1. Physics 207: Lecture 5, Pg 1 Physics 207, Lecture 5, Sept. 17 l Goals:  Solve problems with multiple accelerations in 2- dimensions (including linear, projectile and circular motion)  Discern different reference frames and understand how they relate to particle motion in stationary and moving frames  Recognize different types of forces and know how they act on an object in a particle representation  Identify forces and draw a Free Body Diagram Assignment: HW3, (Chapters 4 & 5, due 9/25, Wednesday) Read through Chapter 6, Sections 1-4

2. Physics 207: Lecture 5, Pg 2 Exercise (2D motion with acceleration) Relative Trajectories: Monkey and Hunter A. go over the monkey? B. hit the monkey? C. go under the monkey? 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 …

3. Physics 207: Lecture 5, Pg 3 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

4. Physics 207: Lecture 5, Pg 4 Uniform Circular Motion (UCM) is common so we have specialized terms Arc traversed s =  r l Tangential velocity v t l Period, T, and frequency, f Angular position,  Angular velocity,  Period (T): The time required to do one full revolution, 360 ° or 2  radians Frequency (f): 1/T, number of cycles per unit time Angular velocity or speed  = 2  f = 2  /T, number of radians traced out per unit time (in UCM average and instantaneous will be the same) r  vtvt s

5. Physics 207: Lecture 5, Pg 5 Example Question (note the commonality with linear motion) 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? l 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must the angular acceleration be? l 7 What is the magnitude and direction of the acceleration after 10 seconds?

6. Physics 207: Lecture 5, Pg 6 Example Question l A horizontal turntable 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 turntable during the initial rotation T (time for one revolution) =  t /# of revolutions/ time = 5 sec / 10 rev = 0.5 s l 2 What is initial angular velocity?  = angular displacement / time = 2  f = 2  / T = 12.6 rad / s l 3 What is the tangential speed of a point on the rim during this initial period? We need more…..

7. Physics 207: Lecture 5, Pg 7 Relating rotation motion to linear velocity l In UCM a particle moves at constant tangential speed v t around a circle of radius r (only direction changes). l Distance = tangential velocity · time Once around… 2  r = v t T or, rearranging (2  /T) r = v t  r = v t Definition: If UCM then  constant 3So v T =  r = 4  rad/s 1 m = 12.6 m/s r  vtvt s 4 A graph of angular displacement (  ) vs. time

8. Physics 207: Lecture 5, Pg 8 Angular displacement and velocity Arc traversed s =  r in time  t then  s =  r so  s /  t = (  /  t) r in the limit  t  0 one gets ds / dt = d  / dt r v t =  r  = d  / dt if  is constant, integrating  = d  / dt, we obtain:  =   +   t Counter-clockwise is positive, clockwise is negative r  vtvt s

9. Physics 207: Lecture 5, Pg 9 Sketch of  vs. time time (seconds)        (radians)  =   +   t  =  +  5  rad  =   +   t  =   rad +  rad 5 Avg. angular velocity =  /  t = 24  /10 rad/s

10. Physics 207: Lecture 5, Pg 10 Next part….. l 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must the “tangential acceleration” be?

11. Physics 207: Lecture 5, Pg 11 Then angular velocity is no longer constant so d  /dt ≠ 0 Define tangential acceleration as a t = dv t /dt = r d  /dt So s = s 0 + (ds/dt) 0  t + ½ a t  t 2 and s =  r We can relate a t to d  /dt  =  o +  o  t +  t 2  =  o +  t l Many analogies to linear motion but it isn’t one-to-one l Note: Even if the angular velocity is constant, there is always a radial acceleration. Well, if  is linearly increasing … atat r 1 2 atat r

12. Physics 207: Lecture 5, Pg 12 Tangential acceleration? Tangential acceleration?  =  o +  o  t +  t 2 (from plot, after 10 seconds) 24  rad = 0 rad + 0 rad/s  t + ½ (a t /r)  t 2 48  rad 1m / 100 s 2 = a t r  vtvt s 1 2 atat r l 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must the “tangential acceleration” be? l 7 What is the magnitude and direction of the acceleration after 10 seconds?

13. Physics 207: Lecture 5, Pg 13 Circular motion also has a radial (perpendicular) component Centripetal Acceleration a r = v t 2 /r Circular motion involves continuous radial acceleration vtvt r arar Uniform circular motion involves only changes in the direction of the velocity vector, thus acceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction. Quantitatively (see text)

14. Physics 207: Lecture 5, Pg 14 Non-uniform Circular Motion For an object moving along a curved trajectory, with non-uniform speed a = a r + a t (radial and tangential) arar atat a r = v2v2 r dt d| | v a t =

15. Physics 207: Lecture 5, Pg 15 Tangential acceleration? Tangential acceleration? a t = 0.48  m / s 2 and  r =  o r + r  t = 4.8  m/s = v t a r = v t 2 / r= 23  2 m/s 2 r  vtvt s atat r l 7 What is the magnitude and direction of the acceleration after 10 seconds? Tangential acceleration is too small to plot!

