1. Relative and Circular Motion Mechanics Lecture 3, Slide 1 a ) Relative motion b) Centripetal acceleration

2. Mechanics Lecture 3, Slide 2

3. What is the speed of Mike relative to the station? A. -1 m/s B. 30 m/s C. 29 m/s D. 31 m/s Mechanics Lecture 3, Slide 3

4. Relative Motion in 1 dimension Mechanics Lecture 3, Slide 4

5. Mechanics Lecture 3, Slide 5 Relative Position and Reference Frames Position of Mike in the ground frame is the vector sum of the position vector of Mike in the train reference frame and the position vector of the train in the ground reference frame.

6. Mechanics Lecture 3, Slide 6 Relative Motion and Reference Frames Differentiate the position vectors to obtain the velocity vectors

7. Mechanics Lecture 3, Slide 7 Relative Motion and Reference Frames

8. Mechanics Lecture 3, Slide 8 Relative Motion and Reference Frames

9. Prelecture 3, Questions 1 again A. B. C. D. Mechanics Lecture 3, Slide 9

10. Mechanics Lecture 3, Slide 10 v belt,ground = 2 m/s v dog,belt = 8 m/s A) 6 m/s B) 8 m/s C) 10 m/s CheckPoint A girl stands on a moving sidewalk that moves to the right at 2 m/s relative to the ground. A dog runs toward the girl in the opposite direction along the sidewalk at a speed of 8 m/s relative to the sidewalk. What is the speed of the dog relative to the ground?

11. Mechanics Lecture 3, Slide 11 v dog, ground = v dog, belt + v belt, ground = (  8 m/s) + (2 m/s) =  6 m/s +x+x v belt,ground = 2 m/s v dog,belt = 8 m/s What is the speed of the dog relative to the ground? A) 6 m/s B) 8 m/s C) 10 m/s

12. Mechanics Lecture 3, Slide 12 About 55% of you got this right – lets try it again. CheckPoint A girl stands on a moving sidewalk that moves to the right at 2 m/s relative to the ground. A dog runs toward the girl in the opposite direction along the sidewalk at a speed of 8 m/s relative to the sidewalk. What is the speed of the dog relative to the girl? v belt,ground = 2 m/s v dog,belt = 8 m/s A) 6 m/s B) 8 m/s C) 10 m/s

13. What is the speed of the dog relative to the girl? A. B. C. Mechanics Lecture 3, Slide 13 v belt,ground = 2 m/s v dog,belt = 8 m/s A) 6 m/s B) 8 m/s C) 10 m/s C) The dog and girl are running towards each other so when you add the two velocities together it would be 8+2. A) Because the girl is actually moving and the two vectors are opposite, so together they make 6 m/s B) Because the girl is not moving relative to the belt, and the dog is going 8 m/s relative to the belt, the dog is also moving 8 m/s relative to the girl..

14. Mechanics Lecture 3, Slide 14 B) Because the girl is not moving relative to the belt, and the dog is going 8 m/s relative to the belt, the dog is also moving 8 m/s relative to the girl. Using the velocity formula: v dog, girl = v dog, belt + v belt, girl =  8 m/s + 0 m/s =  8 m/s What is the speed of the dog relative to the girl? v belt,ground = 2 m/s v dog,belt = 8 m/s A) 6 m/s B) 8 m/s C) 10 m/s

15. Relative Motion in 2 Dimensions Mechanics Lecture 3, Slide 15 Speed relative to shore

16. Relative Motion in 2 Dimensions Mechanics Lecture 3, Slide 16 Direction w.r.t shoreline

17. Relative Motion in 2 Dimensions Mechanics Lecture 3, Slide 17 Caveat !!!! The simple addition of velocities as shown only works for speeds much less than the speed of light…need special relativity at v~c.

