Chapter 10 Spin and orbital motionwdx.cxtc.edu.cn/physics/shuangyu/pdf/ch10.pdf2 10.1 Some concepts...

1

Chapter 10 Spin and orbital motion

Rotation:

All around us: wheels, skaters, ballet, gymnasts, helicopter, rotors, mobile engines, CD disks, …

Atomic world: electrons— “spin”, “orbit”.

Universe: planets spin and orbiting the sun, galaxies spin, …

Chapter 4 kinematicsChapter 10 dynamics

2

§10.1 Some concepts about rotation

1. Spin—describe rotational motion of a system about an axis through its center of mass.2. Rigid body—a system composed of many pointlike particles that maintain fixed distances from each other at all time.

each particle of the spinning rigid body system executes circular motion about the axis through the center of mass.3. Orbital motion—the center of the mass of the system is moving in space from a perspective of a particular reference frame.

The motion of the center of mass must not be circular.


3

4. The orbital angular momentum of a particle

Define: vmrprLrrrrr

×=×=

xy

z

m θrr

pr

o

⊥r

Lr

⊥p

Magnitude:prrprmvL ⊥⊥ === θsin

Direction:

pr rr and

Perpendicular to the plane containing the


Notice:

1 is measured with respect to the origin at O;2 unit: kg·m2/s; 3 whatever the path or trajectory of a particle is straightline , curved path, closed orbital path, ….

Lr

5. The angular momentum of the circular orbital motion of a particle(a) Angular momentum


4

mvrLvmrprL

=×=×=

rrrrr

rvrv

ωω

=×=rrr rr

pr

o m

ωr

ωω 22 mrrmL ==then

(b) Moment of inertia of a particleωr

r2mrL =


2mrI = ωω rrrImrL == 2thendefine

§10.2 The time rate of change of angular momentum and torque

1. The time rate of change of angular momentum for a single particle

tprp

trpr

ttL

dd

dd)(

dd

dd

rrr

rrr

r

×+×=×=

prLrrr

×=

totaldd

dd

0dd

Frtpr

tL

vmvpvptr

rrr

rr

rrrrrr

Q

×=×=∴

=×=×=×

rrFr

θ

om

5

Define: totaltotal Frrrr

×=τ2. torque

is the position vector of the point of application of the force with respect to the chosen origin.rr

Unit of the torque: N·m

Magnitude:

⊥⊥ === rFFrrF θτ sinDirection: Perpendicular to the plane containing the Fr

rr and

rrFr

θ

om

dr =⊥

⊥F


rrFr

θ

om

dr =⊥

⊥F

If ,thecross

,0,or0

total

OFFrr

== πθ

0total =τr


Discussion:

6

3. Dynamics of circular orbital motion of a single particle

ωω rrrQ ImrL == 2

totaltotaldd

dd τr

rrr

rr

=×=×= Frtpr

tL

Example 1: P433 10.5Example 2: P433 10.6

αωωτ rr

rr It

IIt

===∴dd)(

dd

total

Can not be used in noncircular orbital motion.


§10.3 The angular momentum of a system of particles and moment of inertia of rigid body

1. The angular momentum of a system of particles

∑ ∑ ∑ ×=×==i i i

iiiiii vmrprLL rrrrrr

⎩⎨⎧

′+=

′+=

iCMi

iCMi

vvvrrrrrr

rrr

Q θipr

o

CMrr

irr im

irr′C

( )( )

iii

iCMii i

iiiCM

iCMii i

iiiCM

iii

iCM

vmrvmrvmr

vvmrvmr

vmrrL

′×′+×′+×=

′+×′+×=

×′+=

∑∑ ∑∑ ∑

∑

rrrrrr

rrrrr

rrrr∴

7

∑ ×=×i

CMCMiiCM vMrvmrrrrr

First term:

0=×′=×′=×′ ∑∑ CMCMCMi

iiCMii

i vrMvrmvmr rrrrrrSecond term:

The position vector of center of mass with respect to the center of mass

iii

i vmrrr ′×′∑Third term:

is the vector sum of angular momentum of all particles with respect to the center of mass.


