Chapter 11: Rotational Vectors and Angular Momentum Vector (cross) products efinition of vector product and its properties Axis of rotation a b b a sin ˆ b a n b a n ˆ unit vector normal to the plane defined by a and b a b b a 0 a a a a
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Chapter 11: Rotational Vectors and Angular Momentum Vector (cross) products Definition of vector product and its properties Axis of rotation unit vector.
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Chapter 11: Rotational Vectors and Angular Momentum
Vector (cross) products Definition of vector product and its properties
Axis of rotation
a
b
ba
sinˆ banba
n
unit vector normal to the planedefined by a
and b
abba
0
aaaa
Vector (cross) products (cont’d)
Properties of vector product
kji ˆˆˆ
x
y
z
i
jk
kji ˆ,ˆ,ˆ : unit vector in x,y,z direction
ikj ˆˆˆ jik ˆˆˆ Consider three vectors:
),,( zyx aaaa
),,( zyx bbbb
),,( zyx cccc
Then
kbaba
jbabaibaba
kbjbibkajaiabac
xyyx
zxxzyzzy
zyxzyx
ˆ)(
ˆ)(ˆ)(
)ˆˆˆ()ˆˆˆ(
Vector (cross) products (cont’d)
Properties of vector product (cont’d)
x
y
z
i
jk
yzzyx babac zxxzy babac
xyyxz babac
bac
zyx
zyx
bbb
aaa
kji ˆˆˆ
Torque
Case for 1 point-like object of mass m with 1 force
x
y
r
m
F
r
massless rigid rod
• r is constant only tangential componentof causes rotation F
FFt
mr
rrrm
amF
ˆ)ˆˆ(
ˆˆ
2
0
I
mrrFt
2
torque moment of inertiaunit Nm
tF
Torque is a quantitative measure of the tendency of a force to causeor change the rotational motion.
Torque
Case for 1 point-like object of mass m with 1 force
x
y
r
F
r
sinrrt lever arm
Fr
FrFrrF tt
sin
Define Fr
is aligned with the rotation axis
tF
Torque (cont’d)
Case for 2 point-like objects with 2 forces
21 1r 2r
tF2
tF1
2211 rFrF ttnet
0 0
increases
decreases
x
y
Example: A see-saw in balance
M mR r
Mg mg
02211
mgrMgR
rFrF ttnet
rRMm //
Work & energy (I)
A massive body on massless rigid rod
x
y
r
m
tF Work done by the force:
drdFdsFW tt
Also
ddtdIdIW )/(
ddtdddI )/)(/(
1
0
1
0
|)2/1( 2
IdI
KII 20
21 )2/1()2/1(
W=K
Work & energy (I) (cont’d)
Power in rotational motion
Work done by the force:
drdFdsFW tt
ddW
dtddtdW //
PPower :
Correspondence between linear & angular quantities
linear angular
displacement
velocity
acceleration
mass
force
Newton’s law
kinetic energy
work
x dtdxv / dtd / dtdva / dtd /
m 2iirmI
F
ImaF Fr
2)2/1( mvK 2)2/1( IK
FdxW dW
Work & energy (II)
A massive body in rotational & translational motion
Kinetic energy: iii vvmK
)2/1(
Vvv ii
'
Now where 'iv
is the velocity with respect to
V
the center of mass (COM) and the velocity of COM. Then
22'
'22'
'22'
2'2'
''
)2/1()2/1(
/)()2/1()2/1(
/)2/1()2/1(
)2/1()2/1(
)()()2/1(
)2/1(
MVvm
dtrmdVMVvm
dtrdmVmVvm
VmvVmvm
VvVvm
vvmK
ii
iiii
iiiii
iiiii
iii
iii
0
com
'iv
V
Work & energy (II) cont’d
A massive body in rotational & translational motion (cont’d)
22' )2/1()2/1(
)2/1(
MVvm
vvmK
ii
iii
2''' ; iiii rmIvr
22 )2/1()2/1( IMVK
kinetic energy due totranslational motion
kinetic energy dueto rotational motionabout a rotation axisthrough COM
Work & energy (II) cont’d
A massive body in rotational & translational motion (cont’d)
Example: A rolling cylinder (without sliding due to friction)
P P
O O
s
s
COMv
COMv
Rsx
RdtRd
dtdsdtdxvCOM
/
//
View point 1:
COMvv
COMvv
COMvv
COMvv
COMv
COMvv2
0 COMCOM vvv
+ =O O O
COMv
This view point was used in the last few slides
R
rota
tion
tra
nsla
tion
Work & energy (II) cont’d
