Chapter 3: Kinematics in Two Dimensions

Transcription

1 Chapter 3: Kinematics in Two Dimensions Vectors and Scalars A scalar is a number with units. It can be positive, negative, or zero. Time: 100 s Distance and speed are scalars, although they cannot be negative because of the way they are defined. A vector is a quantity which specifies magnitude and direction. Velocity:4 m/s East Acceleration: 10 m/s to the right

2 Notation Vector quantities are represented using bold type or an arrow above the symbol. The magnitude of a vector (a scalar) is represented using the absolute value symbol or without bold type: velocity: v = v = 5 m/s East v = v =5m/s

3 Arrow Representation Arrows are also used to represent vectors themselves. The length of the arrow is proportional to the magnitude. The direction of the arrow indicates the direction of the vector. 4 m/s East m/s North

4 Adding Vectors Only add vectors with the same units. For example: v 3 = v 1 + v is OK, but v 3 = x 1 + v does not make sense Remember that a vector -AA points opposite to A. A A

5 Adding Vectors Graphically To add two vectors A + B, put the tail of B at the tip of faa and connect the tail of faa to the tip of B. Use a ruler and protractor to determine the magnitude and direction of C.

6 Subtracting Vectors, and Multiplication by a Scalar The negative of a vector is a vector having the same length, but pointing in the opposite direction So we can define the subtraction of one vector from another as A vector can also be multiplied by a scalar c. We define this product so that cv has the same direction as V and a magnitude c V.

7 Components of a Vector A vector can be specified either by its magnitude and direction, or by its components in a coordinate system. r,θθ r x,r y

8 Be Careful The components of a vector are not always positive. The angle is not always defined relative to the x axis.

9 Trigonometric Functions SOH CAH TOA opposite side sinθ = = hypotnuse cosθ = tanθ = A A x + A y adjacent side = hypotnuse opposite side = adjacent side A y A A Ax A A A y x = A + A A x θ A A y Be careful with your calculator! Are you using degrees or radians?

10 Example Find the x and y components of a vector of magnitude 5, if its angle relative to the x axis is 00. ( ) A = Acosθ = 5cos 00 = 4.70 x ( ) A = Asinθ = 5sin 00 = 1.71 y

11 Adding Vectors Using Components Break vectors up into components. Add components. Find magnitude and direction of the sum from its components.

12 Example 1. The vectors are shown in the figure below. Their magnitudes are given are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with x axis.

13 Unit Vectors A unit vector is a vector with magnitude equal to 1 and pointing in a certain defined direction. Example: the unit vector in the x direction is usually written: ˆ xˆ or i Now I can write a vector as its components multiplied by the unit vector: A= A xˆ+ A yˆ x or A= Aiˆ+ A ˆj where A x and A y are scalars. x y j y

14 Example Write the vector in the figure in unit vector notation. y θ v v = 10 x θ = 50

15 Position and Displacement Vectors The position vector r points from the origin to a particular location. The displacement vector Δr indicates the change in position: Δr = r f - r i

16 Writing of displacement in terms of components and unit vectors, Δ r = ( x x ) iˆ+ ( y y ) ˆj+ ( z z ) kˆ We defined the instantaneous velocity as v = = Δt Δr dr lim Δ t 0 So we can write the instantaneous velocity as, dx ˆ dy ˆ dz v = i j kˆ v ˆ ˆ ˆ xi vy j vk z dt + dt + dt = + + dt

17 Likewise for acceleration, Δv aav = = Δ t v v Δ t 1 The instantaneous acceleration is given by so a lim v Δ = = dv Δ t 0 Δ t dt dv dv x ˆ y ˆ dvz a = i j kˆ a ˆ ˆ ˆ xi ay j azk + + = + + dt dt dt

18 Acceleration occurs whenever there is a change in velocity. If a car travels in a circle at a constant speed of 50 mi/hr, is it accelerating? YES! Because the direction of the velocity vector is changing.

19 Kinematic Equations for Constant Acceleration in Dimensions x-component y-component 1 x = xo + vxo t + ax t v = v + a t x xo x vx vxo ax x xo y = y + v t+ a t 1 o yo y v = v + a t y yo y = + ( ) v = v + a ( y y ) y yo y o

20 Projectile Motion Motion in the x direction is independent from motion in the y direction. We use the same equations from Chapter, but for each dimension separately. There are not really any new equations in this chapter.

21 The equations you need. 1 x x= x0 + v0xt+ a t 1 y = y0 + v0yt+ ayt NOTE v = v + a t x 0x x v = v + a t y 0 y y ( ) v = v + a x x x 0x + x f i ( ) v = v + a y y y 0 y y f i For projectile motion problems, a x (the horizontal componant of the acceleration) will usually be zero since usually there will be no acceleration in the x direction

22 Projectile Motion Assume that acceleration of gravity is constant, downward and has a magnitude of g = 9.81 m/s Air resistance is ignored The Earth s rotation is ignored Horizontal velocity is constant: because a x = 0 Vertical motion governed by the constant acceleration of gravity

23 Motion of a Projectile Launched Horizontally The dots represents the position of the object every 0.05 s. y 0 = 8 m; x 0 = 0 v 0y = 0; v 0x = 6 m/s a y = m/s ; a x = 0 t = 1 s 1. Verify the position of the object at t = 1s.. What would be the position of the object at t = 1 s if it were dropped (v 0x = 0)?

