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Transcript
Chapter 3 Kinematics in Two Dimensions Vectors 3-1 Vectors & Scalars •A vector has magnitude as well as direction. •Examples: displacement, velocity, acceleration, force, momentum •A scalar has only magnitude •Examples: time, mass, temperature 3-2 Vector Addition – One Dimension A person walks 8 km East and then 6 km East. Displacement = 14 km East A person walks 8 km East and then 6 km West. Displacement = 2 km 3-2 Vector Addition Example 1: A person walks 10 km East and 5.0 km North DR D1 D2 DR D12 D22 DR (10.0 km) 2 (5.0 km) 2 11.2 km D2 sin DR sin 1 ( D2 5.0 km ) sin 1 ( ) 26.5 0 DR 11.2 km Order doesn’t matter 3-2 Graphical Method of Vector Addition Tail to Tip Method V1 VR V2 V3 3-2 Graphical Method of Vector Addition Tail to Tip Method V1 V2 V1 V3 VR V2 V3 Parallelogram Method 3-3 Subtraction of Vectors Negative of vector has same magnitude but points in the opposite direction. For subtraction, we add the negative vector. 3-3 Multiplication by a Scalar A vector V can be multiplied by a scalar c; the result is a vector cV that has the same direction but a magnitude cV. If c is negative, the resultant vector points in the opposite direction. Module 5 - 8 3-4 Adding Vectors by Components Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other. Trigonometry Review Hypotenuse Opposite Adjacent Opposite Hypotenuse Adjacent cos Hypotenuse sin Opposite sin tan Adjacent cos 3-4 Adding Vectors by Components If the components are perpendicular, they can be found using trigonometric functions. sin Opposite Hypotenuse Vy V Adjacent Vx Hypotenuse V Opp sin tan Adj cos cos V y V sin Vx V cos 3-4 Adding Vectors by Components The components are effectively one-dimensional, so they can be added arithmetically: Module 5 - 12 Signs of Components y Rx Rx Ry Ry x Rx Ry Rx Ry 3-4 Adding Vectors by Components Adding vectors: 1. Draw a diagram; add the vectors graphically. 2. Choose x and y axes. 3. Resolve each vector into x and y components. 4. Calculate each component using sines and cosines. 5. Add the components in each direction. 6. To find the length and direction of the vector, use: sin Vy V Example 2 A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes a angle of 1200 with the positive x-axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.00 to the positive x axis. Find the magnitude and direction of the second displacement. B A A 150 cm R A B R 120 0 35 R 140 cm B R A B R A X 0 X X Bx (140cm) cos 35 (150cm) cos120 Bx 190cm B R A y y y By (140cm) sin 35 (150cm) sin 120 By 49.6cm Example 2 Alternative Solution. In the solution below, the angles for vector A are measured from the negative x axis. In this case, we have to assign the signs for the components. The answer is the same. A B R 60 A 150 cm R A B 0 35 R 140 cm B R A B R A X 0 X X Bx (140cm) cos 35 (150cm) cos 60 Bx 190cm B R A y y y By (140cm) sin 35 (150cm) sin 60 By 49.6cm Example 2 Continued A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes a angle of 1200 with the positive x-axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.00 to the positive x axis. Find the magnitude and direction of the second displacement. B B B 2 2 x y B (190cm) 2 (49.6cm) 2 sin 49.6cm 196cm 49.6 sin 196 1 14.6 196 cm 3-5/3-6 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola. Projectile Motion •Neglect air resistance •Consider motion only after release and before it hits •Analyze the vertical and horizontal components separately (Galileo) •No acceleration in the horizontal, so velocity is constant •Acceleration in the vertical is – 9.8 m/s2 due to gravity and thus velocity is not constant. •Object projected horizontally will reach the ground at the same time as one dropped vertically Equations for Projectile Motion Horizontal ax=0 Vertical ay = - g vx= constant