physics 111 lecture 11 angular momentum sj 8th ed.: chap ...janow/physics 111 spring...
TRANSCRIPT
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Physics 111 Lecture 11
Angular Momentum
SJ 8th Ed.: Chap 11.1 –11.4
•Recap and Overview
•Cross Product Revisited
•Torque Revisited
•Angular Momentum
•Angular Form of Newton’s Second Law
•Angular Momentum of a System of Particles
•Angular Momentum of a Rigid Body about a
Fixed Axis
•Conservation of Angular Momentum
11
.1
T
he
Ve
cto
r P
rod
uc
t a
nd
To
rqu
e
11
.2
A
ng
ula
r M
om
en
tum
: T
he
No
n-I
so
late
d S
ys
tem
11
.3
A
ng
ula
r M
om
en
tum
of
a R
ota
tin
g R
igid
Ob
jec
t
11
.4
C
on
se
rva
tio
n o
f A
ng
ula
r M
om
en
tum
: Is
ola
ted
Sys
tem
11
.5
T
he
Mo
tio
n o
f G
yro
sc
op
es
an
d T
op
s
Co
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So far: simple (planar) geometries
Ro
tati
on
al q
ua
nti
ties ∆
θ∆
θ∆
θ∆
θ,
ω,
α,
τ,,
ω,
α,
τ,,
ω,
α,
τ,,
ω,
α,
τ,etc
…re
pre
se
nte
d b
y s
ca
lars
Ro
tati
on
axis
sim
ply
C
CW
or
CW
2 d
ime
nsio
nal
pro
ble
ms,
rota
tio
n a
xis
perp
en
dic
ula
r to
pag
e
∑ ∑∑∑τ τττ
= ===τ τττ
θ θθθ= ===
τ τττ× ×××
= ===τ τττ
i all
in
et
ii
ii
ii
i
)sin
(F
r
Fr
rr
rr
r
Now: more powerful tool for 3D
“C
ross p
rod
uc
t”re
pre
sen
ts r
ota
tio
nal
qu
an
titi
es a
s v
ecto
rs:
•C
ross p
rod
ucts
po
int
alo
ng
in
sta
nta
neo
us a
xes o
f ro
tati
on
•D
irecti
on
s o
f ro
tati
on
axes c
an
be c
alc
ula
ted
an
d s
um
med
up
lik
eo
ther
vecto
rs, e.g
.,
An
gu
lar
mo
men
tum
-n
ew
co
nserv
ed
ro
tati
on
al q
uan
tity
.•
Lik
e l
inear
mo
men
tum
, i
t is
co
nserv
ed
fo
r is
ola
ted
sys
tem
s
•D
efi
nit
ion
:∑ ∑∑∑
= ===θ θθθ
= ==== ===
× ×××= ===
i all
i to
ti
ii
ii
ii
ii
i l
L
)sin
(p
r
l
vm
p
pr
lr
rr
rr
rr
Seco
nd
Law
in t
erm
s o
f co
nserv
ed
qu
an
titi
es:
•L
inear:
•R
ota
tio
nal:
0= ===
τ τττ⇒ ⇒⇒⇒
= ===τ τττ
⇒ ⇒⇒⇒α ααα
= ===τ τττ
ne
tif
co
ns
tan
t
is
ne
tn
et
L
dtL
d
Ir
rr
rr
r
0= ===
⇒ ⇒⇒⇒= ===
⇒ ⇒⇒⇒= ===
net
ifco
nsta
nt
isn
et
net
F
p
dtp
dF
am
Fr
rr
rr
r
Co
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t R
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an
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2012
Angular momentum –concepts & definition
How
muc
h linear
or r
otation
al “s
taying
power”
doe
s a m
oving
obje
ct h
ave
?
-Pr
oduc
t of
an
inert
ia m
easu
re w
ith a
spe
ed m
easu
re.
-
Linear
mom
ent
um:
p = m
v(linear).
-If
obje
ct is
rota
ting
only:
ang
ular
mom
ent
um
-mom
ent
of
inert
ia x
ang
ular
velocity
= I
ω ωωω= L
-line
ar
mom
ent
um x
mom
ent
arm
abou
t so
me a
xis f
or s
imple c
ase
s
inertia
speed
linearrotational
m vI ω ωωω
p=mv
L=I
ω ωωωlinear
momentum
rigid body
angular
momentum
L =
the a
ngular m
om
entu
m o
f a rigid
body relative to a
selecte
d a
xis a
bout
which I a
nd
ω ωωωare
measure
d:
•units: [k
g.m
2/s
] ω ωωω
≡ ≡≡≡r
rI
L
ω ωωωr
Lr
Why bother with angular momentum? It’s conserved for isolated systems.
