Lecture Outlines Chapter 10 Physics, 3rd Edition James S. Walker

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1 Lecture Outlines Chapter 10 Physics, 3rd Edition James S. Walker
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2 Rotational Kinematics and Energy
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 Rotational motion is everywhere in our universe, on every scale of length and time. A galaxy like this one may take millions of years to complete a single rotation about its center.

5 10-1 Angular Position, Velocity, and Acceleration

6 10-1 Angular Position, Velocity, and Acceleration
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 10-1 Angular Position, Velocity, and Acceleration

10 Exercise 10-1(p.284) En gammal LP-skiva gjorde
33 1/3 rpm (varv [revolutions] per minut) 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/s2 Hur lång tid tar det för kvarnen att stanna? Om ursprungsvinkelhastigheten 2,1 rad/s blir Δt = Δω/αav = (ωf – ωi)/αav = (0-2,1)/-0,45 = 4,7 s

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

16 Exercise 10-4 ω0 = - 8,4 rad/s och α = - 2,8 rad/s2
Exercise 10-4 ω0 = - 8,4 rad/s och α = - 2,8 rad/s2. Vad är ω, 1,5 s senare? ω = ω0 + αt ger ω = - 8,4 -2,8•1,5 = - 12,6 rad/s Amass is attached to a string wrapped around a pulley. As the mass falls, it causes the pulley to increase its angular speed with a constant angular acceleration.

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

18 Example 10-1 Thrown for a Curve ω0 = 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 - ω0 )/Δt = (- 1,8 rad/s)/0,595 s = - 3,03 rad/s2 θ = θ0 + ω0t + αt2/2 ger θ - θ0 = 21,4 rad – 0,536 rad = 20,9 rad = 3,33 varv To throw a curve ball, a pitcher gives the ball an initial angular speed of 36.0 rad/s. When the catcher gloves the ball s later, its angular speed has decreased (due to air resistance) to 34.2 rad/s. (a) What is the ball’s angular acceleration, assuming it to be constant? (b) How many revolutions does the ball make before being caught?

19 Example 10-2 Wheel of Misfortune
On a certain game show, contestants spin a wheel when it is their turn. One contestant gives the wheel an initial angular speed of 3.40 rad/s. It then rotates through one-and-one-quarter revolutions and comes to rest on the BANKRUPT space. (a) Find the angular acceleration of the wheel, assuming it to be constant. (b) How long does it take for the wheel to come to rest?

20 Example 10-2 Wheel of Misfortune
Hjulet roterar ett och ett kvarts varv innan det stannar. Vad blir α? Om ursprungsvinkelhastigheten är 3,40 rad/s ger ωf2 = ω02 + 2α(θ – θ0) att den konstanta vinkelaccelerationen α = - 0,736 rad/s2 b) Hur lång tid tar förloppet? t = (ωf – ω0)/α = 4,62 s

21 Active Example 10-1 Find the Time To Rest
A pulley rotating in the counterclockwise direction is attached to a mass suspended from a string. The mass causes the pulley’s angular velocity to decrease with a constant angular acceleration α = rad/s2. (a) If the pulley’s initial angular velocity is ω0 = 5.40 rad/s, how long does it take for the pulley to come to rest? (b) Through what angle does the pulley turn during this time?

22 Active Example 10-1 Find the Time to Rest
Om ursprungsvinkelhastigheten är 5,40 rad/s och den konstanta vinkelaccelerationen = - 2,10 rad/s2 a) Vad blir t? ω = ω0 + αt t = (ωf – ω0)/α = 2,57 s b) Hur stor vinkel har då blocket roterat? θ = θ0 + ω0t + αt2/2 θ - θ0 = 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?
Two children ride on a merry-go-round, with child 1 at a greater distance from the axis of rotation than child 2. Is the angular speed of child 1 (a) greater than, (b) less than, or (c) the same as the angular speed of child 2?

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? acp = rω2 = 4,25 m • (0,838 rad/s)2 = 2,96 m/s2 (0,3 g)

28 Example 10-3 The Microhematocrit
In a microhematocrit centrifuge, small samples of blood are placed in heparinized capillary tubes (heparin is an anticoagulant). The tubes are rotated at 11,500 rpm, with the bottom of the tubes 9.07 cm from the axis of rotation. (a) Find the linear speed of the bottom of the tubes. (b) What is the centripetal acceleration at the bottom of the tubes?

29 Exemple 10-3 The Microhematocrit
Vinkelhastigheten ω = rpm och r = 9,07 cm. Hur stor är v? v = rω = 0,0907 m • • 2π rad/60 s = 109 m/s b) Hur stor är centripetalacceleration? acp = rω2 = 0,0907 m • (1200 rad/s)2 = 1,32 •105 m/s2

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
Suppose the centrifuge in Example 10–3 is starting up with a constant angular acceleration of 95.0 rad/s2. (a) What is the magnitude of the centripetal, tangential, and total acceleration of the bottom of a tube when the angular speed is 8.00 rad/s? (b) What angle does the total acceleration make with the direction of motion?

33 Active example at = r • Δω/Δt = r • α = 0,0907 m • 95,0 rad/s2 = 8,62 m/s2 acp = r • ω2 = 0,0907 m • (8,00 rad/s)2 = 5,80 m/s2 Storleken av totala accelerationen blir då a = (at2 + acp2)1/2 = 10,4 m/s2 och vinkeln blir Φ = arctan acp/at = 33,9º If the angular speed of the merry-go-round is increased, the child will experience two accelerations: (i) a tangential acceleration, at, and (ii) a centripetal acceleration, acp. The child’s total acceleration, a, is the vector sum of at and acp.

