Motion In A Plane PDF

CHAPTER FOUR MOTION IN A PLANE 4.1 INTRODUCTION In the last chapter we developed the concepts of position, displacement, velocity and accelerati...

CHAPTER FOUR MOTION IN A PLANE 4.1 INTRODUCTION In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to 4.1 Introduction describe the motion of an object along a straight line. We 4.2 Scalars and vectors found that the directional aspect of these quantities can be 4.3 Multiplication of vectors by taken care of by + and – signs, as in one dimension only two real numbers directions are possible. But in order to describe motion of an 4.4 Addition and subtraction of object in two dimensions (a plane) or three dimensions vectors — graphical method (space), we need to use vectors to describe the above- 4.5 Resolution of vectors mentioned physical quantities. Therefore, it is first necessary 4.6 Vector addition — analytical to learn the language of vectors. What is a vector ? How to method add, subtract and multiply vectors ? What is the result of 4.7 Motion in a plane multiplying a vector by a real number ? We shall learn this 4.8 Motion in a plane with to enable us to use vectors for defining velocity and constant acceleration acceleration in a plane. We then discuss motion of an object 4.9 Relative velocity in two in a plane. As a simple case of motion in a plane, we shall dimensions discuss motion with constant acceleration and treat in detail 4.10 Projectile motion the projectile motion. Circular motion is a familiar class of 4.11 Uniform circular motion motion that has a special significance in daily-life situations. We shall discuss uniform circular motion in some detail. Summary The equations developed in this chapter for motion in a Points to ponder plane can be easily extended to the case of three dimensions. Exercises Additional exercises 4.2 SCALARS AND VECTORS In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided 66 PHYSICS just as the ordinary numbers*. For example, represented by another position vector, OP′ if the length and breadth of a rectangle are denoted by r′. The length of the vector r 1.0 m and 0.5 m respectively, then its represents the magnitude of the vector and its perimeter is the sum of the lengths of the direction is the direction in which P lies as seen four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = from O. If the object moves from P to P′, the 3.0 m. The length of each side is a scalar vector PP′ (with tail at P and tip at P′) is called and the perimeter is also a scalar. Take the displacement vector corresponding to another example: the maximum and motion from point P (at time t) to point P′ (at time t′). minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of 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 Fig. 4.1 (a) Position and displacement vectors. parallelogram law of addition. So, a vector is (b) Displacement vector PQ and different specified by giving its magnitude by a number courses of motion. and its direction. Some physical quantities that It is important to note that displacement are represented by vectors are displacement, vector is the straight line joining the initial and velocity, acceleration and force. final positions and does not depend on the actual To represent a vector, we use a bold face type path undertaken by the object between the two in this book. Thus, a velocity vector can be positions. For example, in Fig. 4.1b, given the represented by a symbol v. Since bold face is initial and final positions as P and Q, the difficult to produce, when written by hand, a displacement vector is the same PQ for different vector is often represented r by an arrow placed r paths of journey, say PABCQ, PDQ, and PBEFQ. over a letter, say v. Thus, both v and v Therefore, the magnitude of displacement is represent the velocity vector. The magnitude of either less or equal to the path length of an a vector is often called its absolute value, object between two points. This fact was indicated by |v| = v. Thus, a vector is emphasised in the previous chapter also while represented by a bold face, e.g. by A, a, p, q, r,... x, y, with respective magnitudes denoted by light discussing motion along a straight line. face A, a, p, q, r,... x, y. 4.2.2 Equality of Vectors 4.2.1 Position and Displacement Vectors Two vectors A and B are said to be equal if, and To describe the position of an object moving in only if, they have the same magnitude and the a plane, we need to choose a convenient point, same direction.** say O as origin. Let P and P′ be the positions of Figure 4.2(a) shows two equal vectors A and the object at time t and t′, respectively [Fig. 4.1(a)]. B. We can easily check their equality. Shift B We join O and P by a straight line. Then, OP is parallel to itself until its tail Q coincides with that the position vector of the object at time t. An of A, i.e. Q coincides with O. Then, since their arrow is marked at the head of this line. It is tips S and P also coincide, the two vectors are represented by a symbol r, i.e. OP = r. Point P′ is 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. MOTION IN A PLANE 67 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. 