Co
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t R
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2012
Angular momentum –concepts & defining examples
Example: Angular momentum of a rotating hoop about symmetry axisthrough “P”
same as a point particle, for hoop
L =
linear
mom
ent
um X
mom
ent
arm
ω ωωω= ===
IL
p
rm
vr
mr
L= ===
= ===ω ωωω
= ===2
If it’s a hoop:
rv
m
rI
ω ωωω= ===
= ===2
If it’s a disc:
ω ωωω= ===
= ===2
212
21m
rL
m
rI
P
vr
rr
Example: Angular momentum of a point mass moving in a straight line
choose point P as a rotation axis
L =
linear
mom
ent
um X
mom
ent
arm
Note: L = 0 if moment arm =0
L is the same for object at any point along line of v
⊥ ⊥⊥⊥⊥ ⊥⊥⊥
⊥ ⊥⊥⊥⊥ ⊥⊥⊥
ω ωωω= ===
= ==== ===
ω ωωω= ===
ω ωωω= ===
rv
pr
mvr
mr
IL
2
P
vr⊥ ⊥⊥⊥rr
Example: Same as above, but with velocity not perpendicular to r; v raddoes not affect L
r/v
pr
)s
in(
mv
rr
mv
mr
IL
⊥ ⊥⊥⊥⊥ ⊥⊥⊥
= ===ω ωωω
× ×××= ===
φ φφφ= ===
= ===ω ωωω
= ===ω ωωω
= ===r
r2
L =
linear
mom
ent
um X
mom
ent
arm
Note: L = 0 if v is parallel to r (radiallyin or out)
P
⊥ ⊥⊥⊥vr
rr
φ φφφvr
Co
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2012
Example: Calculating
Ang
ular
Mom
ent
um f
or a
Rigid B
ody
Calculate
the a
ngular
mom
ent
um o
f a 1
0 k
g disc
when:
kg
mc
m
s
, /
ra
d
10
m
,
0.0
9
9 r
320
ω= ===
= ==== ===
= ===
Rot
ation
axis is
norm
al to
disc
thro
ugh its
CM
Solut
ion:
dis
k
a fo
r2
21
rm
I
I
L
= ===ω ωωω
= ===
24
21
02
31
09
+ +++− −−−
× ×××× ×××
× ×××× ×××
× ×××= ===
ω ωωω= ===
.1
0
rm
L21
2
21
s/
2m
Kg
13
12.9
6
L
≈ ≈≈≈
= ===
What
ang
ular
speed w
ould a
10 k
g SOLID
SPH
ERE (sa
me d
imens
ions
) have
if
it’s a
ngular
mom
ent
um is
the s
ame a
s abov
e?
kg
mc
m
/s2
m K
g
10
m
,
0.0
9
9 r
,
12.9
6
L
= ==== ===
= ==== ===
sp
he
re
a fo
r2
52
rm
I
I /
L
= ==== ===
ω ωωω
41
0− −−−
× ×××× ×××
× ×××× ×××
= ===ω ωωω
2
259
10
12
.96
Solut
ion:
s
/
rad
00
4
= ===ω ωωω
Co
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Definition: Angular momentum of a single particle
•Extension of linear momentum
•Depends on chosen rotation axis (here along z)
•“Moment of momentum”
vm
ps
= ===
Same picture as for torques
Use moment arm = r sin(
θ θθθ)….or...
