MOTION IN A PLANE 1. Scala Quantities Physical quantities that have only magnitude and no diection ae called scala quantities o scalas. e.g. Mass, time, speed etc. 2. Vecto Quantities Physical quantities that have both magnitude and diection ae called vecto quantities o vectos. e.g. displacement, weight, velocity, foce etc. 3. Repesentation of vectos:-a vecto is epesented by using a staight line with an aowhead at one end. The length of the line epesents the magnitude of the vecto and the aowhead gives its diection. In pint, vecto is epesented by bold lette A 4. Equal vectos Vectos ae said to be equal if they have the same magnitude and same diection 5. Negative vecto A vecto having the same magnitude but diection opposite to that of a given vecto is called negative vecto of the given vecto. Hee, A = -B 6. Coplana vectos: Vectos which ae confined to the same plane. 7. Modulus of a vecto: - Modulus of a vecto is the magnitude of the vecto. Modulus of vecto A is epesented as A If A = A x i + A y j +A z k then, A = Ax 2 + Ay 2 + Az 2 e.g: If A = i 2 j + k then find A. Ax =1 ; Ay= -2; Az=1 A = Ax 2 + Ay 2 + Az 2 = 1 2 + ( 2) 2 + 1 2 = 6 8. Unit vecto:a vecto whose magnitude is unity (one) is called a unit vecto.it is epesented as.a A = A A In Catesian coodinates i, j, k ae the unit vectos along X, Y, and Z axes espectively i = j = k = 1 and i, j, k ae mutually pependicula.
9. Zeo vecto o null vecto It is a vecto, which has zeo magnitude and an abitay diection. It is O epesented by. 3.Zeo vecto / Null vecto: - A + ( - A ) = 0 0 x A = 0 and A + 0 =A e.g:the velocity vecto of a stationay object is a null vecto. ** Addition of vectos. a) When the vectos ae in the same diection. Hee the magnitude of the esultant vecto is equal to the sum of the magnitudes of the two vectos and diection is same as that of the two vectos. R = A + B b) When the vectos ae in opposite diections. Hee the magnitude of the esultant is the diffeence in magnitude of the two vectos and the diection is same as that of the bigge vecto. R = A + B = A - B c) When the vectos ae inclined at an angle. When the vectos ae inclined at an angle, the sum of vectos can be detemined using two methods: (a) Tiangle law of vectos (b) Paallelogam law of vectos. (a).tiangle law of vectos. If two vectos ae epesented by the two sides of a tiangle, both in magnitude and in diection, and in same ode, then the thid side of the tiangle in the evese ode epesents the vecto sum of the vectos. If OA epesents P and AC epesents Q in the antic lock wise ode, then OC in the clockwise ode epesents the vecto sum of P and Q
(b) Paallelogam law of vectos. If two adjacent sides of a paallelogam epesent two vectos in magnitude and diection, then the diagonal of the paallelogam stating fom the point of intesection of the two vectos epesent the vecto sum of the two vectos. Resolution of vectos:- Splitting a vecto along its components (geneally along co-odinate axis) Let us conside a vecto A A = Ax i + Ay j But Ax = A Cosθ & Ay = A Sinθ Ax 2 + Ay 2 = A 2 (Cos 2 θ +Sin 2 θ ) = A 2 Diection is A = Ax 2 + Ay 2 AY A X = A Cosθ A Sinθ = tan θ Θ = tan -1 ( AY A X ) Θ detemines the diection of vecto A. ANALYTICAL METHOD OF VECTOR ADDITION *Find the magnitude and diection of the esultant of two vectos P and Q in tems of thei magnitudes and angle θ between them. Let α be the angle which the esultant makes with the diection of P then R = P + Q Constuct CD OA, In OCD, (OC) 2 = (OD) 2 + (CD) 2 (OC) 2 = (OA+AD) 2 + (CD) 2 (OC) 2 = ( Q + P Cos θ) 2 +(P Sin θ ) 2 (OC) 2 = Q 2 + 2PQ Cos θ + P 2 Cos 2 θ + P 2 Sin 2 θ ( Sin 2 θ + Cos 2 θ =1)
R 2 = Q 2 + 2PQ Cos θ + P 2 ( Cos 2 θ + Sin 2 θ ) R 2 = P 2 + Q 2 + 2PQ Cos θ R = P 2 + Q 2 + 2PQ Cos θ This is known as law of cosines. & Diection of R is given by α CD Fom OCD, tan α = OD = CD OA +AD P Sinθ Q+ P Cosθ ) ie., tan α = P Sinθ Q+ P Cosθ α = tan -1 ( Case (i):- If P & Q ae to each othe. (θ =90) R = P 2 + Q 2 + 2PQ Cos 90 = P 2 + Q 2 α = tan -1 ( P Sin90 Q+ P Cos90 ) Case (ii):- If P & Q ae along the same line. (θ =0) R = P 2 + Q 2 + 2PQ Cos 0 = P+Q α = tan -1 ( P Sin 0 Q+ P Cos 0 ) Case (ii):- If P & Q ae in opposite diection. (θ =180) R = P 2 + Q 2 + 2PQ Cos 180 = P Q α = tan -1 ( P Sin 180 Q+ P Cos 180 )
Laws of vecto addition:- 1. Vecto sum is commutative A + B = B + A 2. Vecto sum is associative A + (B +C ) = ( A + B ) +C 3. Vecto sum is distibutive m (A + B ) = ma + mb Subtaction of two vecto:- Subtaction of a vecto Q fom P is defined as the addition of vecto Q (negative of vecto Q) to the vecto P. Position vecto:- P - Q =P + (-Q ) A vecto to epesent any position of a body. It is epesented by = OP = x i + yj, if x and y ae the coodinates of P. = x 2 + y 2 Multiplication of vectos:- poduct of vecto is not necessaily a vecto quantity. It may be scala o vecto. 1. Dot poduct(scala poduct):- If A and B ae two vectos and θ is the angle between them, taken anti clockwise then scala poduct is given by Chaactesitics:- A. B = A B Cos θ eads A dot B 1. A. B = 0, then two vectos ae pependicula to each othe. (θ= 90 0 ) 2.If A.B = AB, then two vectos ae paallel to each othe. ( i.e. θ = 0 0 ) 3. Scala poduct is commutative A. B = B. A 4. i. i = j. j = k. k =1 (since θ= 0 0 ) and i. j = j. k = k. i = 0 ( since θ = 90 0 ) 5. In Catesian co odinates, If A = A x i + A y j +A z k and B = B x i + B y j +B z k then