4.4 ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD Fig. 4.2 (a) Two equal vectors A and B. (b) Two As mentioned in section 4.2, vectors, by vectors A′ and B′ are unequal though they definition, obey the triangle law or equivalently, are of the same length. the parallelogram law of addition. We shall now describe this law of addition using the graphical as A = B. Note that in Fig. 4.2(b), vectors A′ and method. Let us consider two vectors A and B that B′ have the same magnitude but they are not lie in a plane as shown in Fig. 4.4(a). The lengths equal because they have different directions. of the line segments representing these vectors Even if we shift B′ parallel to itself so that its tail are proportional to the magnitude of the vectors. Q′ coincides with the tail O′ of A′, the tip S′ of B′ To find the sum A + B, we place vector B so that does not coincide with the tip P′ of A′. its tail is at the head of the vector A, as in 4.3 MULTIPLICATION OF VECTORS BY REAL Fig. 4.4(b). Then, we join the tail of A to the head NUMBERS of B. This line OQ represents a vector R, that is, Multiplying a vector A with a positive number λ the sum of the vectors A and B. Since, in this gives a vector whose magnitude is changed by procedure of vector addition, vectors are 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. 4.3(a). Multiplying a vector A by a negative number λ gives a vector λA 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 4.3(b). (c) (d) Fig. 4.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2. Fig. 4.4 (a) Vectors A and B. (b) Vectors A and B (b) Vector A and resultant vectors after added graphically. (c) Vectors B and A multiplying it by a negative number –1 added graphically. (d) Illustrating the and –1.5. associative law of vector addition. 68 PHYSICS arranged head to tail, this graphical method is What is the physical meaning of a zero vector? called the head-to-tail method. The two vectors Consider the position and displacement vectors and their resultant form three sides of a triangle, in a plane as shown in Fig. 4.1(a). Now suppose so this method is also known as triangle method that an object which is at P at time t, moves to of vector addition. If we find the resultant of P′ and then comes back to P. Then, what is its B + A as in Fig. 4.4(c), the same vector R is displacement? Since the initial and final obtained. Thus, vector addition is commutative: positions coincide, the displacement is a “null A+B=B+A (4.1) vector”. The addition of vectors also obeys the associative Subtraction of vectors can be defined in terms law as illustrated in Fig. 4.4(d). The result of of addition of vectors. We define the difference adding vectors A and B first and then adding of two vectors A and B as the sum of two vectors vector C is the same as the result of adding B A and –B : and C first and then adding vector A : A – B = A + (–B) (4.5) (A + B) + C = A + (B + C) (4.2) It is shown in Fig 4.5. The vector –B is added to What is the result of adding two equal and vector A to get R2 = (A – B). The vector R1 = A + B opposite vectors ? Consider two vectors A and is also shown in the same figure for comparison. –A shown in Fig. 4.3(b). Their sum is A + (–A). We can also use the parallelogram method to Since the magnitudes of the two vectors are the find the sum of two vectors. Suppose we have same, but the directions are opposite, the two vectors A and B. To add these vectors, we resultant vector has zero magnitude and is bring their tails to a common origin O as represented by 0 called a null vector or a zero shown in Fig. 4.6(a). Then we draw a line from vector : the head of A parallel to B and another line from the head of B parallel to A to complete a A–A=0 |0|= 0 (4.3) parallelogram OQSP. Now we join the point of Since the magnitude of a null vector is zero, its the intersection of these two lines to the origin direction cannot be specified. O. The resultant vector R is directed from the The null vector also results when we multiply common origin O along the diagonal (OS) of the a vector A by the number zero. The main parallelogram [Fig. 4.6(b)]. In Fig.4.6(c), the properties of 0 are : triangle law is used to obtain the resultant of A A+0=A and B and we see that the two methods yield the λ0=0 same result. Thus, the two methods are 0A=0 (4.4) equivalent. Fig. 4.5 (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. MOTION IN A PLANE 69 Fig. 4.6 (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. Example 4.1 Rain is falling vertically with t 4.5 RESOLUTION OF VECTORS a speed of 35 m s–1. Winds starts blowing Let a and b be any two non-zero vectors in a after sometime with a speed of 12 m s–1 in plane with different directions and let A be east to west direction. In which direction another vector in the same plane(Fig. 4.8). A can should a boy waiting at a bus stop hold be expressed as a sum of two vectors – one his umbrella ? 