…tangential momentum component = p sin(
θ θθθ)
Only the tangential momentum component
contributes
r and ptail-to-tail always form a plane
Lis perpendicular to that plane
⊥ ⊥⊥⊥r
)v
r(m
pr
L
mo
men
tum
lin
ear
x
arm
mo
men
t
rr
rr
r
× ×××= ===
× ×××≡ ≡≡≡= ===
)sin
(
pr
r
p
pr
Lθ θθθ
= ==== ===
= ===⊥ ⊥⊥⊥
⊥ ⊥⊥⊥
⊥ ⊥⊥⊥p
x
y
z
θ θθθ
⊥ ⊥⊥⊥r
⊥ ⊥⊥⊥p
rad
p
r
θ θθθ
line of action
of momentum p
p
P
θ θθθ
90
o
90
o
moment arm
for p
L
Convention:
vector up out
of paper
vector down into
paper (tail)
Co
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2012
Example: Ang
ular
mom
ent
um o
f a p
art
icle in
unifor
m c
ircu
lar
mot
ion
•T
he a
ng
ula
r m
om
en
tum
vec
tor
po
ints
o
ut
of
the
dia
gra
m
•T
he m
ag
nit
ud
e i
s
L=
mvr
sin
(90
o)=
mvr
–s
in (
90
o)is
us
ed
sin
ce v
is
pe
rpe
nd
icu
lar
to r
•A
part
icle
in
un
ifo
rm c
ircu
lar
mo
tio
n h
as
a c
on
sta
nt
an
gu
lar
mo
men
tum
ab
ou
t an
axis
th
rou
gh
th
e c
en
ter
of
its p
ath
•E
xam
ple
s:
sa
tell
ites i
n c
ircu
lar
orb
its
pr
L
L
...
LL
L i
all
ii
i all
i n
2 1
net
∑ ∑∑∑∑ ∑∑∑
× ×××= ===
= ===+ +++
+ ++++ +++
= ===r
rr
vr
r
Superposition: Angular Momentum of a System
p
r - p
r
|L|
22
11
net
⊥ ⊥⊥⊥⊥ ⊥⊥⊥
+ +++= ===
rfo
r th
is c
ase
:
O
Co
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Example: calculating angular momentum for particles
PP
10602-2
3*:
Tw
o o
bje
cts
are
movin
g a
s s
ho
wn in t
he f
igure
. W
hat
is t
heir t
ota
l
angu
lar
mom
entu
m a
bout
po
int
O?
m2
m1
No need to use formal coordinate
system for such a simple problem
For each particle:
mv
rp
rl
⊥ ⊥⊥⊥⊥ ⊥⊥⊥
= ==== ===
Where is the moment arm
⊥ ⊥⊥⊥r
Momentaand displacements from “O”all lie in the plane of the slide,
so angular momentaare perpendicular to the slide (in or out)
z)
(a
lon
gx
xx
x
21
.45
-
31
.25
2.2
6.5
1.5
-
3.6
3.1
2.8
l
l
L= ===
= ===+ +++
= ===2
1
rr
r
CC
W
/sk
g.m
9.8
L
2= ===
Co
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�11.1
. A
car
of
mass 1
000 k
g m
ove
s w
ith
a s
peed
of
50 m
/so
n a
cir
cu
lar
track o
f ra
diu
s 1
00 m
. W
hat
is t
he m
ag
nit
ud
e o
f it
s a
ng
ula
r m
om
en
tum
(i
n k
g •
m2/s
) re
lati
ve t
o t
he c
en
ter
of
the r
ace t
rack (
po
int “P”)
?
Angular momentum for a car
A)
0
B)
5.0
× ×××10
6
C)
2.5
× ×××10
4
D)
2.5
× ×××10
6
E)
5.0
× ×××10
3
)sin
(
pr
r
p
pr
L
θ θθθ= ===
= ==== ===
⊥ ⊥⊥⊥⊥ ⊥⊥⊥
�11.2
. W
hat
wo
uld
th
e a
ng
ula
r m
om
en
tum
ab
ou
t p
oin
t “P”
be i
f th
e c
ar
lea
ves t
he t
rack a
t “A”
an
d e
nd
s u
p a
t p
oin
t “B”
wit
h t
he s
am
e v
elo
cit
y ?
P
A
B
Co
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The Rotational Second Law and Angular Momentum
�The f
orce
caus
es
the linear
mom
ent
um t
o ch
ang
e.
�The n
et
forc
e a
cting
on a
bod
y is
the t
ime r
ate
of
chang
e o
f it’s linear
mom
ent
um
Linear
mot
ion:
App
ly f
orce
to
a p
art
icle.
dtp
dF
ne
t
rr
= ===
an
dτ
∑Lr
r�
to b
e m
easu
red
ab
ou
t th
e s
am
e o
rig
in
�T
he o
rig
in s
ho
uld
no
t b
e a
ccele
rati
ng
(sh
ou
ld b
e a
n in
ert
ial
fram
e)
Rot
ation
al mot
ion:
App
ly t
orqu
e t
o a r
igid b
ody.