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 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? vovansida = 2ωr = 59,2 m/s If the angular speed of the merry-go-round is increased, the child will experience two accelerations: (i) a tangential acceleration, at, and (ii) a centripetal acceleration, acp. The child’s total acceleration, a, is the vector sum of at and acp.

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 10-13 Kinetic energy of a rotating object of arbitrary shape
To calculate the kinetic energy of an object of arbitrary shape as it rotates about an axis with angular speed ω, imagine dividing it into small mass elements, mi. The total kinetic energy of the object is the sum of the kinetic energies of all the mass elements.

40 Exercise 10-7 A dumbbell(=hantel)-shaped object rotating about its (mass)center, figure Massorna kan behandlas som punktmassor. I=?, I = ∑miri2 = mr2 + mr2 = 2mr2

41 Example 10-4 Nose to the Grindstone Om K = 13,0 J vad är då I
Example 10-4 Nose to the Grindstone Om K = 13,0 J vad är då I? K = Iω2/2 så I = 2K/ω2 = 2Kr2/v2 = 4,30 kg•m2 A grindstone with a radius of m is being used to sharpen an ax. If the linear speed of the stone relative to the ax is 1.50 m/s, and the stone’s rotational kinetic energy is 13.0 J, what is its moment of inertia?

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
If the dumbbell-shaped object in Figure 10–14 is rotated about one end, is its moment of inertia (a) more than, (b) less than, or (c) the same as the moment of inertia about its center? As before, assume that the masses can be treated as point masses.

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) A1.20-kg disk with a radius of 10.0 cm rolls without slipping. If the linear speed of the disk is 1.41 m/s, find (a) the translational kinetic energy, (b) the rotational kinetic energy, and (c) the total kinetic energy of the disk.

46 Example 10-5 a) Translationsenergin är mv2/2 = = 1,20 kg • (1,41 m/s)2/2 = 1,19 J b) Rotationsenergin är Iω2/2 = = (Iskiva)•(v/r)2/2 = (mr2/2)•(v/r)2/2 = 0,596 J c) Etot = 1,19 J + 0,596 J = 1,79 J If the angular speed of the merry-go-round is increased, the child will experience two accelerations: (i) a tangential acceleration, at, and (ii) a centripetal acceleration, acp. The child’s total acceleration, a, is the vector sum of at and acp.

47 Figure An object rolls down an incline Punch line: Ju större tröghetsmoment, desto större del “bundet” i rotationsenergi, desto lägre hastighet efter rullningen. An object starts at rest at the top of an inclined plane and rolls without slipping to the bottom. The speed of the object at the bottom depends on its moment of inertia— a larger moment of inertia results in a lower speed.

48 Conceptual Checkpoint 10–4 Which Object Wins the Race?
A disk and a hoop of the same mass and radius are released at the same time at the top of an inclined plane. Does the disk reach the bottom of the plane (a) before, (b) after, or (c) at the same time as the hoop?

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
A ball is released from rest on a no-slip surface, as shown. After reaching its lowest point, the ball begins to rise again, this time on a frictionless surface. When the ball reaches its maximum height on the frictionless surface, is it (a) at a greater height, (b) at a lesser height, or (c) at the same height as when it was released?

51 Example 10-6 Spinning Wheel
A block of mass m is attached to a string that is wrapped around the circumference of a wheel of radius R and moment of inertia I. The wheel rotates freely about its axis and the string wraps around its circumference without slipping. Initially the wheel rotates with an angular speed ω, causing the block to rise with a linear speed v. To what height does the block rise before coming to rest? Give a symbolic answer.

52 Example Translationsenergin är mv2/2 Rotationsenergin är I•ω2/2 = I•(v/R)2/2 Denna (totala) energi mv2/2(1 + I/mR2) har blivit lägesenergin = mgh h = v2/[2g(1+I/mR2)] If the angular speed of the merry-go-round is increased, the child will experience two accelerations: (i) a tangential acceleration, at, and (ii) a centripetal acceleration, acp. The child’s total acceleration, a, is the vector sum of at and acp.

53 Active Example 10-3 Find the Yo-Yo’s Speed
Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo-yo is kg, its moment of inertia is 2.9 x 10-5 kg.m2, and the radius, r, of the axle the string wraps around is m. What is the linear speed, v, of the yo-yo after it has dropped through a height h = 0.50m?

54 Active Example 10-3 Find the Yo-Yo´s speed m = 0,056 kg I = 2,9 •10-5 kg m2 r = 0,0064 m h = 0,50 m Ei = mgh Ef består dels av translationsenergin mv2/2 och dels av rotationsenergin I•ω2/2 = I•(v/r)2/2 Denna energi, [mv2/2](1 + I/mr2) är lika stor som lägesenergin, så att v2 = 2gh/(1+I/mr2) v = ((2•9,81•0,50)/(1+2,9•10-5/(0,056(0,0064)2))1/2 = = [9,81/(1 + 12,6)]1/2 = 0,85 (m/s) If the angular speed of the merry-go-round is increased, the child will experience two accelerations: (i) a tangential acceleration, at, and (ii) a centripetal acceleration, acp. The child’s total acceleration, a, is the vector sum of at and acp.

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.