A. B = Ax Bx +Ay B y +Az B z e.g:1. The Foce F and displacement d of a paticle ae given by F = (2i +3j +5k ) and d = (i +2j +3k ). Evaluate the wok done. Ans : wok done W= F. d = (2i +3j +5k ). (i +2j +3k ) 2+6+15 = 23 Joules. 2. What is the angle between the vectos A= (2i -3j +4k ) and B= (-2i +5j -3k ) A. B = A B Cos θ Cos θ = A.B A B = - 31 (29 x30) = -0.9340 Θ = 159 0 6 3.Find the value of m if the A =(2i +3j +6k ) is pependicula to B= (3i +mj -6k ) Since A B A. B = 0 A. B = (2i+3j-6k) (3i+mj-6k) 6 + 3m 36 =0 m= -10. 2. Vecto poduct (coss poduct):- The coss poduct of two vecto A and B is defined as A x B = AB Sin θ eads A coss B *If A = A x i + A y j +A z k and B = B x i + B y j +B z k then A x B in deteminant fom is given by,
Chaacteistics:- 1. Vecto poduct is not commutative A x B B x A But A x B = - (B x A ) 2. Fo two vectos to be paallel A x B = 0 ( coss poduct is zeo) 3. i xi =j xj = k x k =0 e.g 1. Show that the vectos A = 2i -3j k and B = -6i + 9j +3k ae paallel. We know, if two vectos A and B ae paallel then A x B = 0 Shows A and B ae paallel. Cicula Motion A body is said to be in cicula motion, if it moves in such a way that its distance fom a fixed point always emains the same. The path of the body is called a cicle, fixed point cente of cicle and fixed distance, the adius of cicle. 1)Angula displacement ( ): The angle swept ove by the adius vecto in a given inteval of time iscalled angula displacement ( ). Unit of is adian. 2) Angula velocity ( ): Angula velocity is the ate of change of angula displacement. If θ is the angle tuned in time, then Angula velocity, = θ Unit = adians/second = ad/s = ad s 1
Instantaneous angula velocity is the angula velocity of the paticle at an instant. θ Instantaneous angula velocity, = lim = dθ 0 dt 3) Unifom cicula motion: A cicula motion is said to be unifom, when a paticle moves along a cicula path with a constant speed. 4) Time peiod: It is the time taken by the paticle to complete one evolution. 5) Fequency ( ): It is the numbe of evolutions made by the object in one second. ie. = 1 T ** Also angula velocity = = θ = 2π = 2π t T Relation between linea velocity and angula velocity. Conside a paticle moving ound a cicle of cente O and adius with unifom angula velocity. Let its linea velocity be v. Let t be the time taken by the paticle to go fom A to B. Then Linea velocity, v = AB AB = v t ------(1) Also θ = AB t [ angle = ac adius AB = θ ------------- (2) Fom eqn.(1) and (2); v t = θ. O v= θ i.e. v = t Theefoe, Linea velocity = adius x angula velocity. Angula acceleation (α) Angula acceleation is the ate of change of angula velocity. If is the change in d angula velocity in a time, then angula acceleation, α = lim = 0 dt Unit: ad/s 2. Relation connecting linea acceleation and angula acceleation. We know, v = ; whee v = linea velocity, = adius and = angula velocity. Diffeentiating w..to time, dv = d( ) dt dt = d dt i.e., a = d a = α dt ie., Linea acceleation = adius x angula acceleation. Centipetal Acceleation. When a body moves in a cicula path, its velocity changes continuously at evey point. Hence it expeiences acceleation. The diection of acceleation is towads the cente of the cicle. So it is called centipetal acceleation. The acceleation acting on an object undegoing unifom cicula motion is called centipetal acceleation. ]
1 Conside a paticle executing unifom cicula motion along a cicle of cente O and adius, with unifom speed v and angula velocity. At any instant t, let the paticle be at P. Its velocity v(t) is along the tangent PA and position vecto is along OP. Afte a time, let the paticle each Q. Hee the velocity v (t + ) is along the tangent QB and position vecto is along OQ. The displacement of paticle in time is, PQ = Now, we know v(t) = v(t + t) = v and = 1 = Now conside a vecto diagam on which LM and LN epesent the velocities v(t) and v(t + t) in magnitude and diection. Then MN = v epesents the change in velocity in time. Hee Tiangle s LMN and OPQ ae simila. MN PQ v = v Now, eq(1) = LM OP v a = v v v= v ----------- (1) = v a = v2 (We have v ( centipetal acceleation = a = v2 i.e., a = 2 2 = 2 ------(3) = a & ) ------------(2) eqn.(2) and (3) gives the equations fo centipetal acceleation. Also centipetal foce, F = m a = m v2 = m 2 = v)