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 (4.6) But since OQ is parallel to a, and QP is parallel to b, we can write : Fig. 4.7 OQ = λ a, and QP = µ b (4.7) where λ and µ are real numbers. Answer The velocity of the rain and the wind are represented by the vectors vr and vw in Fig. Therefore, A = λ a + µ b (4.8) 4.7 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 = vr + vw = 35 + 12 ms = 37 m s The direction θ that R makes with the vertical is given by vw 12 tan θ = = = 0.343 Fig. 4.8 (a) Two non-colinear vectors a and b. vr 35 (b) Resolving a vector A in terms of vectors a and b. θ = tan ( 0.343) = 19° -1 Or, 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. t respectively. Using this method one can resolve 70 PHYSICS a given vector into two component vectors along and A2 is parallel to
j , we have : a set of two vectors – all the three lie in the same plane. It is convenient to resolve a general vector A1= Ax
i , A2 = Ay
j (4.11) along the axes of a rectangular coordinate where Ax and Ay are real numbers. system using vectors of unit magnitude. These are called unit vectors that we discuss now. Thus, A = Ax
i + Ay
j (4.12) Unit vectors: A unit vector is a vector of unit magnitude and points in a particular direction. This is represented in Fig. 4.9(c). The quantities It has no dimension and unit. It is used to specify Ax and Ay are called x-, and y- components of the a direction only. Unit vectors along the x-, y- vector A. Note that Ax is itself not a vector, but and z-axes of a rectangular coordinate system A
i is a vector, and so is A
j. Using simple x y are denoted by î , ĵ and k̂ , respectively, as trigonometry, we can express Ax and Ay in terms shown in Fig. 4.9(a). of the magnitude of A and the angle θ it makes Since these are unit vectors, we have with the x-axis : Ax = A cos θ  î  =  ĵ  =  k̂ =1 (4.9) Ay = A sin θ (4.13) These unit vectors are perpendicular to each As is clear from Eq. (4.13), a component of a other. In this text, they are printed in bold face vector can be positive, negative or zero with a cap (^) to distinguish them from other depending on the value of θ. vectors. Since we are dealing with motion in two Now, we have two ways to specify a vector A dimensions in this chapter, we require use of in a plane. It can be specified by : only two unit vectors. If we multiply a unit vector, (i) its magnitude A and the direction θ it makes say n by a scalar, the result is a vector with the x-axis; or (ii) its components Ax and Ay λ = λ n. In general, a vector A can be written as If A and θ are given, Ax and Ay can be obtained A = |A| n (4.10) using Eq. (4.13). If Ax and Ay are given, A and θ can be obtained as follows : where n is a unit vector along A. 2 2 2 2 2 2 We can now resolve a vector A in terms A x + Ay = A cos θ + A sin θ of component vectors that lie along unit vectors = A2 î and
j. Consider a vector A that lies in x-y Or, A= A 2x + Ay2 (4.14) plane as shown in Fig. 4.9(b). We draw lines from the head of A perpendicular to the coordinate Ay −1 Ay axes as in Fig. 4.9(b), and get vectors A1 and A2 tan θ = , θ = tan And (4.15) such that A + A = A. Since A is parallel to
i Ax Ax 1 2 1 Fig. 4.9 (a) Unit vectors
i ,
j and k
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
i and
j. MOTION IN A PLANE 71 So far we have considered a vector lying in B = B x i
+ By
j an x-y plane. The same procedure can be used to resolve a general vector A into three Let R be their sum. We have components along x-, y-, and z-axes in three R=A+B dimensions. If α , β, and γ are the angles * between A and the x-, y-, and z-axes, respectively ( ) ( = A x
i + Ay
j + B x i
+ By
j ) (4.19a) Fig. 4.9(d), we have Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (4.19a) as convenient to us : ( R = ( A x + B x ) i
+ Ay + By
j ) (4.19b) Since R = R x i
+ Ry
j (4.20) we have, R x = A x + B x , R y = Ay + B y (4.21) 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 = A i
+ A
j + A k
x y z B = B x i
+ By
j + B z k
(d) R = A + B = R x i
+ Ry
j + Rz k
Fig. 4.9 (d) A vector A resolved into components along x-, y-, and z-axes with Rx = Ax + Bx A x = A cos α , A y = A cos β , A z = A cos γ (4.16a) Ry = Ay + By In general, we have Rz = Az + Bz (4.22) A = Ax ˆi + Ay ˆj + Az k ˆ (4.16b) This method can be extended to addition and The magnitude of vector A is subtraction of any number of vectors. For A = Ax2 + Ay2 + Az2 (4.16c) example, if vectors a, b and c are given as A position vector r can be expressed as a = a i
+ a
j + a k
x y z r = x i
+ y
j + z k
(4.17) b = b x i
+ by
j + bz k
where x, y, and z are the components of r along x-, y-, z-axes, respectively. c = c x
i + c y
j + c z k
(4.23a) 4.6 VECTOR ADDITION – ANALYTICAL then, a vector T = a + b – c has components : METHOD Tx = a x + b x − c x Although the graphical method of adding vectors Ty = a y + by − c y (4.23b) helps us in visualising the vectors and the Tz = a z + b z − c z. resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors Example 4.2 Find the magnitude and t by combining their respective components. Consider two vectors A and B in x-y plane with direction of the resultant of two vectors A and B in terms of their magnitudes and components Ax, Ay and Bx, By : angle θ between them. A = A x i