A massive body in rotational & translational motion (cont’d)
Example: A rolling cylinder (without sliding) (cont’d)
View point 2: Any point in the cylinder rotates around P
rotation axisP
2)2/1( PIK From the parallel axis theorem
2MRII COMP
22
222
)2/1()2/1(
)2/1()2/1(
COMCOM
COM
MvI
MRIK
rotational translational
RvCOM
Angular momentum & torque
x
y
r
p
r
sinrrt lever arm
tp tt rpprprprL
sinDefine angular momentum as:
prL
Since Fr
and dtpdF /
dtLdnet /
dtpdrvmv
dtpdrpdtrddtLd
/)(
///
0 F
angular momentum
net torque
Angular momentum of a particle
Angular momentum & torque
iiiiii LvrmprL
As before let’s break down vectors into two components
VvvRrr iiii
'' ;
Then
MVRpr
MVRdtrmdRpr
mVRdtrdmRVrmvrm
VRmvRmVrmvrm
VvRrmL
ii
iiii
iiiiiiii
iiiiiiii
iii
''
'''
''''
''''
''
/)(0
/)(
)()(
)()(
ang. mom. about COM + ang. mom. of COM0
position vector for com
velocity of com
Angular momentum of a multi-particle system
Conservation of angular momentum
iiiiii LvrmprL
netii dtLddtLd //
If the net torque is zero, the total angular momentumof the system is conserved.
.0/ constLdtLd
Conservation of angular momentum
Angular momentum & torque
x
z
plane A
plane B
plane A
i i iii
iiiiii
IrmLL
rmrrmL
)(
)(2
2
IL
plane B
1m2m
1r
2r
1L
2L
In general,
IL because of non-zero x/y component of
angular momentum unless the object is symmetric about the
axis of rotation.
If the object is symmetric about the axis of rotation, then
IdtLdIL /,
Angular momentum vector and axis of rotation
Gyroscope Gyroscopic motion
pivot
support
mg
remove
pivot
precession
Gyroscope Principle of gyroscopic
pivotL
wrdtLd
/
r
gmw
zy
x
LrLwr
//,
LLd
Gyroscope
L
LdL
Ld x
y
d
)/()(//)/(/ IwrLdtLLddtd
Precession angular speed:
Assumption: The angular momentum vector is associated only with the spin of the flywheel and is purely vertical.
The procession is much slower than the rotation,
Gyroscope
Gyroscope
Spinning Top
Example
A disk of mass M and radius R rotates around the axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f.
1) f = i 2) f = ½ i 3) f = ¼ i
i
z
f
z
First realize that there are no external torques acting on
the two-disk system. Angular momentum will be conserved!
ii MRL 211 2
1I
i
z2
1
f
z
ff MRL 22211 II
fi MRMR 22
21
fi 21
Example
You are sitting on a freely rotating bar-stool with your arms stretched out and a heavy glass mug in each hand. Your friend gives you a twist and you start rotating around a vertical axis though the center of the stool. You can assume that the bearing the stool turns on is frictionless, and that there is no net external torque present once you have started spinning. You now pull your arms and hands (and mugs) close to your body.
What happens to the angular momentum as you pull in your arms?
1. it increases 2. it decreases 3. it stays the same
L1 L2
CORRECT
1 2
I2 I1
L L
What happens to your angular velocity as you pull in your arms?
1. it increases 2. it decreases 3. it stays the same
CORRECT
What happens to your kinetic energy as you pull in your arms?
1. it increases 2. it decreases 3. it stays the same
CORRECT
1 2
I2 I1
L L
K 12
2I 12
2 2
II 1
22
IL (using L = I )
Example
A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens?
1. She starts to spin CCW.2. She starts to spin CW.3. Nothing
CORRECT
Since there is no net external torque acting on the student
stool system, angular momentum is conserved.
Remember, L has a direction as well as a magnitude!
