Vectors and 2D Kinematics. AIT AP Physics C

Transcription

1 Vectors and 2D Kinematics

2 Coordinate Systems Used to describe the position of a point in space Coordinate system consists of a fixed reference point called the origin specific axes with scales and labels instructions on how to label a point relative to the origin and the axes

3 Cartesian Coordinate System Also called rectangular coordinate system x- and y- axes intersect at the origin Points are labeled (x,y)

4 Polar Coordinate System Origin and reference line are noted Point is distance r from the origin in the direction of angle, ccw from reference line Points are labeled (r, )

5 Polar to Cartesian Coordinates Based on forming a right triangle from r and x = r cos y = r sin

6 Cartesian to Polar Coordinates r is the hypotenuse and an angle tan y x r x y 2 2 must be ccw from positive x axis for these equations to be valid

7 Example 3.1 The Cartesian coordinates of a point in the xy plane are (x,y) = (-3.50, -2.50) m, as shown in the figure. Find the polar coordinates of this point. Solution: From Equation 3.4, r x y and from Equation 3.3, y 2.50 m tan x 3.50 m 216 ( 3.50 m) ( 2.50 m) 4.30 m

8 Vectors and Scalars A scalar quantity is completely specified by a single value with an appropriate unit and has no direction. A vector quantity is completely described by a number and appropriate units plus a direction.

9 Vector Notation When handwritten, use an arrow: When printed, will be in bold print: A When dealing with just the magnitude of a vector in print, an italic letter will be used: A or A The magnitude of the vector has physical units The magnitude of a vector is always a positive number A

10 Vector Example A particle travels from A to B along the path shown by the dotted red line This is the distance traveled and is a scalar The displacement is the solid line from A to B The displacement is independent of the path taken between the two points Displacement is a vector

11 Equality of Two Vectors Two vectors are equal if they have the same magnitude and the same direction A = B if A = B and they point along parallel lines All of the vectors shown are equal

12 Adding Vectors When adding vectors, their directions must be taken into account Units must be the same Graphical Methods Use scale drawings Algebraic Methods More convenient

13 Adding Vectors Graphically Choose a scale Draw the first vector with the appropriate length and in the direction specified, with respect to a coordinate system Draw the next vector with the appropriate length and in the direction specified, with respect to a coordinate system whose origin is the end of vector A and parallel to the coordinate system used for A

14 Adding Vectors Graphically, cont. Continue drawing the vectors tip-to-tail The resultant is drawn from the origin of A to the end of the last vector Measure the length of R and its angle Use the scale factor to convert length to actual magnitude

15 Adding Vectors Graphically, final When you have many vectors, just keep repeating the process until all are included The resultant is still drawn from the origin of the first vector to the end of the last vector

16 Adding Vectors, Rules When two vectors are added, the sum is independent of the order of the addition. This is the commutative law of addition A + B = B + A

17 Adding Vectors, Rules cont. When adding three or more vectors, their sum is independent of the way in which the individual vectors are grouped This is called the Associative Property of Addition (A + B) + C = A + (B + C)

18 Adding Vectors, Rules final When adding vectors, all of the vectors must have the same units All of the vectors must be of the same type of quantity For example, you cannot add a displacement to a velocity

19 Negative of a Vector The negative of a vector is defined as the vector that, when added to the original vector, gives a resultant of zero Represented as A A + (-A) = 0 The negative of the vector will have the same magnitude, but point in the opposite direction

20 Subtracting Vectors Special case of vector addition If A B, then use A+(-B) Continue with standard vector addition procedure

21 Multiplying or Dividing a Vector 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 If the scalar is positive, the direction of the result is the same as of the original vector If the scalar is negative, the direction of the result is opposite that of the original vector

22 Components of a Vector A component is a part It is useful to use rectangular components These are the projections of the vector along the x- and y-axes

23 Vector Component Terminology A x and A y are the component vectors of A They are vectors and follow all the rules for vectors A x and A y are scalars, and will be referred to as the components of A

24 Components of a Vector, 2 The x-component of a vector is the projection along the x-axis A Acos x The y-component of a vector is the projection along the y-axis A Asin y Then, A A x A y

25 Components of a Vector, 3 The y-component is moved to the end of the x-component This is due to the fact that any vector can be moved parallel to itself without being affected This completes the triangle