v0 vx 0 x x0 vx0 t vy vy 0 g t 1 2 y y0 v y 0 t g t 2 v y2 v y20 2 g ( y y0 ) Initial Velocity v y 0 v0 sin vx 0 v0 cos •If the ball returns to the y = 0 point, then the velocity at that point will equal the initial velocity. •At the highest point, v0 y = 0 and v = vx0 Example 3A A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level. v (18.0 m )(cos 50.0 ) 11.6 m s s v (18.0 m )(sin 50.0 ) 13.8 m s s x0 y0 at top: vy vy0 g t 0 v 13.8 m s t 1.41s m g 9.80 s 1 y y v t gt 2 y0 2 2 max 0 yo 1 ymax 0 (13.8 m )(1.41s ) (9.8 m 2 )(1.41s ) 2 s s 2 ymax 9.7m Level Horizontal Range •Range is determined by time it takes for ball to return to ground level or perhaps some other vertical value. •If ball hits something a fixed distance away, then time is determined by x motion •If the motion is on a level field, when it hits: y = 0 1 1 y y0 v y 0 t g t 2 0 0 v yo t g t 2 2 2 2 v y0 Solving we find t g We can substitute this in the x equation to find the range R R x vx 0 t vxo ( 2 v y0 g ) 2 vx 0 v yo g 2 v02 sin 0 cos 0 g Level Horizontal Range We can use a trig identity 2 sin cos sin 2 v02 sin 2 R g •Greatest range: = 450 • = 300 and 600 have same range. ( 450 150 ) Caution– the range formula has limited usefulness. It is only valid when the projectile returns to the same vertical position. Example 3B A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. Calculate the range. Assume that the ball was kicked at ground level and lands at ground level. Assume time down = time up For Range: t (2)(1.41s) 2.82s R x x v t 33m 0 x0 0 (11.6 m )(2.82s) s Could also use range formula v02 sin 2 (18 m / s ) 2 sin (2) ( 500 ) R 33 m 2 g 9.8 m / s Example 4A A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. The football hits a window in a house that is 25.0 m from where it was kicked. How high was the window above the ground. v (18.0 m )(cos 50.0 ) 11.6 m s s v (18.0 m )(sin 50.0 ) 13.8 m s s x0 y0 xv t Time to hit the window: x0 25.0m 11.6 m s x t v x0 2.16s 1 y y v t gt 2 0 2 y0 1 m y 0 (13.8 )( 2.16s) (9.8 m 2 )( 2.16s) 2 s s 2 y 6 .9 m Example 4 B What is the final velocity and angle of the football that hit the window in Example 4 A. t 2.16s v v gt y y0 v y (13.8 m ) (9.8 m 2 ) (2.16s) s s v 11.6 m x s v (11.6 m ) (7.37 m ) s s 2 v tan v 7.37 m 2 13.7 m y x 7.37 tan 1 11.6 32.4 below x axis s s Example 5. (35) A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km /h (69.4 m / s) how far in advance of the recipients (horizontal distance) must the goods be dropped (Fig. 3–37a)? . vy0 0 v x 0 69.4 m / s Coordinate system is 235 m below plane 1 y 235 m 0 g t 2 0 2 ( 2 ) (235 m ) t 6.93 s 2 9.8 m / s x x0 vxo t 0 ( 69.4 m / s ) ( 6.93 s ) x 481 m 3-7 Projectile Motion Is Parabolic In order to demonstrate that projectile motion is parabolic, the book derives y as a function of x. When we do, we find that it has the form: This is the equation for a parabola. 3-8 Relative Velocity •Will consider how observations made in different reference frames are related to each other. A person walks toward the front of a train at 5 km / h (VPT). The train is moving 80 km / h with respect to the ground (VTG). What is the person’s velocity with respect to the ground (VPG)? VPG VPT VTG VPG 5 km / h 80 km / h 85 km / h 3-8 Relative Velocity •Boat is aimed upstream so that it will move directly across. •Boat is aimed directly across, so it will land at a point downstream. •Can expect similar problems with airplanes. Example 6 An airplane is capable of flying at 400 mi/h in still air. At what angle should the pilot point the plane in order for it to travel due east, if there is a wind of speed 50.0 mi/h directed due south? What is the speed relative to the ground? VPA VAG VPG VAG sin VPA mi 50 . 0 1 0 1 VAG h sin 7 . 18 sin 400 mi V PA h North of East Module 6 - 6 VPG ( 400 mi h ) cos ( 7.18 0 ) 397 mi h