Presentation laddar. Vänta.

Presentation laddar. Vänta.

© 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their.

Liknande presentationer


En presentation över ämnet: "© 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their."— Presentationens avskrift:

1 © 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Lecture Outlines Chapter 10 Physics, 3 rd Edition James S. Walker

2 Chapter 10 Rotational Kinematics and Energy

3 Units of Chapter 10 Angular Position, Velocity, and Acceleration Rotational Kinematics Connections Between Linear and Rotational Quantities Rolling Motion Rotational Kinetic Energy and the Moment of Inertia Conservation of Energy

4 Photo 10-1 Galaxy

5 10-1 Angular Position, Velocity, and Acceleration

6 Degrees and revolutions:

7 10-1 Angular Position, Velocity, and Acceleration Arc length s, measured in radians:

8 10-1 Angular Position, Velocity, and Acceleration

9

10 Exercise 10-1(p.284) En gammal LP-skiva gjorde 33 1/3 rpm (varv [revolutions] per minut) a) Vinkelhastigheten = ? ω = - (!) 100/3 2 π rad/60 s = - 3,49 rad/s b) Om en CD roterar 22,0 rad/s vad är det i rpm? 1 rpm = 2π rad/60 s 1 rad/s = 60/2π rpm 22 rad/s = 22 60/2π rpm = 210 rpm

11 10-1 Angular Position, Velocity, and Acceleration

12 Exercise 10-2 (p.284) En någon yngre EP-skiva gjorde 45-varv per minut. T=? ω = 45 2 π rad/60 s T = 2 π/ω = 60 s/45(rad) = 1,3 s

13 10-1 Angular Position, Velocity, and Acceleration

14 Exercise 10-3 (p.286) En väderkvarn saktar ned med en konstant vinkelacceleration = - 0,45 rad/s 2 Hur lång tid tar det för kvarnen att stanna? Om ursprungsvinkelhastigheten 2,1 rad/s blir Δt = Δω/α av = (ω f – ω i )/α av = (0-2,1)/-0,45 = 4,7 s

15 10-2 Rotational Kinematics If the angular acceleration is constant:

16 Exercise 10-4 ω 0 = - 8,4 rad/s och α = - 2,8 rad/s 2. Vad är ω, 1,5 s senare? ω = ω 0 + αt ger ω = - 8,4 -2,81,5 = - 12,6 rad/s

17 10-2 Rotational Kinematics Analogies between linear and rotational kinematics:

18 Example 10-1 Thrown for a Curve ω 0 = 36,0 rad/s och 0,595 s senare så är ω = 34,2 rad/s på grund av luftmotståndet. a) Hur stor är (medel)accelerationen om den antas vara konstant? b) Hur många varv hinner bollen göra i flykten? α = (ω - ω 0 )/Δt = (- 1,8 rad/s)/0,595 s = - 3,03 rad/s 2 θ = θ 0 + ω 0 t + αt 2 /2 ger θ - θ 0 = 21,4 rad – 0,536 rad = 20,9 rad = 3,33 varv

19 Example 10-2 Wheel of Misfortune

20 Hjulet roterar ett och ett kvarts varv innan det stannar. a)Vad blir α? Om ursprungsvinkelhastigheten är 3,40 rad/s ger ω f 2 = ω α(θ – θ 0 ) att den konstanta vinkelaccelerationen α = - 0,736 rad/s 2 a)b) Hur lång tid tar förloppet? t = (ω f – ω 0 )/α = 4,62 s

21 Active Example 10-1 Find the Time To Rest

22 Active Example 10-1 Find the Time to Rest Om ursprungsvinkelhastigheten är 5,40 rad/s och den konstanta vinkelaccelerationen = - 2,10 rad/s 2 a) Vad blir t? ω = ω 0 + αt t = (ω f – ω 0 )/α = 2,57 s b) Hur stor vinkel har då blocket roterat? θ = θ 0 + ω 0 t + αt 2 /2 θ - θ 0 = 13,88 rad – 6,94 rad = 6,94 rad

23 10-3 Connections Between Linear and Rotational Quantities

24 Exercise 10-5 Vilken vinkelhastighet måste en CD ha för att ge en linjär hastighet = 1,25 m/s när lasern belyser skivan a) 2,50 cm b) 6,00 cm från centrum? ω = v/r a) ω(r) = 50,0 rad/s b) ω(r) = 20,8 rad/s

25 Conceptual Checkpoint 10–1 Compare the Speeds Är vinkelhastigheten för barn1 > = < barn2?

26 10-3 Connections Between Linear and Rotational Quantities

27 Vilken centripetalacceleration känner ett barn i en karusell om r = 4,25 m och vinkelhastigheten är = 0,838 rad/s? a cp = rω 2 = 4,25 m (0,838 rad/s) 2 = 2,96 m/s 2 (0,3 g)

28 Example 10-3 The Microhematocrit

29 Exemple 10-3 The Microhematocrit Vinkelhastigheten ω = rpm och r = 9,07 cm. a)Hur stor är v? v = rω = 0,0907 m π rad/60 s = 109 m/s b) Hur stor är centripetalacceleration? a cp = rω 2 = 0,0907 m (1200 rad/s) 2 = 1, m/s 2

30 10-3 Connections Between Linear and Rotational Quantities This merry-go-round has both tangential and centripetal acceleration.