Initially: LLINI = LLW,I
Finally: LLFIN = LLW,F + LLS
LLW,F
LLS
LLW,I LLW,I = LLW,F + LLS
Example
A puck slides in a circular path on a horizontal frictionless table. It is held at a constant radius by a string threaded through a frictionless hole at the center of the table. If you pull on the string such that the radius decreases by a factor of 2, by what factor does the angular velocity of the puck increase?
(a) 2 (b) 4 (c) 8
Since the string is pulled through a hole at the center of rotation, there is no torque: Angular momentum is conserved.
L1 = I11 = mR21
mR
mR21 = m R2241
41
1 = 2 2 = 41
Example A uniform stick of mass M and length D is pivoted at the center. A
bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.
What is the angular speed F of the stick after the collision? (Ignore gravity)
v1 v2
M
F
initial final
mD
D/4
Set Li = Lf using
v1 v2
M
F
initial finalm
DD/4
I 1
122MD
mvD
mvD
MD F1 22
4 4
1
12 F
m
MDv v
31 2
ExercisesProblem 1
Find the acceleration of an object of mass m.
Solution
m
R
M
h
x
y
T
Mg
n
T
mg
For the object, Newton’s 2nd law gives:
yy maTmgF
For the cylinder, the total torque is:
zzz MRIRT 2)2/1(
The tangential acceleration of the cylinderis equal to that of the object:
zy Raa tan
(1)
(2)
(3)
(2)+(3):
yMaT )2/1( (4)
ay>0 in –y direction
ExercisesProblem 1 (cont’d)
Solution (cont’d)
m
R
M
h
x
y
T
Mg
n
T
mg
(1)+(4):
yy maMamg )2/1(
)]2/(1/[ mMgay (5)
(4)+(5):
)/21/( MmmgmamgT y The final velocity of the object when itwas at rest initially:
hahavv yy 2220
2
)]2/(1/[22 mMghhav y
ExercisesProblem 2
Solution
m
R
M
h
x
y
T
Mg
n
T
mg
)]/21/()3[(
)]/21/([
MmmMg
MmmMgMgTn
(a) The normal force on the cylinder is:
(b) Compare n with (m+M)g:
As the suspended mass is accelerating downthe tension is less than mg. Thereforen is less than the total weight (m+M)g.
(c) If the cylinder is initially rotating clockwise and so that the object gets initial velocity upward, what effect does this have on T and n?
As long as the cable remains taut, T and n remain the same.
ExercisesProblem 3
Solution
T1
T2
RMm1
m2
A glider of mass m1 slides without friction ona horizontal air track. It is attached to anobject of mass m2 by a massless string.The pulley is a thin cylindrical shell withmass M and radius R, and the string turnsthe pulley without slipping or stretching.Find the acceleration of each body, theangular acceleration of the pulley, and thetension in each part of the string.
m1
n1 T1
m1g
m2
m2g
T2T1
T2
Mgn2
glider hanging obj. pulley
glider: xx amTF 111
object: yy amTgmF 2222
pulley:
zzz MRIRTRT )( 212
no stretching and slipping:
Zyx Raa 21
y
x (1)
(2)
(3)
(4)
glider
hangingobject
pulley
+
y xy
x
Exercises
Solution cont’d
Problem 3 (cont’d)
(1)-(4):
Mmm
gmMmT
Mmm
gmmT
21
212
21
211
)(;
Exercises
Solution
Problem 4
A primitive yo-yo is made by wrapping a stringseveral times around a solid cylinder with massM and radius R. You hold the end of the stringstationary while releasing the cylinder with noinitial motion. The string unwinds but does notslip or stretch as the cylinder drops and rotates.Find the speed vcm of the center of mass of thesolid cylinder after it has dropped a distance h.
h
1
22
2,cmv
0
0
1
1,
cmv
2211 UKUK
RvMRI
IMvKK
UMghU
cm
cm
/,)2/1(
)2/1()2/1(,0
0,
2,22
22
22,21
21
Energy conservation
0)4/3(0 22, cmMvMgh ghvcm )3/4(2,
Exercises
Solution
Problem 5
B
A
C
(Atwood’s machine)
Find the linear accelerations of blocks A and B,the angular acceleration of the wheel C, and the tension in each side of the cord if there isno slipping between the cord and the surface ofthe wheel.