spinorbital LL

vmrvMrL iii

iCMCMrr

rrrrr

+=

′×′+×= ∑then

2. Spin angular momentum of a rigid body about a axis through the center of mass

iv ′r

o

ωrz

o′im

ir ′r⊥′irr

//ir ′r

iii

i vmrLrrr

Q ′×′= ∑spin

⊥⊥ ′+′=′′×=′ iiiii rrrrvrrrrrr

//ω

)()( //spin ⊥⊥ ′××′+′=∴ ∑ ii

iii rrrmLrrrrr

ω


8


∑∑∑

∑

⊥

⊥

⊥⊥

⊥

′+

′−=

′××′+

′××′=

iii

iiii

iii

i

iii

i

rm

rrm

rrm

rrmL

ω

ω

ω

ω

r

r

rrr

rrrr

2

//

//spin

)(

)(

iv ′r

o

ωrz

o′im

ir ′r⊥′irr

//ir ′r

The vector is an involved vector summation.The rotation of an oddly shaped object about any axis of rotation is beyond the scope of this course.

iv ′r

o

ωrz

o′im

ir ′r⊥′irr

//ir ′r

3. The moment of inertia or rotational inertia of a rigid body about a fixed axis through center of mass

∑∑ ⊥⊥ ′−=′××′i

iiiiii

i rrmrrm rrrrωω )(//

If 1the rigid body is symmetry about the axis; 2the axis is fixed.

This term has no effect. Then

∑∑

⊥

⊥⊥

′=

′××′=

iii

iii

i

rm

rrmL

ω

ωr

rrrr

2

spin )(


9

iv ′r

o

ωrz

o′im

ir ′r⊥′irr

//ir ′r

∑ ⊥′=i

iiCM rmI 2Define:

This is the moment of inertia or rotational inertia of a rigid body about a fixed axis through center of mass

The spin angular momentum of a rigid body

ωω rrr)( 2

spin ∑ ⊥==i

iiCM rmIL


4. The moment of inertia of various rigid bodies

(a) Point particle 2mrI =r –a distance from the axis of rotation

(b) Collection of point particles

∑ ⊥=i

i irmI 2

--the perpendicular distance of each mass mi from the axis of rotation

⊥ir


(c) Rigid body of distributive mass

mrI d2∫ ⊥=

10

--the perpendicular distance of each mass dm from the axis of rotation

⊥r

=mdλλ ：densitylineardlσσ ：densitysurfacedS

ρρ ：densityvolumdV

The element of mass:


l

lll

A

m m2 m3

m4

m52

2

22

32)2)(54(

)2(32

mllmm

lmmlI

=++

+=

Example1: 5 particles are connected by 4 light staffs as shown in figure. Find the moment of the system with respect to the axis through point A, and perpendicular to the paper plane.

Solution: 2⊥∑= i

ii rmI


11

Example 2: There is a light thin staff of mass m and length L. Find the moment with respect to different axis.

mx

mrI

d

d2

2

∫∫

=

= o xmd x

2L2

L−

233

32

2

2

121

8831

2

231d

mLLLLm

L

L

xLmx

Lmx

L

L

=⎟⎟⎠

⎞⎜⎜⎝

⎛+=

−== ∫−

Solution:


L

mdo

xx

23

0

222

31

031

ddd

mLL

xLm

xLmxmxmrI

L

==

=== ∫ ∫ ∫


12


Some rotational inertias

Summary of the previous sections

1. The orbital angular momentum of a particle

2. The time rate of angular momentum for asingle particle

For a circular orbital motion of a particle

ωω rrrImrL == 2

totaltotaldd

dd τr

rrr

rr

=×=×= Frtpr

tL

Define: totaltotal Frrrr

×=τ

xy

z

m θ

rr

pro⊥r

Lr

⊥p

rr

totalFr

θ

om

vmrprL rrrrr×=×=Define:

Rotational inertiarr

pr

o m

ωr

13

For a circular orbital motion of a particle

αωωτr

rrr I

tII

t===

dd)(

dd

total

3. The angular momentum of a system of particles

spinorbital LL

vmrvMrL iii

iCMCM

rr

rrrrr

+=

′×′+×= ∑


( ) ( )iCMii

iCMiii

i vvmrrvmrL ′+×′+=×= ∑∑ rrrrrrr

θipr

o

CMrr

irr im

irr′C

4. The angular momentum of a rigid body spinningabout a fixed axis through center of mass

ωωrrr

spin2

spin )( IrmLi

ii == ∑ ⊥


ivr

o

ωrz

o′im

irr⊥irr

//irr

∫= mrI d2spin

∑ ⊥=i

ii rmI 2spin

The rotational inertia

It represents the inertia of rotating body.

14

Some rotational inertias


1. The time rate of change of the angular momentum of a system of particlesA group of particles with masses ,nmmm ,,, 21 L

and angular momentums nLLLr

Lrr

,,, 21

∑=+++=i

iLLLLLrr

Lrrr

321

The total angular momentum of the system is

totaltotaldd

dd τr

rrrr

=×== ∑∑i

iii

i Frt

LtLThen

§10.4 The dynamics of rigid body with a fixed axis

totali,totaldd

τrrr

r

=×= iii Fr

tLRecall

15

1m

2m

3m4m

5m

6m

2. The torques due to internal and external forces

2Fr

1Fr

1rr

2rro

∑∑∑ +==i

ii

ii

i exintotal ττττ rrrrA system bounded with orange colour

0)(

)(in121

212121in

=×−=

×+×=Frr

FrFrrrr

rrrrrτ

For any two element mass of the system m1 and m2

∑ ≡i

in 0iτr

therefore

12Fr

21Fr


∑∑∑ =×==×=i

extexttotaltotaldd

iii

iii

i FrFrtL ττ rrrrrrr

3. The dynamics of rigid body with a fixed axis--The rotational counterpart of Newton’s second law of a rigid body with fixed axis

tL

i dd

iexttotal

rrr

== ∑ττ

ωrrr

spinspin ILL ==

αω

τr

rrr

spinspinexttotal dd

dd I

tI

tL

===


The torque due to the internal forces do not change the angular momentum of a system.

zzz I

ttL

αω

τ spinz dd

dd

===

amtvm

tpF

rrrr===

dd

dd

exttotal

16

Example 1: As shown in figure a uniform disk, with mass M=2.5 kg and radius R=20 cm, mounted on a fixed horizontal axle. A block with mass m=1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. The cord does not slip, and there is no friction at the axle.


gmr

gMr

NrSolution:

2

2

rad/s242.08.4

N0.621

m/s8.42

2

−=−

==

=−=

−=+

−=

Ra

MaTmM

mga

α

)3(

)2(21

)1(2

α

α

Ra

MRRT

mamgT

=

=−

=−

++


17

Example 2: Find the acceleration of the falling block by direct application of .tL ddtotal

rr=τ

Solution: RmvILz )(+= ω

mRaRaMR

mRaItvmR

tI

mvRIt

Rmg

+=

+=

+=

+=

2

21

dd

dd

)(dd)(

α

ω

ω

mMmga

22+

=


gmr

Nr

gr

Example 3: As shown in figure, one block has mass M=500 g, the other has mass m=460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls h=75.0 cm in 5.00 s ( without the cord slipping ). (a) What is the magnitude of the blocks? (b) What are the tensions of the cord in both sides? (c) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia?

