Physics 201: Lecture 5, Pg 1 Lecture 5 l Goals: (Chapter 4.1-3)  Introduce position, displacement, velocity and acceleration in 2D  Address 2D motion in systems with constant acceleration (i.e. both magnitude and acceleration)  Discuss horizontal range (special) Key point 1: In many case motion occurs 2D with constant acceleration (typically on the surface of a planet) Key point 2: The “superposition principle” allows us to discuss the x & y motion individually

Physics 201: Lecture 5, Pg 2 Decomposing vectors Any vector can be resolved into components along the x and y axes  tan -1 ( y / x ) y x (x,y)  r ryry rxrx vyvy vxvx (v x,v y )  v vyvy vxvx ayay axax (a x,a y )  a ayay axax  tan -1 ( v y / v x )  tan -1 ( a y / a x )

Physics 201: Lecture 5, Pg 3 Dynamics: Motion along a line but with a twist (2D dimensional motion, magnitude and directions) l Particle motions involve a path or trajectory l In 2D the position of a particle is r = x i + y j and this vector is dependent on the origin. (i, j unit vectors ) ri’ri’ rf’rf’ O’O’ Displacement,  r, is independent of origin

Physics 201: Lecture 5, Pg 4 Motion in 2D l Position l Displacement l Velocity (avg)

Physics 201: Lecture 5, Pg 5 Instantaneous Velocity As  t  0  r shrinks and becomes tangent to the path l The direction of the instantaneous velocity is along a line that is tangent to the path of the particle’s direction of motion. v

Physics 201: Lecture 5, Pg 6 Average Acceleration l The average acceleration of particle motion reflects changes in the instantaneous velocity vector (divided by the time interval during which that change occurs). l Instantaneous acceleration

Physics 201: Lecture 5, Pg 7 Instantaneous Acceleration l The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path l Changes in a particle’s path may produce an acceleration  The magnitude of the velocity vector may change  The direction of the velocity vector may change  Both may change simultaneously (depends: path vs time)

Physics 201: Lecture 5, Pg 8 2 D Kinematics l Position, velocity, and acceleration of a particle: r = x i + y j v = v x i + v y j (i, j unit vectors ) a = a x i + a y j

Physics 201: Lecture 5, Pg 9 2 D Kinematics (special case) l If a x and a y are constant then l In many case one of the a’s is zero

Physics 201: Lecture 5, Pg 10 Trajectory with constant acceleration along the vertical What do the velocity and acceleration vectors look like? x t = 0 4 y x vs y Position Special Case: a x =0 & a y = -g v x (t=0) = v 0 v y (t=0) = 0 v y (t) = – g t x(t) = x 0  v 0 t y(t) = y 0  - ½ g t 2

Physics 201: Lecture 5, Pg 11 Trajectory with constant acceleration along the vertical What do the velocity and acceleration vectors look like? Velocity vector is always tangent to the curve! x t = 0 4 y x vs y Velocity

Physics 201: Lecture 5, Pg 12 Trajectory with constant acceleration along the vertical What do the velocity and acceleration vectors look like? Velocity vector is always tangent to the curve! Acceleration may or may not be! x t = 0 4 y x vs y Acceleration Position Velocity

Physics 201: Lecture 5, Pg 13 Another trajectory x vs y t = 0 t =10 Can you identify the dynamics in this picture? How many distinct regimes are there? 0 < t < 3 3 < t < 77 < t < 10  I.v x = constant = v 0 ; v y = 0  II.v x = v y = v 0  III. v x = 0 ; v y = constant < v 0 y x What can you say about the acceleration?

Physics 201: Lecture 5, Pg 14 Exercises 1 & 2 Trajectories with acceleration l A rocket is drifting sideways (from left to right) in deep space, with its engine off, from A to B. It is not near any stars or planets or other outside forces. l Its “constant thrust” engine (i.e., acceleration is constant) is fired at point B and left on for 2 seconds in which time the rocket travels from point B to some point C  Sketch the shape of the path from B to C. l At point C the engine is turned off.  Sketch the shape of the path after point C (Note: a = 0)

Physics 201: Lecture 5, Pg 15 Exercise 1 Trajectories with acceleration A. A B. B C. C D. D E. None of these B C B C B C B C A C B D From B to C ?

