Chapter 11: Rotational Vectors and
Angular Momentum
Vector (cross) products
 Definition
 
a b
of vector product and its properties
 
 
ˆ a b sin 
a b  n
Axis of rotation
unit vector normal
tothe plane

defined by a and b

b
n̂

 
 
a  b  b  a

a

 
 
a  a  a  a  0
Vector (cross) products (cont’d)
Properties of vector product
z
iˆ, ˆj , kˆ
ĵ
k̂
y
: unit vector in x,y,z direction
iˆ  ˆj  kˆ
ˆj  kˆ  iˆ
kˆ  iˆ  ˆj
Consider three vectors:
x
iˆ
Then

a  (a x , a y , a z )
b  (bx , by , bz )
c  ( c x , c y , cz )
  
c  a  b  (a xiˆ  a y ˆj  az kˆ)  (bxiˆ  by ˆj  bz kˆ)
 (a y bz  az by )iˆ  (az bx  a x bz ) ˆj
 (a x by  a y bx )kˆ
Vector (cross) products (cont’d)
Properties of vector product (cont’d)
z
ĵ
k̂
y
x
iˆ
cx  a y bz  a z by
c y  a z bx  a x bz
cz  a x by  a y bx
iˆ
ˆj
kˆ
bx
by
bz
  
c  a  b  ax a y az
Torque
Torque is a quantitative measure of the tendency of a force to cause
or change the rotational motion.
Case for 1 point-like object of mass m with 1 force
y

F 
ˆ

r

• r is constant
only tangential component
of F causes rotation

 ˆ
ˆ
F    ma  
r̂
m
Ft

Ft  F  ˆ
 m( rˆ  r 2 rˆ)  ˆ
 mr
2
Ft r  mr 
x
massless rigid rod
  I
torque
unit Nm
moment of inertia
0
Torque
Case
 for 1 point-like object of mass m with 1 force
F
y
Ft

r

  Ft r  Frt  Fr sin 
r̂
 
 r F

x
rt  r sin 
lever arm
Define





 r F
is aligned with the rotation axis
Torque (cont’d)
Case for 2 point-like objects with 2 forces
 net  F1t r1  F2t r2
y
  0 
  0 
F2 t
F1t

r1

r2
increases
decreases
Example: A see-saw in balance
  1  2
 net  F1t r1  F2t r2
x
M
R
Mg
r
m
 MgR  mgr  0
mg
m/ M  R/ r
Work & energy (I)
 A massive body on massless rigid rod
y

Ft

r
Work done by the force:

W   Ft ds   Ft rd   d
m
Also
x
W=K
W   Id   I (d / dt )d
  I (d / d )( d / dt )d
1
  Id  (1 / 2) I |0
2 1
0
 (1 / 2) I  (1 / 2) I0  K
2
1
2
Work & energy (I) (cont’d)
 Power in rotational motion
Work done by the force:
W   Ft ds   Ft rd   d
dW  d
dW / dt  d / dt
Power :
P  
Correspondence between linear & angular
quantities
displacement
velocity
acceleration
mass
force
Newton’s law
kinetic energy
work
linear
angular
v  dx / dt
a  dv / dt
  d / dt
x
m

F
F  ma
K  (1 / 2)mv2
W   Fdx

  d / dt
I   mi ri2
  
  r F
  I
K  (1 / 2) I 2
W   d
Work & energy (II)
 A massive body in rotational & translational motion
Kinetic energy:
Now
 
K   (1 / 2)mi vi  vi
 ' 
vi  vi  V
'
where vi
the center of mass (COM) and
com

V
is the velocity with respect to
the velocity of COM. Then
 
K   (1 / 2)mi vi  vi
'  ' 
 (1 / 2) mi (vi  V )  (vi  V )
'
 '
vi
'2
 (1 / 2) mi vi   miV  vi  (1 / 2) miV 2
  (1 / 2) m v ' 2  (1 / 2)V 2 m  V  m dr ' / dt
 ii
 i  i i
V

'
'2
2
 (1 / 2) mi vi  (1 / 2) MV  V  d (  mi ri ) / dt
 (1 / 2) m v  (1 / 2) MV 2
'2
i i
0
Work & energy (II) cont’d
 A massive body in rotational & translational motion (cont’d)
 
