Conservation of Momentum Results 1 Conservation of Momentum 2.5 p - - PowerPoint PPT Presentation

▶
conservation of momentum results 1
SMART_READER_LITE
LIVE PREVIEW

Conservation of Momentum Results 1 Conservation of Momentum 2.5 p - - PowerPoint PPT Presentation

Feb 24, 2023 197 likes •769 views

Conservation of Momentum Results 1 Conservation of Momentum 2.5 p / p =- 10 7 2.0 1.5 N 1.0 0.5 0.0 - 25 - 20 - 15 - 10 - 5 0 5 p / p 100 Jerry Gilfoyle A Star Is Born! 1 / 44 Some changes to Lab 28

slide-1
SLIDE 1

Conservation of Momentum Results 1

〈Δp/〈p〉〉=-10 ± 7

  • 25
  • 20
  • 15
  • 10
  • 5

5 0.0 0.5 1.0 1.5 2.0 2.5 Δp/〈p〉 ×100 N Conservation of Momentum

Jerry Gilfoyle A Star Is Born! 1 / 44

slide-2
SLIDE 2

Some changes to Lab 28 2

1 For Activities 3b and 3d do NOT make a video of your own. Go to

the course webpage here or enter the address below.

https://facultystaff.richmond.edu/~ggilfoyl/genphys.html

2 Then under General Physics I (Physics 131) navigate to General

Physics I links and then to the item Videos for Activities 3.b and 3.d in Momentum Conservation and Center-of-Mass.

3 There will be links for the videos for Activities 3b and 3d. Analyze

these.

Jerry Gilfoyle A Star Is Born! 2 / 44

slide-3
SLIDE 3

Center-of-Mass Motion - Equal Massses 3

Jerry Gilfoyle A Star Is Born! 3 / 44

slide-4
SLIDE 4

Center-of-Mass Motion - Equal Massses 4

Jerry Gilfoyle A Star Is Born! 3 / 44

slide-5
SLIDE 5

Center-of-Mass Motion - Different Massses 5

Jerry Gilfoyle A Star Is Born! 4 / 44

slide-6
SLIDE 6

Center-of-Mass Motion - Different Massses 6

Jerry Gilfoyle A Star Is Born! 4 / 44

slide-7
SLIDE 7

Center-of-Mass Motion - Different Massses 7

Jerry Gilfoyle A Star Is Born! 4 / 44

slide-8
SLIDE 8

Center-of-Mass Motion - Different Massses 8

CM = mH xH + mL xL mH + mL 0.0 0.2 0.4 0.6 0.8

  • 0.3
  • 0.2
  • 0.1

0.0 0.1 0.2 Time (s) Center of Mass (m)

Jerry Gilfoyle A Star Is Born! 4 / 44

slide-9
SLIDE 9

Center-of-Mass 9

Jerry Gilfoyle A Star Is Born! 5 / 44

slide-10
SLIDE 10

Center-of-Mass 10

Jerry Gilfoyle A Star Is Born! 6 / 44

slide-11
SLIDE 11

Center-of-Mass 11

Jerry Gilfoyle A Star Is Born! 7 / 44

slide-12
SLIDE 12

Center-of-Mass 12

Blue - x Red - y 1 2 3 4 5 6 7

  • 0.4
  • 0.2

0.0 0.2 0.4 Time Position Flying Tennis Racket Jerry Gilfoyle A Star Is Born! 8 / 44

slide-13
SLIDE 13

Center-of-Mass 13

Blue - x Red - y 1 2 3 4 5 6 7

  • 0.4
  • 0.2

0.0 0.2 0.4 Time Position Flying Tennis Racket

  • 0.4
  • 0.2

0.0 0.2 0.4

  • 0.4
  • 0.2

0.0 0.2 0.4 x Position y Position Flying Tennis Racket

Jerry Gilfoyle A Star Is Born! 8 / 44

slide-14
SLIDE 14

A Star Is Born! 14

Jerry Gilfoyle A Star Is Born! 9 / 44

slide-15
SLIDE 15

A Star Is Born! 15

Jerry Gilfoyle A Star Is Born! 9 / 44

slide-16
SLIDE 16

A Star Is Born! 16

The photograph below shows a cloud of molecules called Bernard 68 (B68). It is located about 300 light-years (2.8 × 1015 km) away from us in the constellation Ophiuchus and is about 1.6 trillion kilometers across. It is made of molecules like CS, N2H, H2, and CO and is slowly rotating (ω = 9.4 × 10−14 rad/s). The internal gravitational attraction

  • f B68 may make the molecular cloud collapse far enough so it will ignite the nuclear

fires and B68 will begin to shine.