�The t
orqu
e c
aus
es
the a
ngular
mom
ent
um t
o ch
ang
e.
�The n
et
torq
ue a
cting
on a
bod
y is
the t
ime r
ate
of
chang
e o
f it’s a
ngular
mom
ent
um.
dtL
d
net
rr
= ===τ τττ
�The a
ngular
mom
ent
um is
cons
tant
if
the n
et
torq
ue =
zero
.
�The m
oment
um is
cons
tant
if
the n
et
forc
e =
zero
.
Co
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Rotational Second Law applies to single particles,
systems, and rigid bodies:
BU
T…
inte
rnal
torq
ues i
n t
he s
um
can
ce
l in
3rd
law
pair
s.
On
ly E
xte
rnal
To
rqu
es c
on
trib
ute
to
Ls
ys
net
exte
rnalto
rque
on
the s
yst
em
net
iext
,isys
d
t
Ld
τ τττ= ===
τ τττ= ===∑ ∑∑∑
r
r •S
am
e r
efe
ren
ce
ax
is f
or
L’s
an
d τ τττ
’s•
Ho
lds
ab
ou
t a
ny r
ota
tio
n a
xis
, a
nd
if
I is
ch
an
gin
g•
Mu
st
us
e m
as
s c
en
ter
as
ori
gin
if
it (
cm
) is
ac
ce
lera
tin
g(o
the
rwis
e n
on
-in
ert
ial
eff
ec
ts s
ho
w u
p)
•τ τττ i
= n
et
torq
ue
on
pa
rtic
le “
i”•
Inte
rna
l to
rqu
e p
air
s a
re i
nc
lud
ed
i
sys
LL
∑ ∑∑∑= ===
rr
•a
ll a
bo
ut
sa
me
ori
gin
∑ ∑∑∑∑ ∑∑∑
τ τττ= ===
= ===∴ ∴∴∴
ii
isys
dtL
d
dt
Ld
r
rr
dtL
di
i
:b
od
y
sin
gle
a
Fo
r
r
v= ===
τ τττ
To
tal
an
gu
lar
mo
men
tum
of
a s
yste
mo
f b
od
ies:
Syste
m is c
alled
“IS
OL
AT
ED
”if
fn
et
torq
ue o
n it
is z
ero
……
then
an
gu
lar
mo
men
tum
of
syste
m is c
on
sta
nt
Co
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COMPARISON
Translation
For
ce
Linear
Mom
ent
um
Kinetic
Ene
rgy
Fr
vm
pr
v= ===
2
21m
vK
= ===
Rotation
Tor
que
Ang
ular
Mom
ent
um
Kinetic
Ene
rgy
Fr
rr
r× ×××
= ===τ τττ
pr
lr
rv
× ×××= ===
2
21ω ωωωΙ ΙΙΙ
= ===K
Linear
Mom
ent
umcm
iv
Mp
Pr
rr
= ==== ===∑ ∑∑∑
Seco
nd
Law
dtP
dF n
et
rr
= ===
Ang
ular
Mom
ent
umi
ii
L
L∑ ∑∑∑
∑ ∑∑∑ω ωωω
Ι ΙΙΙ= ===
= ===r
rr
for
rig
id b
od
ies
ab
ou
t c
om
mo
n f
ixe
d a
xis
Seco
nd
Law
dt
Ld
sys
net
r
v= ===
τ τττ
Momentum conservation -for closed, isolated systems
Systems and Rigid Bodies
co
nsta
nt
P
sys
= ===r
co
nsta
nt
L
sys
= ===r
Ap
ply
se
pa
rate
ly t
o x
, y,
z a
xe
s
Co
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Example: A non-isolated system
Ma
ss
es
co
nn
ec
ted
by a
lig
ht
co
rd
Fin
d t
he
lin
ea
r a
cc
ele
rati
on
a.
•
Us
e a
ng
ula
r m
om
en
tum
ap
pro
ac
h•
No
fri
cti
on
be
twee
n m
2a
nd
ta
ble
•M
ec
ha
nic
al
en
erg
y i
s c
on
sta
nt
(no
no
n-
co
ns
erv
ati
ve
fo
rce
s),
bu
t w
e i
gn
ore
th
at
•B
loc
k,
pu
lle
y a
nd
sp
he
re a
re n
on
-is
ola
ted
sys
tem
in
th
at
ne
t to
rqu
e i
s n
ot
ze
ro.