+ Ay
j (4.18) * Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. 72 PHYSICS t Example 4.3 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. 4.10 the water current are shown in Fig. 4.11 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. 4.10). 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 cos θ (4.24a) 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, = (4.24b) sin θ sin α Fig. 4.11 Similarly, PM = A sin α = B sin β We can obtain the magnitude of R using the Law A B of cosine : or, = (4.24c) sin β sin α Combining Eqs. (4.24b) and (4.24c), we get R = v 2b + v c2 + 2v bv c cos120o R A B = = (4.24d) = 252 + 102 + 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. (4.24d), we get: R vc vc B = or, sin φ = sin θ sin α = sin θ (4.24e) sin θ sin φ R R where R is given by Eq. (4.24a). 10 × sin120  10 3 = = ≅ 0.397 SN B sin θ 21.8 2 × 21.8 or, tan α = = (4.24f) OP + PN A + B cos θ  φ ≅ 23.4 t Equation (4.24a) gives the magnitude of the resultant and Eqs. (4.24e) and (4.24f) its direction. 4.7 MOTION IN A PLANE Equation (4.24a) is known as the Law of cosines In this section we shall see how to describe and Eq. (4.24d) as the Law of sines. t motion in two dimensions using vectors. MOTION IN A PLANE 73 4.7.1 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. 4.12(b)]. Then, the displacement is : reference frame (Fig. 4.12) is given by ∆r = r′ – r (4.25) and is directed from P to P′. r = x i
+ y
j We can write Eq. (4.25) in a component form: where x and y are components of r along x-, and y- axes or simply they are the coordinates of the object. ∆r ( ) ( = x'
i + y'
j − x i
+ y
j ) =
i∆x +
j∆y where ∆x = x ′ – x, ∆y = y′ – y (4.26) Velocity The average velocity ( v ) of an object is the ratio of the displacement and the corresponding time interval : ∆r ∆x
i + ∆y
j ∆x
∆y v= = = i
+j (4.27) ∆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. 4.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : ∆ r dr v = lim = (4.28) ∆t d t ∆t → 0 The meaning of the limiting process can be easily understood with the help of Fig 4.13(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. 4.12 (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. 4.13 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. 74 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, ( ∆ t 1 > ∆ t 2 > ∆ t 3 ). As ∆ t → 0, ∆ r → 0 and is along the tangent to the path [Fig. 4.13(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. 4.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that dr v= vx = v cos θ, vy = v sin θ. dt  ∆x ∆y  Or, a = a x i
+ a y
j. = lim  i
+
j  (4.29) ∆t → 0 ∆t ∆t  (4.31b) The acceleration (instantaneous acceleration) ∆x
∆y = i
lim + j lim is the limiting value of the average acceleration ∆t → 0 ∆t ∆t → 0 ∆t as the time interval approaches zero : dx
dy ∆v Or, v = i
+j = v x i
+ v y
j. a = lim (4.32a) dt dt ∆t → 0 ∆t dx dy where vx = , vy = (4.30a) Since ∆v = ∆v x
i + ∆v y
j, we have dt dt So, if the expressions for the coordinates x and ∆v x ∆vy a = i
lim +
j lim y are known as functions of time, we can use ∆t → 0 ∆t ∆t → 0 ∆t these equations to find vx and vy. The magnitude of v is then Or, a = a x i
+ a y
j (4.32b) v = v x + vy 2 2 (4.30b) dv x dv y where, ax = , ay = (4.32c)* and the direction of v is given by the angle θ : dt dt As in the case of velocity, we can understand vy v  graphically the limiting process used in defining −1  y  tanθ = , θ = tan (4.30c) acceleration on a graph showing the path of the   vx vx  object’s motion. This is shown in Figs. 4.15(a) to (d). P represents the position of the object at vx, vy and angle θ are shown in Fig. 4.14 for a time t and P1, P2, P3 positions after time ∆t1, ∆t2, velocity vector v. ∆t3, respectively (∆t 1> ∆t 2>∆t 3). The velocity Acceleration vectors at points P, P1, P2, P3 are also shown in Figs. 4.15 (a), (b) and (c). In each case of ∆t, ∆v is The average acceleration a of an object for a obtained using the triangle law of vector addition. time interval ∆t moving in x-y plane is the change in velocity divided by the time interval : By definition, the direction of average acceleration is the same as that of ∆v. We see a= ∆v = ( ∆ v x
i + v y
j ) = ∆v x i
+ ∆v y