16. Physics 207: Lecture 5, Pg 16 Angular motion, signs l Also note: if the angular displacement, velocity and/or accelarations are counter clockwise then this is said to be positive. l Clockwise is negative

17. Physics 207: Lecture 5, Pg 17 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.

18. Physics 207: Lecture 5, Pg 18 Circular Motion l UCM enables high accelerations (g’s) in a small space l Comment: In automobile accidents involving rotation severe injury or death can occur even at modest speeds. [In physics speed doesn’t kill….acceleration does (i.e., the sudden change in velocity).]

19. Physics 207: Lecture 5, Pg 19 Mass-based separation with a centrifuge How many g’s? BeforeAfter a r =v t 2 / r and f = 10 4 rpm is typical with r = 0.1 m and v t =  r = 2  f r bb5 ca. 10000 g’s

20. Physics 207: Lecture 5, Pg 20 g’s with respect to humans l 1 gStanding l 1.2 gNormal elevator acceleration (up). l 1.5-2g Walking down stairs. l 2-3 gHopping down stairs. l 1.5 gCommercial airliner during takeoff run. l 2 gCommercial airliner at rotation l 3.5 gMaximum acceleration in amusement park rides (design guidelines). l 4 gIndy cars in the second turn at Disney World (side and down force). l 4+ gCarrier-based aircraft launch. l 10 gThreshold for blackout during violent maneuvers in high performance aircraft. l 11 gAlan Shepard in his historic sub orbital Mercury flight experience a maximum force of 11 g. l 20 gColonel Stapp’s experiments on acceleration in rocket sleds indicated that in the 10-20 g range there was the possibility of injury because of organs moving inside the body. Beyond 20 g they concluded that there was the potential for death due to internal injuries. Their experiments were limited to 20 g. l 30 gThe design maximum for sleds used to test dummies with commercial restraint and air bag systems is 30 g.

21. Physics 207: Lecture 5, Pg 21 A bad day at the lab…. l In 1998, a Cornell campus laboratory was seriously damaged when the rotor of an ultracentrifuge failed while in use. l Description of the Cornell Accident -- On December 16, 1998, milk samples were running in a Beckman. L2-65B ultracentrifuge using a large aluminum rotor. The rotor had been used for this procedure many times before. Approximately one hour into the operation, the rotor failed due to excessive mechanical stress caused by the g-forces of the high rotation speed. The subsequent explosion completely destroyed the centrifuge. The safety shielding in the unit did not contain all the metal fragments. The half inch thick sliding steel door on top of the unit buckled allowing fragments, including the steel rotor top, to escape. Fragments ruined a nearby refrigerator and an ultra-cold freezer in addition to making holes in the walls and ceiling. The unit itself was propelled sideways and damaged cabinets and shelving that contained over a hundred containers of chemicals. Sliding cabinet doors prevented the containers from falling to the floor and breaking. A shock wave from the accident shattered all four windows in the room. The shock wave also destroyed the control system for an incubator and shook an interior wall.

22. Physics 207: Lecture 5, Pg 22 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

23. Physics 207: Lecture 5, Pg 23 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)

24. Physics 207: Lecture 5, Pg 24 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)

25. Physics 207: Lecture 5, Pg 25 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

26. Physics 207: Lecture 5, Pg 26 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)

27. Physics 207: Lecture 5, Pg 27 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 v y 1 m/s m/s y x 2m/s river’s frame

28. Physics 207: Lecture 5, Pg 28 Home Exercise 2 l The time taken to swim straight across = (distance across) / (v y ) time = 50 m / ( 2 m/s cos  50/3 ½ seconds Where do you land if the river flows at 2 m/s while swimming at the same heading in the river (i.e.,  = asin ½) ? y x Dist in river = v x t = -2 m/s sin  t = -2 (50/3½) 1/2 m = -29 m (upstream) 2 m/s v y 1 m/s m/s y x 2 m/s river’s frame  l Dist river flows = v r t = 2 m/s t = -2 (50/3½) m = 58 m l Final position = -29 m + 58 m = 29 m down the shore.

29. Physics 207: Lecture 5, Pg 29 What causes motion? (Actually changes in motion) What are forces ? What kinds of forces are there ? How are forces and motion related ?

30. Physics 207: Lecture 5, Pg 30 Physics 207, Lecture 5, Sept. 17 Assignment: HW3, (Chapters 4 & 5, due 9/25, Wednesday) Read Chapter 5 through Chapter 6, Sections 1-4