001 Newtonian Mechanics - George Mason Universitycomplex.gmu.edu/ Newtonian Mechanics.pdfNewtonian...
Transcript of 001 Newtonian Mechanics - George Mason Universitycomplex.gmu.edu/ Newtonian Mechanics.pdfNewtonian...
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
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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
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Conservation Theorems
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
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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.
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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
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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
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11
2 212 2 1
2
12
2d vm mW dt d v
dt
W m v v
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Conservation Theorems
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
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Conservation Theorems
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
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Conservation Theorems
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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Mechanics of a System of Particles
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
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ir
Mechanics of a System of Particles
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
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ir
O
( )1
eF
( )2
eFR
O
M
exttotF
Mechanics of a System of Particles
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.
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2
2exttot
dMdt
RF gives
0exttot F totp
Mechanics of a System of Particles
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
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i i ii
d mdt
r v
Mechanics of a System of Particles
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
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i j jipairs
r F
Mechanics of a System of Particles
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
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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
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Notes on the 3rd law
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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Notes on the 3rd law
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)
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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
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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
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' '
' ' 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
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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
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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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