j (4.31a) that as ∆t decreases, the direction of ∆v changes and consequently, the direction of the ∆t ∆t ∆t ∆t acceleration changes. Finally, in the limit ∆t g0 * In terms of x and y, ax and ay can be expressed as 2 2 d dx d x d dy d y ax = = , ay = = dt dt 2 dt dt 2 dt dt MOTION IN A PLANE 75 x (m) Fig. 4.15 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. Fig. 4.15(d), the average acceleration becomes  vy  −1  4  ° the instantaneous acceleration and has the θ = tan -1   = tan   ≅ 53 with x-axis. direction as shown.  vx  3 Note that in one dimension, the velocity and the acceleration of an object are always along t the same straight line (either in the same di- rection or in the opposite direction). However, 4.8 MOTION IN A PLANE WITH CONSTANT for motion in two or three dimensions, veloc- ACCELERATION ity and acceleration vectors may have any angle between 0° and 180° between them. Suppose that an object is moving in x-y plane t and its acceleration a is constant. Over an Example 4.4 The position of a particle is interval of time, the average acceleration will given by equal this constant value. Now, let the velocity r = 3.0t ˆi + 2.0t 2ˆj + 5.0 kˆ of the object be v0 at time t = 0 and v at time t. where t is in seconds and the coefficients Then, by definition have the proper units for r to be in metres. v − v0 v − v0 (a) Find v(t) and a(t) of the particle. (b) Find a= = t−0 t the magnitude and direction of v(t) at t = 1.0 s. Or, v = v 0 + at (4.33a) In terms of components : Answer v x = vox + a x t v( t ) = dr dt = d dt ( 2 3.0 t i
+ 2.0t
j + 5.0 k
) vy = v oy + a y t (4.33b) Let us now find how the position r changes with = 3.0
i + 4.0t
j time. We follow the method used in the one- dv dimensional case. Let ro and r be the position a (t) = = +4.0
j vectors of the particle at time 0 and t and let the dt velocities at these instants be vo and v. Then, a = 4.0 m s–2 along y- direction over this time interval t, the average velocity is At t = 1.0 s, v = 3.0ˆi + 4.0ˆj (vo + v)/2. The displacement is the average velocity multiplied by the time interval : -1 It’s magnitude is v = 3 + 4 = 5.0 m s 2 2 v + v0 ( v0 + at ) + v 0 and direction is r − r0 = t= t 2 2 76 PHYSICS Given x (t) = 84 m, t = ? 1 2 = v0t + at 2 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s At t = 6 s, y = 1.0 (6)2 = 36.0 m 1 2 Or, r = r0 + v 0t + at (4.34a) dr 2 Now, the velocity v = = (5.0 + 3.0 t ) ˆi + 2.0 t ˆj dt It can be easily verified that the derivative of dr At t = 6 s, v = 23.0i
+ 12.0
j Eq. (4.34a), i.e. gives Eq.(4.33a) and it also dt speed = v = 232 + 122 ≅ 26 m s−1. t satisfies the condition that at t=0, r = r o. Equation (4.34a) can be written in component 4.9 RELATIVE VELOCITY IN TWO form as DIMENSIONS 1 The concept of relative velocity, introduced in x = x 0 + v ox t + a xt 2 2 section 3.7 for motion along a straight line, can be easily extended to include motion in a plane 1 or in three dimensions. Suppose that two objects y = y0 + v oy t + ayt 2 (4.34b) 2 A and B are moving with velocities vA and vB One immediate interpretation of Eq.(4.34b) is that (each with respect to some common frame of the motions in x- and y-directions can be treated reference, say ground.). Then, velocity of object independently of each other. That is, motion in A relative to that of B is : a plane (two-dimensions) can be treated as two vAB = vA – vB (4.35a) separate simultaneous one-dimensional and similarly, the velocity of object B relative to motions with constant acceleration along two that of A is : perpendicular directions. This is an important vBA = vB – vA result and is useful in analysing motion of objects Therefore, vAB = – vBA (4.35b) in two dimensions. A similar result holds for three dimensions. The choice of perpendicular and, v AB = v BA (4.35c) directions is convenient in many physical t Example 4.6 Rain is falling vertically with situations, as we shall see in section 4.10 for a speed of 35 m s–1. A woman rides a bicycle projectile motion. with a speed of 12 m s–1 in east to west t direction. What is the direction in which Example 4.5 A particle starts from origin she should hold her umbrella ? at t = 0 with a velocity 5.0 î m/s and moves Answer In Fig. 4.16 vr represents the velocity in x-y plane under action of a force which of rain and vb , the velocity of the bicycle, the produces a constant acceleration of ( ) woman is riding. Both these velocities are with 3.0i + 2.0 $ j m/s 2. (a) What is the respect to the ground. Since the woman is riding y-coordinate of the particle at the instant a bicycle, the velocity of rain as experienced by its x-coordinate is 84 m ? (b) What is the speed of the particle at this time ? Answer The position of the particle is given by 1 2 r (t ) = v 0 t + at 2 ( = 5.0ˆi t + (1/2 ) 3.0ˆi + 2.0ˆj t 2) ( ) = 5.0 t + 1.5 t 2 ˆi + 1.0 t 2 ˆj Therefore, x (t ) = 5.0 t + 1.5 t 2 Fig. 4.16 her is the velocity of rain relative to the velocity y (t ) = +1.0 t 2 of the bicycle she is riding. That is vrb = vr – vb MOTION IN A PLANE 77 This relative velocity vector as shown in Fig. 4.16 makes an angle θ with the vertical. It is given by vb 12 tan θ = = = 0.343 vr 35 Or, θ ≅ 19  Therefore, the woman should hold her umbrella at an angle of about 19° with the vertical towards the west. Note carefully the difference between this Example and the Example 4.1. In Example 4.1, the boy experiences the resultant (vector Fig 4.17 Motion of an object projected with velocity sum) of two velocities while in this example, vo at angle θ0. the woman experiences the velocity of rain relative to the bicycle (the vector difference If we take the initial position to be the origin of the reference frame as shown in Fig. 4.17, we of the two velocities). t have : 4.10 PROJECTILE MOTION xo = 0, yo = 0 As an application of the ideas developed in the Then, Eq.