1 Class exercise sheet - Solution 1.1 Generalized coordinates

1
1.1
Class exercise sheet - Solution
Generalized coordinates
We first recall that to describe a system with N particles we may use 3N coordinates. The
number of units required to describe the position of the system is called degrees of freedom,
these don’t have to be the Cartesian coordinates of the particles, each s units q1 , q − 2, ..., qs
define the position of the system uniquely with s degrees of freedom and are called generalized coordinates.
We start by asking how many degrees of freedom we have in the problem. So, for each
”particle” we always start with 3 degrees of freedom (were we doing General Relativity there
would be 4). Degrees of freedom can be reduced by using constraints. example of a constraint is x2 + y 2 = R for a circle. we will talk about constraints in more detail later in the
course.
Exercise 1.1: A pendulum attached to a spring
Consider a mass m hanging from a pendulum of length l, attached to a mass M which in
turn is attached to a spring of spring constant k that moves horizontally. Find appropriate
generalized coordinates and describe the kinetic and potential energy in terms of these.
Solution:
Each particle has 3 degrees of freedom but they are forced to move on the x−y plane meaning
they each have 2 degrees of freedom. The mass, which is connected to the spring, is forced to
move solely on the x axis so it has only 1 degree of freedom.
Now moving to the pendulum, it moves on the x − y axis, so we would think that it has 2
degrees of freedom, but since it is moving in a circular manner we will have another constraint
1
reducing the degrees of freedom to 1 as well.
so the generalized coordinates for the mass which is connected to a spring are
x1 = a + x̃
(1)
where a is the loose length of the spring.
For the pendulum the generalized coordinates are
x2 = x1 + l sin q
y2 = −l cos q
We can now write the kinetic and potential energies of the system
1
1
1
1
T =
m ẋ22 + ẏ22 + M ẋ21 = (m + M ) ẋ2 + ml2 q̇ 2 + mẋq̇l cos q
2
2
2
2
1 2
U =
kx̃ − mgl cos q
2
1.2
(2)
(3)
(4)
(5)
Calculus of variations
Exercise 1.2: Snell’s law
You are standing at point (x1 , y1 ) on the beach and you want to get to a point (x2 , y2 )
in the water, a few meters offshore. The interface between the beach and the water lies at
x = 0. What path results in the shortest travel time?
Solution:
We want the fastest route, meaning the shortest time. So we can write that the total time is
p
p
x21 + (y − y1 )2
x22 + (y − y2 )2
|~r1 | |~r2
t=
+
=
+
(6)
v1
v2
v1
v2
2
So to find the shortest time we need to minimize t(y)
y − y1
dt
y − y2
=p 2
+p 2
=0
dy
x1 + (y − y1 )2 v1
x2 + (y − y2 )2 v2
(7)
we now go to the proper generalized coordinates
y − y1 = r1 sin θ1
x1 = r1 cos θ1
y − y2 = r2 sin θ2
x2 = −r2 cos θ2
(8)
(9)
and plugin them in we get
r1 sin θ1 r2 sin θ2
−
=0
v1 r1
v2 r2
sin θ1
v1
=
sin θ2
v2
(10)
(11)
which is snell’s law.
Exercise 1.3: brachistochrone
Find the path between (x1 , y1 ) and (x2 , y2 ) which a beed sliding without friction under
constant gravitational acceleration will traverse in the shortest time.
1. Express the time as a functional of the curve y(x)
2. Show that:
∂L
−L
∂y 0
(12)
1 + y 02
2g(y1 − y)
(13)
H = y0
where
s
L=
is conserved, namely dH/dx = 0.
3. Use the conservation of H to find an integral equation for y(x).
4. Solve the equation to obtain the path.
3
Solution:
1. The functional we want is for the time, so the minimal action will be for time.
Z
Z
ds
I = dt =
v
(14)
so we now need to find what is v. We know the energy is conserved and we assume
that the beed starts sliding with no starting velocity.
1 2
mv + mgy = mgy1
2
⇒
v=
p
2g (y1 − y)
(15)
we can now write
Z
ds
=
v
Z
ds
p
2g (y1 − y)
2. We need to verify that
dH
dx
=
=
=
=
n
o Z
p
2
2
= ds = dx + dy =
s
dx2 + dy 2
=
2g (y1 − y)
Z s
(1 + y 02 ) dx
(16)
2g (y1 − y)
dH
dx
= 0, so we start,
d
dy 0 ∂L
∂L
d
0 ∂L
0 d
y 0 −L =
+
y
−
(L) =
dx
∂y
dx ∂y 0
dx ∂y 0
dx
∂L 0 ∂L 00 ∂L
d
(L) =
y + 0y +
=
dx
∂y
∂y
∂x
∂L
∂L 0 ∂L 00 ∂L
00 ∂L
0 d
y
+y
y − 0y +
=
−
0
0
∂y
dx ∂y
∂y
∂y
∂x
d ∂L
∂L
0
y
−
=0
dx ∂y 0
∂y
(17)
(18)
(19)
(20)
we notice that what we received is nothing but the Euler-Lagrange equation and is just
zero, so we have proved that H is a constant in respect to x.
3. We start by plugin in L into H
H = y0
02
∂L
y
−L = p
−
0
∂y
2g (y1 − y) (1 + y 02 )
4
s
1 + y 02
2g (y1 − y)
(21)
1 = H 2 2g (y1 − y) + H 2 2g (y1 − y) y 02
s
1 − H 2 2g (y1 − y)
y0 =
H 2 2g (y1 − y)
s
H 2 2g (y1 − y)
dy
dx =
1 − H 2 2g (y1 − y)
4. To solve the equation we use the following change of variables
1
θ
dy
1
2
(y1 − y) =
sin
;
=
sin θ
2
2H g
2
dθ
4H 2
(22)
(23)
(24)
(25)
this gives us
1
dx =
2H 2 g
Z
0
θ1
Z θ1
θ
1
sin
(1 − cos θ) dθ
dθ =
2
4H 2 g 0
2
(26)
which has a parametric solution
x − x1 =
1
(θ1 − sin θ1 )
4H 2 g
and
5
y − y1 =
1
(cos θ1 − 1)
4H 2 g
(27)