31 10-3 Connections Between Linear and Rotational Quantities

32 Active Example 10-2 (p.293) Find the Acceleration

33 Active example 10-2 a t = r Δω/Δt = r α = 0,0907 m 95,0 rad/s 2 = 8,62 m/s 2 a cp = r ω 2 = 0,0907 m (8,00 rad/s) 2 = 5,80 m/s 2 Storleken av totala accelerationen blir då a = (a t 2 + a cp 2 ) 1/2 = 10,4 m/s 2 och vinkeln blir Φ = arctan a cp /a t = 33,9º

34 10-4 Rolling Motion If a round object rolls without slipping, there is a fixed relationship between the translational and rotational speeds:

35 10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion:

36 Exercise 10-6 En bil med hjulradien 32 cm kör med hastigheten 24,6 m/s (55 mi/h). a) Hur stor är vinkelhastigheten? ω = v/r = 24,6 m/s/ 0,32 m = 77 rad/s b) Vad är hastigheten i en punkt på däckets ovansida? v ovansida = 2ωr = 59,2 m/s

37 10-5 Rotational Kinetic Energy and the Moment of Inertia For this mass,

38 10-5 Rotational Kinetic Energy and the Moment of Inertia We can also write the kinetic energy as Where I, the moment of inertia, is given by

39 Figure Kinetic energy of a rotating object of arbitrary shape

40 Exercise 10-7 A dumbbell(=hantel)-shaped object rotating about its (mass)center, figure Massorna kan behandlas som punktmassor. I=?, I = ∑m i r i 2 = mr 2 + mr 2 = 2mr 2

41 Example 10-4 Nose to the Grindstone Om K = 13,0 J vad är då I? K = Iω 2 /2 så I = 2K/ω 2 = 2Kr 2 /v 2 = 4,30 kgm 2

42 10-5 Rotational Kinetic Energy and the Moment of Inertia Moments of inertia of various regular objects can be calculated: (hoop = rull/tunnband, rim = rand, kant, [fälg])

43 Conceptual Checkpoint 10–2 Compare the Moments of Inertia

44 10-6 Conservation of Energy The total kinetic energy of a rolling object is the sum of its linear and rotational kinetic energies: The second equation makes it clear that the kinetic energy of a rolling object is a multiple of the kinetic energy of translation.

45 Example 10-5 Like a Rolling Disk Bestäm a) translationsenergin b) rotationsenergin och c) totala rörelseenergin, då skivan rullar utan att slira (m=1,20 kg)

46 Example 10-5 a) Translationsenergin är mv 2 /2 = = 1,20 kg (1,41 m/s) 2 /2 = 1,19 J b) Rotationsenergin är Iω 2 /2 = = (I skiva )(v/r) 2 /2 = (mr 2 /2)(v/r) 2 /2 = 0,596 J c) E tot = 1,19 J + 0,596 J = 1,79 J

47 större Figure An object rolls down an incline Punch line: Ju större tröghetsmoment, desto större del “bundet” i rotationsenergi, desto lägre hastighet efter rullningen.

48 Conceptual Checkpoint 10–4 Which Object Wins the Race?

49 10-6 Conservation of Energy If these two objects, of the same mass and radius, are released simultaneously, the disk will reach the bottom first – more of its gravitational potential energy becomes translational kinetic energy, and less rotational.

50 Conceptual Checkpoint 10–5 Compare Heights

51 Example 10-6 Spinning Wheel

52 Example 10-6 Translationsenergin är mv 2 /2 Rotationsenergin är Iω 2 /2 = I(v/R) 2 /2 Denna (totala) energi mv 2 /2(1 + I/mR 2 ) har blivit lägesenergin = mgh h = v 2 /[2g(1+I/mR 2 )]

53 Active Example 10-3 Find the Yo-Yo’s Speed

54 Active Example 10-3 Find the Yo-Yo´s speed m = 0,056 kg I = 2, kg m 2 r = 0,0064 m h = 0,50 m E i = mgh E f består dels av translationsenergin mv 2 /2 och dels av rotationsenergin Iω 2 /2 = I(v/r) 2 /2 Denna energi, [mv 2 /2](1 + I/mr 2 ) är lika stor som lägesenergin, så att v 2 = 2gh/(1+I/mr 2 ) v = ((29,810,50)/(1+2, /(0,056(0,0064) 2 )) 1/2 = = [9,81/(1 + 12,6)] 1/2 = 0,85 (m/s)

55 Summary of Chapter 10 Describing rotational motion requires analogs to position, velocity, and acceleration Average and instantaneous angular velocity: Average and instantaneous angular acceleration:

56 Summary of Chapter 10 Period: Counterclockwise rotations are positive, clockwise negative Linear and angular quantities:

57 Summary of Chapter 10 Linear and angular equations of motion: Tangential speed: Centripetal acceleration: Tangential acceleration:

58 Summary of Chapter 10 Rolling motion: Kinetic energy of rotation: Moment of inertia: Kinetic energy of an object rolling without slipping: When solving problems involving conservation of energy, both the rotational and linear kinetic energy must be taken into account.


Ladda ner ppt "© 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their."

Liknande presentationer


Google-annonser