The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions. 4.2 SCALARS AND VECTORS. In physics
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CHAPTER FOURMOTION IN A PLANE4.1 INTRODUCTIONIn the last chapter we developed the concepts of position,displacement, velocity and acceleration that are needed todescribe the motion of an object along a straight line. Wefound that the directional aspect of these quantities can betaken care of by + and Œ signs, as in one dimension only twodirections are possible. But in order to describe motion of anobject in two dimensions (a plane) or three dimensions (space), we need to use vectors to describe the above-mentioned physical quantities. Therefore, it is first necessaryto learn the language of vectors. What is a vector? How toadd, subtract and multiply vectors ? What is the result ofmultiplying a vector by a real number ? We shall learn thisto enable us to use vectors for defining velocity andacceleration in a plane. We then discuss motion of an objectin a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail the projectile motion. Circular motion is a familiar class ofmotion that has a special significance in daily-life situations.We shall discuss uniform circular motion in some detail.The equations developed in this chapter for motion in aplane can be easily extended to the case of three dimensions.4.2 SCALARS AND VECTORSIn physics, we can classify quantities as scalars orvectors. Basically, the difference is that a direction isassociated with a vector but not with a scalar. A scalarquantity is a quantity with magnitude only. It is specifiedcompletely by a single number, along with the properunit. 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 forcombining scalars are the rules of ordinary algebra.Scalars can be added, subtracted, multiplied and divided4.1Introduction4.2Scalars and vectors4.3Multiplication of vectors byreal numbers4.4Addition and subtraction of vectors Š graphical method4.5Resolution of vectors4.6Vector addition Š analyticalmethod4.7Motion in a plane4.8Motion in a plane with constant acceleration4.9Relative velocity in two dimensions4.10Projectile motion4.11Uniform circular motionSummaryPoints to ponder Exercises Additional exercises
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PHYSICS66just as the ordinary numbers*. For example,if the length and breadth of a rectangle are1.0 m and 0.5 m respectively, then itsperimeter 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 10Œ3 m3 (a scalar)and its density is 2.7´103 kg mŒ3 (a scalar).A vector quantity is a quantity that has botha magnitude and a direction and obeys thetriangle law of addition or equivalently theparallelogram law of addition. So, a vector isspecified 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 typein this book. Thus, a velocity vector can berepresented by a symbol v. Since bold face isdifficult to produce, when written by hand, a vector is often represented by an arrow placedover a letter, say rv. Thus, both v and rvrepresent the velocity vector. The magnitude ofa vector is often called its absolute value, indicated by |v| = v. Thus, a vector isrepresented by a bold face, e.g. by A, a, p, q, r, x, y, with respective magnitudes denoted by lightface A, a, p, q, r, x, y.4.2.1 Position and Displacement VectorsTo describe the position of an object moving ina plane, we need to choose a convenient point,say O as origin. Let P and P¢ be the positions ofthe object at time t and t¢, respectively [Fig. 4.1(a)].We join O and P by a straight line. Then, OP isthe position vector of the object at time t. Anarrow is marked at the head of this line. It isrepresented by a symbol r, i.e. OP = r. Point P¢ isrepresented by another position vector, OP¢denoted by r¢. The length of the vector rrepresents 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¢, thevector PP¢ (with tail at P and tip at P¢) is calledthe displacement vector corresponding tomotion from point P (at time t) to point P¢ (at time t¢).Fig. 4.1(a) Position and displacement vectors.(b) Displacement vector PQ and different courses of motion.It is important to note that displacementvector is the straight line joining the initial andfinal positions and does not depend on the actualpath undertaken by the object between the twopositions. For example, in Fig. 4.1(b), given theinitial and final positions as P and Q, thedisplacement vector is the same PQ for differentpaths of journey, say PABCQ, PDQ, and PBEFQ.Therefore, the magnitude of displacement iseither less or equal to the path length of anobject between two points. This fact wasemphasised in the previous chapter also while discussing motion along a straight line.4.2.2 Equality of VectorsTwo vectors A and B are said to be equal if, andonly if, they have the same magnitude and thesame direction.**Figure 4.2(a) shows two equal vectors A andB. We can easily check their equality. Shift Bparallel to itself until its tail Q coincides with thatof A, i.e. Q coincides with O. Then, since theirtips 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 multiplyand 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 ofapplication of a vector is important. Such vectors are called localised vectors.
