Short Version :
11. Rotational Vectors & Angular Momentum
11.1. Angular Velocity & Acceleration Vectors
Right-hand rule
Angular acceleration vector:
α  lim
t  0
 //   
ω
dω

t
dt
 //   
 change
direction
11.2. Torque & the Vector Cross Product
τr
τF
  r F sin 
τˆ  r
τˆ  F
Right hand rule
τ  r F
cross product
Cross Product
Cross product C of vectors A & B:
C  A B
C  A B sin 
Ĉ
Given by right-hand rule,
Dot product C of vectors A & B:
= area of A-B parallelogram .
is a vector  A-B plane.
C  A  B  A B cos 
Properties of cross product :
1. Distributive
A   B  C  A  B  A  C
2. Anti-commutative
A  B  B  A
3. NOT associative
A   B  C   A  B   C
A   A  B

A A  0
B
is a vector in the A-B plane and  A.
A
 A  A  B  0
A× (A× B )
xˆ
A  B  Ax
Bx
yˆ
Ay
By
zˆ
Az
Bz
  Ay Bz  Az B y  xˆ   Az Bx  Ax Bz  yˆ   Ax B y  Ay Bx  zˆ
xˆ
A  B  Ax
Bx
yˆ
Ay
By
zˆ
0
0
  Ax B y  Ay Bx  zˆ
  AB sin   zˆ
11.3. Angular Momentum
Linear momentum:
Angular momentum:
pmv
L  rp  m r v
Iω
 I ω
In terms of I defined in chap 10 :
ˆ L  I 
L//  ω
particle
rigid body with axis of rotation
along principal axis
general case, I a tensor.
 L &  can have different directions.
v  ωr
 y2  z2

I   dV   r    x y
 xz

x y
x2  z2
yz
xz 

yz 
x 2  y 2 
Example 11.1. Single Particle
A particle of mass m moves CCW at speed v around a circle of radius r in the x-y plane.
Find its angular momentum about the center of the circle,
express the answer in terms of its angular velocity.
L  m rv
 m r v kˆ
 m r 2  kˆ
 m r2 ω
Iω
I  m r2
Torque & Angular Momentum
L   Li
System of particles:
  ri  pi
i

d Li
dL
 dr
dp 

   i  pi  ri  i 
dt
dt
dt 
i
i  dt
  ri 
i
d pi
dt
  ri  Fi
i

i
dL
τ
dt
d ri
 pi  v i  m v i  0
dt
  τi
i
rotational analog of 2nd law.
11.4. Conservation of Angular Momentum
Rotating Stool with Weights
Conceptual Example 11.1. Playground
A merry-go-round is rotating freely when a boy runs straight toward the center & leaps on.
Later, a girl runs tangentially in the same direction as the merry-go-round also leaps on.
Does the merry-go-round’s speed increase, decrease, or stays the same in each case?
Boy
Girl
Lb = 0   L = 0
I = Im + Ib
 
 L = Lg
I = Im + Ig
?
Making the Connection
A merry-go-round of radius R = 1.3 m has rotational inertia I = 240 kg m2
& is rotating freely at 1 = 11 rpm.
A boy of mass mb = 28 kg runs straight toward the center at vb = 2.5 m/s & leaps on.
At the same time, a girl of mass mg = 32 kg, running tangentially at speed vg = 3.7 m/s
in the same direction as the merry-go-round also leaps on.
Find the new angular speed 2 once both children are seated on the rim.
L0  I 1  mg R vg
Before :
After :
L  I 2  mb R2 2  mg R2 2
L  L0
 2 
I 1  mg R vg
I   mb  mg  R 2
 1

2
240
kg
m
11
rpm

32
kg
1.3
m
3.7
m
/
s
rev / rad   60 s / min 









 2

 12 rpm
2 
2
2
 240 kg m    28 kg  32 kg 1.3 m 
Demonstration of Conservation of Angular Momentum
Rotating Stool & Bicycle Wheel
11.5. Gyroscopes & Precession
Gyroscope: spinning object whose rotational axis is fixed in space.
External torque required to change axis of rotation
 Higher spin rate  larger L  harder to change orientation
Usage:
• Navigation
• Missile & submarine guidance.
• Cruise ships stabilization.
• Space-based telescope like Hubble.
Gyroscopic Stability
Precession
Precession: Continuous change of direction of rotation axis,
which traces out a circle.
dL
τ
dt
r L
 r  Fg

L  L
Gyroscope with
Adjustable Weights
Rate of Precession
Precession occurs if   L.
dL

 τ  r    mg zˆ  
zˆ  Lˆ
dt
sin 
  rmg sin 
r̂ Lˆ
z
 L precesses CCW around z.
For L constant:

ˆ
dL

ˆ

zˆ  L
d t L sin 

x
ˆ   sin  cos  , sin  sin  , cos  
L
ˆ    sin  sin  , sin  cos  , 0
zˆ  L
   const

    sin  sin  ,   sin  cos  , 0     sin  , cos  , 0 
L
Rate of precession :
 

L sin 
y
“Torqueless” Precession
L
r
L 
L//
v
v
r
L is conserved.
Only L// is conserved.
Precession is due to torque caused by centripetal force.
Torqueless Precession
Earth’s Precession
Earth’s precession (period ~ 26,000 y )
The equatorial bulge is highly exaggerated.
Perfect sphere
=0
=0
Oblate spheroid
 <

The equatorial bulge is highly exaggerated.
GOT IT? 11.3.
You push horizontally at right angles to the shaft of a spinning gyroscope.
Does the shaft move
(a)upward,
(b)downward,
(c)in the direction you push,
(d)opposite the direction you push?
Looking down at bike.
Bicycling
Direction of
bike’s motion
L+  t
wheel
L

τ  r  Fg
Wheel
turns
Biker leans
points into paper
L //   wheel turns to biker’s left