MOTION IN A PLANE - Lecture Notes

Motioninqaupanlatnityep→hyssic.caalatr quantity scalar quantity → onlnyicemtaogonl itqudueantity eg : - speed , distance time etc , vector quantity → both magnitude and direction eg :- Velocity , displacement , force etc A nice for is me presented by straight line with an arrow mheaagdn;itulednegtohf of the straight line gives the the vector and arrow head gives the direction of the vector . tail \"→ Head . if it is a vector quantity • then it is represented by A→ Magnitude of ¥ is 1*1 on A unit nice tool vectors with magnitude one . If it is unit rector then it is represented \", by A

* ¥÷, Unit vector along ✗ axis F along Yaxisj along Tvaxis Ñ . A→=1A→lÑ nlcctorformofa = Magnitude of * lsnitniector physical quantity that vector in that direction; Typcsofniectors -☒ Equal vectors -☒-☒ Collinear → Hector :S ←_* Negative vectors → vectors =☒\" Parallel vectors coplanar *- Antiparallel vectors ¥☒ _☒ Coinitial rectors .

Ñengu1artawofvectooaddition_ f- two vectors are arranged as the adjacent side of a triangle, then the third side taken in the opposite direction will represent their resultant - If ☒☒ Ñ=A→+B→ BE -A→☒ arallelo grain law of vector addition If two vectors are arranged as the adjacent side of a parallelogram , then diagonal passing through the point of intersection of two vectors will represent their resultant - ☒- - - - pu→=A→+B→ A→ \"É agnitude of the resultant É If R ☒ B B-) ¥-1k - IQ .. - A A→

I coSO=x/B YIBSino __ }yJmmÉBy=BSinQm✓☐ BI ← a- → BECA -1×7+1912 Rf=fA -11305012+1135117012 122=11-2+211-131050 -1134050 -1133hPa 1122=11-2+2ABCOSO -1132×1 B=✓¥ÉABTso cased fA→andB→ insane direction 0=00 Cos @ =L 12=9-+13 if¥%÷p5> are in opposite direction 0=1800 COS 0=-1 R=A - B cases - F) and B-' are perpendicular to eachother 0=90 [ 050=0 R=fÉB

Vector subtraction Ñ=A→_B→=A→+fB→) 7 IF ¥7 B* A→ ☒# \" →* \"ñ¥☒ Magnitude of put F- x 4: % }y Cos 0=04,3 Sino -- 9113 \" x=BCosoMh✓y=BSinomW 11292=111--047-1912 1Bc12 os') o:t-(fBSAI.n-o/2lRY2=A2-2ABCoso-B2cosZe-B2sinZe 11242=11-2-2 ABCOSO -1132×1 R' =jA¥ABTso

Resolving of a vector Process of splitting a single vector into two ox more vector is known as resolving of a Hector . ☒µ-§ from triangular law -I¥ A→=A→×+Ñy \" A→=A×i^ + Ayj - Ay 10 Ax 11¥ 1T¥Cosa = Since = AyAx = Acoso =/1- since AF+AyÉA7o5o + A}in2o = A2 (1) = A2 = 1¥12 I # f- jAF+Ay

Jot product / scalar product ) II. 151-11++7115> lcoso \" i^=j^•j^=Ñ•k^=1 • F. Ñ=j^•Ñ=i^•j^=o If A→=A×F+Ayj^+AaÑ B→=B×i^+Byj^+BaK^ then A→•B→=A×B×tAyBy+AzBz cross product /scalar product ) A-'✗ 131-1571157 since Ñ it is a unit vector perpendicular to A→andB→ i^xj=k^ j×T=-k' i' ✗ i^=jxj=k^xÉ=o j^xk^=F k^×j^= F K - i¥° É×i=j' i'×ñ=-j ifÑ=A×i^tAyj+AzK^ 11+-1×13-7=1*11137 since B→=B×i^tByj+BzÑ Ñ×Ñ=|A× (find A→×B→ i Iñ = ((AA×YBB,z--BBy Aa ) Ag Az _ j ootz) + k^fA✗By - BxAg) Be By B2

telatiiicnielocitym TEA - Velocity of A nip - Melo city of B IIB - Velocity of A writ B III. =¥ - YI → - VBeWloc.iTrtipy.t-lo4/fT¥ ii3A-- BA River crossing problem * ' \" \"\" ; ÷ 11 → Net of man in Stillwater 1) → Velocity of river d → width of the river ✗ → drift . since time taken to cross the river = ¥6drift -1111050 + b) ofoles .net man 11M = Vittcosa

