Lecture 8 Outline - End-o-2
•Chapter 2 outtro:
•Energy function (Section 2.5,2.6,2.7)
•Friction (Section 1.5, 2.5)
•Conservation of momentum/energy (Sections 2.6,2.7)
•(if time) Electrical circuits (Section 2.5)
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Lecture 8: Energy, Friction, Conservation of Momentum & Energy with Lagrange Multiplier, Study notes of Physics
The concepts of energy function, friction, conservation of momentum and energy, and the use of lagrange multiplier in physics. It covers topics such as the energy function in sections 2.5, 2.6, and 2.7, friction in sections 1.5 and 2.5, and the conservation of momentum and energy in sections 2.6 and 2.7. It also discusses electrical circuits as an additional topic. Explanations and examples of these concepts, including the example of a particle falling down a hemisphere and the use of lagrangian l(q, ˙q, t) = l0(q, t) + l1(q, ˙q, t) + l2(q, ˙q, t) to find the energy function h(q, ˙q, t).
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Lecture 8 Outline - End-o-
Chapter 2 outtro:
Energy function (Section 2.5,2.6,2.7)
Friction (Section 1.5, 2.5)
Conservation of momentum/energy (Sections 2.6,2.7)
(if time) Electrical circuits (Section 2.5)
example III: Lagrange Multiplier
Particle falling down hemisphere...
spare
Sometimes
h = T + V = E
Part I
Sometimes
h ( q,
˙q, t
) is the total energy of the system
Recall that
T = ∑ i m i
2
|~ v i | 2 = T 0 +
T
1
(^) T 2 in gen.coords
T = ∑ i m i
j
r i
∂q
j ˙q (^) +
r i
∂t
2
M
0 (^) +
(^) ∑ j
M
j (^) ˙q j (^) +
(^) ∑ j,k
M
jk
˙q j (^) ˙q k
eg.
T
2 is quadratic in gen.coords
(if constraints indep of time then only have
T
2 )
consider
L
L
0 ( q, t
L
1 ( q,
˙q, t
) +
L
2 ( q,
˙q, t
)
(as most problems do conform to!) then
h
∑ j
j^ ˙q ∂ L
(^) ˙q j − L
j
(
˙q j ∂ L 2
(^) ˙q j
- ˙
q j ∂ L 1
(^) ˙q j )
− (^) ( L 2 (^) +
(^) L
1 (^) +
(^) L
0 )
Velocity Dependent Potentials (1.5)
if
L
contains only conservative forces
and
Q
j represents forces not from potential
dt d
L
q k −
L
∂q
k
Q k
Consider frictional forces such as
F f x = − k x v x.
These come from Rayleigh’s
dissipation
function:
F
∑ i ( k x v
xi 2
(^) k y v yi 2
(^) k z (^) v zi 2 )
where
F
f xi
=
− (^) ∂ F
∂v
xi
for
i particles. Or we can write
F~
f i =
−∇
v i F
The work done
against
friction is then
dW
f
− F~ f (^) .d~
r
− ( F~ f (^) .~ v ) dt
(^) = (
k x v x 2 (^) +
(^) k y v y 2
(^) k z (^) v z 2 (^) ) dt
F
Fraction too much friction
F
2 1 ∑ i ( k x v
xi 2 (^) +
(^) k y v yi 2
(^) k z (^) v zi 2 )
Generalised force was (originally) given by
Q
j
∑ i
F ~ f i . ∂~ r i
∂q
j
− (^) ∑ i ∇ v i F.
r i
∂q
j
− (^) ∑ i ∇ v i F. ∂
(^) ˙r i
(^) ˙q j =
∂ F
(^) ˙q j
With the Lagrange Eqn’s of motion are:
dt d
∂ L
q j −
L
∂q
j
∂ F
(^) ˙q j = 0
note:
F
(^) ˙q j ≡
λ ∂ ∂f
˙ q k
so two scalars functions are reqd...(
L
F
eg. Stokes Law: sphere (radius
a ) moving at velocity
~v
in medium viscosity
η experiences drag
F~
j
=
− 6 πηa~
v
Final Note on Dissipation
If Frictional Forces derivable from dissipation
F
ie, the Lagrange’s Equation’s with dissipation
dt d
(
∂^ L
(^) ˙q j )
−
∂ L
∂q
j
∂ F
(^) ˙q j = 0
dt^ dh we can then show that the energy function relation
L
∂t
dt d
( ∑ j
j^ ˙q ∂ L
(^) ˙q j − L
L
∂t
j
F
(^) ˙q j ˙q j
since
F
is a homogeneous function of
dq j
dt
’s (degree
dt^ dh
F −
L
∂t
If
L
not explicitly depend on
t , and
h
E ,
then we have
dtdE
=
− 2 F
ie. the rate of energy loss
Electrical Circuits
are a simple dissipative system
Generalised Momentum
That’s energy conservation, what about momentum....
Consider a system with forces from potentials
V
q ) then
L
(^) ˙x i =
∂ ∂T (^) ˙x i − ∂^ ∂V (^) ˙x i =
∂ ∂T (^) ˙x i =
(^) ˙x i ∑ i
m
i
˙x^ i 2
- ˙
y i 2
- ˙
z i 2 ) =
m i ˙x i =
p ix
w.r.t
q i generalised (canonical) momentum
p j =
L
(^) ˙q j
Note:
p j not always has units of linear momentum
if
V (^) ( ˙ q ) then
p j not equal to mechanical momentum
If
L
indep of
q j (ie.
q j is a
cyclic
coordinate)
dt d
∂ L
(^) ˙q j −
L
(^) ˙q j =
dt d
∂ L
(^) ˙q j =
dp
j
dt
ie. the generalised momentum
p j
is conserved
Translation and Rotation of System
I’m skipping section 2.6 on symmetry (read for HW)
a system Translation or RotationWhich looks at how conservation works given
Consider a cyclic coordinate
q j where Lagrangian
L
q j (^) )
(i)
Translation of cyclic coordinate
q j , the generalised force
∂q^ ∂V
j ≡
Q j = 0
which is really just conservation of linear momentum.
(ii) if change cyclic coordinate
q j means system rotation
then the generalised force
Q
j is the component of Torque
about rotation axis, and the generalised momentum
p j is
the angular momentum along the same axis.