§10.4 The time rate of change of the spin angular momentum and dynamics of rigid body

18

Solution:

gMrgm

r

1Tr

1Tr

2Tr

2Tr

+

+

2

21

2

1

21

)(

ath

RaIRTT

mamgTMaTMg

=

==−

=−=−

αα

22

22

1

22

mkg1038.11.2rad/s

N54.4N87.4

m/s100.6

⋅×====

×=

−

−

I

TTa

α

§10.4 The time rate of change of the spin angular momentum and dynamics of rigid body

1. The angular impulse-momentum theorem

totaltotaldd

dd τr

rrrr

=×== ∑∑i

iii

i Frt

LtLRecall

tL dd totalτrr=

∫∫ ==∆f

i

f

i

t

t

L

LtLL dd totalτr

rr r

r

The angular impulse-momentum theorem

Compare with the impulse-momentum theorem

totaldd F

tp rr

= ItFpp f

i

f

i

t

t

p

p

rrrr r

r ===∆ ∫∫ dd total

§10.5 The angular impulse-momentum theoremand the procession of a rapidly spinning top

19

§10.5 The angular impulse-momentum theoremand The procession of a rapidly spinning top

Example 4:Two cylinder having radii R1 and R2 and rotational inertias I1 and I2 respectively, are supported by axes perpendicular to the plane of the screen. The two cylinders are initially rotating with angular speeds , respectively. The small cylinder is moved to the right until it touches the large one and the two cylinders rotate at constant rates in opposite directions. Find the final angular speeds of the two cylinders in terms of I1, I2, R1, R2, .

2010 andωω

2010 andωω

2R2I

10ω

1R 1I1O2O20ω

2R2I1ω

1R 1I1O2O2ω

)3()2(d

)1(d

2211

2022222

1011111

RRIItRf

IItRf

ωωωω

ωω

=−=

−=−

∫∫

Solution:

According to the angular impulse-momentum theorem

Free body diagram:


We can get

1222

21

2021210211

2 IRIRRIRRI

++

=ωω

ω1

222

21

2021210221

1 IRIRRRIRI

++

=ωω

ω

12 ffrr

−=

1fr

222

111

RfRf

=−=

ττ

20

2. The procession of a rapidly spinning top0spin =L

rIf the spinning angular momentum is zero, i.e.

jMgrgMrcˆsintotal φτ =×=

rrrThe torque of the gravity with respect to the point O


It will make the top fall down.

If the spinning angular momentum is not zero, i.e.

0spin ≠Lr

tLtL dd,

dd

totaltotal ττ rrr

r==

According to the angular impulse-momentum theorem

Crr

gM r

spinLr

x

z


ttL ∆==∆ ∫ totaltotald ττ rrr

The change of the angular momentum is

21

The result is that the top moves in the direction perpendicular to the plane of symmetry axis and the local vertical direction.

LgMrc

rrrr⊥×=totalτbecause

LLLLLrrrrr

=′+=′ ∆

The total angular momentum


φ

φ

Crr

Crr

φ

φθd

Lr

∆

Lr

∆

3. The angular speed of the procession

φθ

θφ

sindd

dsind

spin

spin

LL

LL

=

=


φ

φθd

spinspin

spinprocs

sinsin

sind/d

dd

LMgr

LMgr

LtL

t

==

==

φφ

φθω

22

4. The application of the procession of top

ωo

c

ωr

rr

vr

fr

gm r


Think! why does the bicycle wheel behave as shown in the following video film?

§10.6 Simultaneous spin and orbital motion

2. Rotational kinetic energy of a system

1. The kinetic energy of a spinning system

∑∑ ⊥×==i

iiiii

rmvmKE 22total )(

21

21 rrω

for spinning motion ⊥= ii rv ω222

rot 21

21 ωω CM

iii IrmKE == ∑ ⊥

for orbital motion of the center of mass2

21

CMMvKE =

23

Total kinetic energy22

spin 21

21

CMCM MvIKE += ω


Example 4: A uniform spherical shell of mass M and radius R rotates about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertial I and radius r, and is attached to a small object of mass m. There is no friction on pulley’s axle; the cord does not slip on the pulley. What is the speed of the object after it falls a distance h from rest?