K   (1 / 2)mi vi  vi
 (1 / 2) m v  (1 / 2) MV 2
'2
i i
ri'  vi' ; I   mi ri' 2
K  (1 / 2) MV 2  (1 / 2) I 2
kinetic energy due to
translational motion
kinetic energy due
to rotational motion
about a rotation axis
through COM
Work & energy (II) cont’d
 A massive body in rotational & translational motion (cont’d)
Example: A rolling cylinder (without sliding due to friction)
O

vCOM
s
P
rotation
O


v  vCOM
translation
View point 1:
+
R O
 vCOM
vCOM  dx / dt  ds / dt
 Rd  / dt  R
P
s
 
v  vCOM
x  s  R
 
v  vCOM

vCOM
O
 
v  vCOM


v  2vCOM

vCOM
=
O



v  vCOM  vCOM  0
This view point was used in the last few slides
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
K  (1 / 2) I P
2
From the parallel axis theorem
P
rotation axis
I P  I COM  MR 2
K  (1 / 2) I COM   (1 / 2) MR 
2
vCOM  R
2
2
2
 (1 / 2) I COM  2  (1 / 2) MvCOM
rotational
translational
Angular momentum & torque
 Angular
 momentum of a particle
y
p

Define angular momentum as:






r̂
L  r pt  r p sin   r  p  p rt
pt

  
r
angular momentum
Lrp



x


rt  r sin  Since   r  F and F  dp / dt
lever arm


  
dL / dt  dr / dt  p  r  dp / dt

  
 v  (mv )  r  dp / dt

0
F

net torque

 net  dL / dt
Angular momentum & torque
 Angular momentum of a multi-particle system


 
 
L   ri  pi   mi ri  vi   Li
As before let’s break down vectors into two components
'
'
velocity of com
i
i
i
i
Then
position vector for com
     
r  r  R ; v  v V

' 
' 
L   mi ( ri  R )  ( vi  V )
 '
 
' '
' 
  mi ri  vi  (  mi ri )  V   mi R  vi  (  mi R )  V
 
' '
'  
'
  mi ri  vi  (  mi ri )  V  R   mi dri / dt  R  V  mi
 
'  '  
'
  ri  pi  0  R  d (  mi ri ) / dt  R  VM
'  '  
  ri  pi  R  VM
0
ang. mom. about COM + ang. mom. of COM
Conservation of angular momentum


 
 
L   ri  pi   mi ri  vi   Li


 
dL / dt   dLi / dt   i   net
If the net torque is zero, the total angular momentum
of the system is conserved.



dL / dt  0  L  const.
Conservation of angular momentum
Angular momentum & torque
 Angular momentum vector and axis of rotation
z


plane A

L2
L1
2
Li  mi (ri )ri  mi ri 
m2 m1
plane B
2
L  i Li  (i mi ri )  I


r2


L  I
r1
x
plane A
plane B


In general, L  I 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

 


L  I, dL / dt    I
 Gyroscopic motion
Gyroscope
remove
support
pivot

mg
precession
pivot

Gyroscope
 Principle of gyroscopic

L
pivot



w  mg
z

r



 
  dL / dt  r  w

   
 
r  w, L // r    L
 
dL  L
y
x
y
Gyroscope
Assumption:
The angular momentum vector
is associated only with the spin
of the flywheel and is purely
vertical.

L

dL
d


L  dL
x
The procession is much slower
than the rotation,   
Precession angular speed:
 
  d / dt  ( dL / L ) / dt   / L  ( wr ) /( I )
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
z
z
i
f
First realize that there are no external torques acting on
the two-disk system. Angular momentum will be conserved!
1
Li  I1 1  MR 2 i
2
z
2
1
i
L f  I1 1  I 2 2  MR  f
1
2
2
MR i  MR  f
2
1
i   f
2
2
z
f
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
CORRECT
3. it stays the same
L1
L
2
What happens to your angular velocity as you
pull in your arms?
1. it increases
CORRECT
2. it decreases
3. it stays the same
2
1
I2
I1
L
L
What happens to your kinetic energy as you pull
in your arms?
CORRECT
1. it increases
2. it decreases
3. it stays the same
K
1
1 2 2
I 2 
I 
2
2I