Jerry Gilfoyle A Star Is Born! 9 / 44

slide-17
SLIDE 17

A Star Is Born 17

The molecular cloud B68 in the constellation Ophiuchus is rotating with an angular speed ω = 9.4 × 10−14 rad/s. The gravitational attraction among the atoms in the cloud may make it collapse until the core is hot enough to ignite nuclear reactions and B68 will begin to shine. If the final properties of B68 are the same as our Sun, i.e., the same mass and size, then what will be its final angular velocity and period? Assume the lost mass carries away very little angular momentum. Compare this with the angular velocity of the Sun. Is your result reasonable? Why or why not? MB68 = 6.04 × 1030 kg IB68 = 2.7 × 1054 kg − km2 MSun = 1.989 × 1030 kg RSun = 6.96 × 105 km TSun = 25.4 d

Jerry Gilfoyle A Star Is Born! 10 / 44

slide-18
SLIDE 18

Rotational Quantities 18

Jerry Gilfoyle A Star Is Born! 11 / 44

slide-19
SLIDE 19

Linear → Rotational Quantities 19

Linear Rotational Quantity Connection Quantity s s = rθ θ = s

r

vT vT = rω ω = vT

r = dθ dt

aT aT = rα α = aT

r = dω dt

KE = 1

2mv2

KER = 1

2Iω2

  • F = m

a τ = rF⊥

  • τ = I

α

  • p = m

v

  • L =

r × p

  • L = I

ω

  • rcm =

mi ri mi

Jerry Gilfoyle A Star Is Born! 12 / 44

slide-20
SLIDE 20

How Fast Will the Star Spin? 20

The pulsar in the Crab nebula has a period T0 = 0.033 s and this period has been

  • bserved to be increasing by ∆T = 1.26 × 10−5 s each year. Assuming constant angular

acceleration what is the expression for the angular displacement of the pulsar? What are the values of the parameters in that expression? What is the torque exerted on the pulsar? mC = 3.4 × 1030 kg rC = 25 × 103 m

Jerry Gilfoyle A Star Is Born! 13 / 44

slide-21
SLIDE 21

A Pulsar 21

Jerry Gilfoyle A Star Is Born! 14 / 44

slide-22
SLIDE 22

Linear → Rotational Quantities 22

Linear Rotational Quantity Connection Quantity s s = rθ θ = s

r

vT vT = rω ω = vT

r = dθ dt

a aT = rα α = aT

r = dω dt

KE = 1

2mv2

KER = 1

2Iω2

  • F = m

a τ = rF⊥

  • τ = I

α

  • p = m

v

  • L =

r × p

  • L = I

ω

  • rcm =

mi ri mi

Jerry Gilfoyle A Star Is Born! 15 / 44

slide-23
SLIDE 23
  • F ∝

a → F = m a 23

Force and Motion 1

Jerry Gilfoyle A Star Is Born! 16 / 44

slide-24
SLIDE 24

Torque - Rotational Equivalent of Force 24

F

Jerry Gilfoyle A Star Is Born! 17 / 44

slide-25
SLIDE 25

Torque on a Point Particle 25

F ω r

  • rigin

m F

c

Consider a point particle at a fixed dis- tance from the origin (attached by the famed massless rod or a string) that moves in a circle.

Jerry Gilfoyle A Star Is Born! 18 / 44

slide-26
SLIDE 26

Torque on a Point Particle 26

F ω r

  • rigin

m F

c

  • F = m

a → τ = r F⊥ Consider a point particle at a fixed dis- tance from the origin (attached by the famed massless rod or a string) that moves in a circle.

