Classical Mechanics Lecture 7

Transcription

1 Classical Mechanics Lecture 7 Today s Concepts: Work & Kine6c Energy Mechanics Lecture 7, Slide 1

2 Karate Will not do Session 3 of Unit 8. It is a Karate thing. We will only mark Session 2 of unit 8. You can do it as a project in Unit 14 instead of the assigned project on the pendulum.

3 Which Job? All three jobs pay the same amount of money. Which one would you choose for your crew? A. Job 1 B. Job 2 B C. Job 3 A B C

4 Joke of the day B Don't drink and DERIVE!

5 Stuff you asked about: WHAT IS GOING ON WITH ALL THE INTEGRALS?????? HELP :( I hate springs, but we should talk about them. sta6c fric6on Everything is fine except the work done by Gravity far from earth The slides had a lot of informa6on that was confusing and it flew by. I didn't have 6me to understand what was going on. The dot product is ocnfusing Predy much everything. This prelecture sort of went over my head. I'll definitely have to watch it again. The concept of posi6ve and nega6ve work is very confusing. please explain Mechanics Lecture 7, Slide 2

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7 Your comments and questions Will we have to use integrals to calculate work? clarification on under what circumstances the dot product should be used would be helpful. Are you going to test us on the dot product or is it just a concept you want us to know?

8 The Dot Product Mechanics Lecture 7, Slide 3

9 Dot Product using Unit Vectors î î =1 ˆ ˆ =1 ˆk ˆk =1 î ˆ =0 ˆ ˆk =0 ˆk ˆ =0 etc. ~A =A x î + A y ˆ + A zˆk ~B =B x î + B y ˆ + B zˆk

10 Dot Product in Rectangular Coordinates ~A ~B =(A x î + A y ˆ + A zˆk) (B x î + B y ˆ + B zˆk) =(A x B x î î + A x B y î ˆ + + A z B yˆk ĵ + A z B zˆk ˆk) =(A x B x 1 + A x B y A z B y 0 + A z B z 1) ~A ~B =(A x B x + A y B y + A z B z )

11 Finding the angle between vectors If we know the xyz components of two vectors we can find the angle between them: ~A ~B =(A x B x + A y B y + A z B z ) = ~A ~B cos therefore = arccos 0 B@ ~A ~B 1 CA ~A ~B

12 Example ~A = 3î + 4ˆ ~B = 4î + 3ˆ A = B = 5 A B = = A B = arccos B@ CA A B = arccos 24! = arccos =0.28 rad = 16

13 Work-Kinetic Energy Theorem The work done by the net force F as it acts on an object that moves between posi6ons r 1 and r 2 is equal to the change in the object s kine6c energy: W = Z ~r 2 ~r 1 ~F d~l Mechanics Lecture 7, Slide 4

14 Work-Kinetic Energy Theorem: 1-D Example If the force is constant and the direc6ons aren t changing then this is very simple to evaluate: car F d W = Z ~r 2 In this case ~F d~l = ~F ~d ~r 1 = Fd since cos(0)=1 0 Z ~r 2 1 d~l = ~d B@ CA ~r 1 This is probably what you remember from High School. (The nota6on may be confusing though.) Mechanics Lecture 7, Slide 5

15 Clicker Question A lighter car and a heavier van, each ini6ally at rest, are pushed with the same constant force F. Aier both vehicles travel a distance d, which of the following statements is true? (Ignore fric6on) car F d van F d A) They will have the same velocity B) They will have the same kine6c energy C) They will have the same momentum Mechanics Lecture 7, Slide 6

16 Z ~r 2 ~r 1 ~F d~l = K Deriva6on not so important Concept very important A force pushing over some distance will change the kine6c energy. θ Mechanics Lecture 7, Slide 7

17 Work done by gravity near the Earth s surface mg path Mechanics Lecture 7, Slide 8

18 Work done by gravity near the Earth s surface = m~g d~l 1 + m~g d~l m~g d~l N 2 dl N dl 1 mg dl 2 dl 1 dy 1 dx 1 mg Mechanics Lecture 7, Slide 9

19 Work done by gravity near the Earth s surface = m~g d~l 1 + m~g d~l m~g d~l N dl N Δy mg dl 1 dl 2 Mechanics Lecture 7, Slide 10

20 Work-Kinetic Energy Theorem If there are several forces ac6ng then W is the work done by the net (total) force: You can just add up the work done by each force Mechanics Lecture 7, Slide 12

21 CheckPoint Three objects having the same mass begin at the same height, and all move down the same ver6cal distance H. One falls straight down, one slides down a fric6onless inclined plane, and one swings on the end of a string. H Free Fall Fric6onless incline String In which case does the object have the biggest net work done on it by all forces during its mo6on? A) Free Fall B) Incline C) String D) All the same Mechanics Lecture 7, Slide 13