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MOTION IN A PLANE67as A = B. Note that in Fig. 4.2(b), vectors A¢ andB¢ have the same magnitude but they are notequal because they have different directions.Even if we shift B¢ parallel to itself so that its tailQ¢ coincides with the tail O¢ of A¢, the tip S¢ of B¢does not coincide with the tip P¢ of A¢.4.3MULTIPLICATION OF VECTORS BY REALNUMBERSMultiplying a vector A with a positive number lgives a vector whose magnitude is changed by the factor l but the direction is the same as thatof A :½l A½ = l ½A½ if l > 0.For example, if A is multiplied by 2, the resultantvector 2A is in the same direction as A and hasa magnitude twice of |A| as shown in Fig. 4.3(a).Multiplying a vector A by a negative number-l gives another vector whose direction isopposite to the direction of A and whosemagnitude is l times |A|.Multiplying a given vector A by negativenumbers, say Œ1 and Œ1.5, gives vectors as shown in Fig 4.3(b).The factor l by which a vector A is multipliedcould be a scalar having its own physical dimension. Then, the dimension of l A is theproduct of the dimensions of l and A. Forexample, if we multiply a constant velocity vector by duration (of time), we get a displacement vector.4.4ADDITION AND SUBTRACTION OF VECTORS Š GRAPHICAL METHODAs mentioned in section 4.2, vectors, by definition, obey the triangle law or equivalently,the parallelogram law of addition. We shall nowdescribe this law of addition using the graphicalmethod. Let us consider two vectors A and B thatlie in a plane as shown in Fig. 4.4(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 thatits tail is at the head of the vector A, as inFig. 4.4(b). Then, we join the tail of A to the headof B. This line OQ represents a vector R, that is,the sum of the vectors A and B. Since, in thisprocedure of vector addition, vectors areFig. 4.2 (a) Two equal vectors A and B. (b) Twovectors A¢ and B¢ are unequal though theyare of the same length.Fig. 4.3(a) Vector A and the resultant vector aftermultiplying A by a positive number 2.(b) Vector A and resultant vectors aftermultiplying it by a negative number Œ1and Œ1.5.(c)(d)Fig. 4.4 (a) Vectors A and B. (b) Vectors A and Badded graphically. (c) Vectors B and Aadded graphically. (d) Illustrating theassociative law of vector addition.
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PHYSICS68arranged head to tail, this graphical method iscalled the head-to-tail method. The two vectorsand their resultant form three sides of a triangle, so this method is also known as triangle methodof vector addition. If we find the resultant ofB + A as in Fig. 4.4(c), the same vector R isobtained. Thus, vector addition is commutative:A + B = B + A(4.1)The addition of vectors also obeys the associative law as illustrated in Fig. 4.4(d). The result of adding vectors A and B first and then addingvector C is the same as the result of adding Band C first and then adding vector A :(A + B) + C = A + (B + C)(4.2)What is the result of adding two equal and opposite vectors ? Consider two vectors A andŒA shown in Fig. 4.3(b). Their sum is A + (ŒA).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 zerovector :A Œ A = 0 |0|= 0(4.3)Since the magnitude of a null vector is zero, its direction cannot be specified.The null vector also results when we multiplya vector A by the number zero. The mainproperties of 0 are :A + 0 = Al 0 = 00 A = 0 (4.4)Fig. 4.5 (a)Two vectors A and B, Œ B is also shown. (b) Subtracting vector B from vector A Œ the result is R2. Forcomparison, 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. 4.1(a). Now suppose that an object which is at P at time t, moves toP¢ and then comes back to P. Then, what is itsdisplacement? Since the initial and final positions coincide, the displacement is a finull vectorfl.Subtraction of vectors can be defined in termsof addition of vectors. We define the differenceof two vectors A and B as the sum of two vectorsA and ŒB :A Œ B = A + (ŒB)(4.5)It is shown in Fig 4.5. The vector ŒB is added tovector A to get R2 = (A Œ B). The vector R1 = A + Bis also shown in the same figure for comparison.We can also use the parallelogram method tofind the sum of two vectors. Suppose we have two vectors A and B. To add these vectors, webring their tails to a common origin O as shown in Fig. 4.6(a). Then we draw a line fromthe head of A parallel to B and another line fromthe head of B parallel to A to complete aparallelogram OQSP. Now we join the point ofthe intersection of these two lines to the origin O. The resultant vector R is directed from thecommon origin O along the diagonal (OS) of theparallelogram [Fig. 4.6(b)]. In Fig.4.6(c), thetriangle law is used to obtain the resultant of Aand B and we see that the two methods yield thesame result. Thus, the two methods areequivalent.
