2D - 1 1/18/2009 University of Colorado at Boulder Motion in 2D Velocity and acceleration are vectors. They can have any direction. When we are considering motion in the xy plane, these vectors can point anywhere in the plane. A common example of motion in 2D is... Projectile motion Consider a projectile fired from a cannon, with an initial velocity 0v with a direction of above the horizontal. 0x 00y 0v v cosv v sin Acceleration is a vector, and can have any direction. But in the special case of acceleration due solely to gravity, the acceleration is always straight down. Review of 1D motion: d v d xa , vd t d t, From these two equations, we can derive, for the special case a = constant, (a) ov v a t (b) 2oox x v t (1/ 2)a t   (c) 22oov v 2a(x x )   (d) ovvv2 xo , vo = initial position, initial velocity x, v = position, velocity at time t y x ax = 0 ay = –g a y vo x vx vy2D - 2 1/18/2009 University of Colorado at Boulder Suppose that a = 0 . In this case v = constant, and 0xv v v constantt 00 0 0 0 0xxv , v t x x , x x v tt If a ≠ 0 then 21002position if a = 0how much more (a > 0) or less (a < 0) distance yougo if a 0x x v t a t End of 1D motion review. Now, 2D Motion yxxydvdvdva a , adt dt dt and xyd r d x d yv v , vd t d t d t Special case: xya constant a constant, a constant This is exactly like the 1D motion case, except now we have separate equations for x-motion and y-motion. We can treat the x-motion and y-motion separately: These are the x- and y-components of the vector equations 21oo2r r v t a t     ov v a t Example: Horizontal Rifle. A rifle bullet is fired horizontally with vo = 100 m/s from an initial height of y0 = 2.0 m. Assume no air resistance. How long is the bullet in flight? How far does the bullet go before it hits the ground? Key idea in all projectile motion problems: treat x- and y-motions separately! y x vox = 100 m/s 2 m 21o ox x2x ox x21o oy y2y oy yx x v t a tXv v a ty y v t a tYv v a t    2D - 3 1/18/2009 University of Colorado at Boulder The motion along the y-direction (vertical motion) is completely independent of the motion along the x-direction (horizontal motion). 0oxxX: x 0v 100m/sa0 0oy2yY: y 2mv0a g 9.8 m/s The time to hit the ground is entirely controlled by the y-motion: 2 2 21 1 1o oy 0 02 2 20022000y y v t g t 0 y g t , y g t2y 2y2(2)2y g t , t , t 0.64sg g 9.8           Now we look at the x-equations to see how far along the x-direction the bullet traveled in 0.64 s. x x ox21o ox x ox200a 0 v constant = v 100 m/sx x v t a t v t 100(0.64) 64m         Why vx = constant ? The force of gravity is straight down. There is no sideways force to change vx (assuming no air resistance). Another question: What is the speed of the bullet as it falls? x ox oy oy y0gv constant v vv v a t g t As the bullet travels, its vx remains constant, while |vy | grows larger and larger. y x vox = vo vx vy v vx |vy | v2D - 4 1/18/2009 University of Colorado at Boulder speed = magnitude of velocity = 22 2 2x y oxv v v v g t The speed is a minimum at t = 0 when vy = 0 (the moment when the bullet leaves the gun). The speed is maximum when vy is maximum, just before the bullet reaches the ground. Don’t forget that we are assuming no air resistance. For a real rifle fired in real air, the bullet’s speed is usually maximum when leaving the barrel, and then air resistance slows the bullet down as it travels. Example: A projectile is fired on an airless world with initial speed vo at an angle above the horizontal. What is the minimum speed of the projectile? Answer: vox = vo cos . Proof: x x ox oy y oya 0 v constant = v v cosa g v v g t Review of acceleration: 1D: dvadt 2D: 21vvd v vad t t t 21v v v   means 21v v v   v is the vector you add to 1v to get 2v. The direction of a is the same as the direction of v (since a v positive number (1/ t ) x V V2 V1 The direction of the acceleration of gravity is the direction of V: straight down! V1 V2 y x vy ( vy = 0 ) voy vox vox vox Here, the speed is minimum at the top of the trajectory, where vy = 0. v1 vv22D - 5 1/18/2009 University of Colorado at Boulder "Shoot the Monkey" Experiment: A hunter aims a rifle at a monkey hanging in a tree. The rifle fires at the same instant that the monkey lets go and drops. Does the bullet hit the poor monkey? Answer: Yes! First, consider the situation with no gravity : If no gravity, the bullet goes in a straight line, and the monkey does not fall. So the monkey is hit. The height of the bullet (with ay = 0) is 0 0y 00y y v t v sin t Now, turn on gravity. The height of the bullet is now: 210 0y2y y v t g t. With gravity on, the bullet falls below the straight-line path by a distance (1/2)g t2 , which is exactly the same distance that the monkey falls. So the monkey falls into the path of the bullet. Poor monkey! Circular Motion and Acceleration Circular motion: consider an object moving in a circle of radius r, with constant speed v. T = period = time for 1 complete revolution, 1 cycle y x (1/2)g t2 y x r v2D - 6 1/18/2009 University of Colorado at Boulder 2rdistancespeed v vtime T An object moving in a circle is accelerating, because its velocity is changing -- changing direction. Recall the definition of acceleration: 21dvv v vad t t t   , velocity v can change is two ways: Magnitude can change or direction can change: For circular motion with constant speed, we will show that 1) the magnitude of the acceleration is 2vaar 2) the direction of the acceleration is always towards the center of motion. This is centripetal acceleration. "centripetal" = "toward center" Notice that the direction of acceleration vector is always changing, therefore this …