3 Home exercise sheet 3.1 Conservation Laws and Neother’s theorem

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3.1
Home exercise sheet
Conservation Laws and Neother’s theorem
Exercise 3.1: Disk rolling down an inclined plane
A hoop of mass m and radius R rollsl without slipping down an inclined plane. The
inclined plane has opening angle α and mass M , and itself slides frictionlessly along a horizontal surface.
Find the EOM of the system and solve them.
Solution:
We start by writing the Lagrangian in the most basic coordinates:
L=
1
1
1 2
ẋ + ẏ 2 + I θ̇2 + Ẋ 2 − mgy
2
2
2
(1)
We now can write the constraints of the problem
x = X + s cos α − a cos α
y = s sin α + a sin α
s = aθ
(2)
(3)
(4)
We now take the derivative of the constraint equations and plug them into the Lagrangian (It’s
obvious that if we choose to use Lagrange multipliers we would still get the same equations
of motion)
1
I
1
2
m + 2 ṡ2 + m cos αẊ ṡ − mgs sin α
(5)
L = (m + M ) Ẋ +
2
2
a
1
First thing we notice is that Ẋ is a cyclic coordinate which tells us there is a conserved
quantity which is the linear momentum.
PX =
∂L
= (m + M ) Ẋ 2 + m cos αṡ
∂ Ẋ
(6)
Using E-L for s we find
I
m+ 2
a
s̈ + m cos αẌ = −mg sin α
(7)
So we can now use the conserved momentum to find Ẍ (s̈) which we can plug into our E-L
equation for s and get
mg sin α
2 cos2 α ≡ as
m + aI2 − mm+M
m cos α
mg sin α
Ẍ =
2 cos2 α ≡ aX
m + M m + aI2 − mm+M
s̈ = −
(8)
(9)
Both these accelerations are constant which will give us the simplest EOM solutions
1
X = X0 + Ẋ0 t + aX t2
2
1
s = s0 + ṡ0 t + as t2
2
(10)
(11)
Exercise 3.2: Bead on a stick
A stick is pivoted at the origin and is arranged to swing around in a horizontal plane at
constant angular speed ω. A bead of mass m slides frictionlessly along the stick. Let r be
the radial position of the bead. Find the conserved quantity E given by
E=
X ∂L
i
∂ q̇i
q̇i − L
Explain why this quantity is not the energy of the bead.
2
(12)
Solution:
There is no potential energy here, so the Lagrangian consists of just the kinetic energy, T ,
which comes from the radial and tangential motions:
1
1
L = T = mṙ2 + mr2 ω 2
2
2
(13)
And the energy is given by eq. (1) giving us
1
1
E = mṙ2 + mr2 ω 2
2
2
(14)
Since ∂L
= 0 we say that this quantity is conserved. But it is not the energy of the bead,
∂t
due to the minus sign in the second term. The point here is that in order to keep the stick
rotating at a constant angular speed, there must be an external force acting on it. This force
in turn causes work to be done on the bead, thereby increasing its kinetic energy. The kinetic
energy T is therefore not conserved. From our equations we see that E = T − mr2 ω 2 is the
quantity that is constant in time.
Exercise 3.3: Spherical Pendulum
A bead is free to slind along a frictionless hoop of radius R. The hoop rotates with
constant angular velocity ω around a vertical diameter.
1. Find the EOM for the angle θ shown.
2. What are the equilibrium positions?
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3. What is the frequency of small oscillations about the stable equilibrium?
4. There is one value of ω that is rather special; what is it, and why is it special?
Solution:
1. To solve those it would be wise to move to spherical coordinates and plug in the constraints of the problem r = R and φ̇ = ω into the Lagrangian to receive
1 (15)
L = m R2 θ̇2 + R2 sin2 θω 2 − mgR (1 − cos θ)
2
Which gives us the following EOM
g
θ̈ = sin θ cos θω 2 −
R
2. We are now looking for equilibrium, this means that θ̈ = 0 meaning
g
0 = sin θ cos θω 2 −
R
(16)
(17)
This has three solutions: θ = 0, θ = π and cos θ = ω2gR .
Notice that there is a condition for cos θ = ω2gR because cos θ can only go between −1
and 1 meaning that ω 2 ≥ Rg , this means we have two cases:
(a) The case where ω 2 ≤ Rg : This means that the two extrema are θ = 0 and θ = π.
It is very intuitive that the stable extrema is θ = 0 but one can also check by taking
θ → π + δ if we expand for small δ we receive
g
2
δ
(18)
δ̈ = ω +
R
This solution is exponential which means that this point is unstable, making θ = 0
the stable equilibrium point.
So to find the frequency of small oscillations we need to solve for small amplitude,
giving us
g
θ̈ = −
− ω2 θ
(19)
R
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(b) The case where ω 2 ≥ Rg : The extremas are the same as item (a.) and cos θ0 = ω2gR .
This time around θ = 0 is also unstable (this is obvious because Rg − ω 2 is
negative now giving us an exploding solution). So the stable point is now cos θ0 =
g
.
ω2 R
To find the frequency of small oscillations we expand around θ = θ0 + δ which will
give us
r
g2
ω sin θ0 = ω 2 − 2 2
(20)
ω R
p
3. The special frequency is ω = Rq , this is the critical frequency above which the mass
wants to move away from the bottom of the hoop.
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