1. Physics 207: Lecture 8, Pg 1 Lecture 5 l Goals: (finish Ch. 4)   Address 2D motion in systems with acceleration (includes linear, projectile and circular motion)  Discern differing reference frames and understand how they relate to particle motion 

2. Physics 207: Lecture 8, Pg 2 Kinematics in 2 D l The position, velocity, and acceleration of a particle moving in 2-dimensions can be expressed as: r ij v ij a ij r = x i + y j  v = v x i + v y j  a = a x i + a y j aj Special Case: a x =0 & a y = -g or a = -g j 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 

3. Physics 207: Lecture 8, Pg 3 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) v y (  t) = – g  t x(  t) = x 0  v 0  t y(  t) = y 0  - ½ g  t 2

4. Physics 207: Lecture 8, Pg 4 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

5. Physics 207: Lecture 8, Pg 5 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

6. Physics 207: Lecture 8, Pg 6 Concept Check E1. You drop a ball from rest, how much of the acceleration from gravity goes to changing its speed? A. All of it B. Most of it C. Some of it D. None of it E2. A hockey puck slides off the edge of a horizontal table, just at the instant it leaves the table, how much of the acceleration from gravity goes to changing its speed? A. All of it B. Most of it C. Some of it D. None of it

7. Physics 207: Lecture 8, Pg 7 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)?

8. Physics 207: Lecture 8, Pg 8 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

9. Physics 207: Lecture 8, Pg 9 Exercise 3 Relative Trajectories: Monkey and Hunter A. go over the monkey. B. hit the monkey. C. go under the monkey. All free objects, if acted on by gravity, accelerate similarly. A hunter sees a monkey in a tree, aims his gun at the monkey and fires. At the same instant the monkey lets go. Does the bullet …

10. Physics 207: Lecture 8, Pg 10 Schematic of the problem x B (  t) = d = v 0 cos   t y B (  t) = h f = v 0 sin   t – ½ g  t 2 x M (  t) = d y M (  t) = h – ½ g  t 2 Does y M (  t) = y B (  t) = h f ? Does anyone want to change their answer ? (x 0,y 0 ) = (0,0) (v x,v y ) = (v 0 cos , v 0 sin  Bullet  v 0 g (x,y) = (d,h) hfhf What happens if g=0 ? How does introducing g change things? Monkey

11. Physics 207: Lecture 8, Pg 11 Relative motion and frames of reference l Reference frame S is stationary Reference frame S ’ is moving at v o This also means that S moves at – v o relative to S ’ l Define time t = 0 as that time when the origins coincide

12. Physics 207: Lecture 8, Pg 12 Relative motion and frames of reference l Reference frame S is stationary, A is stationary Reference frame S ’ is moving at v o A moves at – v o relative to an observer in S ’ Define time t = 0 as that time (O & O ’ origins coincide) r ’= r- v 0 t O,O ’ t=0 -v0-v0 r A -v 0 t O’O’ r t later A

13. Physics 207: Lecture 8, Pg 13 Relative Velocity The positions, r and r ’, as seen from the two reference frames are related through the velocity, v o, where v o is velocity of the r ’ reference frame relative to r  r ’ = r – v o  t l The derivative of the position equation will give the velocity equation  v ’ = v – v o l These are called the Galilean transformation equations l Reference frames that move with “constant velocity” (i.e., at constant speed in a straight line) are defined to be inertial reference frames (IRF); anyone in an IRF sees the same acceleration of a particle moving along a trajectory.  a ’ = a (dv o / dt = 0)

14. Physics 207: Lecture 8, Pg 14 Central concept for problem solving: “x” and “y” components of motion treated independently. l Example: Man on cart tosses a ball straight up in the air. l You can view the trajectory from two reference frames: Reference frame on the ground. Reference frame on the moving cart. y(t) motion governed by 1) a = -g y 2) v y = v 0y – g  t 3) y = y 0 + v 0y – g  t 2 /2 x motion: x = v x t Net motion : R = x(t) i + y(t) j (vector)

15. Physics 207: Lecture 8, Pg 15 Relative Velocity l Two observers moving relative to each other generally do not agree on the outcome of an experiment (path) l For example, observers A and B below see different paths for the ball

16. Physics 207: Lecture 8, Pg 16 Acceleration in Different Frames of Reference l Again: The derivative of the velocity equation will give the acceleration equation  v ’ = v – v 0  a ’ = a l The acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer moving at a constant velocity relative to the first frame.

