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Chapter 11: Angular Momentum
11.1: The Vector Product and Torque: vector
product, cross product
11.4: Conservation of Angular Momentum
11.2: Angular Momentum
11.5: The Motion of Gyroscopes and Tops: precessional
motion & frequency
11.3: Angular Momentum of a Rotating Rigid Object
11.6: Angular Momentum as a Fundamental Quantity
11.1 Vector Product & Torque
Right Hand rule
The vector product A x B is a third vector C having a
magnitude AB sinθ equal to the area of the
parallelogram described by AxB perpendicular the
plane of A x B.
The direction of the Torque will be centered
along the axis of rotation (the +z axis).
Of course the rod will accelerate in the
general direction of the applied force.
Example of Cross Product
Lets say that a 3 meter rod is oriented in the x-y
plane as seen to the right (pointing to the forward
and right position so that, r = 2i + 3j + 0k).
A 5 Newton force is applied in the same plane, but
with only a small component in the x-plane, F = 1i +
4j + 0k.
What is the resultant torque where  = r x F?
r
i (+)
j (-)
k (+)
2
3
0


= +( 3*0 - 4*0 ) i - ( 2*0 - 1*0 ) j + (2*4 - 1*3 ) k
F
1
4
0




= 0 i - 0j + 5 k
= 5 k (positive means counter clockwise)
Some Properties of Vector Products
1. A x B = -B x A
2. If A is || to B then A x B = 0 (AxA=0 also)
3. If A is ┴ to B then |A x B| = AB
4. A x (B + C) = A X B + A X C
5. d (A x B) / dt
= A x dB/dt + dA/dt x B
11.2 Angular Momentum of a Particle
Torque = r x F
or
Σ = r x dp/dt
Let's add ZERO to the above equation
Σ = r x dp / dt
 = Fd = F sinφ r
Let 0 = dr/dt x p
Angular momentum, L, is defined as r x p
r x p  off centered momentum will cause rotation
L=rxp
L = r x (mv)
L = mv r sinφ
(both r & p are vectors)
F Δt = Δp
Δp = F Δt
dp = F dt

Impulse
dr/dt = v ll p
Σ = r x dp/dt + dr/dt x p
d (A x B)/dt = A x dB/dt + dA/dt x B
apply Rule 5 which yields
Σ = d(r x p) / dt
Σ dt = d(r x p)
Σ dt = dL
Thus net force on a particle is the time rate of
change of the linear momentum
ΣF = dp / dt
+ 0
Σ = dL / dt
by def  L = r x p
Σ = r x ΣF
Σ = r x dp/dt
Demo: Fallling Bottle and keys: ME-Q-FB
11.3 Angular Momentum of a Rotating Rigid Object
L
=rxp
L
= r x (mv)
Li
= miri2ω
L
=I ω
dL/dt = I dω/dt
If you recall, we introduced this formula
previously (Ch 10) and related it to
Newton’s 2nd law, we now can see the proof.
v = ωr
I = mr2
F =m a
where I is constant for a rigid object
Σ = I α
Example
The angular momentum of a bowling ball
spinning at 60 rpm is about ____.
Σ = I
α
mbowlingball ≈ 4 kg
rbowlingball ≈ 10 cm
Lz = Iω
Table 10.2  I = 2/5 MR2
Lz = 2/5 MR2 ω
60 rpm = 1 rev / sec
Lz = 2/5(4)(0.1)2 (1 rev/sec)(2π / rev)
Lz = 0.10 Nm2
11.4 Conservation of Angular Momentum
Σext = dLtot / dt
Σext = 0
Ltot = constant so Linit = Lfinal
Demo: Euler’s Disk: ME-M-ED
Example of exploding galaxy
What was the angular velocity of our galaxy
I
½mri2
ω
ω
=
If
= ½mrf2
ωf
2/T
when it was (a) 1014 meters in diameter
(b) 1010 m in diameter?
(a)
Need to look up the following information online
1011 stars in a galaxy
mstar = 2 x 1030 kg
rgalaxy = 50,000 light years = 5x1020 m
Tgalaxy = 2x108 years = 6x1015 sec
v = r = 2r/T
 f = 2/T
Example 11.9 The spinning bicycle wheel
Use the right hand rule...
L=rxp
The wheel is spinning in the horizontal
plane,
so L = Linitial
Rotate the wheel through its center
by 180°. Now inverted wheel has
Linvert = - Linitial
No external torques have been applied
(only one torque internal to the system) so
Angular momentum should be conserved.
(b)
assuming a disk
½m(5x1013)2 ω = ½m(5x1020)2 2/6x1015
ω = (5x1020)2/(5x1013)2 * 2/6x1015
ω = 1014 * 2/6x1015
ω = 0.1 rad / sec
ω = (5x1020)2/(5x109)2 * 2/6x1015
ω = 107 rad / sec
The wheel started at +Linit to the inverted -Linit.
The student on the stool (the only other thing internal to the
system) went from 0 to 2 Li (to conserve angular momentum).
Lecture Example
During lecture I spin three
times around on the
platform in 3 seconds. In
each arm is a 2 kg mass.
These are brought in to the
center of my body.
1st point: calc the inertia at the
center of rotation and when
extended to 1 meter.
At the center:
I = mr2 = (2(0.02m)2 )
Icenter = 0.0008 kg m2
Remember I’ve two arms!!!
Extended:
I = mr2 = 2(1)2
Iextend = 2 kg m2
Given: f
I

Then I extend my arms in to
(I+ Icenter)

1 meters…and now 3
(I+ 2(0.0008)) 
rotations take about 6
2I
seconds. What is my
I ≈ 4 kg m2
inertia?
= ½
(same angular distance in half the time)
= If
f
= (I+ Iextend) ½
= (I+ 2(2)) ½
≈ (I+ 4)
Demo: Angular Momentum Platform: ME-Q-AM
11.5 Gyroscopes
As a Gyroscope is spinning it wobbles (precesses) at a
certain frequency, ω.
The weight vector (mg) of the gyroscope is off
centered (this perpendicular distance is the lever arm, r)
from the pivot point so that,
 = r x mg
The Torque produces a change in dL in the direction
of the Torque, ( = dL / dt).
Just like , dL must also be perpendicular to L.
Since dL is perpendicular to L, we know that dL is
independent of L

Note: dL & only change in the x-y plane. The z component
is constant during the rotation
The gyroscope is precessing about the pivot
point with an angular velocity, ω, so that L =
Iω.
From the above diagram
sinφ
dφ
= ΔL
= dL
dφ
= d dt
dφ
= mgr dt
dφ/dt
= mgr
dω
= mgr
ωprec = mgr / I ω
/L
/L
/L
/L
/Iω
/Iω
(φ << 10°)
Demo: Bicycle Wheel, Gyroscope: ME-Q-GB
Demo: Gyroscope: ME-Q-GY
Demo: Gyroscopic Stabilizer: ME-Q-GS
11.6 Angular Momentum as a Fundamental Quantity
L
=ħ
ħ = 1.0546 x 10-34 J s
ICMω = ħ
ħ = h / 2π
ω = ħ /ICM
Example
What is the angular velocity of a N2 molecule?
mN-atom = 14 amu
mamu = 1.66 x 10-27 kg
dN-molecule = 3.4 Angstroms
ω
ω
ω
ω
ω
=ħ
/
ICM
2
=ħ
/ (mr + mr2) (two nitrogen atoms)
=ħ
/2
m
r2
= 10-34 / 2 (14)( 1.66x10-27) (1.7 x 10-10)2
= 7.7 x 1010 rad / sec