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Vectors and Two-dimensional Motion. PHY231. 1. Vectors. In 1-d motion, we have used vectors But, because their components could be written with a single number, we didn’t bother introducing a new notation In 2-d a clearer notation for vectors becomes important
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Vectors and Two-dimensional Motion PHY231 PHY231 1
Vectors In 1-d motion, we have used vectors But, because their components could be written with a single number, we didn’t bother introducing a new notation In 2-d a clearer notation for vectors becomes important From now on, vectors will be written using either arrows like or bold letters like A or u PHY231
Vector properties Vectors are different than scalars because they have a magnitude and a direction Two vectors are equal if they have same magnitude and direction Vectors can be added, subtracted or multiplied and divided by scalars Addition and subtraction can be done graphically PHY231
Vector addition - graphically Continue drawing the vectors “tip-to-tail” The resultant is drawn from the origin of to the end of the last vector The order in which the vectors are added doesn’t affect the result PHY231
Vector subtraction - graphically Special case of vector addition Add the negative of the subtracted vector Continue with standard vector addition procedure PHY231
Vector basis (2-D) In the following, vectors will be used in a context where a coordinate system is defined In this case vectors are expressed w.r.t. a basis In 2-D, a vector basis is defined by two unit vectors. For example (ex, ey) PHY231
Vectors components (I) Each vector has 2 components For example V=(Vx, Vy) or Assuming the basis vectors are ex and ey V = Vxex+ Vyey Bound or fixed vectors are vectors whose initial point is the origin (e.g. position vector) Free vectors are vectors whose initial point is not necessarily the origin (e.g. velocity, acceleration vectors) Usually, the context determines if a vector is bound or free PHY231
Vector components (II) Basis (ex, ey) ex=(1,0) ex=(0,1) Let’s consider the bound vector V=(2,3) V = 2ex+ 3ey We can see V as the sum V = VX + VY VX = (2,0) = 2ex VY = (0,3) = 3ey PHY231
Vector components (III) A vector can also be parameterized using its magnitude (I.e. length) and its direction (I.e. angle w.r.t. the coordinate system) Consider The magnitude of V and angle w.r.t. the x-axis are One also has PHY231
Vector components (IV) It may be convenient to use a coordinate system other than horizontal and vertical Choose axes that are perpendicular to each other Adjust the components accordingly PHY231
Reminder !! Tanq = y/x q = tan-1 (y/x) If x>=0 your calculator will tell you the truth If x<0 your calculator may lie to you ! x<0 and y>0 add π to calc. result x<0 and y<0 subtract π from calc. result tan-1( -1/2 / -√3/2) Calculator = π/6 wrong !! Correct answer = π/6– π = -5π/6 PHY231
In 2-D motion it is usual to have a coordinate system with One direction parallel to the direction of the motion One direction perpendicular to the direction of the motion Typical coordinate system Fx=Fsin Fy=Fcos F y • Choices for coordinate system • Second choice with x parallel and y perpendicular to the direction of motion is better y v x u x PHY231
Vector between two points A vector is sometimes named using its two extreme points, for example u = v = w = And one has for example Two vectors are equal if they have equal components. u = v = w since PHY231
Vector multiplication by a scalar The result of the multiplication or division is a vector The magnitude of the vector is multiplied or divided by the scalar Consider Multiplication and division In particular PHY231
Vector addition (2-D basis) Consider Then with C=A+B C=B+A PHY231
Vector subtraction (2-D basis) • Consider • Then • and C=A-B PHY231
Relative motion • Relative velocity is about relating the measurements of two different observers • Sometimes, it can be useful to use a moving frame to analyze the motion • Motion can appear very different in different frames !! PHY231 17
Relative motion • The following pattern of subscripts can be useful when solving relative motion problems • rAE = Position of A as measured in referential E • rBE = Position of B as measured in referential E • rAB = Position of A as measured in referential attached to B PHY231 18
Relative motion - train • A girl is in a train. The train is moving at 100 km/h and the girl walks at 5 km/h toward the head of the train. • What is the velocity of the girl for a cow standing and watching the train pass by? • What about if the girl is walking toward the tail of the train instead? • Girl: G, Train: T, Cow:C PHY231 19
Relative motion - 2 planes • Two planes are leaving an airport at the same time • The first flies south-east at 600 km/h • The second flies west at 500 km/h • What is the position of the second plane w.r.t. the first plane after 30 minutes? G = Ground 1 = 1st plane 2 = 2nd plane y2 yG v2G x2 xG y1 x1 v1G PHY231 20
Solution – 2 planes • Let’s solve the problem in a general manner • Let’s apply the result to our case PHY231 21
Solution – 2 planes • Plane #1 • Plane #2 • Let’s visualize the problem graphically in ground frame • Initial instant • After 30 minutes yG yG xG xG • Ground frame PHY231 22
Solution – 2 planes • Plane #1 • Plane #2 • Let’s visualize the problem graphically in plane#1 frame • Initial instant • After 30 minutes y1 y1 x1 x1 • Plane#1 frame PHY231 23
Relative motion – boat/river • A boat is moving on a river • The boat speed in still water is 20 km/h • The river flows north to south at 10 km/h • The boat is crossing the river west to east • What angle q should the boat be directed toward to go straight across? • A) -60º • B) -30º • C) 0º • D) 30º • E) 60º yG xG PHY231 24