24 1 x = 0 + v0x t y = y0 + v0yt + ayt x 1 = 6 1 = 6 m y = ( 1) = 3.1 m 1 ( ) y = = 3.1 m x = 0

25 Example Pitcher s mounds are raised to compensate for the vertical drop of the ball as it travels 18 m to the catcher. (a) If a pitch is thrown horizontally with an initial speed of 3 m/s (71 mi/hr), how far does it drop by the time it reaches the catcher? (b) If the speed of the pitch is increased, does the drop distance increase, decrease or stay the same? Explain. (c) If this baseball game were to be played on the moon, would (c) If this baseball game were to be played on the moon, would the drop distance increase, decrease, or stay the same? Explain.

26 ( a) x= v t 0x 18 = 3t t = 0.56 s reach catcher reach catcher 1 1 y = y ( ) 0 + v0yt 9.8t y y0 = = 1.54 m y0 y = m (b) If the speed of the pitch increases, the time gets smaller, so the drop distance is less (c) On the moon g is less than 9.8 m/s, so the drop distance is less.

27 General Launch Angle Consider an object launched from the origin at an angle θ with respect to the horizontal. y v o θ x What are the x and y components of the initial velocity vector?

28 Example: On a hot summer day a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is.55 m/s at an angle of 35.0 above the horizontal. If she is in flight for 1.60 s, how high above the water was she when she let go of the rope? What is the girl s minimum speed during her flight? What is her acceleration at the top of her trajectory?

29 1 y = y0 + v0sinθt 9.81t 1 0 = y0 +.5sin 35 t 9.81t 1 0 = y0 +.5sin y 0 = m ( ) ( ) Girl s minimum speed is at the maximum in her trajectory! Can you see why? Girl s acceleration is 9 81 m/s at the top of her trajectory Girl s acceleration is 9.81 m/s at the top of her trajectory. What is her acceleration at other points on her trajectory?

30 Range The range R of a projectile is the horizontal distance it travels before landing. R v = 0 sin θ g (BE CAREFUL!, This equation only works when the initial and final elevation are the same) You should try to derive this equation! It s not difficult. Just remember that x=r when the projectile hits the ground. What angle θ results in the maximum range? Remember 0 θ 90. What if we do not ignore air resistance?

31 Example A projectile is fired at 00 m/s at an angle of 45 degrees above the horizontal. How far does it travel? R v = 0 sin θ g (00m/s) R o = sin((45 )) = 98m/s 9.8m/s 4081m

32 Maximum Height The maximum height (and therefore the hang time )ofa projectile depends only on the vertical component of its initial velocity. At y max, the vertical velocity v y is zero. v 0 y y = = v max v 0 0 y sin v = + aδy 0 θ + ( g) y sin θ g max

33 Example From the previous example, how high does the projectile go? y v = o max sin θ g ( 00m / s) o (45 ) y max = = sin (9.8) 100m

34 Uniform Circular Motion Consider a particle moving in a circle. v 1 Is the particle accelerating? lim v Δ a = = dv Δ t 0 Δ t dt - The speed of the particle is constant - The direction of the ball is changing v

35 What is the direction of the acceleration?

36 Magnitude of Radial Acceleration This type acceleration is called centripetal or radial acceleration. The magnitude of the acceleration is given by, a R = v r

37 Period and Frequency The amount of time that it takes the particle to make one full revolution is called the period, T. The number of times that the particle goes around the circle per unit time is called the frequency, y,f. T 1 = f The speed of a particle moving around in a circle is given by, distance π r v= = = π rf time T

38 Example 33 A h t tt th th th h t( 73k ) ith 33. A shotputter throws throws the shot (mass = 7.3 kg) with an initial speed of 14.0 m/s at a 40 angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete s hand at a height of. m above the ground.

39 Example 18. The position of a particular particle as a function of time is given by, ˆ ˆ r = (7.60 ti j t k ˆ ) m (a) Determine the particle s velocity and acceleration as a function of time. (b) Describe the motion in each direction.

40 Example Chapter #40. A space vehicle accelerates uniformly from 65 m/s at t=0 to 16 m/s at 10.0 s. How far did it move bt between t=.0s 0 and dt= s?

41 Example A baseball is hit at an angle of 50 relative to the horizontal with an initial speed of 10 miles/hour. ()H hi hd it? (a) How high does it go? (b) How long does it spend in the air? (c) How far does it travel?