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.
§10.1 Some concepts about rotation
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
§10.1 Some concepts about rotation
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
§10.1 Some concepts about rotation
4
mvrLvmrprL
=×=×=
rrrrr
rvrv
ωω
=×=rrr rr
pr
o m
ωr
ωω 22 mrrmL ==then
(b) Moment of inertia of a particleωr
r2mrL =
§10.1 Some concepts about rotation
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
§10.2 The time rate of change of angular momentum and torque
rrFr
θ
om
dr =⊥
⊥F
If ,thecross
,0,or0
total
OFFrr
== πθ
0total =τr
§10.2 The time rate of change of angular momentum and torque
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.
§10.2 The time rate of change of angular momentum and torque
§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.
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
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
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
8
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
∑∑∑
∑
⊥
⊥
⊥⊥
⊥
′+
′−=
′××′+
′××′=
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 )(
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
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
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
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
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
(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:
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
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
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
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:
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
L
mdo
xx
23
0
222
31
031
ddd
mLL
xLm
xLmxmxmrI
L
==
=== ∫ ∫ ∫
§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
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§10.3 The angular momentum of a system of particles and moment of inertia of rigid body
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
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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
+=
′×′+×= ∑
Summary of the previous sections
( ) ( )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 == ∑ ⊥
Summary of the previous sections
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
Summary of the previous sections
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
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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
§10.4 The dynamics of rigid body with a fixed axis
∑∑∑ =×==×=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
===
§10.4 The dynamics of rigid body with a fixed axis
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.
§10.4 The dynamics of rigid body with a fixed axis
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
=
=−
=−
++
§10.4 The dynamics of rigid body with a fixed axis
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+
=
§10.4 The dynamics of rigid body with a fixed axis
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:
§10.5 The angular impulse-momentum theoremand The procession of a rapidly spinning top
We can get
1222
21
2021210211
2 IRIRRIRRI
++
=ωω
ω1
222
21
2021210221
1 IRIRRRIRI
++
=ωω
ω
12 ffrr
−=
1fr
222
111
RfRf
=−=
ττ
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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
§10.5 The angular impulse-momentum theoremand The procession of a rapidly spinning top
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
§10.5 The angular impulse-momentum theoremand The procession of a rapidly spinning top
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
§10.5 The angular impulse-momentum theoremand The procession of a rapidly spinning top
φ
φ
Crr
Crr
φ
φθd
Lr
∆
Lr
∆
3. The angular speed of the procession
φθ
θφ
sindd
dsind
spin
spin
LL
LL
=
=
§10.5 The angular impulse-momentum theoremand The procession of a rapidly spinning top
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
§10.5 The angular impulse-momentum theoremand The procession of a rapidly spinning top
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
§10.6 Simultaneous spin and orbital motion
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?
§10.6 Simultaneous spin and orbital motion
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++
=++
=
§10.6 Simultaneous spin and orbital motion
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
§10.6 Simultaneous spin and orbital motion
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
§10.6 Simultaneous spin and orbital motion
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.
§10.6 Simultaneous spin and orbital motion
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
§10.6 Simultaneous spin and orbital motion
(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
§10.6 Simultaneous spin and orbital motion
+ =
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
§10.6 Simultaneous spin and orbital motion
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
§10.6 Simultaneous spin and orbital motion
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
−×−=∑
§10.6 Simultaneous spin and orbital motion
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
§10.6 Simultaneous spin and orbital motion
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
§10.6 Simultaneous spin and orbital motion
§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
§10.7 Conservation of angular momentum
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 =
§10.7 Conservation of angular momentum
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
§10.7 Conservation of angular momentum
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.
§10.7 Conservation of angular momentum
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
§10.7 Conservation of angular momentum
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?
§10.7 Conservation of angular momentum
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, ==
§10.7 Conservation of angular momentum
o60cos34
34 222
iiff mdvMdMdMd −=+ ωωω
rad/s80.0)60cos4(51
−=−= oiif vd
dωω
iball,its,fball,fts, LLLLrrrr
+=+
+
§10.7 Conservation of angular momentum
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.
§10.7 Conservation of angular momentum
Solution:
22
31
2
mdMd
ghmd
+=ω
§10.7 Conservation of angular momentum
)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.
§10.7 Conservation of angular momentum
§10.7 Conservation of angular momentum
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