001 Newtonian Mechanics - George Mason Universitycomplex.gmu.edu/ Newtonian Mechanics.pdfNewtonian Mechanics: Basic Description r O f 1 The influence of the external world on this

PHYS 705: Classical MechanicsNewtonian Mechanics

1

Quick Review of Newtonian Mechanics

Basic Description:

-An idealized point particle or a system of point particlesin an inertial reference frame

-r is a position vector of the particle relative to O

[Rigid bodies (ch. 5 later)]

3

2

1

distance bet 1 and 2 1 2i ii

r r

r rparticle (m)

r

Oddt

rv

-Define the linear momentum vector as the product of the particle mass m and v:

mp v

2

-We are in 3D Euclidean space, i.e.,

- is its velocity vector

Newtonian Mechanics: Basic Description

r

O

f1

f2The influence of the external world on this particle is encoded as forces (vectors) fi acting on it

The dynamics of the particle is describe by the Newton’s second law of motion:

ddt

pF p where is the net sum (vector sum) of

all forces acting on the particlei

iF f

If m is constant in time,

d d dm m mdt dt dt

p vF v a

If m is not constant in time, differentiation will involve m e.g., rocket problem

3

Conservation Theorems

If 0, then 0 consta1 nt. ddt

pF p Conservation of Linear

Momentum

Define angular momentum by: L r p Recall that vector cross product is not commutative, i.e., A B B A

Now define torque as: N r F

In terms of these angular variables, the 2nd law can be written as:

d d d mdt dt dt

p pF r F r r v

d mdt

N r v

4


If 0, then 0 consta2 nt. ddt

LN L Conservation of Angular

Momentum

Notice that:

zero !

Plug this into the previous equation for the 2nd law, we have:

d d d dm m mdt dt dt dt

d mdt

m

L rr v r v

v v

v

r v

d dm mdt dt

N r v r v ddt

LN 2nd law for rotation

5

Conservation TheoremsDefinition of work:

is the work done by a force F along a particular path between 1 to 2 and it can be +, -, or zero.

2

121

W d F s

for a constant mass m

F

ds

1

2

We have taken F as being the net force acting on the particle,

And, now expanding the two separate terms in the integral,

d dm mdt dt

vF v d dts v

6

Conservation TheoremsThis gives:

So, we can write the above eq as:

2

121

ddt

W m dt v v

Side note:

2d d d dvdt dt dt dt

v vv v v v

212

d vt dtd

d

v v

2 222

12

11

2 212 2 1

2

12

2d vm mW dt d v

dt

W m v v

7


Net work done KE The Work-Energy Theorem

Now, define as the kinetic energy,212

T mv

12 2 1W T T

Then, we have,

A conservative force is one such that the work done in going from 1 to 2 depends only on the end points and not on the particular path taken.

- as an example, fiction is not a conservative force while gravity is.

Equivalent statement: work done around a closed path will be zero!

F

ds

1

2

8


Recall from vector calculus (Stoke’s Theorem):

Mathematically, if we have a conservative F on a closed path

The right hand side is zero if there exist a scalar function U such that,

where the surface is defined with the path as its boundary and dnis the normal on the surface.

Note that one can add an arbitrary constant to U without affecting this result . zero level for U is not important and only is physically relevant.

0d F s

surface

d d F s F n

ds

dn

U F ( : 0)note U

U is called the potential or potential energy.

U

9


So, for any conservative systems, we can write,

From our previous discussion, we also know that the net work from this conservative force is also equal to,

So,

2 2

12 1 21 1

W d dU U U F s

12 2 1W T T

2 1 1 2T T U U

1 1 2 23. T U T U Conservation of Mechanical Energy

Then, rewriting,

12W

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Mechanics of a System of Particles

- For a system of particles, one needs to distinguish between :

“external forces” acting on the entire system and“internal forces” acting within the system

- 2nd law for the ith particle is:

internal force on i from j

( )eji i i

j

F F p

ir

O

net external force on i

ji ij F F

We further assume the weak form of Newton’s 3rd law applies such that,

Note: there is a stronger form about the direction of forces (later)

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Summing over all particles (LHS):

= 0 by 3rd law

,ji

i j j i F ( )e ext

i toti

F F

We now define the center of mass vector R as,

Summing over all particles (RHS): (also assume mi are constant in time)

2

2i i i

i i i i ii i i i i

d d dd dm m mdt dt dt dt dt

p v rp r

i i i ii i

iii

i

m mwith M m

m M

r rR weighted avg of by the

particle’s mass

12

ir


Then, LHS = RHS gives,

The center of mass moves as if it was a point particle with the total mass of the entire system and all the external forces acting directly on it.

2 2

2

2

2 2i ii

exttot

dMd

d dm Md dt tt

R RF r( )e

i ii i

F p

13

ir

O

( )1

eF

( )2

eFR

O

M

exttotF


Now calculate the total linear momentum :

itot i i i i

i i i

d d dm m Mdt dt dt

r Rp p r

Putting this in our previous red equation for the Center of Mass,

exttot totF p

Conservation of Linear Momentum for a system of particles:

If total external force , then the total linear momentum is

conserved.