24

Solution:

mghPErvRv

mvIMRKE

=

=

=

++=

pull

sph

22pull

2sph

2

21

21)

32(

21

ω

ω

ωω

According to the conservation law of mechanical energy PE=KE

mMmrIgh

MrIm

mghv3/2/1

2

321

21 2

2++

=++

=


3. The total angular momentum(a) Spinning motion

spinspin ωrr

CMIL =

(b) Orbital motion of center of mass

orbit2

orbit ωrrrrrr

⊥=×=×= MrvMrprL CMCMCMCM

(c) Total angular momentum of the system

orbit2

spinspinorbital ωω rrrrr⊥+=+= MrILLL CM

Example : P450 10.10


25

4. Synchronous rotationIf the spin angular velocity and the orbital angular velocity are parallel, and have the same magnitude, then, the motion is called synchronous rotation.

ωωω rrr== orbitspin

5. Parallel axis theoremThe total angular momentum for a synchronous rotation

2orbit

2spintotal

MdIIIMdIL

CM

CM

+==+= ωωω rrrr


The parallel axis theorem:

2MdII CM +=

The moment of inertia I of the system about an axis parallel to the symmetry axis of the system and separated from it by a perpendicular distance d is I.


6. Rolling motion without slipping(a) The spin and orbital axes are parallel and separated by the radius of the circularly shaped system undergoing rolling motion.

without slipping—ABS system?

26

(b) Rolling constrains:

αω

ωθθ

Rt

Raat

v

Rvt

Rts

Rs

t ====

==

=

dd

dd

ordd

dd

centercenter

center

(c) Velocity

rrr ′+=rrr

center

vvv ′+=rrr

center

rr

centerrrr ′r

o


(d) Total kinetic energy

CM

CM

CMCM

ImRI

IImRKE

IKEandmvKE

+=

=+=

==

2

2222total

2rot

2trans

21

21

21

21

21

ωωω

ω

CMv

CMv

ωR

ωR

CMv2•C CC

T TT

B BB

CMv CMv


+ =

27

Example 1: P459 10.12, find the acceleration and the tension in the cord.

R

Solution:method 1: choose the center of mass of the body as the reference point.

αα

RaITR

maTmgCM

==

=−

CM

CM

CM ImRmgITg

ImRmRa

+=

+= 22

2


Tr

gmr

oR

+

method 2: choose the point P on the rim of the body as the reference point.

R

Ra

mRIIImgR

CM

=

+==

α

α2

CM

CM

CM ImRmgITg

ImRmRa

+=

+= 22

2

The same results can be obtained


Tr

gmr

oR

+

P

28

6. A general discussion about the angular momentum and the torque(§10.17)

(a) Angular momentum about a point P(When both spin and orbital motion exist)

ir ′r

Po

im

Prrirr

PiiiPi rrrrrrrrrrrr

−=′′+= or

ii

iiP vrmL ′×′= ∑ rrr

PiiPii vvvt

rtr

tr rrr

rrr

−=′⇒−=′

dd

dd

dd

)()( PiPi

iiP vvrrmLrrrrr

−×−=∑


Pi

iiP

Pi

iii

ii

PP

iii

iiP

ii

PP

iii

iP

iii

PiPi

ii

PiPi

iiP

arm

armtpr

tvrrm

tvmrr

tvrrm

tvrrm

vvt

rrm

vvrrt

mt

L

rrr

rrr

r

rrr

rrr

rrr

rrr

rrrr

rrrrr

×′−=

×′−×′=

×−−×−=

×−−×−=

−×−+

−×−=

∑∑∑

∑∑

∑∑

∑

∑

)(

)(dd

dd)(

d)d()(

dd)(

dd)(

)](dd[)(

)()](dd[

dd

τ

(b) The time rate of change of the angular momentum about point P


29

if 1P is in an inertial reference frame;2P is the center of mass of the system;3P has an acceleration parallel or antiparallel to the vector locating the center of mass.

thent

LPP d

dr

r=τ

For a rigid body rotating about the axis through the center of mass

αττ rrr

r It

LPP === totald

d


§10.7 Conservation of angular momentum

∑=i

LL itotal

rr1. The total angular momentum of a system of particles

2. The time rate of change of the angular momentum

exttotaltotal

dd τrr

=t

L

3. Conservation of angular momentum

If 0exttotal =τr then vectorconstanttotal =Lr

30


The diver’s angular momentum is constant

The student decreasing his inertia to increase his angular speed. The angular momentum is constant.