C
on
str
ain
ts:
ar
ar
dv/d
t
R
a
/d
td
Rv
masses
F
or
pu
lley
F
or
sp
here
an
db
lock
fo
r
sa'
an
d
sv'
Eq
ual
= ===α ααα
= ===
ω ωωω= ===
α αααω ωωω
= ===
•Ig
no
re i
nte
rna
l fo
rce
s,
co
ns
ide
r e
xte
rna
l fo
rce
s o
nly
•N
et
ex
tern
al
torq
ue
on
sys
tem
:
•A
ng
ula
r m
om
en
tum
of
sys
tem
:(n
ot
co
ns
tan
t)ω ωωω
+ ++++ +++
= ===I
vR
m
vR
m
Ls
ys
21
•U
se
Se
co
nd
La
w (
dif
fere
nti
ate
):
gR
m )
Rm
Rm
(I I
aR
m
aR
m
dt
Ld
ne
t2
1s
ys
1
22
21
= ===τ τττ
= ===α ααα
+ ++++ +++
= ===α ααα
+ ++++ +++
= ===
w
he
el
of
c
en
ter
ab
ou
t
n
et
gR
m1
= ===τ τττ
)R
mR
m(I
gR
m
21
221
+ ++++ +++
= ===α ααα
∴ ∴∴∴sa
me r
esu
lt f
ollowed f
rom e
arlier
meth
od
using
3 F
BD’s
& 2
ndlaw
I
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Angular Momentum Conservation…
…for systems and rigid bodies
sta
te
init
ial
is
i""
fin
al,
is
f""
wh
ere
if
L
L
rr
= ===
•L
is c
on
serv
ed
sep
ara
tely
fo
r x,
y,
z d
irecti
on
s•
An
gu
lar
mo
men
tum
co
nserv
ati
on
is a
s i
mp
ort
an
t as e
nerg
y a
nd
lin
ear
mo
men
tum
co
nserv
ati
on
L
L f
ffi
na
lfi
na
l0
init
ial
init
ial
ω ωωωΙ ΙΙΙ
= ==== ===
ω ωωωΙ ΙΙΙ
= ===∑ ∑∑∑
∑ ∑∑∑0
Typ
ical p
rob
lem
s m
ay in
vo
lve...
•Is
ola
ted
syste
ms,
so
∆ ∆∆∆
L=
0
•R
igid
bo
die
s r
ota
tin
g a
rou
nd
a c
om
mo
n r
ota
tio
n a
xis
•S
om
e e
ven
t ch
an
ges m
om
en
t o
f in
ert
ia,
sh
ap
e,
co
llis
ion
…
Wh
en
so
me e
ven
t ch
an
ges t
he s
tate
of
an
iso
late
d s
yste
m:
Wh
en
th
e n
et
exte
rnal to
rqu
e o
n a
syste
m is z
ero
:
net
dtL
d0
= ===τ τττ
= ===
r
co
nsta
nt
L
= ===r
syst
em is
“iso
late
d”
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: A
part
icle o
n a s
tring
rota
ting
aro
und a
peg
Str
ing
wra
ps a
rou
nd
peg
as m
ass
sp
irals
in
ward
….W
hat
yo
u s
ee:
•R
ad
ius
r d
ec
rea
se
s
•A
ng
ula
r s
pe
ed
ω ωωωin
cre
as
es
•T
an
ge
nti
al
sp
ee
d v
ch
an
ge
s
Init
ial co
nd
itio
ns:
At
t =
0:
m
= 0
.2 k
g,
v0
=1
m/s
,
r 0=
0.5
m,
ω ωωω0
= v
0/r
0=
2 r
ad
/s
Fin
d:
ω ωωωf, v
fw
hen
rad
ius r
f=
0.2
m
m
v0
Meth
od
:•
Sys
tem
is
is
ola
ted
(W
hy?
