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:
We can ignore the potential energy here (even if there was a potential energy nothing would
change - see if you understand why), 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:
Consider a particle with kinetic energy
1
T = m ẋ2 + ẏ 2 − ż 2
2
(15)
and constrain it to the hyperboloid
x2 + y 2 − z 2 = −1
z>0
1. Choose the convenient generalized coordinates to solve this problem.
2. Write the Lagrangian
3. Show that the kinetic energy is positive and find the constants of motion.
Solution:
3
(16)
1. We have to choose the appropriate coordinates, hyperbolic ones!
x = r sinh ν cos θ ;
y = r sinh ν sin θ ;
z = r cosh ν
;
ẋ = ṙ sinh ν cos θ + rν̇ cosh ν cos θ − rθ̇ sinh ν sin θ (17)
ẏ = ṙ sinh ν sin θ + rν̇ cosh ν sin θ + rθ̇ sinh ν cos θ (18)
ż = ṙ cosh ν + rν̇ sinh ν
(19)
2. The Lagrangian is
1 2
2
2 2
2
2
L = m ṙ + r ν̇ + r sinh ν θ̇ − λ (r − 1)
2
(20)
3. It is easy to see that T > 0. We have 2 constants of motion
∂L
= 0
∂t
∂L
= 0
∂ θ̇
⇒
E = Const
(21)
⇒
J = mr2 sinh2 ν θ̇
(22)
Exercise 3.4: Motion on a movable inclined plane
Consider a wedge, of mass M , which is free to move in the x direction. On the wedge there
is a small object of the mass m which is free so slide on the wedge.
1. Find the Lagrangian in the generalized coordinates of the problem.
2. What are the constants of motion.
3. Find the equations of motion, solve them for different cases and explain the meaning
of the results.
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Solution:
1. We choose as generalized coordinates the distance x of the vertical side of the wedge
from our coordinate system and the distance l of the center of mass of the small object
from the top of the wedge. If we denote by xCM and yCM the distances along the two
axis of the center of mass of the wedge from its bottom-left corner, then
xM = x + xcm ; ẋM = ẋ
yM = ycm ; ẏm = 0
(23)
(24)
The coordinates of the center of mass of the small object are
xm = x + l cos α ;
ym = h − l sin α ;
ẋm = ẋ + l˙ cos α
l˙ sin α
(25)
(26)
The Lagrangian of the system is
L=
1
1
(M + m) ẋ2 = ml˙2 + mẋl˙ cos α + mgl sin α
2
2
(27)
2. Since x is a cyclic coordinate we know that it’s momentum is the constant of motion.
Also the Lagrangian is t independent so the energy is also a constant of motion.
Px = (m + M ) ẋ + ml˙ cos α
(28)
3. We want to find the equation of motion. We start with l
m¨l + mẍ cos α = mg sin α
(29)
We can use the constant of motion to find ẍ which we can plug into the l EOM and
get
2
¨l 1 − m cos α
= g sin α
(30)
m+m
sin α
¨l =
g
(31)
cos2 α
1 − mm+m
m sin α cos α
ẍ = −
g
(32)
m (1 − cos2 α) + M
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These two equations show that, unless α = 0, the horizontal motion of the wedge as
well as the motion of the small mass on the surface of the wedge are characterized by
constant accelerations. When α = 0, the wedge is completely horizontal and neither
of the two objects move. On the other hand, when the mass of the wedge is much
larger the mass of the small object, i.e., when M >> m, the two expression reduce to
¨l = g sin α and ẍ = 0.
This is the familiar result that the wedge remains fixed and the small mass is accelerated
by the component of the force that is parallel to the inclined plane.
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