Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Lecture 19

Chapter 11

Angular momentumVector product.

04.09.2014Physics I

Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

Lecture Capture: http://echo360.uml.edu/danylov2013/physics1spring.html

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Chapter 11

Angular Momentum Vector Cross Product Conservation of Ang. Mom. Ang. Mom. of point particle Rigid Objects

Outline

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Let’s introduce

Vector Cross Product

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

A

B

C

Vector Cross Product

C

A

B

If we have two vectors

C A

B ABsinMagnitude

Direction: perp. to both A and B (right hand rule)

Then the vector product is

A

B

B

AOrder matters:

A

B

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Cross product

The cross product vector increases from 0 to AB as θ increases from 0 to 90

A

B

C

A

B

C A

B ABsin

0BA

θ=30

θ=0

A

B

ABBA 21

θ=90A

B

ABBA

The vector product is zero when vectors are parallel

The vector product increases The vector product is max when vectors are perpendicular

ABBA

ABBA 21

ConcepTest 1 Vector product

A)

B)

C)

For the unit vectorsFind the following vector products

kji ˆ,ˆ,ˆ

?ˆˆ)2 ji

?ˆˆ)1 ii

0ˆˆ iikji ˆˆˆ

iii ˆˆˆ)1

0ˆˆ)2 ji

0ˆˆ)1 iikji ˆˆˆ)2

x

y

z

0ˆˆ)1 iijji ˆˆˆ)2

C A

B ABsin

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

A Axi Ay j Azk

Vector Cross Product

B Bxi By j Bzk

i i 0 j j 0 k k 0 i j k j k i k i j

A

B (Axi Ay j Azk) (Bxi By j Bzk) AxBx (i i ) AxBy (i j) AxBz (i k)

AyBx ( j i ) AyBy ( j j) AyBz ( j k)

AzBx (k i ) AzBy (k j) AzBz (k k)

(AyBz AzBy )i (AzBx AxBz ) j (AxBy AyBx )k

00ˆˆˆˆ Siniiii 190ˆˆˆˆ Sinjiji

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

What is the vector cross product of the two vectors:

A 1i 2 j 4k

A

B 14i 9 j 1k

Vector Cross Product. Example

A

B [(21) (43)]i [(42) (11)] j [(13) (22)]k

A

B (AyBz AzBy )i (AzBx AxBz ) j (AxBy AyBx )k

B 2i 3 j 1k

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Write Torque as the Cross Product

r

F

Axis of rotation

F

r

Let’s look at a door top view:

Applied force F produces torque

sinrFNow, with vector product notation we can rewrite torque as

Torque direction – out of page (right hand rule)

Direction – out of the page

Direction – into the pageNotation convention:

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Angular Momentum

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Angular momentum is the rotational equivalent of linear momentum

?L

vmp

For translational motion we needed the concepts of

force, Flinear momentum, p

mass, m

For rotational motion we needed the concepts of

torque, angular momentum, L

moment of inertia, I

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Angular Momentum of a single particle

L

r p rpSinL

x

z

y

O

r pm

L

r p Suppose we have a particle with-linear momentum -positioned at r

p

Then, by definition: Angular momentum of a particle about point O isO

Carefull: Let’s calculate angular momentum of m about point O’ prL

sinpr

r

,since pr

0

0sin,0 so

Thus, angular momentum of m 0OL but 0OL Cont.

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Angular Momentum is not an intrinsic property of a particle.

It depends on a choice of origin So, never forget to indicate which origin is being used

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Example: projectile motion

L

r p

Let’s calculate the angular momentum of a particle about point O.

What is the angular momentum of a free particle of mass m moving near the surface of the Earth?

r p

sin0prLL

0

1) L at point O:

O

By definition:

x

y 0p A

00

2) L at point A: sinprLL

So, the angular momentum is changing.

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

m

Example: the Earth around the Sun.

L

r pLet’s calculate the angular momentum of a particle about point O.

What is the angular momentum of a particle of mass m moving with speed v=constin a circle of radius r in a counterclockwise direction?

L

r p

rp

L

θ=90sinrpLL

rmv const

The direction of the angular momentum is perpendicular to the plane of circle(right-hand rule)

The magnitude is

O

But, again, L calculated relative to O’ is obviously not a constant.It depends on a choice of origin

By definition:

So, the angular momentum is constant.

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Angular Momentum of a rigid body

L I

points towardsL

For the rotation of a symmetrical object about the symmetry axis, the angular momentum and the angular velocity are related by (without a proof)

IL

IL

IL

I – moment of inertia of a body

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Two definitions of Angular Momentum

L

r

p

L

L I

L

r p

Rigid symmetrical bodySingle particle

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Rotational N. 2nd law

L

r p

dtLd

dtLd

Let’s find relationship between angular momentum and torque for a point particle:

0

dtpdFlawndN

2.

vmp

dtLd

Torque causes the particle’s angular momentum to change

Rotational N. 2nd lawwritten in terms of L.

p

dtrd

dtpdr

vmv Fr

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Rotational N. 2nd law

dtLd

We got exactly the same expression

Let’s show that this rotational N. 2nd law is the same to the one presented in Lecture 18

I

I

dtdI

dtId )(

dtLd

amF

dtpdF

I

Translational N.2nd law Rotational N.2nd law

dtLd

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Example:What is the angular momentum (about the origin) of an object of mass m dropped from rest.

ConcepTest 2 traffic light/car

A car of mass 1000 kg drives away from a traffic light h=10 m high, as shown below, at a constant speed of v=10 m/s. What is the angular momentum of the car with respect to the light?

A) B) C)

skgmk 2 )ˆ(000,100

x

y

z

h

skgmi 2 ˆ000,100

v

prL

skgmk 2 )ˆ(000,10

r

)ˆ)(( krSinmv )ˆ( kmvh )ˆ(000,100 k

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Conservation of Angular Momentum

I11 I22

netdtLd

Angular momentum is an important concept because, under certain conditions, it is conserved.

If the net external torque on an object is zero, then the total angular momentum is conserved.

0,0 dtLdthenIf net

constL

IL For a rigid body 21 LL

Department of Physics and Applied Physics95.141, Spring 2014, Lecture 19

Thank youSee you on Monday