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

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Example: Vector algebra

If and = 12 find .

Let

Then

Finally:

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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:

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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

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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

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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

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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:

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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

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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