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MOTION IN A PLANE69tExample 4.1 Rain is falling vertically witha speed of 35 m sŒ1. Winds starts blowingafter sometime with a speed of 12 m sŒ1 ineast to west direction. In which directionshould a boy waiting at a bus stop hold his umbrella ?Fig. 4.7Answer The velocity of the rain and the windare represented by the vectors vr and vw in Fig.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 shownin the figure. The magnitude of R isRvv r2w2=+=+= — 3512 m s37 m s 221 1The direction q that R makes with the verticalis given by12 tan 0.343 35 wrv vq=== Or, () q==° tan -.1034319 Therefore, the boy should hold his umbrellain the vertical plane at an angle of about 19owith the vertical towards the east. tFig. 4.6(a) Two vectors A and B with their tails brought to a common origin. (b) The sum A + B obtained usingthe parallelogram method. (c) The parallelogram method of vector addition is equivalent to the trianglemethod.4.5 RESOLUTION OF VECTORSLet a and b be any two non-zero vectors in aplane with different directions and let A beanother vector in the same plane(Fig. 4.8). A canbe expressed as a sum of two vectors Š oneobtained by multiplying a by a real number andthe other obtained by multiplying b by anotherreal number. To see this, let O and P be the tail and head of the vector A. Then, through O, drawa straight line parallel to a, and through P, astraight line parallel to b. Let them intersect atQ. Then, we haveA = OP = OQ + QP(4.6)But since OQ is parallel to a, and QP is parallelto b, we can write :OQ = l a, and QP = µ b(4.7)where l and µ are real numbers.Therefore,A = l a + µ b(4.8)Fig. 4.8 (a) Two non-colinear vectors a and b.(b) Resolving a vector A in terms of vectorsa and b.We say that A has been resolved into twocomponent vectors l a and m b along a and brespectively. Using this method one can resolve
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PHYSICS70Fig. 4.9 (a)Unit vectors i , j and klie along the x-, y-, and z-axes. (b) A vector A is resolved into itscomponents Ax and Ay along x-, and y- axes. (c) A1 and A2 expressed in terms of i and j.a given vector into two component vectors alonga set of two vectors Œ 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. Aunit 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- andz-axes of a rectangular coordinate system aredenoted by i, j and ‹k, respectively, as shownin Fig. 4.9(a).Since these are unit vectors, we have½‹i ½ = ½‹j½ = ½‹k½=1(4.9)These unit vectors are perpendicular to eachother. In this text, they are printed in bold facewith 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 ‹n by a scalar, the result is a vectorll ll l = l‹n. In general, a vector A can be written asA = |A|‹n(4.10)where ‹nis a unit vector along A.We can now resolve a vector A in termsof component vectors that lie along unit vectorsi‹ and j. Consider a vector A that lies in x-yplane as shown in Fig. 4.9(b). We draw lines fromthe head of A perpendicular to the coordinateaxes as in Fig. 4.9(b), and get vectors A1 and A2such that A1 + A2 = A. Since A1 is parallel to iand A2 is parallel to j, we have :A1= Ax i, A2 = Ay j(4.11)where Ax and Ay are real numbers.Thus, A = Ax i+ Ay j (4.12)This is represented in Fig. 4.9(c). The quantitiesAx and Ay are called x-, and y- components of thevector A. Note that Ax is itself not a vector, butAx i is a vector, and so is Ay j. Using simpletrigonometry, we can express Ax and Ay in termsof the magnitude of A and the angle q it makeswith the x-axis :Ax = A cos qAy = A sin q(4.13)As is clear from Eq. (4.13), a component of avector can be positive, negative or zero depending on the value of q.Now, we have two ways to specify a vector Ain a plane. It can be specified by : (i)its magnitude A and the direction q it makeswith the x-axis; or(ii)its components Ax and AyIf A and q are given, Ax and Ay can be obtainedusing Eq. (4.13). If Ax and Ay are given, A and qcan be obtained as follows :AAAA x2y22222 +=+ cossin qq = A2Or,AAA x2y2=+ (4.14)Andtan,tan qq == -AAAAyxyx1 (4.15)
