p = × = ∑ i i n i 1 The torque or moment of force on a system of n particles about the origin is = × ∑r F i i 1 τ The force Fi acting on the i th 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 τ int = 0 and mechanism is neglected, there is no external torque about the axis of rotation of the chair and hence Iω is constant. Stretching the arms increases I about the axis of rotation, resulting in decreasing the angular speed ω. Bringing the arms closer to the body has the opposite effect. A circus acrobat and a diver take advantage of this principle. Also, skaters and classical, Indian or western, dancers performing a pirouette (a spinning about a tip–top) on the toes of one foot display ‘mastery’ over this principle. Can you explain?