(4.47b) becomes : previous sections, we consider the motion of a x = vox t = (vo cos θo ) t projectile. An object that is in flight after being thrown or projected is called a projectile. Such and y = (vo sin θo ) t – ( ½ )g t2 (4.38) a projectile might be a football, a cricket ball, a The components of velocity at time t can be baseball or any other object. The motion of a obtained using Eq.(4.33b) : projectile may be thought of as the result of two vx = vox = vo cos θo separate, simultaneously occurring components of motions. One component is along a horizontal vy = vo sin θo – g t (4.39) direction without any acceleration and the other Equation (4.38) gives the x-, and y-coordinates along the vertical direction with constant of the position of a projectile at time t in terms of acceleration due to the force of gravity. It was two parameters — initial speed vo and projection Galileo who first stated this independency of the angle θo. Notice that the choice of mutually horizontal and the vertical components of perpendicular x-, and y-directions for the projectile motion in his Dialogue on the great analysis of the projectile motion has resulted in world systems (1632). a simplification. One of the components of In our discussion, we shall assume that the velocity, i.e. x-component remains constant air resistance has negligible effect on the motion throughout the motion and only the of the projectile. Suppose that the projectile is y- component changes, like an object in free fall launched with velocity vo that makes an angle in vertical direction. This is shown graphically θo with the x-axis as shown in Fig. 4.17. at few instants in Fig. 4.18. Note that at the point of maximum height, v y = 0 and therefore, After the object has been projected, the vy acceleration acting on it is that due to gravity θ = tan −1 =0 which is directed vertically downward: vx a = −g
j Equation of path of a projectile Or, ax = 0, ay = – g (4.36) What is the shape of the path followed by the The components of initial velocity vo are : projectile? This can be seen by eliminating the vox = vo cos θo time between the expressions for x and y as voy= vo sin θo (4.37) given in Eq. (4.38). We obtain: 78 PHYSICS y = ( tan θ o ) x − g x2 ( v 0 sin θ 0 ) 2 hm = 2 (v o cosθo ) 2 (4.40) Or, (4.42) 2g Now, since g, θo and vo are constants, Eq. (4.40) Horizontal range of a projectile is of the form y = a x + b x2, in which a and b are The horizontal distance travelled by a projectile constants. This is the equation of a parabola, from its initial position (x = y = 0) to the position i.e. the path of the projectile is a parabola where it passes y = 0 during its fall is called the (Fig. 4.18). horizontal 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 2 v0 sin 2θ 0 Or, R= (4.43a) g Equation (4.43a) 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, 2 v0 Rm = (4.43b) g t Example 4.7 Galileo, in his book Two new sciences, stated that “for elevations which Fig. 4.18 The path of a projectile is a parabola. exceed or fall short of 45° by equal Time of maximum height amounts, the ranges are equal”. Prove this statement. 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 Answer For a projectile launched with velocity Eq. (4.39): vo at an angle θo , the range is given by vy = vo sinθo – g tm = 0 Or, tm = vo sinθo /g (4.41a) v02 sin 2θ0 R= The total time Tf during which the projectile is g in flight can be obtained by putting y = 0 in Now, for angles, (45° + α ) and ( 45° – α), 2θo is Eq. (4.38). We get : (90° + 2α ) and ( 90° – 2α ) , respectively. The Tf = 2 (vo sin θo )/g (4.41b) values of sin (90° + 2α ) and sin (90° – 2α ) are the same, equal to that of cos 2α. Therefore, Tf is known as the time of flight of the projectile. ranges are equal for elevations which exceed or We note that Tf = 2 tm , which is expected fall short of 45° by equal amounts α. t because of the symmetry of the parabolic path. Maximum height of a projectile t Example 4.8 A hiker stands on the edge The maximum height h m reached by the of a cliff 490 m above the ground and projectile can be calculated by substituting throws a stone horizontally with an initial t = tm in Eq. (4.38) : speed of 15 m s-1. Neglecting air resistance, find the time taken by the stone to reach  v sinθ  g  v sinθ 2 ( ) the ground, and the speed with which it y = hm = v0 sinθ 0  0 0  0 0  −   hits the ground. (Take g = 9.8 m s-2 ).  