Relative motion – boat/river • A boat is moving on a river • The boat speed in still water is 20 km/h • The river flows north to south at 10 km/h • The boat is crossing the river west to east • What angle q should the boat be directed toward to go straight across? • A) -60º • B) -30º • C) 0º • D) 30º • E) 60º yG xG PHY231 25
Explanation – boat/river • Let’s solve the problem in a general manner • Let’s apply the result to our case • We want the boat to go straight across, thus the vertical component of the velocity should be zero G = Ground B = Boat W = Water PHY231 PHY231 26 26
Relative motion – boat/river (2) L • A boat is moving on a river • The boat speed in still water is 20 km/h • The river flows north to south at 10 km/h • The boat is crossing the river west to east • What angle q should the boat be directed toward to across as fast as possible? • A) -60º • B) -30º • C) 0º • D) 30º • E) 60º yG xG PHY231 27
Relative motion – boat/river (2) L • A boat is moving on a river • The boat speed in still water is 20 km/h • The river flows north to south at 10 km/h • The boat is crossing the river west to east • What angle q should the boat be directed toward to across as fast as possible? • A) -60º • B) -30º • C) 0º • D) 30º • E) 60º yG xG PHY231 28
Explanation – boat/river • Let’s apply the result to our case • The time to get across is given when the x-component is equal to the river width L. Dt minimum when G = Ground B = Boat W = Water PHY231 PHY231 29 29
2-D motion - displacement The position of an object is described by its position vector r Displacement is the change in position, SI unit (m) PHY231
2-D motion – average velocity Average velocity during time interval Dt is SI unit (m/s) Dt=2s Dt is a scalar so <v> is colinear to Dr PHY231
2-D motion – instantaneous velocity Limit of the average velocity as Dt goes to zero SI unit (m/s) • Tangent to the path of the particle • In the direction of motion vi vf PHY231
2-D motion average acceleration Average acceleration during time interval Dt is, SI unit (m/s) SI unit (m/s2) vf aav Dv vi vi vf PHY231
2-D motion – instantaneous acceleration Limit of the average acceleration as Dt goes to zero SI unit (m/s2) PHY231
1d-motion, reminder • Notations • Below, Dt is time ellapsed from initial instant • xi, vi and ai refer to quantities at initial instant • xf, vf and af refer to quantities at final instant • Constant Acc (CA) • Other useful equations CA1 CA2 NEED 2-D EQUIVALENT EQUATIONS !! CA3 CA4 CA5 PHY231 PHY231 35
2-D = 1-D + 1-D For our purpose, the 2-D motion will always be decomposed in two independents 1-D problems !! Basically (a, v, r) becomes (a, v, r) and 1-D equations apply to the components of these vectors 2-D equations 1-D equations CA1 CA2 CA3 PHY231
Other relations The equations on the previous slides are the 2-D equivalent of the 1-D equations CA-1, 2 and 3 of Chapter2. Here are the equivalent of Ca-4 and 5: 2-D equations 1-D equations PHY231
Projectile motion We will ignore air friction rotation of the earth With these assumptions, an object in projectile motion will follow a parabolic path The x- and y-directions of motion are completely independent of each other The x-direction is uniform motion ax = 0 The y-direction is free fall ay = -g The initial conditions can be broken down into its x- and y-components PHY231
Projectile motion equations The initial conditions can be broken down into its x- and y-components The equations of motion become PHY231
The train and the canon A little canon is embarked on a train rolling at constant velocity w.r.t the ground. The canon shoots a ball straight in the air w.r.t the train. Where will the ball falls back? (assume no friction for the train and the ball) A) Behind the train B) Right back on the train C) In front of the train VBT VTG PHY231
The train and the canon A little canon is embarked on a train rolling at constant velocity w.r.t the ground. The canon shoots a ball straight in the air w.r.t the train. Where will the ball falls back? (assume no friction for the train and the ball) A) Behind the train B) Right back on the train C) In front of the train yT xT PHY231
The train and the canon - explanation Only thing you need to solve is the motion of the ball in the referential of the train Initial conditions • The equation for the position tells us that the ball is always at the same xT than the train. • The ball will move up, decelerate due to the gravity and falls right back on the train Motion PHY231
Falling dice Two dice are falling from a same height H. The first one is at rest initially. The second one has a horizontal velocity initially. Which one gets to the ground the fastest? A) First one B) Second one C) Same time for both y x PHY231
Falling dice Two dice are falling from a same height H. The first one is at rest initially. The second one has a horizontal velocity initially. Which one gets to the ground the fastest? A) First one B) Second one C) Same time for both y x PHY231
Falling dice - explanation Both dice are following the projectile motion eq. They both reach the ground when y=0 Y(t) is the same equation for both dice They reach the ground at the same time! PHY231
Falling dice Trajectory for both dice H y x PHY231
Canon firing A canon is at rest on a hill at height h=20 m It fires at 45º and the speed of the shell is 30m/s initially What is the speed of the shell when it hits the ground? How long does the shell flies in the air? How far did the projectile go? V=30 m/s 45º y 1 h=20m x X=185m
Canon firing – general solution • Initial conditions change in velocity and position • The shell hits the ground when Dy=-h. Using the y-component in the previous equation we obtain the quadratic equation for the time of impact • Solving and keeping the physical solution
Canon firing – general solution • The velocity and position of the shell at the time of impact are • The speed at impact is found by calculating the magnitude of the velocity vector
Canon firing – example solved • In our example we had initial conditions • Time, velocity, position, speed and angle of impact are found applying the previous equations