)•
Ce
ntr
ipe
tal
forc
e o
n m
ex
ert
s z
ero
to
rqu
e a
bo
ut
pe
g,
sin
ce
r X
F=
0,
so
…•
Re
pre
se
nt
an
gu
lar
mo
me
ntu
m:
s
inc
ep
) r
(
p r
pr
L⊥ ⊥⊥⊥
= ===× ×××
= ===r
rr
rv
m
v,
pω ωωω
= ==== ===
2
I
r
m
L
ω ωωω= ===
ω ωωω= ===
∴ ∴∴∴
•A
ng
ula
r m
om
en
tum
is
co
ns
tan
t (c
on
se
rve
d)
as
co
rd w
rap
s u
p:
ch
an
ge
s)
in
ert
ia
of
(mo
me
nt
f
ff
00
I
L
I
L
ω ωωω= ===
= ===ω ωωω
= ===∴ ∴∴∴
0Theme:
L is
cons
tant
while m
oment
of
inert
ia c
hang
es,
( ((() )))
)
(
ra
d/s
12.5
x
m
r
mr
II
0
mu
ch
.2.5
0
f
0
ff
ω ωωω> >>>
= ==== ===
ω ωωω= ===
ω ωωω= ===
ω ωωω2
2
22 00
)v
(
m
/s
2.5
r
v
0m
uch
f f
f> >>>
= ===ω ωωω
= ===
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Restr
icti
on
:F
or
rota
tio
n o
f a r
igid
bo
dy a
rou
nd
a
SY
MM
ET
RY
AX
IS,
or
rota
tio
n o
f a f
lat
bo
dy i
n t
he x
-yp
lan
e a
bo
ut
z-a
xis
ω ωωω= ===
rr
IL
Examples:
L=I
ω ωωωve
rsus
L=rx
pfo
r 2 d
imens
iona
l mot
ion
•S
ing
le p
oin
t m
as
s•
Cir
cu
lar
pa
th i
n x
-yp
lan
e•
Fin
d L
ab
ou
t c
en
ter
of
pa
th
x
y
z
vrrr
ω ωωωvv,
Lk̂ )
sin
(90
m
vr
v
r m
p
r L
o= ===
× ×××= ===
× ×××= ===
rr
rr
r
r
vω ωωω
= ===
ω ωωω= ===
ω ωωω= ===
ω ωωω= ===
rr
I
k̂ I
k̂ m
r
L2
Isc
aler
vect
or
L &
ω ωωωbot
h
along
Z-axis
For
L=I
ω ωωω
To
wor
k:•
Ne
ed
an
oth
er
eq
ua
l m
as
s s
ym
me
tric
all
y a
cro
ss
fro
m m
, s
am
e m
oti
on
•
Ad
ds
to
Lz,
bu
t c
an
ce
ls c
om
po
ne
nt
of
L n
orm
al
to z
•Same m
otion:
circu
lar
path
para
llel to
x-y
plane
Shift
Origin:
| p
r|
LI
xy
z
so
....
rr
× ×××= ===
= ===ω ωωω
x
y
z
pr
rrω ωωωv
Lv
xy
rr
k̂
p
pla
ne
,
y-
x to
ara
lle
lp
pla
ne
in
isω ωωω
= ===ω ωωωr
r
pk̂
mr
pr
L
k̂L
L
zx
yx
yz
rr
rr
r× ×××
+ +++× ×××
= ===+ +++
= ===
along
z
in x
-y
plane
•A
ng
ula
r m
om
en
tum
ha
s c
om
po
ne
nts
alo
ng
an
d n
orm
al
to ω ωωω
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
�11.3
. A
bo
wlin
g b
all i
s r
ota
tin
g a
s s
ho
wn
ab
ou
t it
s m
ass c
en
ter
axis
.
Fin
d i
t’s a
ng
ula
r m
om
en
tum
ab
ou
t th
at
axis
, in
kg
.m2/s
Angular momentum of a bowling ball
A)
4
B)
½C
) 7
D)
2
E
) ¼
I
L
ω ωωω= ===
ω=
4 r
ad/s
M =
5 k
gr
= ½
mI =
2/5
MR
2
�11.4
. S
up
po
se t
he r
ota
tio
n a
xis
is s
hif
ted
to
b
e t
an
gen
t to
th
e s
ph
ere
an
d p
ara
llel
to t
he z
–
axis
in
th
e p
ictu
re.
