Classical Mechanics Relative and Circular Motion

Transcription

1 Classical Mechanics Relative and Circular Motion Today s Concepts: a) Rela3ve mo3on b) Centripetal accelera3on Mechanics Lecture 4, Slide 1

2 Stuff you asked about:! Discuss centrifugal force.! Everything made sense. Also: What did the nuclear physicist have for lunch?! Fission chips.! I like physics.! Reference frames of moving objects, like the prelecture example of the red and blue ball queseon.! Can you please go over Non- acceleraeng reference frames. Describe/show the difference between non- acceleraeng and acceleraeve frames.! I do not 100% understand how you know that the acceleraeon vector in a case of centripetal acceleraeon is pointed towards the center of the circle of it's rotaeon.! How you got from Fc = ma c to a = (v 2 )/r. Or do we need to know that?! Further disussing angular velocity! A review of the formula's would be a great refresher on this concept. Just for fun, a "pracecal" example of circular moeon. hxps:// watch?v=zjjsktesuws Mechanics Lecture 4, Slide 2

3 How familiar are you with the concepts of relaeve moeon and centripetal acceleraeon from your high school course. A) I already know this stuff B) It seems familiar, but I need a review C) We didn't learn this in high school StudentLogForQuestion Mechanics Lecture 4, Slide 3

4 Train Demo Clicker Question A flatbed railroad car is moving along a track at constant velocity. A passenger at the center of the car throws a ball straight up. NeglecEng air resistance, where will the ball land? A) Forward of the center of the car B) At the center of the car C) Backward of the center of the car correct v train car Ball and car start with same x posieon and x velocity, Since a = 0 they always have same x posieon. Demo - train Mechanics Lecture 2, Slide 16

5 Relative Motion What you just did Mechanics Lecture 4, Slide 4

6 PreLecture 4, Question 1 Mechanics Lecture 4, Slide 5

7 PreLecture 4, Question 1 Recall that the velocity of a with respect to c = velocity of a with respect to b + velocity of b with respect to c so velocity of green with respect to red = velocity of green with respect to Lab + velocity of Lab with respect to red. = velocity of green with respect to Lab velocity of red with respect to Lab v gr = v gl + v Lr = v gl v rl + 6 m/s ( 2 m/s) = +8 m/s Mechanics Lecture 4, Slide 5

8 CheckPoint A girl stands on a moving sidewalk that moves to the right at 2 m/s relaeve to the ground. A dog runs toward the girl in the opposite direceon along the sidewalk at a speed of 8 m/s relaeve to the sidewalk. What is the speed of the dog relaeve to the ground? v dog,belt = 8 m/s v belt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s Mechanics Lecture 4, Slide 6

9 What is the speed of the dog relative to the ground? v dog,belt = 8 m/s v belt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s +x v dog, ground = v dog, belt + v belt, ground = ( 8 m/s) + (2 m/s) = 6 m/s Mechanics Lecture 4, Slide 7

10 CheckPoint A girl stands on a moving sidewalk that moves to the right at 2 m/s relaeve to the ground. A dog runs toward the girl in the opposite direceon along the sidewalk at a speed of 8 m/s relaeve to the sidewalk. What is the speed of the dog relaeve to the girl? v dog,belt = 8 m/s v belt,ground = 2 m/s Most of you got this. A) 6 m/s B) 8 m/s C) 10 m/s Mechanics Lecture 4, Slide 8

11 What is the speed of the dog relative to the girl? v dog,belt = 8 m/s v belt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s 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 relalve to the belt, and the dog is going 8 m/s relalve to the belt, the dog is also moving 8 m/s relalve to the girl.. C) The dog and girl are running towards each other so when you add the two velociles together it would be 8+2. Mechanics Lecture 4, Slide 9

12 What is the speed of the dog relative to the girl? v dog,belt = 8 m/s v belt,ground = 2 m/s A) 6 m/s B) 8 m/s C) 10 m/s B) Because the girl is not moving relalve to the belt, and the dog is going 8 m/s relalve to the belt, the dog is also moving 8 m/s relalve 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 Mechanics Lecture 4, Slide 10

13 Clicker Question A man starts to walk along the doxed 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 relaeve 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 1 m/s Moving sidewalk 2 m/s 4 m Mechanics Lecture 4, Slide 12

14 If the sidewalk wasn t moving: 1 m/s Time to get across: Δt = distance / speed = 4m / 1m/s = 4 s 4 m Mechanics Lecture 4, Slide 13

15 Just the sidewalk: 2 m/s 4 m Mechanics Lecture 4, Slide 14

16 Combination of motions: 1 m/s 2 m/s 4 m Mechanics Lecture 4, Slide 15

17 1 m/s D 2 m/s 4 m D = (speed of sidewalk) (time to get across) = (2 m/s) (4 s) = 8 m Mechanics Lecture 4, Slide 16

18 Clicker Question Three swimmers can swim equally fast relaeve to the water. They have a race to see who can swim across a river in the least Eme. RelaEve 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 Ann Beth Carly y x Mechanics Lecture 4, Slide 17

19 Look at just water & swimmers Time to get across = D / V y D A B 30 o 30 o C y x V y,beth = V o V y,ann = V o cos 30 o V y,carly = V o cos 30 o Mechanics Lecture 4, Slide 18

20 Clicker Question Three swimmers can swim equally fast relaeve to the water. They have a race to see who can swim across a river in the least Eme. RelaEve 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 Ann Carly y x Mechanics Lecture 4, Slide 19

21 Navigating straight across ~v river,ground = 0.1m/s, west

22 Navigating upriver ~v river,ground = 0.1m/s, west θ

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24 CheckPoint (& Demo) 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 brakes at the instant shown, which of the arrows best represents the resullng path of the rock? A B C D ATer the string breaks, the rock will have no force aclng on it, so it cannot accelerate. Therefore, it will maintain its velocity at the Lme of the break in the string, which is directed tangent to the circle. Top view looking down Mechanics Lecture 4, Slide 20

25 Show Prelecture Mechanics Lecture 4, Slide 21

26 Centripetal Acceleration Mechanics Lecture 4, Slide 22

27 v = ωr ω is the rate at which the angle θ changes: θ Once around: v = Δx / Δt = 2πR / T ω = Δθ / Δt = 2π / Τ Mechanics Lecture 4, Slide 23

28 v = ωr Another way to see it: dθ R v dt=r dθ v = Rω Mechanics Lecture 4, Slide 24

29 We can ignore this acceleration due to Earth's rotation since its small 1700 km/h Mechanics Lecture 4, Slide 25

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