22 Clicker Question Three objects having the same mass begin at the same height, and all move down the same ver6cal distance H. One falls straight down, one slides down a fric6onless inclined plane, and one swings on the end of a string. What is the rela6onship between their speeds when they reach the bodom? H v f Free Fall Fric6onless incline Pendulum v i v p A) v f > v i > v p B) v f > v p > v i C) v f = v p = v i Mechanics Lecture 7, Slide 14

23 CheckPoint / Clicker Question H Free Fall Fric6onless incline String A) v f > v i > v p B) v f > v p > v i C) v f = v p = v i Only gravity will do work: W g = Δ K W g = mgh Δ K = 1 / 2 mv 2 2 Mechanics Lecture 7, Slide 15

24 CheckPoint A car drives up a hill with constant speed. Which statement best describes the total work W TOT done on the car by all forces as it moves up the hill? A) W TOT = 0 B) W TOT > 0 C) W TOT < 0 v About 50% got this right Mechanics Lecture 7, Slide 16

25 Clicker Question A car drives up a hill with constant speed. How does the kine6c energy of the car change as it moves up the hill? v A) It increases B) It stays the same C) It decreases Mechanics Lecture 7, Slide 17

26 CheckPoint A car drives up a hill with constant speed. Which statement best describes the total work W TOT done on the car by all forces as it moves up the hill? v A) W TOT = 0 B) W TOT > 0 C) W TOT < 0 A) The change in kine6c energy is zero because the velocity is constant. By the work-kine6c energy theorem, we know that work must also be zero. B) The force of fric6on between the car's wheels and the ramp exert a force up the ramp over a par6cular distance. W=Fd, so the work is posi6ve. C) gravity is downward and car moves up the hill. Mechanics Lecture 7, Slide 18

27 CheckPoint A box sits on the horizontal bed of a moving truck. Sta6c fric6on between the box and the truck keeps the box from sliding around as the truck drives. a µ S The work done on the box by the sta6c fric6onal force as the truck moves a distance D is: A) Posi6ve B) Nega6ve C) Zero About 60% got this right Mechanics Lecture 7, Slide 21

28 From Before A box sits on the horizontal bed of a moving truck. Sta6c fric6on between the box and the truck keeps the box from sliding around as the truck drives. a µ S If the truck moves with constant accelera6ng to the lei as shown, which of the following diagrams best describes the sta6c fric6onal force ac6ng on the box: A B C Physics 211 Lecture 7, Slide 22

29 CheckPoint a F µ S D The work done on the box by the sta6c fric6onal force as the truck moves a distance D is: A) Posi6ve B) Nega6ve C) Zero A) because the direc6on of the sta6c fric6onal force and the direc6on the box travels are all same. B) Since the fric6on force always has opposite direc6on from the movement direc6on, the work done is nega6ve C) Fric6on keeps the box from sliding, thus D=0, and work done is 0. Mechanics Lecture 7, Slide 23

30 Work done by a Spring Mechanics Lecture 7, Slide 24

31 I am confused about the posi6ve work and nega6ve work and also the posi6ve and nega6ve forces for the spring problems. Use the formula to get the magnitude of the work Use a picture to get the sign (look at direc6ons) In this example the spring does ve work since F and Δx are in opposite direc6on. The axes don t mader. Mechanics Lecture 7, Slide 25

32 Clicker Question A box adached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new loca6on. D During this mo6on, the spring does: A) Posi6ve Work B) Nega6ve Work C) Zero work Mechanics Lecture 7, Slide 26

33 Clicker Question A box adached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new loca6on. D During this mo6on, your hand does: A) Posi6ve Work B) Nega6ve Work C) Zero work Mechanics Lecture 7, Slide 27

34 Clicker Question A box adached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new loca6on. During this mo6on, the total work done on the box is: A) Posi6ve B) Nega6ve C) Zero D Mechanics Lecture 7, Slide 28

35 Gravity W = Z ~r 2 ~r 1 ~F(~r) d~r = Z r 2 r 1 GM e m r 2 dr = GM e m 1! 1 r 2 r 1 Mechanics Lecture 7, Slide 29

36 Clicker Question In Case 1 we send an object from the surface of the earth to a height above the earth surface equal to one earth radius. In Case 2 we start the same object a height of one earth radius above the surface of the earth and we send it infinitely far away. In which case is the magnitude of the work done by the Earth s gravity on the object biggest? A) Case 1 B) Case 2 C) They! are the same W = GM e m 1 r 2 1 r 1 Case 1 Case 2 Mechanics Lecture 7, Slide 30

37 Case 1: Case 2: Clicker Question Solution W = GM e m 1 2R E 1 R E W = GM e m R E!! = GM em 2R E = GM em 2R E Same! R E Case 1 Case 2 2R E Mechanics Lecture 7, Slide 31