2D and 3D Motion. with constant (uniform) acceleration

Transcription

1 2D and 3D Motion with constant (uniform) acceleration

2 1 Dimension 2 or 3 Dimensions x x v : position : position : displacement r : displacement : velocity v : velocity a : acceleration a r : acceleration r = xiˆ + yj ˆ+ zkˆ r xi yj zkˆ Unit vector = ˆ+ ˆ+ notation r v = vxiˆ + v ˆ yj+ vzkˆ r a= aiˆ + a ˆj + akˆ r r r r x y z

3 Magnitude and Direction Magnitude of a vector: z y x A A A A + + = Angle between 2 vectors: Use the dot product B A AB B A B A B A AB B A B A B A z z y y x x z z y y x x + + = = + + = θ θ cos cos r r

4 Relationships between x, v, and a for changing acceleration Derivatives x v a = = dx dt dv dt = 2 d x 2 dt Integrals a v= x= adt vdt const. = v o const. = x o

5 Problem A baseball outfielder throws a long ball. The components of the position are x = (30 t) m and y = (10 t 4.9t 2 ) m a) Write vector expressions for the ball s position, velocity, and acceleration as functions of time. Use unit vector notation! b) Write vector expressions for the ball s position, velocity, and acceleration at 2.0 seconds. r tiˆ 2 = 30 + (10t 4.9t ) ˆj r r dr v = = 30ˆ i+ (10 9.8t) ˆj dt r r dv a= = 9.8 ˆj dt r (2) r v(2) r a(2) = 30(2)ˆ i+ (10(2) 4.9(2) = 60ˆ i+ 0.4 ˆj = 30ˆ i+ (10 9.8(2)) ˆj = 30ˆ i+ 0.4 ˆj = 9.8 ˆj 2 ) ˆj

6 Problem A particle undergoing constant acceleration changes from a velocity of 4i 3j to a velocity of 5i + j in 4.0 seconds. What is the acceleration of the particle during this time period? v o = 4iˆ 3 ˆj v= 5 iˆ + ˆj v (5 4)ˆ i + (1 ( 3)) ˆj 1 a= = = iˆ ˆ 4 + j t 4 What is its displacement during this time period? x x acceleration is constant! x= vot+ = = 1 2 at ( i 3 ˆ)( ) ( 1 1 4ˆ j 4s + iˆ + ˆj )( 4s) ˆ 2 2 ( 16+ 2) i+ ( 3+ 8) j= i+ j 4 ˆ 18ˆ 2 5 ˆ

7 Trajectory of Projectile g g g g g This shows the parabolic trajectory of a projectile fired over level ground. Acceleration points down at 9.8 m/s 2 for the entire trajectory.

8 Trajectory of Projectile v x v y v x v x v y v y v x v y v x The velocity can be resolved into components all along its path. Horizontal velocity remains constant; vertical velocity is accelerated.

9 Position graphs for 2-D projectiles. Assume projectile fired over level ground. y y x x t t

10 Velocity graphs for 2-D projectiles. Assume projectile fired over level ground. V y V x t t

11 Acceleration graphs for 2-D projectiles. Assume projectile fired over level ground. a y a x t t

12 Remember To work projectile problems resolve the initial velocity into components. V o V o,y = V o sin θ θ V o,x = V o cos θ

13 Problem A soccer player kicks a ball at 15 m/s at an angle of 35 o above the horizontal over level ground. How far horizontally will the ball travel until it strikes the ground?