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PHYSICS72tFig. 4.10Answer Let OP and OQ represent the two vectorsA and B making an angle q (Fig. 4.10). Then,using the parallelogram method of vectoraddition, OS represents the resultant vector R : R = A + BSN is normal to OP and PM is normal to OS.From the geometry of the figure,OS2 = ON2 + SN2butON = OP + PN = A + B cos qSN = B sin qOS2 = (A + B cos q)2 + (B sin q)2or,R2 = A2 + B2 + 2AB cos qRAB2AB 22 =++ cos q(4.24a)In D OSN, SN = OS sina = R sina, andin D PSN, SN = PS sin q = B sin qTherefore, R sin a = B sin qor, RB sin sin qa =(4.24b)Similarly, PM = A sin a = B sin bor, AB sin sin ba =(4.24c)Combining Eqs. (4.24b) and (4.24c), we get RA sin sin sin qba == B(4.24d)Using Eq. (4.24d), we get: sin sin aq =BR(4.24e)where R is given by Eq. (4.24a).or, sin tan cos SNB OPPNAB qaq== ++ (4.24f)Equation (4.24a) gives the magnitude of theresultant and Eqs. (4.24e) and (4.24f) its direction. Equation (4.24a) is known as the Law of cosinesand Eq. (4.24d) as the Law of sines. tExample 4.3 A motorboat is racingtowards 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 velocityof the motorboat and the vector vc representingthe water current are shown in Fig. 4.11 in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in thefigure.Fig. 4.11We can obtain the magnitude of R using the Lawof cosine :Rvvvv =b2c2bc 2cos120 ++ o=251022510-1/222 km/h 22 ++´´ () @To obtain the direction, we apply the Law of sinesRv csin sin qf = or, sin fq =vR csin =10sin120 21.8 103 221.8 0.397 ´=´@f@ 23.4 t4.7 MOTION IN A PLANEIn this section we shall see how to describemotion in two dimensions using vectors.
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MOTION IN A PLANE734.7.1 Position Vector and DisplacementThe position vector r of a particle P located in aplane with reference to the origin of an x-yreference frame (Fig. 4.12) is given byrij =+ x y where x and y are components of r along x-, andy- axes or simply they are the coordinates ofthe object.(a)(b)Fig. 4.12(a) Position vector r. (b) Displacement Dr andaverage velocity v of a particle.Suppose a particle moves along the curve shownby the thick line and is at P at time t and P¢ attime t¢ [Fig. 4.12(b)]. Then, the displacement is :Dr = r¢ Œ r(4.25)and is directed from P to P¢.We can write Eq. (4.25) in a component form:Dr ()()=+-+ x’y’x ijij y =+ ij DD xy whereDx = x ¢ Œ x, Dy = y¢ Œ y(4.26)VelocityThe average velocity ()v of an object is the ratioof the displacement and the corresponding timeinterval :vrij ij == +=+ D DDD DDDDDtx y txtyt (4.27)Or,‹xy vv =+ v ij Since vr=DDt, the direction of the average velocityis the same as that of Dr (Fig. 4.12). The velocity(instantaneous velocity) is given by the limitingvalue of the average velocity as the time interval approaches zero :vrr == ®lim tt tDDD0dd(4.28)The meaning of the limiting process can be easilyunderstood with the help of Fig 4.13(a) to (d). In these figures, the thick line represents the pathof an object, which is at P at time t. P1, P2 andP3 represent the positions of the object aftertimes Dt1,Dt2, and Dt3. Dr1, Dr2, and Dr3 are thedisplacements of the object in times Dt1, Dt2, andFig. 4.13As the time interval Dt approaches zero, the average velocity approaches the velocity v. The directionof v is parallel to the line tangent to the path.