26 Components of a Vector, 4 The previous equations are valid only if θ is measured with respect to the x-axis The components are the legs of the right triangle whose hypotenuse is A A A A and tan x y May still have to find θ with respect to the positive x-axis A A y x

27 Components of a Vector, final The components can be positive or negative and will have the same units as the original vector The signs of the components will depend on the angle

28 Unit Vectors A unit vector is a dimensionless vector with a magnitude of exactly 1. Unit vectors are used to specify a direction and have no other physical significance

29 Unit Vectors, cont. The symbols î, ĵ, and kˆ represent unit vectors They form a set of mutually perpendicular vectors

30 Unit Vectors in Vector Notation A x is the same as A x and A y is the same as A y ĵ etc. The complete vector can be expressed as A A ˆi A ˆj A kˆ x y z î

31 Adding Vectors Using Unit Vectors Using R = A + B Then A ˆ ˆ ˆ ˆ x Ay Bx By R i j i j R ˆi ˆj R A B A B R x x x y y R y and so R x = A x + B x and R y = A y + B y R R R x y tan R R y x

32 Trig Function Warning The component equations (A x = A cos and A y = A sin ) apply only when the angle is measured with respect to the x-axis (preferably ccw from the positive x-axis). The resultant angle (tan = A y / A x ) gives the angle with respect to the x-axis. You can always think about the actual triangle being formed and what angle you know and apply the appropriate trig functions

33 Adding Vectors with Unit Vectors

34 Adding Vectors Using Unit Vectors Three Directions Using R = A + B A ˆ ˆ ˆ ˆ ˆ ˆ x Ay Az Bx By Bz R i j k i j k A B A B A B R ˆi ˆj kˆ R x x y y z z R R R x y z R x = A x + B x, R y = A y + B y and R z = A z + B z R R R R x y z x tan R R x etc.

35 Example 3.5: Taking a Hike A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40.0 km in a direction 60.0 north of east, at which point she discovers a forest ranger s tower.

36 Example 3.5 (A) Determine the components of the hiker s displacement for each day. Solution: We conceptualize the problem by drawing a sketch as in the figure above. If we denote the displacement vectors on the first and second days by A and B respectively, and use the car as the origin of coordinates, we obtain the vectors shown in the figure. Drawing the resultant R, we can now categorize this problem as an addition of two vectors.

37 Example 3.5 We will analyze this problem by using our new knowledge of vector components. Displacement A has a magnitude of 25.0 km and is directed 45.0 below the positive x axis. From Equations 3.8 and 3.9, its components are: A A x y Acos( 45.0 ) (25.0 km)(0.707) = 17.7 km Asin( 45.0 ) (25.0 km)( 0.707) 17.7 km The negative value of A y indicates that the hiker walks in the negative y direction on the first day. The signs of A x and A y also are evident from the figure AIT AP Physics above. C

38 Example 3.5 The second displacement B has a magnitude of 40.0 km and is 60.0 north of east. Its components are: B B x y Bcos60.0 (40.0 km)(0.500) = 20.0 km Bsin60.0 (40.0 km)(0.866) 34.6 km

39 Example 3.5 (B) Determine the components of the hiker s resultant displacement R for the trip. Find an expression for R in terms of unit vectors. Solution: The resultant displacement for the trip R = A + B has components given by Equation 3.15: R x = A x + B x = 17.7 km km = 37.7 km R y = A y + B y = km km = 16.9 km In unit-vector form, we can write the total displacement as R = (37.7 î ĵ) km

40 Example 3.5 Using Equations 3.16 and 3.17, we find that the vector R has a magnitude of 41.3 km and is directed 24.1 north of east. Let us finalize. The units of R are km, which is reasonable for a displacement. Looking at the graphical representation in the figure above, we estimate that the final position of the hiker is at about (38 km, 17 km) which is consistent with the components of R in our final result. Also, both components of R are positive, putting the final position in the first quadrant of the coordinate system, which is also consistent with the figure.

