Lecture Outlines Chapter 10 Physics, 3

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Lecture Outlines Chapter 10

Physics, 3^rd Edition James S. Walker

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Chapter 10 Rotational Kinematics and

Energy

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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

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Photo 10-1 Galaxy

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10-1 Angular Position, Velocity, and

Acceleration

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10-1 Angular Position, Velocity, and Acceleration

Degrees and revolutions:

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10-1 Angular Position, Velocity, and Acceleration

Arc length s, measured in radians:

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10-1 Angular Position, Velocity, and

Acceleration

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10-1 Angular Position, Velocity, and

Acceleration

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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

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10-1 Angular Position, Velocity, and

Acceleration

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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

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10-1 Angular Position, Velocity, and

Acceleration

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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/α_av = (ω_f – ω_i)/α_av = (0-2,1)/-0,45 = 4,7 s

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10-2 Rotational Kinematics

If the angular acceleration is constant:

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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

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10-2 Rotational Kinematics

Analogies between linear and rotational kinematics:

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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

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Example 10-2 Wheel of Misfortune

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Example 10-2 Wheel of Misfortune

Hjulet roterar ett och ett kvarts varv innan det stannar.

a)Vad blir α? Om ursprungsvinkelhastigheten är 3,40 rad/s ger

ω_f²= ω₀² + 2α(θ – θ₀)

att den konstanta vinkelaccelerationen α = - 0,736 rad/s²

a)b) Hur lång tid tar förloppet?

t = (ω_f – ω₀)/α = 4,62 s

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Active Example 10-1 Find the Time To Rest

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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 = (ω_f – 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

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10-3 Connections Between Linear and

Rotational Quantities

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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

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

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10-3 Connections Between Linear and

Rotational Quantities

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Vilken centripetalacceleration känner ett barn i en karusell om r = 4,25 m och vinkelhastigheten är

= 0,838 rad/s?

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

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Example 10-3

The Microhematocrit

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Exemple 10-3 The Microhematocrit

Vinkelhastigheten ω = 11 500 rpm och r = 9,07 cm.

a)Hur stor är v?

v = rω = 0,0907 m • 11500 • 2π rad/60 s = 109 m/s b) Hur stor är centripetalacceleration?

a_cp = rω² = 0,0907 m • (1200 rad/s)² = 1,32 •10⁵ m/s²

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10-3 Connections Between Linear and Rotational Quantities

This merry-go-round

has both tangential and centripetal

acceleration.

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10-3 Connections Between Linear and

Rotational Quantities

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

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Active example 10-2

a_t= r • Δω/Δt = r • α = 0,0907 m • 95,0 rad/s² = 8,62 m/s²

a_cp = r • ω²= 0,0907 m • (8,00 rad/s)² = 5,80 m/s²

Storleken av totala accelerationen blir då a = (a_t² + a_cp²)^1/2 = 10,4 m/s²

och vinkeln blir

Φ = arctan a_cp/a_t = 33,9º

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10-4 Rolling Motion

If a round object rolls without slipping, there is a fixed relationship between the

translational and rotational speeds:

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10-4 Rolling Motion

We may also consider rolling motion to be a combination of pure rotational and pure

translational motion:

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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

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10-5 Rotational Kinetic Energy and the Moment of Inertia

For this mass,

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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

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Figure 10-13

Kinetic energy of a rotating object of arbitrary shape

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Exercise 10-7 A dumbbell(=hantel)-shaped object rotating about its

(mass)center, figure 10-14. Massorna kan behandlas som punktmassor.

I=?, I = ∑m_ir_i² = mr² + mr² = 2mr²

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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

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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])

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Conceptual Checkpoint 10–2 Compare the Moments of Inertia

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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.

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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)

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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

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Figure 10-17 An object rolls down an incline

Punch line: Ju större tröghetsmoment, desto större del “bundet” i större rotationsenergi, desto lägre hastighet efter rullningen.

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Conceptual Checkpoint 10–4 Which Object Wins the Race?

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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.

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Conceptual Checkpoint 10–5 Compare Heights

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Example 10-6 Spinning Wheel

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Example 10-6

Translationsenergin är mv

²

/2

Rotationsenergin är I•ω

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

)]

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

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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 E_i = mgh

E_fbestår 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)

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Summary of Chapter 10

• Describing rotational motion requires analogs to position, velocity, and acceleration

• Average and instantaneous angular velocity:

• Average and instantaneous angular acceleration:

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Summary of Chapter 10

• Period:

• Counterclockwise rotations are positive, clockwise negative

• Linear and angular quantities:

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Summary of Chapter 10

• Linear and angular equations of motion:

Tangential speed:

Centripetal acceleration:

Tangential acceleration:

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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.