14 Problem A cannon is fired at 40 m/s at a 15 o angle above the horizontal from the top of a 120 m high cliff. How long will it take the cannonball to strike the plane below the cliff? How far from the base of the cliff will it strike? 40m/s 15 40cos15 m/s = 38.6 m/s x o x v o v a + 40sin15 m/s = 10.4 m/s x y +

15 Chain Rule If displacement is a function of a variable other than t, you need to use the chain rule: dy = dt dy dx dx dt Find the y-component of velocity if x-component of velocity is constant, and the y-component of displacement is described by y= x 2 + 3x dy dydx dx v y = = = 3 dt dxdt dt ( 2 x+ 3) = ( 2x+ ) v x

16 Chain rule problem An object moves with a horizontal velocity of 4 m/s and a vertical motion described by y=2x 2-3. Find the velocity of the object at 5 seconds. v v v v x y y y = = = (5s) 4m/s dy dt = dy dx dx dt ( 4x)( 4m/s) d = dx = 16x ( 2 2x 3)( v ) m/s = (16)(5s) = 80 m/s x v = v 2 x + v 2 y = 2 2 ( 4m/s) + ( 80m/s) = 80.1m/s

17 Review of Uniform Circular Motion

18 Uniform Circular Motion Occurs when an object moves in a circle without changing speed. Velocity (a vector) is continually changing; therefore, the object must be accelerating. Centripetal acceleration = the acceleration vector, which is pointed toward the center of the circle

19 Vectors in Uniform Circular Motion v a a v a = v 2 / r v a a v

20 Problem The Moon revolves around the Earth every 27.3 days. The radius of the orbit is 382,000,000 m. What is the magnitude and direction of the acceleration of the Moon relative to Earth?

21 Tangential acceleration Sometimes the speed of an object in circular motion is not constant (it s not uniform circular motion). An acceleration component is tangent to the path, aligned with the velocity. This is called tangential acceleration. The centripetal acceleration component causes the object to continue to turn while the tangential component causes the radius or speed to change.

22 Tangential Acceleration tangential component (a T ) v If tangential acceleration exists, the orbit is not stable. a radial or centripetal component (a r or a c )

23 Problem Given the figure at right rotating at constant radius, find the radial and tangential acceleration components if θ = 30 o and a has a magnitude of 15.0 m/s 2. What is the speed of the particle? How is it behaving? θ 5.00 m a

24 Problem Suppose you attach a ball to a 60 cm long string and swing it in a vertical circle. The speed of the ball is 4.30 m/s at the highest point and 6.50 m/s at the lowest point. Find the acceleration of the ball at the highest and lowest points.

25 Relative Motion When observers are moving at constant velocity relative to each other, we have a case of relative motion. The moving observers can agree about some things, but not about everything, regarding an object they are both observing.

26 Consider two observers and a particle. Suppose observer B is moving relative to observer A. P particle A observer v rel B observer

27 Also suppose particle P is also moving relative to observer A. P v A A observer particle In this case, it looks to A like P is moving to the right at twice the speed that B is moving in the same direction. v rel B observer

28 However, from the perspective of observer B P v B v A particle it looks like P is moving to the right at the same speed that A is moving in the opposite direction, and this speed is half of what A reports for P. vrel -v rel A observer B observer

29 The velocity measured by two observers depends upon the observers velocity relative to each other. P v B v A particle v B = v A v rel v A = v B + v rel -v rel A observer v rel B observer

30 Problem Now show that although velocity of the observers is different, the acceleration they measure for a third particle is the same provided v rel is constant. Begin with v B = v A -v rel

31 Galileo s Law of Transformation of Velocities If observers are moving but not accelerating relative to each other, they agree on a third object s acceleration, but not its velocity!

32 Inertial Reference Frames Frames of reference which may move relative to each other but in which observers find the same value for the acceleration of a third moving particle. Inertial reference frames are moving at constant velocity relative to each other. It is impossible to identify which one may be at rest. Newton s Laws hold only in inertial reference frames, and do not hold in reference frames which are accelerating.

33 Problem How long does it take an automobile traveling in the left lane at 60.0km/h to pull alongside a car traveling in the right lane at 40.0 km/h if the cars front bumpers are initially 100 m apart?

34 Problem A pilot of an airplane notes that the compass indicates a heading due west. The airplane s speed relative to the air is 150 km/h. If there is a wind of 30.0 km/h toward the north, find the velocity of the airplane relative to the ground.