Fin
d t
he a
ng
ula
r m
om
en
tum
ab
ou
t th
at
axis
, in
kg
.m2/s
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Demonstration: Spinning Professor
co
nsta
nt
axis
- z
ab
ou
t n
et
L
= ===⇒ ⇒⇒⇒
= ===τ τττ
rr
0
ff
fin
al
0in
itia
l
I
I
L ω ωωω
= ===ω ωωω
= ===∑ ∑∑∑
∑ ∑∑∑0
r Moment of inertia changes
Isolated
System
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
How fast should the professor spin?
L is
cons
tant
…while m
oment
of
inert
ia c
hang
es,
Th
e p
rofe
sso
r is
ro
tati
ng
(n
o f
ricti
on
) w
ith
an
gu
lar
sp
eed
1.2
re
v/s
.
•A
rms a
re o
uts
tretc
hed
wit
h a
bri
ck i
n e
ach
han
d.
•T
he m
om
en
t o
f in
ert
ia o
f th
e s
yste
m c
on
sis
tin
g o
f th
e p
rofe
sso
r, t
he
bri
cks,
an
d t
he p
latf
orm
ab
ou
t th
e c
en
tral
axis
is 6
.0 k
g·m
2.
By l
ow
eri
ng
th
e b
ricks t
he m
om
en
t o
f in
ert
ia o
f th
e s
yste
m d
ecre
as
es t
o 2
.0
kg
·m2.
(a)
what
is t
he r
esu
lting
ang
ular
speed o
f th
e p
latf
orm?
(b)
what
is t
he r
atio
of t
he s
yst
em’s n
ew k
inetic
ene
rgy t
o th
e o
rigina
l kine
tic
ene
rgy?
(c)
In
part
(b), w
hat
acc
ount
s fo
r th
e d
iffe
renc
e (added K
E) ?
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
ff
0
a
xis
fi
xe
d
aa
bo
ut
...
init
ial
fin
al
to
rqu
e
ex
tern
al
Z
ero
I
I
L
L
L
L
ω ωωω= ===
ω ωωω= ===
= ==== ===
⇒ ⇒⇒⇒
0
r
L is
cons
tant
…while m
oment
of
inert
ia c
hang
es,
How fast should the professor spin?
I 0=
6 k
g-m
2
ω ωωω0
= 1
.2 r
ev/s
I f=
2 k
g-m
2
ω ωωωf=
? r
ev/s
Solut
ion
(a):
rev/s
ra
d/s
6
fI 0I
f
3.6
1.2
x
= ===
= ===ω ωωω
= ===ω ωωω
20
KE h
as
incr
ease
d!!
Solut
ion
(b):
30
2 00
2 022
0
2 00
2
0
II
I
I
II
KK
f
fII
f
21
ff
21
f= ===
= ===ω ωωω
ω ωωω
= ===ω ωωωω ωωω
= ===
Solut
ion
(c):
The e
xtr
a K
E c
ame f
rom w
ork
don
e w
hen
pulling
the w
eights
in
and
/or
reduc
ing
pote
ntial ene
rgy b
y low
ering
the a
rms.
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Controlling spin (
ω ωωω) by changing I (moment of inertia)
In
the a
ir,
τ τττ net
= 0
L is
cons
tant
ff
0
II
Lω ωωω
= ===ω ωωω
= ===0
Chang
e I
by c
urling
up
or s
tretc
hing
out
-sp
in r
ate
ω ωωωmus
t adju
st
Moment of inertia changes
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Internal torques do not change total angular momentum...
... it is redistributed within the isolated system
co
nsta
nt
sys
axis
- zab
ou
t n
et
L
= ===⇒ ⇒⇒⇒
= ===τ τττ
rr
0
Inte
rnal
torq
ues n
ot
0th
ey r
eve
rse L
wh w
heel
fin
al
pro
f,L
2L
rr
= ===
Sp
acecra
ft m
an
eu
ve
rs b
y
sp
inn
ing
th
e f
lyw
heel
the c
raft
co
un
ter-
rota
tes
flyw
heel
cra
ft
tot
LL
0 L
rr
r+ +++
= ==== ===
Demonstration
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
�11.5
. T
wo
astr
on
au
ts e
ach
wit
h m
ass M
are
co
nn
ecte
d b
y a
ma
ssle
ss
rop
e o
f le
ng
th d
. T
hey a
re i
so
late
d i
n s
pace,
orb
itin
g t
heir
cen
ter
of
mass a
t id
en
tical
sp
eed
s v
.