Jerry Gilfoyle A Star Is Born! 18 / 44

slide-27
SLIDE 27

Torque - Rotating a Rigid Body (a door) 27 F ω hinge

i torque

r

Jerry Gilfoyle A Star Is Born! 19 / 44

slide-28
SLIDE 28

Torque - Rotating a Rigid Body (a door) 28 F ω m i hinge ri chunks rtorque

∆

Jerry Gilfoyle A Star Is Born! 20 / 44

slide-29
SLIDE 29

Moments of Inertia 29

Jerry Gilfoyle A Star Is Born! 21 / 44

slide-30
SLIDE 30

Linear → Rotational Quantities 30

Linear Rotational Quantity Connection Quantity s s = rθ θ = s

r

vT vT = rω ω = vT

r = dθ dt

a aT = rα α = aT

r = dω dt

KE = 1

2mv2

KER = 1

2Iω2

  • F = m

a τ = rF⊥

  • τ = I

α

  • p = m

v L = rp⊥

  • L = I

ω

  • rcm =

mi ri mi

Jerry Gilfoyle A Star Is Born! 22 / 44

slide-31
SLIDE 31

Torque - Rotational Force 31

The shield door at a neutron test fa- cility at Lawrence Livermore Labora- tory is possibly the world’s heaviest hinged door. It has a mass m = 44, 000 kg, a rotational inertia about a vertical axis through its hinges of I = 8.7 × 104 kg − m2, and a (front) face width of w = 2.4 m. A steady force Fa = 73 N, applied at its outer edge and perpendicular to the plane

  • f the door, can move it from rest

through an angle θ = 90◦ in ∆t = 75 s. What is the torque exerted by the friction in the hinges?

Jerry Gilfoyle A Star Is Born! 23 / 44

slide-32
SLIDE 32

Moments of Inertia 32

Jerry Gilfoyle A Star Is Born! 24 / 44

slide-33
SLIDE 33

Rotational Form of F = m a in lab 33

Applied torque Rotator Disk

Jerry Gilfoyle A Star Is Born! 25 / 44

slide-34
SLIDE 34

Which One Wins? 34

A wooden disk and a metal ring have the same mass m and radius r, start from rest, and roll down identical inclined planes (see figure). Which one wins?

h

Jerry Gilfoyle A Star Is Born! 26 / 44

slide-35
SLIDE 35

Which One Wins? 35

Jerry Gilfoyle A Star Is Born! 27 / 44

slide-36
SLIDE 36

Moments of Inertia 36

Jerry Gilfoyle A Star Is Born! 28 / 44

slide-37
SLIDE 37

Rolling Down an Incline 37

ri

cm

r v

d

h

Jerry Gilfoyle A Star Is Born! 29 / 44

slide-38
SLIDE 38

Rolling Down an Incline - 1 38

Jerry Gilfoyle A Star Is Born! 30 / 44

slide-39
SLIDE 39

Rolling Down an Incline - 2 39

Jerry Gilfoyle A Star Is Born! 31 / 44

slide-40
SLIDE 40

Rolling Down an Incline - 3 40

  • 40
  • 20

20 40

  • 40
  • 20

20 40 x (pixels) y (pixels) Rolling Down an Incline Jerry Gilfoyle A Star Is Born! 32 / 44

slide-41
SLIDE 41

Which One Wins? 41

A wooden disk and a metal ring have the same mass m and radius r, start from rest, and roll down identical inclined planes (see figure). Which one wins?

h

Jerry Gilfoyle A Star Is Born! 33 / 44

slide-42
SLIDE 42

Linear → Rotational Quantities 42

Linear Rotational Quantity Connection Quantity s s = rθ θ = s

r

vT vT = rω ω = vT

r = dθ dt

a aT = rα α = aT

r = dω dt

KE = 1

2mv2

KER = 1

2Iω2

  • F = m

a τ = rF⊥

  • τ = I

α

  • p = m

v L = rp⊥

  • L = I

ω

  • rcm =

mi ri mi

Jerry Gilfoyle A Star Is Born! 34 / 44

slide-43
SLIDE 43

Moments of Inertia 43

Jerry Gilfoyle A Star Is Born! 35 / 44

slide-44
SLIDE 44

A Star Is Born 44

The molecular cloud B68 in the constellation Ophiuchus is rotating with an angular speed ω = 9.4 × 10−14 rad/s. The gravitational attraction among the atoms in the cloud may make it collapse until the core is hot enough to ignite nuclear reactions and B68 will begin to shine. If the final properties of B68 are the same as our Sun, i.e., the same mass and size, then what will be its final angular velocity and period? Assume the lost mass carries away very little angular momentum. Compare this with the angular velocity of the Sun. Is your result reasonable? Why or why not? MB68 = 6.04 × 1030 kg IB68 = 2.7 × 1054 kg − km2 MSun = 1.989 × 1030 kg RSun = 6.96 × 105 km TSun = 25.4 d