Examples: 2211 ωω II =


An idealized spacecraft containing a flywheel. If the flywheel is made to rotate clockwise as shown, the spacecraft itself will rotate counterclockwise. When the flywheel is braked to a stop, the spacecraft will also stop rotating but will be reoriented by the angle ∆θsc.

02211 =+ ωω rr II

31


As shown in figure, the bicycle wheel whose rotational inertia about its central axis is Iwh, the wheel is rotating at an angular speed ωwh counterclockwise. When the wheel is inverted, the student, the stooland the wheel’s center rotate together as a composite bodyabout the stool’srotation axis, with rotational inertia Ib. With what angular speed and in what direction does the composite body rotate after the inversion of the wheel.


Exercise 1:

32

Solution:The angular momentum of the system is conserved

b

whwhb

bbwhwhbf

whwhwhwhbf

whibiwhfbf

II

IkILkIkIL

LLLL

ωω

ωωωω

2

ˆ2

ˆ0ˆ

=

==

+=−

+=+

r

r

rrrr


Exercise 2: In the overhead view of the figure, four thin, uniform rods, each of mass M and length d= 0.50 m, are rigidly connected to a vertical axle to form a turnstile. The turnstile rotates clockwise about the axle, which is attached to a floor, with initial angular velocity ω i=2.0 rad/s. A mud ball of mass m=M/3 and initial speed vi=12 m/sis thrown along the path shown and sticksto the end of one rod. What is the final angularvelocity ωf of the ball-turnstile system?


33

Solution: Mechanical energy, linear momentum and angular momentum, which is conserved?

iball,its,fball,fts, LLLLrrrr

+=+

2ts

22ts

tsfts,tsits,

34or

])2

(121[4

MdI

dMMdI

ILIL fi

=

+=

== ωω

o

oo

60cos)60180sin(iball,

dmvdmvL

i

i

−=−−=

+

ff mdIL ωω 2ballfball, ==


o60cos34

34 222

iiff mdvMdMdMd −=+ ωωω

rad/s80.0)60cos4(51

−=−= oiif vd

dωω

iball,its,fball,fts, LLLLrrrr

+=+

+


34

Example 3:The particle of mass m in figure slides down the frictionless surface through height h and collides with the uniform vertical rod of mass M and length d, sticking to it. The rod pivots about point Othrough the angle θbefore momentarily stopping. Find θ.


Solution:

22

31

2

mdMd

ghmd

+=ω


)cos22

()cos(

)(21

)2

(121

)(21

22rod

22rod

2rod

2

θθ

ω

ω

ddMgddmg

mdI

dMMdI

mdImvd

mvmgh

−+−

=+

+=

+=

=

])3)(2(

61[cos2

1

MmMmdhm

++−= −θ

35

Exercise 4:during a jump to his partner, an aerialist is to make a quadruple somersault lasting a time t=1.87 s. For the first and last quarter revolution, he is in the extended orientation shown in figure, with rotational inertia I1=19.9 kg m2 around his center of mass. During the rest of the flight he is in a tight tuck, with rotational inertia I2=3.93kg m2. What must be his angular speed around his center of mass during the tuck.



Solution:There is no net external torque about his center of mass, his angular momentum about his center of mass is conserved.

1

2212211 I

III ωωωω ==

First and last quarter -revolutionrev5.0111 == tωθ

rev5.3222 == tωθThe rest of time

36

)(1

s87.1

22

11

22

2

22

11

2

2

1

121

θθωω

θωθ

ωθ

ωθ

+=+=

=+=+=

II

IIt

ttt

Then we have

10.14rad/srev/s23.32 ==ω


Chapter 10 Spin and orbital motionwdx.cxtc.edu.cn/physics/shuangyu/pdf/ch10.pdf2 10.1 Some concepts...

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Transcript of Chapter 10 Spin and orbital motionwdx.cxtc.edu.cn/physics/shuangyu/pdf/ch10.pdf2 10.1 Some concepts...