Lecture 5

1. Lecture 5 • 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. Trajectory with constant acceleration along the vertical Special Case: ax=0 & ay= -g vx(t=0) = v0 vy(t=0) vy(Dt) = – g Dt x(Dt) = x0 + v0Dt y(Dt) = y0- ½ g Dt2 Position xvsy t = 0 4 y What do the velocity and acceleration vectors look like? x

3. Trajectory with constant acceleration along the vertical What do the velocity and acceleration vectors look like? Velocity vector is always tangent to the curve! Velocity t = 0 xvsy 4 y x

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! Acceleration may or may not be! Position Velocity Acceleration xvsy t = 0 4 y x

5. 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

6. 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/s2 & no air resistance)?

7. 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/s2 & no air resistance)? • at t=0 the knife is at (0.0 m, 1.0 m) with vy=0 • after Dt the kinfe is at (x m, 0.0 m) • x = x0 + vxDt and y = y0 – ½ g Dt2 So x = 10 m/s Dt and 0.0 m = 1.25 m – ½ g Dt2  2.5/10 s2 = Dt2 Dt = 0.50 s x = 10 m/s 0.50 s = 5 .0 m

8. Exercise 3Relative Trajectories: Monkey and Hunter 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 … • go over the monkey. • hit the monkey. • go under the monkey.

9. Schematic of the problem • xB(Dt) = d = v0cosqDt • yB(Dt) = hf = v0sinqDt – ½ g Dt2 • xM(Dt) = d • yM(Dt) = h – ½ g Dt2 • Does yM(Dt) = yB(Dt) = hf? (x,y) = (d,h) Monkey Does anyone want to change their answer ? What happens if g=0 ? How does introducing g change things? hf g v0 q Bullet (x0,y0) = (0 ,0) (vx,vy) = (v0cosq, v0sinq)

10. Relative motion and frames of reference • Reference frame S is stationary • Reference frame S’ is moving at vo This also means that S moves at – vo relative to S’ • Define time t = 0 as that time when the origins coincide

11. t later t=0 -v0 -v0 t A A r r O’ O,O’ Relative motion and frames of reference • Reference frame S is stationary, A is stationary • Reference frame S’ is moving at vo A moves at – vo relative to an observer in S’ • Define time t = 0 as that time (O & O’ origins coincide) r’= r-v0t

12. Relative Velocity • The positions, r and r’, as seen from the two reference frames are related through the velocity, vo, where vo is velocity of the r’ reference frame relative to r • r’ = r – voDt • The derivative of the position equation will give the velocity equation • v’ = v – vo • These are called the Galilean transformation equations • 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 (dvo/ dt = 0)

13. y(t) motion governed by 1) a = -gy 2) vy = v0y – gDt 3) y = y0 + v0y – g Dt2/2 Reference frame on the moving cart. x motion: x = vxt Reference frame on the ground. Net motion: R = x(t) i + y(t) j(vector) Central concept for problem solving: “x” and “y” components of motion treated independently. • Example: Man on cart tosses a ball straight up in the air. • You can view the trajectory from two reference frames:

14. 2 m/s 50m 1 m/s Exercise,Relative Motion • You are swimming across a 50. mwide river in which the current moves at 1.0 m/s with respect to the shore. Your swimming speed is 2.0m/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? Home exercise: What is your apparent speed if you swim across?

15. Case 2: Uniform Circular MotionCircular motion has been around a long time

16. Circular Motion (UCM) is common so specialized terms • Angular position q (CCW + CW -) q

17. Circular Motion (UCM) is common so specialized terms • Angular position q (CCW + CW -) • Radius is r r q

18. Circular Motion (UCM) is common so specialized terms • Angular position q (CCW + CW -) • Radius is r • Arc traversed s = rq (if starting at 0 deg.) s r q

19. Circular Motion (UCM) is common so specialized terms • Angular position q (CCW + CW -) • Radius is r • Arc traversed s = rq (if starting at 0 deg.) • Tangential “velocity” vT = ds/dt s vT r q

20. Circular Motion (UCM) is common so specialized terms • Angular position q (CCW + CW -) • Radius is r • Arc traversed s = rq • Tangential “velocity” vT = ds/dt • Angular velocity, w ≡dq/dt (CCW + CW -) • vT = ds/dt = r dq/dt = r w s vT r q

21. Dq v Uniform Circular Motion (UCM) has only radial acceleration UCM changes only in the direction of 1. Particle doesn’t speed up or slow down! 2. Velocity is always tangential, acceleration perpendicular !

22. vT ar r Again 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. Centripetal Acceleration ar = vT2/r = w2r Circular motion involves continuous radial acceleration

23. Angular motion, signs • If angular displacement velocity accelerations are counter clockwise then sign is positive. • If clockwise then negative

24. Recap • Assignment: HW2, (Chapters 2,3 & some 4) due Wednesday • Read through Chapter 6, Sections 1-4