On
e o
f th
em
pu
lls o
n t
he r
op
e,
sh
ort
en
ing
th
e d
ista
nce b
etw
een
th
em
to
d
/2.
Wh
at
are
th
e n
ew
to
tal an
gu
lar
mo
men
tum
L’an
d s
peed
s v’?
Tethered Astronauts
A)
L’=
mvd
/2,
v’=
v/2
B)
L’=
mvd
, v
’=
2v
C)
L’=
2m
vd
, v
’=
vD
)L
’=
2m
v’d
, v
’=
v/2
E)
L’=
mvd
, v
’= v
/2
I
L
ω ωωω= ===
rv
m L
T
= ===
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Conservation of Angular Momentum:
Merry-Go-Round Problems
•T
he m
om
en
t o
f in
ert
ia o
f th
e
syste
m=
–
the m
om
en
t o
f in
ert
ia o
f th
e
pla
tfo
rm p
lus…
–th
e m
om
en
t o
f in
ert
ia o
f th
e
pers
on
(a p
art
icle
).
•A
s t
he p
ers
on
mo
ve
s t
ow
ard
th
e c
en
ter
of
the r
ota
tin
g
pla
tfo
rm t
he m
om
en
t o
f in
ert
ia d
ecre
ases.
•T
he a
ng
ula
r sp
eed
mu
st
incre
ase s
inc
e t
he a
ng
ula
r m
om
en
tum
is c
on
sta
nt.
or
•A
pers
on
ru
ns t
an
gen
t to
th
e
ed
ge o
f th
e p
latf
orm
an
d t
hen
ju
mp
s o
n.
•T
he t
ota
l an
gu
lar
mo
men
tum
o
f th
e s
yste
m (
pers
on
+
pla
tfo
rm)
is c
on
sta
nt.
•T
he a
ng
ula
r velo
cit
y o
f th
e
pla
tfo
rm c
han
ges.
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: A m
err
y-go
-ro
und p
roblem
A 40-kg child running at 4.0 m/sjumps tangentially onto a stationary
circular merry-go-round platform whose radius is 2.0 m and whose
moment of inertia is 20 kg-m2. There is no friction.
a)Find the angular velocity of the platform after the child jumpson.
b)Find the change in the total kinetic energy of the system.
c)Find the change in the kinetic energy of the child alone.
Mo
men
t o
f in
ert
ia c
han
ges
To
tal
L i
s c
on
sta
nt
ff
0
tot
I
I
L ω ωωω
= ===ω ωωω
= ===∑ ∑∑∑
∑ ∑∑∑0
r
•S
ys
tem
= c
hild
+ m
err
y-g
o r
ou
nd
, b
efo
re a
nd
aft
er
co
llis
ion
•S
ys
tem
is
iso
late
d (
zero
ne
t exte
rnal to
rqu
e).
•In
ela
sti
c c
olls
ion
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: A m
err
y-go
-ro
und p
roblem-
Solut
ion
ff
0
tot
I
I
L ω ωωω
= ===ω ωωω
= ===∑ ∑∑∑
∑ ∑∑∑0
r
I =
20 k
g.m
2
vT
= 4
.0 m
/sm
c=
40 k
gr
= 2
.0 m
ω ωωω0
= 0
fto
tT
c
0
tot
I
rv
m L
ω ωωω= ===
= ===+ +++
2r
m I
I
c
tot
+ +++= ===
4 x
40
20
2 x
4 x
40
r
m I
rv
m
cTc
f
+ +++= ===
+ +++= ===
ω ωωω2
rad
/s
1.7
8
a)
f
= ===ω ωωω
∴ ∴∴∴
0320
4284
22
..
vm
I
K
KK
T
c21
fto
t21
0
f− −−−
= ===− −−−
ω ωωω= ===
= ===∆ ∆∆∆
− −−−
Jo
ule
s
.K
b)
6
35
− −−−= ===
∆ ∆∆∆∴ ∴∴∴
22
2
Tc
21f
cm
210
cf
ch
ild
vm
r
K
KK
− −−−
ω ωωω= ===
= ===∆ ∆∆∆
− −−−
Jo
ule
sch
ild
.
K
c)
5
66
− −−−= ===
∆ ∆∆∆∴ ∴∴∴
inelast
ic c
ollision
KE d
ecr
ease
s