Jerry Gilfoyle A Star Is Born! 36 / 44

slide-45
SLIDE 45

The Shape It’s In 45

The plot below shows the ‘obscuration’ in the angular area around B68 based on measurements of background stars. The light in the center is 1014 dimmer than outside the edge of the cloud. To make life simple we will treat the mass distribution of B68 as three, rigid, uniform spheres that lie along the axis shown in the figure and rotate with ω = 9.4 × 10−14 rad/s. The spheres do NOT rotate independently of the rest of the

  • cloud. The origin is at the center of the central lobe. What is the moment of inertia of

the cloud? Lobe Radius (km) Mass (kg) central Rc = 1.0 × 1012 mc = 6.0 × 1030 inner Ri = 2.0 × 1011 mi = 4.6 × 1028

  • uter

Ro = 1.7 × 1011 mo = 2.9 × 1028

  • rigin
  • inner

cloud center li = 1.4 × 1012 km

  • rigin
  • uter

cloud center lo = 2.0 × 1012 km

  • li

l

Jerry Gilfoyle A Star Is Born! 37 / 44

slide-46
SLIDE 46

Angular Momentum 46

Jerry Gilfoyle A Star Is Born! 38 / 44

slide-47
SLIDE 47

Angular Momentum 47

| L| = rp⊥ = Iω

Jerry Gilfoyle A Star Is Born! 38 / 44

slide-48
SLIDE 48

Linear → Rotational Quantities 48

Linear Rotational Quantity Connection Quantity s s = rθ θ = s

r

vT v = rω ω = v

r = dθ dt

a a = rα α = a

r = dω dt

  • F = m

a = d

p dt

τ = rF⊥

  • τ = I

α = d

L dt

KE = 1

2mv2

KER = 1

2Iω2

  • p = m

v L = rp⊥

  • L = I

ω

  • rcm =

mi ri mi I =

  • mir2

i

Jerry Gilfoyle A Star Is Born! 39 / 44

slide-49
SLIDE 49

Conservation of Momentum From FBA = − FAB 49

  • FAB = −

FBA mA aA = −mB aB mA d vA dt = −mB d vB dt dmA vA dt = −dmB vB dt d pA dt = −d pB dt d pA dt + d pB dt = 0 d dt ( pA + pB) = 0 ∴ pA + pB = const

Jerry Gilfoyle A Star Is Born! 40 / 44

slide-50
SLIDE 50

Angular Momentum Conservation 50

Jerry Gilfoyle A Star Is Born! 41 / 44

slide-51
SLIDE 51

Angular Momentum Conservation 51

d Rd

w

r

Jerry Gilfoyle A Star Is Born! 41 / 44

slide-52
SLIDE 52

Moments of Inertia 52

Jerry Gilfoyle A Star Is Born! 42 / 44

slide-53
SLIDE 53

A Star Is Born 53

The molecular cloud B68 in the constellation Ophiuchus is rotating with an angular speed ω = 9.4 × 10−14 rad/s. The gravitational attraction among the atoms in the cloud may make it collapse until the core is hot enough to ignite nuclear reactions and B68 will begin to shine. If the final properties of B68 are the same as our Sun, i.e., the same mass and size, then what will be its final angular velocity and period? Assume the lost mass carries away very little angular momentum. Compare this with the angular velocity of the Sun. Is your result reasonable? Why or why not? MB68 = 6.04 × 1030 kg IB68 = 2.7 × 1054 kg − km2 MSun = 1.989 × 1030 kg RSun = 6.96 × 105 km TSun = 25.4 d

Jerry Gilfoyle A Star Is Born! 43 / 44

slide-54
SLIDE 54

Conservation of Angular Momentum Results 54

No outliers: 33±29 Outliers: 130±230 100 200 300 400 500 600 0.0 0.5 1.0 1.5 2.0 2.5 3.0

αth - αexp αexp ×100

Counts Newton's 2nd Law for Rotation, 2019

Jerry Gilfoyle A Star Is Born! 44 / 44

slide-55
SLIDE 55

Conservation of Angular Momentum Results 55

No outliers: 33±29 Outliers: 130±230 100 200 300 400 500 600 0.0 0.5 1.0 1.5 2.0 2.5 3.0

αth - αexp αexp ×100

Counts Newton's 2nd Law for Rotation, 2019

Nature volume 575, pages 147-150, Nov 6 (2019)

Jerry Gilfoyle A Star Is Born! 44 / 44