14

2

2exttot

dMdt

RF gives

0exttot F totp


Similarly for angular momentum, starting again with the 2nd law,

Crossing r on both sides and summing over all particles,

( )eji i i

j

F F p

( )

( )

,

ei ji i i i

i j i

ei ji i i i i i

i j j i i i

d mdt

r F F r p

r F r F r v

itot

i

ddt

L Lext

i toti

N N

for the ith particle

d dm mdt dt

m r v vv r v

15

i i ii

d mdt

r v


Looking back at the first term on the LHS:

Rewrite the sum in terms of pairs of particles:

,

i jii j j i

r F

,i ji i ji j ij

i j j i pairs

i j jipairs

r F r F r F

r r F

O

ir

jr

ijr

(by the weak form of the 3rd law)

By defining,

ij i j r r r

16

i j jipairs

r F


Now, assume the strong version of the 3rd law as follows:

In addition to being equal and opposite, Fij also lie along the line connecting the ith and jth particle

,

0i ji i j jii j j i pairs

r F r FThen, the previous sum term will vanish,

Putting everything together, we then have,

exttottot

ddt

L N

Conservation of Angular Momentum for a system of particles:

if total external torque = 0, then the total angular momentum is conserved.

The system of particles again acts like a single particle

17

O

ir

jr

ijrijF

jiF

Notes on the 3rd law

1. Conservation of Linear Momentum weak form of the 3rd law2. Conservation of Angular Momentum strong form of the 3rd law

Let digress in considering a moving charges in an EM field,

Recall Biot-Savert law: a point charge q moving in the direction given by v gives rise to B at a position r away,

RHR

q

r

B (out of page)

For a system of particles,

034

qr

B r v r

v

18


Now, we will give two examples where the 3rd law does not hold (footnote on p.8):

q F E v B

Recall also the Lorenz force equation: a charge q with velocity v in

an E + B field feels a force F given by,

a) two +charges moving || in a plane

1

v2

B

2

F21

r21v1

BF12

The two internal forces F12

and F21 are equal and

opposite (weak) but are

not directed along r21

joining them (not strong).

Consider the magnetic forces acting on each others :

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1

v2

B

2

F21

v1

B=0 F12=0

The two internal forces

F12 and F21 are not even

equal and opposite here

(not weak nor strong).

ii. two +charges moving in a plane

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General Motion of a System of ParticlesWe have shown that for a system of particles, we generally have:

For Linear translational motion of a system of particles

can be described as a single particle at CM with its total mass M

concentrated at that point.

And, for rotational motion of a system of particles, a similar statement in relation to the CM can be made:

tot ang mom Ltot = (ang mom of M as if concentrated at CM) +

(ang mom of rotation about the CM)

21

totdMdt

Rp ptot = (total mass) (velocity of CM)

General Motion of a System of ParticlesTo show the rotational motion claim, consider :

Total angular momentum L for all particles with respect to O is:

''

i i

i i

R r rV v v

O

irR

'ir CMi i

i L r p

' '

' ' ' '

i i ii

i i ii

i i i i i i i ii

m

m

m m m m

r v

R r V v

R V R v r V r v

' ' ' 'i i i i i i i ii i i i

m m m m

R V r v R v r V

22

General Motion of a System of ParticlesNotice that the last two terms are zero by def. of CM:

Taking time derivative of above,

' ' ''i i i ii i

i i ii

Mm mm RR r r vL V v V

' 0i ii

m r V

' '

' 0

i i i i i ii i i

i ii

M m m M m

m

R r R r R r

r Last term

' ' 0i i i ii i

d m mdt

r v ' 0i ii

m R v3rd term

So, we have:

ang mom of CM

ang mom of rot about the CM

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General Motion of a System of ParticlesNext, we can obtain a similar separation for KE:

To see this, start with,

Notice that

212 i i

iT m v we also have,

KE of a rotating system = (KE of M @ CM) + (KE of rotation about the CM)

''

i i

i i

r R rv V v

2

22

' '

' ' 2 '

' 2 '

i i i i i

i i i

i iV v

v v v V v V v

V V v v V v

V v

24

General Motion of a System of Particles

Putting this into the expression for T, we have,

(def of CM)

221 1 ' '2 2i i i i i

i i iT mV m v m V v

So, this is our desired separation,

KE of sys as if concen@CM

KE of motion about the CM

' 0i ii

m V v

2 21 '2

12 i i

imV vT M

25

General Motion of a System of ParticlesLastly, this separation can also be applied to potential energy of a system of particles,

Assumptions:

,

12i ij

i i j j iU U U

-All forces are derivable from a potential function (conservative)-Internal forces also satisfy the strong form of the 3rd law

Total PE of system = (external PE) + (internal PE)

(the ½ is here since the sum double-counts pairs)

Note: in order for the internal forces to satisfy the strong 3rd law, PE (Uij) derivable from internal forces must also be a function of rij distance between particles only.

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001 Newtonian Mechanics - George Mason Universitycomplex.gmu.edu/ Newtonian Mechanics.pdfNewtonian Mechanics: Basic Description r O f 1 The influence of the external world on this

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