1 2
L
2I
(using L = I )
2
1
I2
I1
L
L
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: LINI = LW,I
Finally: LFIN = LW,F + LS
LS
LW,I
LW,I = LW,F + LS
LW,F
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 = I11 = mR21
mR21 = m
1 2
R 2
4
1
1 = 2
4
m
R
2 =
41
1
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)
M
D
m
F
D/4
v1
v2
initial
final
Set Li = Lf using
mv1
D
D 1
 mv 2  MD 2  F
4
4 12
I
1
MD 2
12
F 
3m
v1  v 2 
MD
M
D
m
v1
initial
F
D/4
final
v2
Exercises
Problem 1
Find the acceleration of an object of mass m.
n
y
Solution
R
For the object, Newton’s 2nd law gives:
ay>0 in –y direction
Fy  mg  T  may
(1)
T
For the cylinder, the total torque is:

2


RT

I


(
1
/
2
)
MR
z
z
z
(2)
The tangential acceleration of the cylinder
is equal to that of the object:
a  a  R (3)
y
tan
z
(2)+(3):
T  (1 / 2) Ma y
M
Mg
T
m
h
mg
x
(4)
Exercises
Problem 1 (cont’d)
Solution (cont’d)
n
(1)+(4):
y
mg  (1 / 2) Ma y  ma y
a y  g /[1  M /( 2m)]
R
(5)
T
(4)+(5):
M
T  mg  ma y  mg /(1  2m / M )
Mg
T
The final velocity of the object when it
was at rest initially:
v 2  v02  2a y h  2a y h
v  2a y h  2 gh /[1  M /( 2m)]
m
h
mg
x
Exercises
Problem 2
Solution
n
(a) The normal force on the cylinder is:
y
n  T  Mg  g[ M  m /(1  2m / M )]
R
 g[( M  3m) /(1  2m / M )]
(b) Compare n with (m+M)g:
T
M
As the suspended mass is accelerating down
T
the tension is less than mg. Therefore
n 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
h
upward, what effect does this have on T
and n?
Mg
m
mg
As long as the cable remains taut, T and n remain the same.
x
Exercises
Problem 3
m1
T1
pulley
M
R
A glider of mass m1 slides without friction on
a horizontal air track. It is attached to an
object of mass m2 by a massless string.
The pulley is a thin cylindrical shell with
T2
glider
mass M and radius R, and the string turns
the pulley without slipping or stretching.
m2
Find the acceleration of each body, the
angular acceleration of the pulley, and the
hanging
tension in each part of the string.
object
Solution
y
x
x
y
glider:  Fx  T1  m1a1x
(1)
+
x
n1
object:  Fy  m2 g  T2  m2a2 y (2)
T
1
T1
y
T2
pulley:
m1
m2
2
T2  z  T2 R  T1R  I z  ( MR ) z
m1g
m2 g
(3)
Mg
no stretching and slipping:
n2
a1 x  a2 y  R Z
(4)
glider
hanging obj.
pulley
Exercises
Problem 3 (cont’d)
Solution cont’d
(1)-(4):
m1m2 g
( m1  M )m2 g
T1 
; T2 
m1  m2  M
m1  m2  M
Exercises
Problem 4
A primitive yo-yo is made by wrapping a string
several times around a solid cylinder with mass
M and radius R. You hold the end of the string
stationary while releasing the cylinder with no
initial motion. The string unwinds but does not
slip or stretch as the cylinder drops and rotates.
Find the speed vcm of the center of mass of the
solid cylinder after it has dropped a distance h.
Solution
K1  U1  K2  U2
vcm ,1  0
1  0
1
h
2
2
vcm , 2
Energy conservation
U1  Mgh,U 2  0
2
2
K1  0, K 2  (1 / 2) Mvcm

(
1
/
2
)
I

,2
2
I  (1 / 2) MR 2 , 2  vcm , 2 / R
2
0  Mgh  (3 / 4) Mvcm
,2  0
vcm ,2  (4 / 3) gh
Exercises
Problem 5
(Atwood’s machine)
R
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 is
no slipping between the cord and the surface of
the wheel.
I
m
C
mB
AA
B
Solution
The accelerations of blocks A and B will have the same magnitude a.
As the cord does not slip, the angular acceleration of the pulley will be
  a / R. If we denote the tensions in cord as T and T , the equations
A
B
of motion are:
I
m A g  TA  m Aa ; TB  mB g  mB a ; TA  TB 
ag
R
2
a (  I )
m A  mB
a
m A  mB




