Download - Chapt. 10 Angular Momentum
Chapt. 10Angular Momentum
Definition of angular momentum
Vector nature of torque
04/21/23 1Phys 201, Spring 2011
Angular rotation using vectors• Angular quantities in vector notation
ωur
F
t
τr=r
r× F
ur
τF r sin ϕ = F L
04/21/23 2Phys 201, Spring 2011
The vector product
• It follows:
Non commutative:
• And distribution rule:
Unit vectors:
Cur
=Aur
×Bur
=ABsinΦn$
Aur
×Aur
=0
Aur
×Bur
=−Bur
×Aur
Aur
× Bur
+Cur
( ) =Aur
×Bur
+ Aur
×Cur
04/21/23 3Phys 201, Spring 2011
Example: Vector algebra
If and = 12 find .
Let
Then
Finally:
04/21/23 4Phys 201, Spring 2011
Linear momentum --> Angular momentum
• The linear momentum
• The angular momentum with respect to the axis of rotation must scale with r:
m
p =mvr
L =rp = r m v = m r2 ω = I ω
04/21/23 5Phys 201, Spring 2011
Linear momentum --> Angular momentum
• The angular momentum,
the vector nature:
04/21/23 6Phys 201, Spring 2011
i
j
Angular momentum of a rigid bodyabout a fixed axis:
• Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momenta of each particle:
rr1
rr3
rr2
m2
m1
m3
ϕ vv2
vv1
vv3
We see that LL is in the z direction.
Using vi = ω ri , we get
(since ri and vi are
perpendicular)
rL =
rri
i∑ ×
rpi = mi
rri ×
rvi
i∑ = mirivi
i∑ k̂
rL = miri
2ωi∑ k̂
rL =I
rω
04/21/23 7Phys 201, Spring 2011
Angular momentum of a rigid bodyabout a fixed axis:
• In general, for an object rotating about a fixed (z) axis we can write LZ = I ω
• The direction of LZ is given by theright hand rule (same as ω).
• We will omit the axis (Z) subscript for simplicity, and write L = I ω
ω
z
LZ =Iω
04/21/23 8Phys 201, Spring 2011
Rotational and Linear
Quantity Linear Angular
Position
Velocity
Acceleration
Time
Inertia
Dynamics
Momentum
Kinetic energy
rθrωrαt
Irτ = I
rα
rL = I
rω
K =1
2Iω 2
04/21/23 9Phys 201, Spring 2011
The rotational analogue of force F F is torque
Define the rotational analogue of momentum pp to be
angular momentum
The 2nd Law in rotation:• Translational (linear) motion for a system of particles
• What is the rotational version of this??
rFEXT =
drp
dt
rτ =
rr ×
rF
rL =
rr ×
rp
F = m a
04/21/23 10Phys 201, Spring 2011
Definitions & Derivations...• First consider the rate of change of LL:
drL
dt=
ddt
rr ×
rp( )
d
dt
rr ×
rp( ) =
drr
dt×
rp⎛
⎝⎜⎞⎠⎟+
rr ×
drp
dt⎛⎝⎜
⎞⎠⎟
=rv × m
rv( )
= 0
drL
dt=
rr ×
drp
dtSo:
rFEXT =
drp
dt
drL
dt=
rr ×
rFEXTRecall:
τEXT
rτ EXT =
drL
dtFinally:
rFEXT =
drp
dtDirect analogue:
04/21/23 11Phys 201, Spring 2011
What does it mean?
• where and
In the absence of external torquesIn the absence of external torques
Total angular momentum is conservedTotal angular momentum is conserved
rτ EXT =
drL
dt
rFEXT =
drp
dt rτ EXT =
rr ×
rFEXT
rτ EXT =
drL
dt= 0
04/21/23 12Phys 201, Spring 2011
rτ EXT = τ i = 0 + Rm1g + R∑ m2g
rL = miri
2ωi∑ k̂
L =L1 + L2 + Iω=m1R
2ω + m2R2ω + Iω
rτ EXT =
drL
dt
τr=r
rcm × g
r
τr=
dLur
dt
ΔLur
= τrΔt
L (and therefore the wheel)moves in a horizontalcircle around O: “precession”
Gyroscope:
τr=r
rcm × g
r
The precession frequency
Let’s calculate the precession frequency
ωPr ecession =ΔθPr ecession
Δt=
ΔL
LΔt =τ g
L=
rcmMg
IωWheel
ΔθPr ecession =ΔL
L
L forms a circular motion: ΔL = LΔθPr ecession
τr=
dLur
dt
(Last figure)
τr=r
rcm × g
r
τr=
dLur
dt
Question: We repeat the gyroscope experiment on the moon (g_moon = 1/6 g_Earth) but with an angular velocity double from the one in the lecture.Will the precession of the wheel be faster, slower or the same?
ωPr ecession =ΔθPr ecession
Δt=
ΔL
LΔt
rcmg
IωWheel
=rcmM
I⋅
g
ωWheel
ωP,Moon
ωP,Earth
=1
6
ωWheel ,moon
ωWheel ,lecture
=1
3