I
mA
mB
R
The accelerations of blocks A and B will have the same magnitude .As the cord does not slip, the angular acceleration of the pulley will be . If we denote the tensions in cord as and , the equationsof motion are:
Ra /
a
AT BT
)(;;2
IaR
ITTamgmTamTgm BABBBAAA
RIRmRm
mmg
R
a
RImm
mmga
BA
BA
BA
BA
// 2
2
2
/
/2)(
RImm
RImmmgagmT
BA
ABAAA
2
2
/
/2)(
RImm
RImmmgagmT
BA
BABBB
Exercises
Solution
Problem 6
h=50.0 m
rough
smooth
A solid uniform spherical boulderstarts from rest and rolls down a50.0 m high hill. The top half of thehill is rough enough to cause theboulder to roll without slipping, butthe lower half is covered with ice andthere is no friction. What is thetranslational speed of the boulderwhen it reaches the bottom of the hill?
1st half (rough):
12
222
221
)7/10(
)/]()5/2)[(2/1()2/1(
)2/1()2/1(
ghv
RvmRmv
Imvmgh
2nd half (smooth):
smghghv
vghgh
KmvKmvmgh
B
B
rotBrot
/0.292)7/10(
)2/1(])7/10)[(2/1(
)2/1()2/1(
21
212
222
Exercises
Solution
Problem 7
Occasionally, a rotating neutron star undergoes a sudden andunexpected speedup called a glitch. One explanation is that aglitch occurs when the crust of the neutron star settles slightly,decreasing the moment of inertia about the rotation axis. Aneutron star with angular speed 0=70.4 rad/s underwent sucha glitch in October 1975 that increased its angular speed to, where =2.01x10-6. If the radius of the neutronstar before the glitch was 11 km, by now how much did itsradius decrease in the starquake? Assume that the neutronstar is a uniform sphere.
Conservation of angular momentum: 20
2000 RRII
20000
20
02
0020
2
)()(
RRRR
RRRcmRR 1.1)/)(2/( 00
Exercises
Solution
Problem 8
A small block with mass 0.250 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block isoriginally revolving in a circle with a radius of 0.800 m about thehole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radiusof the circle in which the block revolves.The breaking strength of the string is 30.0N. What is the radius of the circle when thestring breaks. The tension on the string: )/()/()(/ 32322 mrLmrmvrrmvT The radius at which the string breaks is obtained from:
mr
Nkg
msmkg
mTrmvmTLr
440.0
)]0.30)(250.0[(
/)]800.0)(/00.4)(250.0[(
)/()()/(2
max2
00max23
ExercisesProblem 9
pivot
pivotcatcher
v
r
M,I
A ball catcher whose mass is M and momentof inertia is I is hung by a frictionless pivot.A ball with a speed v and mass m is caught by thecatcher. The distance between the pivot point andthe ball is r which is much greater that the radius ofthe ball.
(a)Find the angular speed of the catcher right after it catches the ball.(b) After the ball is caught, the catcher – ball system swings up as high as h. Find the angular speed at the maximum height h.
Exercises
Solution
Problem 10pivot
pivotcatcher
v
r
M,I
(a)The initial angular momentum with respect to the pivot is: The final total moment of inertia is
mvr
2mrI
)/( 2 Imrmvr
(b) The kinetic energy after the collision is:
ghmMImrK )()()2/1( 22
,IL Since
potential energy at height h
h
)(
)(22 Imr
ghMm
ExercisesProblem 11
h= 58.0 m
d= 42.0 cm
g/cmA 42.0 cm diameter wheel, consisting ofa rim and six spokes, is constructed froma thin rigid plastic material having a linearmass density of 25.0 g/cm. This wheel isreleased from rest at the top of a hill 58.0 mhigh.(a)How fast is it rolling when it reaches the bottom of the hill?(b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?
No sliding
Exercises
Solution
Problem 11 (cont’d)
h= 58.0 m
d= 42.0 cm
g/cm
(a) Conservation of energy:
2211 KUKU
)3/(6
)2/1()2/1(
6,0,0,
22
222
211
rmrmIII
IMvK
mmMUKMghU
spokerimspokesrim
spokesrim
)3(2,2 rMrmrm spokerim
322
2
)]3/(62)[2/1(
))(3)(2)(2/1()3(2
rrrr
rrghR
sradRgh /124)]2(/[])3[( 2
(b) Doubling the density would have no effect! As , doubling the diameter would reduce the angular velocity by half. But would be unchanged.