Physics 201: Lecture 5, Pg 16 Exercise 2 Trajectories with acceleration A. A B. B C. C D. D E. None of these C C C C A C B D After C ?

Physics 201: Lecture 5, Pg 17 Kinematics in 2 D; Projectile Motion l The position, velocity, and acceleration of a particle moving in 2-dimensions can be expressed as: r = x i + y j  v = v x i + v y j  a = a x i + a y j Special Case: a x =0 & a y = -g v x (t) = v 0 cos  v y (t) = v 0 sin  – g t x(t) = x 0  v 0 cos  t y(t) = y 0  v 0 sin  t - ½ g t 2 

Physics 201: Lecture 5, Pg 18 Kinematics in 2 D; Horizontal Range Given v 0 and  how far will and object travel horizontally?  Let y 0 = 0 = y initial = y final x 0 =0 Again: a x =0 & a y = -g 1. v x (t) = v 0 cos  2. v y (t) = v 0 sin  – g t 3. x(t) = 0  v 0 cos  t = R (range) 4. y(t) = 0 = 0  v 0 sin  t - ½ g t 2 4 gives: 0 = t (v 0 sin  - ½ g t)  t = 0, 2 v 0 sin  g R = v 0 cos  2 v 0 sin  g = v 0 2 sin 2  / g Maximum when dR/d  = 0  cos 2  = 0 or 45 ° 

Physics 201: Lecture 5, Pg 19 Parabolic trajectories ( v=10 m/s, g = - 10 m/s 2 ) 90 ° R:0.0m H:5.0m t=2.00s 75 ° R:5.0m H:4.7m t=1.93s 60 ° R:8.7m H:3.7m t=1.73s 45 ° R:10.0m H:2.5m t=1.41s 30 ° R:8.7m H:1.2m t=1.00s 15 ° R:5.0m H:0.3m t=0.52s

Physics 201: Lecture 5, Pg 20 Example Problem A medieval soldier is at the top of a castle wall. There are two cannon balls. The 1 st one he fires from the cannon and it lands 200 m away. Simultaneously the 2 nd cannon ball is dislodged and falls directly down. The 2 nd cannon ball lands after 2.0 seconds and the fired cannon ball lands 2.0 seconds later. The ground is completely level around the castle. At what angle from horizontal did he fire the cannon (if g= -10 m/s 2 & no air resistance)?

Physics 201: Lecture 5, Pg 21 Example Problem A medieval soldier is at the top of a castle wall. There are two cannon balls. The 1 st one he fires from the cannon and it lands 200 m away. Simultaneously the 2 nd cannon ball is dislodged and falls directly down. The 2 nd cannon ball lands after 2.0 seconds and the fired cannon ball lands 2.0 seconds later. The ground is completely level around the castle. At what angle from horizontal did he fire the cannon (if g= -10 m/s 2 & no air resistance)? Find height of wall from 2 nd cannon ball 0 = h + 0 – ½ 10 t 2  h = 5(4) m = 20 m Find angle 0 = h + v sin  (t+2) – ½ 10 (t+2) 2  0 = 20 m + 4v sin  - 80 m R = 200 m = v cos  (t+2)  4v = 200 / cos  Combining 60 m = 200 m tan   16.7 degrees

Physics 201: Lecture 5, Pg 22 Example Problem Example Problem: If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel be for it hit the level ground (if g= -10 m/s 2 & no air resistance)?

Physics 201: Lecture 5, Pg 23 Example Problem If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel before it hits the level ground (assume g= -10 m/s 2 & no air resistance)? l at t=0 the knife is at (0.0 m, 1.0 m) with v y =0 after  t the kinfe is at (x m, 0.0 m) x = x 0 + v x  t and y = y 0 – ½ g  t 2 So x = 10 m/s  t and 0.0 m = 1.25 m – ½ g  t 2  2.5/10 s 2 =  t 2  t = 0.50 s x = 10 m/s 0.50 s = 5.0 m

Physics 201: Lecture 5, Pg 24 Fini l For Thursday, Read all of Chapter 4