g
m A  mB  I / R 2
R
m A R  mB R  I / R
2mAmB  mA I / R 2
TA  mA ( g  a )  g
mA  mB  I / R 2
2mB mA  mB I / R 2
TB  mB ( g  a )  g
mA  mB  I / R 2
Exercises
Problem 6
A solid uniform spherical boulder
starts from rest and rolls down a
50.0 m high hill. The top half of the
hill is rough enough to cause the
boulder to roll without slipping, but
h=50.0 m
the lower half is covered with ice and
there is no friction. What is the
translational speed of the boulder
when it reaches the bottom of the hill?
Solution
mgh1  (1 / 2)mv 2  (1 / 2) I 2
1st half (rough):
rough
smooth
 (1 / 2)mv 2  (1 / 2)[( 2 / 5)mR2 ]( v / R ) 2
v 2  (10 / 7) gh1
2nd half (smooth):
mgh2  (1 / 2)mv2  K rot  (1 / 2)mvB2  K rot
gh2  (1 / 2)[(10 / 7) gh1 ]  (1 / 2)vB2
vB  (10 / 7) gh1  2 gh2  29.0 m / s
Exercises
Problem 7
Occasionally, a rotating neutron star undergoes a sudden and
unexpected speedup called a glitch. One explanation is that a
glitch occurs when the crust of the neutron star settles slightly,
decreasing the moment of inertia about the rotation axis. A
neutron star with angular speed 0=70.4 rad/s underwent such
a glitch in October 1975 that increased its angular speed to
0, where /0=2.01x10-6. If the radius of the neutron
star before the glitch was 11 km, by now how much did its
radius decrease in the starquake? Assume that the neutron
star is a uniform sphere.
Solution
Conservation of angular momentum:
R020  ( R0  R)2 (0   )
 R 0  2 R0R0  R 
2
0
2
0
I 00  I  R020  R 2
R  ( R0 / 2)(  / 0 )  1.1 cm
Exercises
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 is
originally revolving in a circle with a radius of 0.800 m about the
hole with a tangential speed of 4.00 m/s. The string is then pulled
slowly from below, shortening the radius
of the circle in which the block revolves.
The breaking strength of the string is 30.0
N. What is the radius of the circle when the
string breaks.
Solution
2
2
3
2
3
The tension on the string: T  mv / r  (mvr) /( mr )  L /( mr )
The radius at which the string breaks is obtained from:
r 3  L2 /( mTmax )  ( mv0 r0 ) 2 /( mTmax )
 [( 0.250 kg )( 4.00 m / s )( 0.800 m )]2 /
[( 0.250 kg )( 30.0 N )]
r  0.440 m
Exercises
Problem 9
A ball catcher whose mass is M and moment
of inertia is I is hung by a frictionless pivot.
A ball with a speed v and mass m is caught by the
r
catcher. The distance between the pivot point and
the ball is r which is much greater that the radius of
the ball.
v
(a) Find the angular speed of the catcher
right after it catches the ball.
pivot
(b) After the ball is caught, the catcher –
catcher
ball system swings up as high as h.
Find the angular speed at the maximum
height h.
pivot
M,I
Exercises
Problem 10
pivot
Solution
(a) The initial angular momentum with
respect to the pivot is:
The final total moment of inertia is
mvr
I  mr
Since
2
r
v
h
L  I,
  mvr /( mr2  I )
(b) The kinetic energy after the collision
is:
pivot
catcher
K  (1 / 2) 2 (mr2  I )  ( M  m) gh

2( m  M ) gh
( mr 2  I )
M,I
potential energy at height h
Exercises
Problem 11
A 42.0 cm diameter wheel, consisting of
a rim and six spokes, is constructed from
a thin rigid plastic material having a linear
mass density of 25.0 g/cm. This wheel is
released from rest at the top of a hill 58.0 m
high.
(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?
d= 42.0 cm
l 25.0 g/cm
No sliding
h= 58.0 m
Exercises
d= 42.0 cm
Problem 11 (cont’d)
l 25.0 g/cm
Solution
(a) Conservation of energy:
No sliding
U1  K1  U2  K2
U1  Mgh , K1  0,U 2  0, M  mrim  6mspokes
K 2  (1 / 2) Mv 2  (1 / 2) I 2
h= 58.0 m
I  I rim  I spokes  mrimr 2  6(mspoker 2 / 3)
mrim  l 2r, mspoke  lr  M  2rl (  3)
2 Rl (  3) gh  (1 / 2)( 2rl )(  3)( r)2 
(1 / 2)[ 2rlr  6(lrr / 3)]
2
2
3
  [(  3) gh ] /[ R 2 (  2)]  124 rad / s
(b) Doubling the density would have no effect! As   1/ r, doubling the
diameter would reduce the angular velocity by half. But v  r would
be unchanged.