41 Problem Solving Strategy Adding Vectors Select a coordinate system Try to select a system that minimizes the number of components you need to deal with Draw a sketch of the vectors Label each vector

42 Problem Solving Strategy Adding Vectors, 2 Find the x and y components of each vector and the x and y components of the resultant vector Find z components if necessary Use the Pythagorean theorem to find the magnitude of the resultant and the tangent function to find the direction Other appropriate trig functions may be used

43 Projectile Motion An object may move in both the x and y directions simultaneously The form of two-dimensional motion we will deal with is called projectile motion

44 Assumptions of Projectile Motion The free-fall acceleration g is constant over the range of motion And is directed downward The effect of air friction is negligible With these assumptions, an object in projectile motion will follow a parabolic path This path is called the trajectory

45 Projectile Motion Diagram

46 Verifying the Parabolic Trajectory Reference frame chosen y is vertical with upward positive Acceleration components a y = -g and a x = 0 Initial velocity components v xi = v i cos and v yi = v i sin

47 Verifying the Parabolic Trajectory Displacements x f = v xi t = (v i cos t y f = v yi t + ½a y t 2 = (v i sin t - ½ gt 2 Combining the equations gives: g y i x x 2 2 2vi cos i 2 tan This is in the form of y = ax bx 2 which is the standard form of a parabola

48 Projectile Motion Implications The y-component of the velocity is zero at the maximum height of the trajectory The accleration stays the same throughout the trajectory

49 Analyzing Projectile Motion Consider the motion as the superposition of the motions in the x- and y-directions The x-direction has constant velocity a x = 0 The y-direction is free fall a y = -g The actual position at any time is given by: r f = r i + v i t + ½gt 2

50 Non-Symmetric Projectile Motion Follow the general rules for projectile motion Break the y-direction into parts up and down or symmetrical back to initial height and then the rest of the height May be non-symmetric in other ways

51 Projectile Motion Vectors r f = r i + v i t + ½ g t 2 The final position is the vector sum of the initial position, the position resulting from the initial velocity and the position resulting from the acceleration

52 Range and Maximum Height of a Projectile When analyzing projectile motion, two characteristics are of special interest The range, R, is the horizontal distance of the projectile The maximum height the projectile reaches is h

53 Height of a Projectile, equation The maximum height of the projectile can be found in terms of the initial velocity vector: h 2 2 v i sin i 2g This equation is valid only for symmetric motion

54 Range of a Projectile, equation The range of a projectile can be expressed in terms of the initial velocity vector: 2 sin 2 R v i i This is valid only for symmetric trajectory g

55 More About the Range of a Projectile

56 Range of a Projectile, final The maximum range occurs at i = 45 o Complementary angles will produce the same range The maximum height will be different for the two angles The times of the flight will be different for the two angles

57 Projectile Motion Problem Solving Hints Select a coordinate system Resolve the initial velocity into x and y components Analyze the horizontal motion using constant velocity techniques Analyze the vertical motion using constant acceleration techniques Remember that both directions share the same time

58 Uniform Circular Motion Uniform circular motion occurs when an object moves in a circular path with a constant speed An acceleration exists since the direction of the motion is changing This change in velocity is related to an acceleration The velocity vector is always tangent to the path of the object

59 Changing Velocity in Uniform Circular Motion The change in the velocity vector is due to the change in direction The vector diagram shows Dv = v f - v i

60 Centripetal Acceleration The acceleration is always perpendicular to the path of the motion The acceleration always points toward the center of the circle of motion This acceleration is called the centripetal acceleration

61 Centripetal Acceleration, cont The magnitude of the centripetal acceleration vector is given by a C v r 2 The direction of the centripetal acceleration vector is always changing, to stay directed toward the center of the circle of motion

62 Period The period, T, is the time required for one complete revolution The speed of the particle would be the circumference of the circle of motion divided by the period 2 r Therefore, the period is T v

63 Tangential Acceleration The magnitude of the velocity could also be changing In this case, there would be a tangential acceleration

64 Relative Velocity Two observers moving relative to each other generally do not agree on the outcome of an experiment For example, observers A and B below see different paths for the ball

65 Relative Velocity, generalized Reference frame S is stationary Reference frame S is moving at v o This also means that S moves at v o relative to S Define time t = 0 as that time when the origins coincide

66 Relative Velocity, equations The positions as seen from the two reference frames are related through the velocity r = r v o t The derivative of the position equation will give the velocity equation v = v v o These are called the Galilean transformation equations

67 Acceleration in Different Frames of Reference The derivative of the velocity equation will give the acceleration equation The acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer moving at a constant velocity relative to the first frame.