17. Physics 207: Lecture 8, Pg 17 Example (with frames of reference) Vector addition An experimental aircraft can fly at full throttle in still air at 200 m/s. The pilot has the nose of the plane pointed west (at full throttle) but, unknown to the pilot, the plane is actually flying through a strong wind blowing from the northwest at 140 m/s. Just then the engine fails and the plane starts to fall at 5 m/s 2. What is the magnitude and directions of the resulting velocity (relative to the ground) the instant the engine fails? A B Calculate: A + B A x + B x = -200 + 140 x 0.71 and A y + B y = 0 – 140 x 0.71 x y BxBx ByBy

18. Physics 207: Lecture 8, Pg 18 Exercise, Relative Motion l You are swimming across a 50. m wide river in which the current moves at 1.0 m/s with respect to the shore. Your swimming speed is 2.0 m/s with respect to the water. You swim perfectly perpendicular to the current, how fast do you appear to be moving to an observer on shore? 2 m/s 1 m/s50m Home exercise: What is your apparent speed if you swim across?

19. Physics 207: Lecture 8, Pg 19 Home Exercise, Relative Motion l You are swimming across a 50 m wide river in which the current moves at 1 m/s with respect to the shore. Your swimming speed is 2 m/s with respect to the water. You swim across in such a way that your path is a straight perpendicular line across the river. l How many seconds does it take you to get across? 2m/s 1m/s50m a) b) c) d)

20. Physics 207: Lecture 8, Pg 20 Home Exercise, l The time taken to swim straight across is (distance across) / (v y ) Choose x axis along riverbank and y axis across river y x l Since you swim straight across, you must be tilted in the water so that your x component of velocity with respect to the water exactly cancels the velocity of the water in the x direction: 1m/s m/s y x 2m/s

21. Physics 207: Lecture 8, Pg 21 Case 2: Uniform Circular Motion Circular motion has been around a long time

22. Physics 207: Lecture 8, Pg 22 Circular Motion (UCM) is common so specialized terms Angular position  (CCW + CW -) 

23. Physics 207: Lecture 8, Pg 23 Circular Motion (UCM) is common so specialized terms Angular position  (CCW + CW -) l Radius is r  r

24. Physics 207: Lecture 8, Pg 24 Circular Motion (UCM) is common so specialized terms Angular position  (CCW + CW -) l Radius is r Arc traversed s = r  (if starting at 0 deg.)  r s

25. Physics 207: Lecture 8, Pg 25 Circular Motion (UCM) is common so specialized terms Angular position  (CCW + CW -) l Radius is r Arc traversed s = r  (if starting at 0 deg.) l Tangential “velocity” v T = ds/dt  r vTvT s

26. Physics 207: Lecture 8, Pg 26 Circular Motion (UCM) is common so specialized terms Angular position  (CCW + CW -) l Radius is r Arc traversed s = r  l Tangential “velocity” v T = ds/dt Angular velocity,  ≡ d  /dt (CCW + CW -) v T = ds/dt = r d  /dt = r  r  vTvT s

27. Physics 207: Lecture 8, Pg 27 Uniform Circular Motion (UCM) has only radial acceleration UCM changes only in the direction of v 1. Particle doesn’t speed up or slow down! 2. Velocity is always tangential, acceleration perpendicular ! 

28. Physics 207: Lecture 8, Pg 28 Again Centripetal Acceleration a r = v T 2 /r =  2  r Circular motion involves continuous radial acceleration vTvT r arar Uniform circular motion involves only changes in the direction of the velocity vector Acceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction.

29. Physics 207: Lecture 8, Pg 29 Home Example Question l A horizontally mounted disk 2 meters in diameter spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds. l 1 What is the period of the initial rotation? l 2 What is initial angular velocity? l 3 What is the tangential speed of a point on the rim during this initial period? l 4 Sketch the angular displacement versus time plot. l 5 What is the average angular velocity?

30. Physics 207: Lecture 8, Pg 30 Angular motion, signs l If angular displacement velocity accelerations are counter clockwise then sign is positive. l If clockwise then negative

31. Physics 207: Lecture 8, Pg 31 Home Exercise A Ladybug sits at the outer edge of a merry-go-round, and a June bug sits halfway between the outer one and the axis of rotation. The merry-go-round makes a complete revolution once each second. What is the June bug’s angular velocity? J L A.half the Ladybug’s. B.the same as the Ladybug’s. C.twice the Ladybug’s. D.impossible to determine. B. Same

32. Physics 207: Lecture 8, Pg 32 Recap l Assignment: HW2, (Chapters 2,3 & some 4) due Wednesday l Read through Chapter 6, Sections 1-4