18. Moving Sidewalk Question A. B. C. D. Mechanics Lecture 3, Slide 18 A man starts to walk along the dotted line painted on a moving sidewalk toward a fire hydrant that is directly across from him. The width of the walkway is 4 m, and it is moving at 2 m/s relative to the fire-hydrant. If his walking speed is 1 m/s, how far away will he be from the hydrant when he reaches the other side? A)2 m B)4 m C)6 m D)8 m Moving sidewalk 2 m/s 1 m/s 4 m

19. Mechanics Lecture 3, Slide 19 Time to get across:  t = distance / speed = 4m / 1m/s = 4 s 1 m/s 4 m If the sidewalk wasn’t moving:

20. Mechanics Lecture 3, Slide 20 4 m 2 m/s Just the sidewalk:

21. Mechanics Lecture 3, Slide 21 1 m/s 4 m 2 m/s Combination of motions:

22. Mechanics Lecture 3, Slide 22 D = (speed of sidewalk) ∙ (time to get across) = (2 m/s) ∙ (4 s) = 8 m 1 m/s 4 m 2 m/s D

23. Swim Race A. B. C. Mechanics Lecture 3, Slide 23 Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Beth swims perpendicular to the flow, Ann swims upstream at 30 degrees, and Carly swims downstream at 30 degrees. Who gets across the river first? A) Ann B) Beth C) Carly x y Ann Beth Carly

24. Mechanics Lecture 3, Slide 24 A B C V y,Beth = V o 30 o V y,Ann = V o cos(30 o ) V y,Carly = V o cos(30 o ) Time to get across = D / V y D Look at just water & swimmers x y

25. Mechanics Lecture 3, Slide 25 x y Clicker Question Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Beth swims perpendicular to the flow, Ann swims upstream at 30 degrees, and Carly swims downstream at 30 degrees. Who gets across the river second? A) Ann B) Carly C) Both same AnnCarly

26. Mechanics Lecture 3, Slide 26 Accelerating (Non-Inertial) Frames of Reference Accelerating Frame of Reference Confusing due to the fact that the acceleration can result in what appears to be a “push or pull”.

27. Accelerated Frames of Reference Mechanics Lecture 3, Slide 27 Accelerating Frame of Reference Accelerometer can detect change in velocity

28. Inertial Frames of Reference Mechanics Lecture 3, Slide 28 Inertial Frames of Reference Non-accelerating frames of reference in a state of constant, rectilinear motion with respect to one another. An accelerometer moving with any of them would detect zero acceleration.

29. Mechanics Lecture 3, Slide 29 A girl twirls a rock on the end of a string around in a horizontal circle above her head as shown from above in the diagram. If the string breaks at the instant shown, which of the arrows best represents the resulting path of the rock? A B C D Top view looking down After the string breaks, the rock will have no force acting on it, so it cannot accelerate. Therefore, it will maintain its velocity at the time of the break in the string, which is directed tangent to the circle. CheckPoint

30. Mechanics Lecture 3, Slide 30 Show Prelecture https://www.smartphysics.com/Course/PlaySlideshow?unitItemID=192821

31. Mechanics Lecture 3, Slide 31 Rotating Reference Frames Direction Changing Speed is constant. Acceleration

32. Mechanics Lecture 3, Slide 32 Rotating Reference Frames

33. Mechanics Lecture 3, Slide 33 Rotating Reference Frames

34. Mechanics Lecture 3, Slide 34 Rotating Reference Frames

35. Mechanics Lecture 3, Slide 35 Rotating Reference Frames

36. Mechanics Lecture 3, Slide 36 Centripetal Acceleration Constant speed in circular path Acceleration directed toward center of circle What is the magnitude of acceleration? Proportional to: 1.Speed 1.time rate of change of angle or angular velocity

37. Mechanics Lecture 3, Slide 37 Centripetal Acceleration

38. Mechanics Lecture 3, Slide 38 Centripetal Acceleration

39. Mechanics Lecture 3, Slide 39 Centripetal Acceleration

40. Mechanics Lecture 3, Slide 40 v =  R Once around:  /  t = 2  /  v  x /  t = 2  R / T  is the rate at which the angle  changes: 

41. Mechanics Lecture 3, Slide 41 dd v dt R =R d  Another way to see it: v =  R v = R 

42. Mechanics Lecture 3, Slide 42 Centripetal Acceleration-Example

43. Mechanics Lecture 3, Slide 43 Centripetal Acceleration due to Earth’s rotation

44. Mechanics Lecture 3, Slide 44