laselilm-ocrossthcriverinshortest imc-O-qoot.dk , ✗ =D dlv \"M=V%2 casdiil-Tocros5s1t1h-c1%riv1crinshortestpatho.co t=y÷ X=o 11m __✓¥r

ofMotion a projectile can be spitted into horizontal motion and vertical motion 1) •÷e Horizonta-otsioonveotica-mboyti=onHhxS-inlct.eco a×=o ay g= - 14=11×+9 ✗ t yy= Hy + ayt ' S ✗ = 11×1-+1=0×1-2 Sy = lsyttyzayt \" 11×2=11×2+29×51 Hy ? , Hy -12945g Time of flight / T ) It is the Time taken by projectile to reach back to the ground . Syat t=T, -_ o Sy = Hyt +112dg th 0 = Hsin of -11+1-2 f- g) 1- 2 129T¥ Hsin OT

-1=2115%71 Range of the projectile CR) 1) •÷e ←R- at t=T , s×=R s×=u×t+÷a×t ' IÉ÷¥%,•My{R= HCOSOIT) -10 R= HCOSO ✗ T 2HSgi#R = HCOSO ✗ R=1IÉl g D= ITS in 20 g- Note if angle of Maximum range will be obtained projection is 450

Maximum range is given by Rina ✗ = V2 g- \" angle of projection is o or ao - o , then range will be same . angle of projection = 0 Mange , R=Ñsing2O_ angleofprojec=_ Ñ=Ygorange , 1=11%111180--201 = liking g = Ry so range is same in both the cases . Maximum Height reached by projectile At topmost point of the projectile Ny = 0 Sy = It Hy? ' -129g Sy Hy

lot ? ( Hsin of-121 - g) It 2g H= is}in2o H=ÑsI¥ Velocity of projectile at any time time-toÉÉ¥☒\" µ ☒ ✗- - - -- ✗ time -0µA. .- - Projectile is projected with an initial velocity velocityis . At any time 't ' its is 11 Which can Vybe resolved into Vx and . I'=V×i^tVyj^ 14=11×+9 ✗ t 11=1171 = ¥+11,2 = HCOSO -10 ✗ t •\" \" =✓¥iÑgtVy== HCOSQ Then velocity of projectile at isytayt = Hsiao +f- g) t any time is gt= Hsiao - \" =j¥+sino-gtT

the above case angle made by projectile Keith horizontal is d -_Y÷then 1-and tana-usi.g ?ta=tan-'fs i':-I-) velocity of projectile at any height \";t☒\" H •¥& Iii\" ✗ A projectile is projected with an initial velocity is at an angle 0 with horizontal . When it reaches at a height h' ' from the ground , its velocity is 11 and it can be resolved Vyinto Vx and ^ 17-11×1^+1/yj^

11=117 / = ✓¥4.2 11×2=11×1+29×5 ✗ \"= fÉsiñosgh OF-11J COS -12×0×5✗ = licorice \"F- \"= ✓iigh ' by -12 aysy = Hsin 05+21-g) h v=fÉgh = ifs into -2g h \" = ✓¥1 coEo-siño=1mI Trajectory of projectile YX p H Fix, y) → ☒ Aa Y ④. 1010) K A projectile is projected with an initial velocity is at an angle 0 with horizontal . When it ,

reaches point , its coordinate is Cx•y ) . iesx-xsy-yuxt-l-za.it?-xHyt-1z-ayt2=yHcosot0=x EyHsiao t -11-2 f- g) 1- 1) [OSO t =K Using,÷go-1-29 ,?÷§o= 1- =L HCOSCE ÷÷é→octane - y=xtan•-÷÷a Note y=xtano-gx2-xt-anoaicos-otano-xt.no - ;÷¥é÷•) = ✗ tano - gÑtano_ 112251170 Cosa =xtan•→¥:•

;÷÷÷+an • - , x4qgo_= seta no - of )= x fan 1 - Xlr of %)y fan°1- •. =x

circular motion Ang display → Angle covered by radius vector in given interval of tin unit → radian Ang velocity → fate of change of an g. disp Cw ) Unit → road Is 4¥Wang = -_&¥wins ,, Ang acceleration → rate of change onang.net (d) ✗avg __%t , Winst =d%t w v=RuTWrelation btw v and W=2ñfm① R → radius of circular path 1-→ Time period

Typesnoffsinger motion → 1) Uniform circular motion → speed constant a) Nonuniform circular motion → speed varying Types of forces in circular motion 1) Centripetal force direction of body → changes 2) Tangential force speed of the body → changes centripetal acceleration q=W2R 09 a[ = 11413 tangential acceleration datat -_ 09 9-1=124 In uniform circular motion tangential force is zero .