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PHYSICS74Dt3, respectively. The direction of the averagevelocity v is shown in figures (a), (b) and (c) forthree decreasing values of Dt, i.e. Dt1,Dt2, and Dt3,(Dt1 > Dt2 > Dt3). As Dt ®® ®® ® 0, Dr ®® ®® ® 0and is along the tangent to the path [Fig. 4.13(d)].Therefore, the direction of velocity at any pointon the path of an object is tangential to thepath at that point and is in the direction ofmotion.We can express v in a component form :vr=ddt =+æèçöø÷®lim xtyttDDDDD0 ij (4.29)=+ ®®ijlim xtlim ytttDDDDDD00Or,vijij =+=+ ddddxtytvv xy .wherevxtvytxy== dddd, (4.30a)So, if the expressions for the coordinates x andy are known as functions of time, we can usethese equations to find vx and vy.The magnitude of v is thenvvv x2y2=+ (4.30b)and the direction of v is given by the angle q :tan tan 1qq == æèççöø÷÷-vvvvyxyx (4.30c)vx, vy and angle q are shown in Fig. 4.14 for avelocity vector v at point p.AccelerationThe average acceleration a of an object for atime interval Dt moving in x-y plane is the changein velocity divided by the time interval : ()avij ij == +=+ DDDDDDDDtvv tvtvtxy xy (4.31a)Or,aij =+ aa xy . (4.31b)The acceleration (instantaneous acceleration)is the limiting value of the average accelerationas the time interval approaches zero :av=®lim ttDDD0 (4.32a)SinceDDD v=+ vv, xy ij we haveai j=+®®lim vtlim vttxtyDDDDDD00Or,aij =+ aa xy (4.32b)where,avt, a vtxxyy== dddd (4.32c)*As in the case of velocity, we can understandgraphically the limiting process used in defining acceleration on a graph showing the path of the object™s motion. This is shown in Figs. 4.15(a) to (d). P represents the position of the object at time t and P1, P2, P3 positions after time Dt1, Dt2,Dt3, respectively (Dt 1> Dt2>Dt3). The velocity vectorsat points P, P1, P2, P3 are also shown in Figs. 4.15(a), (b) and (c). In each case of Dt, Dv is obtainedusing the triangle law of vector addition. Bydefinition, the direction of average acceleration is the same as that of Dv. We see that as Dtdecreases, the direction of Dv changes andconsequently, the direction of the acceleration changes. Finally, in the limit Dt g0 [Fig. 4.15(d)],the average acceleration becomes the instantaneous acceleration and has the direction as shown.Fig. 4.14The components vx and vy of velocity v andthe angle q it makes with x-axis. Note thatvx = v cos q, vy = v sin q.*In terms of x and y, ax and ay can be expressed as
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MOTION IN A PLANE75t x (m)Note that in one dimension, the velocity andthe acceleration of an object are always along the same straight line (either in the samedirection 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 4.4 The position of a particle isgiven by r i j k =++ 3.0 t‹.‹.‹2050 2twhere t is in seconds and the coefficientshave the proper units for r to be in metres.(a) Find v(t) and a(t) of the particle. (b) Findthe magnitude and direction of v(t) att = 1.0 s.Answer() ()vri jk tttt t 2== ++ dddd3.0 2.05.0 =+ 3.0.0 ij 4t () a vjtt=dd=+4.0 a = 4.0 m sŒ2 along y- directionAt t = 1.0 s, ‹‹ 3.04.0 v=i+j It™s magnitude is 22 1-=345.0 m s v+= and direction is-1 14=tantan53 3yxvvq- °=@ with x-axis.t4.8MOTION IN A PLANE WITH CONSTANTACCELERATIONSuppose that an object is moving in x-y planeand its acceleration a is constant. Over aninterval 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 definitionavvvv00=–=-tt0Or,vva0=+t(4.33a)In terms of components :vvatxoxx =+vvatyoyy =+(4.33b)Let us now find how the position r changes withtime. We follow the method used in the one-dimensional case. Let ro and r be the positionvectors of the particle at time 0 and t and let thevelocities at these instants be vo and v. Then,over this time interval t, the average velocity is(vo + v)/2. The displacement is the averagevelocity multiplied by the time interval :rr vv vav 0000-= +æèçöø÷=+() +æèçöø÷22 tttFig. 4.15The average acceleration for three time intervals (a) Dt1, (b) Dt2, and (c) Dt3, (Dt1> Dt2> Dt3). (d) In thelimit Dt g0, the average acceleration becomes the acceleration.
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