g  2 g  MOTION IN A PLANE 79 Answer We choose the origin of the x-,and y- Neglecting air resistance - what does axis at the edge of the cliff and t = 0 s at the the assumption really mean? instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity While treating the topic of projectile motion, and the positive direction of y-axis to be the we have stated that we assume that the vertically upward direction. The x-, and y- air resistance has no effect on the motion components of the motion can be treated of the projectile. You must understand what the statement really means. Friction, force independently. The equations of motion are : due to viscosity, air resistance are all x (t) = xo + vox t dissipative forces. In the presence of any of y (t) = yo + voy t +(1/2) ay t2 such forces opposing motion, any object will Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2, lose some part of its initial energy and vox = 15 m s-1. consequently, momentum too. Thus, a The stone hits the ground when y(t) = – 490 m. projectile that traverses a parabolic path – 490 m = –(1/2)(9.8) t2. would certainly show deviation from its This gives t =10 s. idealised trajectory in the presence of air The velocity components are vx = vox and resistance. It will not hit the ground with vy = voy – g t the same speed with which it was projected so that when the stone hits the ground : from it. In the absence of air resistance, the vox = 15 m s–1 x-component of the velocity remains voy = 0 – 9.8 × 10 = – 98 m s–1 constant and it is only the y-component that Therefore, the speed of the stone is undergoes a continuous change. However, in the presence of air resistance, both of v 2x + vy2 = 152 + 982 = 99 m s−1 t these would get affected. That would mean that the range would be less than the one t Example 4.9 A cricket ball is thrown at a given by Eq. (4.43). Maximum height speed of 28 m s–1 in a direction 30° above attained would also be less than that the horizontal. Calculate (a) the maximum predicted by Eq. (4.42). Can you then, height, (b) the time taken by the ball to anticipate the change in the time of flight? return to the same level, and (c) the In order to avoid air resistance, we will distance from the thrower to the point have to perform the experiment in vacuum where the ball returns to the same level. or under low pressure, which is not easy. When we use a phrase like ‘neglect air resistance’, we imply that the change in parameters such as range, height etc. is Answer (a) The maximum height is given by much smaller than their values without air (v0 sinθo )2 (28 sin 30°)2 resistance. The calculation without air hm = = m 2g 2 ( 9.8 ) resistance is much simpler than that with air resistance. 14 × 14 = = 10.0 m 2 × 9.8 4.11 UNIFORM CIRCULAR MOTION (b) The time taken to return to the same level is When an object follows a circular path at a Tf = (2 vo sin θo )/g = (2× 28 × sin 30° )/9.8 constant speed, the motion of the object is called = 28/9.8 s = 2.9 s uniform circular motion. The word “uniform” (c) The distance from the thrower to the point refers to the speed, which is uniform (constant) where the ball returns to the same level is throughout the motion. Suppose an object is (v sin 2θ ) 2 moving with uniform speed v in a circle of radius o o 28 × 28 × sin 60o R as shown in Fig. 4.19. Since the velocity of the R= = = 69 m g 9.8 object is changing continuously in direction, the object undergoes acceleration. Let us find the t magnitude and the direction of this acceleration. 80 PHYSICS Fig. 4.19 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. 4.19(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. 4.19(a1). v, v′ and ∆v are similar (Fig. 4.19a). Therefore, ∆v is obtained in Fig. 4.19 (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 4.19(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 In Fig. 4.19(c), ∆t Ž 0 and the average ∆t → 0 ∆t ∆ t → 0 R∆ t R ∆ t → 0 ∆t 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, a = lim ∆t → 0 ∆t ∆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. MOTION IN A PLANE 81 v v 2 2 ω R 2 ac =   v = v2/R (4.44) ac = = 2 =ω R R  R R Thus, the acceleration of an object moving with 2 ac = ω R (4.47) speed v in a circle of radius R has a magnitude 2 v /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 acceleration (a term proposed by Newton). A is known as its time period T and the number of thorough analysis of centripetal acceleration was revolution made in one second is called its first published in 1673 by the Dutch scientist frequency ν (=1/T ). However, during this time the Christiaan Huygens (1629-1695) but it was distance moved by the object is s = 2πR. probably known to Newton also some years earlier. Therefore, v = 2πR/T =2πRν (4.48) “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 (4.49) pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector. t Example 4.10 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. 4.19) 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 (4.45) ω = 2π/T = 2π × 7/100 = 0.44 rad/s Now, if the distance travelled by the object The linear speed v is : 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. Since but ∆s = R ∆θ. Therefore : this direction changes continuously, ∆θ acceleration here is not a constant vector. v=R =Rω ∆t However, the magnitude of acceleration is v= Rω (4.46) constant: We can express centripetal acceleration ac in a = ω2 R = (0.44 s–1)2 (12 cm) terms of angular speed : = 2.3 cm s-2 t 82 PHYSICS 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 0A=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 =λa +µb where λ and µ are real numbers. 9. A unit vector associated with a vector A has magnitude one and is along the vector A: A n̂ = A 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 A = A i
+ A
j x y where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle θ Ay with the x-axis, then Ax = A cos θ, Ay=A sin θ and A = A = A x2 + Ay2 , tanθ =. Ax 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 : R = Rx i
+ Ry
j , where, Rx = Ax + Bx, and Ry = Ay + By 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 = ( x ′ − x ) i
+ (y ′ − y )
j = ∆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 MOTION IN A PLANE 83 as ∆t tends to zero : ∆r dr v = lim =. It can be written in unit vector notation as : ∆t → 0 ∆t dt dx dy dz v = v x i
+ v y
j + v z k
where vx = ,v = ,v = dt y dt 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 : lim ∆v dv a= = ∆t → 0 ∆t dt In component form, we have : a = a i
+ a
j + a k x
y z dv x dvy dvz where, a x = , ay = , az = 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 ot + 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 x = x o + vox t + ax t 2 2 1 y = yo + voy t + ay t 2 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 : gx 2 y = ( tanθ0 ) x – 2 (v o cos θo ) 2 The maximum height that a projectile attains is : 84 PHYSICS 2 (v o sin o ) hm = 2g The time taken to reach this height is : vo sinθ o tm = 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 : vo2 R= sin 2θo 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 MOTION IN A PLANE 85 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 (4.33a) and (4.34a) 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 4.1 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. 4.2 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. 4.3 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. 4.4 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. 4.5 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. 4.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| 86 PHYSICS (c) − b| < |a| + |b| |a− (d) − b| > |a− ||a| − |b|| When does the equality sign above apply? 4.7 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 ? 4.8 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. 4.20. What is the magnitude of the displacement vector for each ? For Fig. 4.20 which girl is this equal to the actual length of path skate ? 4.9 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. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 4.21 4.10 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. 4.11 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 ? 4.12 Rain is falling vertically with a speed of 30 m s-1. A woman rides a bicycle with a speed of 10 m s-1 in the north to south direction. What is the direction in which she should hold her umbrella ? 4.13 A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his MOTION IN A PLANE 87 strokes normal to the river current? How far down the river does he go when he reaches the other bank ? 4.14 In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ? 4.15 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 ? 4.16 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 ? 4.17 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 ? 4.18 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. 4.19 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 4.20 The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 k ˆ 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 ? 4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0
j m/s and moves in ( ) 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 ? 4.22
i and
j are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors
i +
j , and
i −
j ? What are the components of a vector A= 2 i
+ 3
j along the directions of
i +
j and
i −
j ? [You may use graphical method] 4.23 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) 4.24 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. 4.25 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 apart is 30°, what is the speed of the aircraft ? 88 PHYSICS Additional Exercises 4.26 A vector has magnitude and direction. Does it have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer. 4.27 A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ? 4.28 Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ? Explain. 4.29 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to be fixed, and neglect air resistance. 4.30 A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s-1 to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m s-2 ). 4.31 A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ? 4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by  v0y − gt  θ (t ) = tan -1    vox  (b) Shows that the projection angle θ0 for a projectile launched from the origin is given by  4h  θ 0 = tan-1  m   R  where the symbols have their usual meaning.