2 Class exercise sheet 2.1 Equations of motion and constraints systems

2
Class exercise sheet
2.1
Equations of motion and constraints systems
We define the Lagrangian as L = T − V , where T is the kinetic energy of the system and V
is the potential energy of the system. ∂L
− ∂q
= 0 we receive the equations of motion.
Utilizing the Euler-Lagrange equation dtd ∂∂L
q̇i
i
In the case where our system is constrained we can do one of two things:
1. We can either implement the constraint at the Lagrangian level.
2. We can you use Lagrange multipliers. The use of Lagrange multipliers requires us to
find a constraint equations in the form of G(q1 , ..., qn , t) = 0, which will be implemented
in the Lagrangian as L = T − V + λG(q1 , ..., qn , t).
The merits of using the Lagrange multipliers is that they represent the constraining force
(assuming that the constraint equation is in the units of length) and will allow us to find
the constraining forces if we wish.
It is important to notice that the meaning of λ is given by it’s multiplied constraint equation
i.e. if G(q1 , ...qn , t) = x2 − y 2 then λ is not the force but the force per length.
Exercise 2.1: Frictionless motion along a curve
A skier moves frictionlessly under the influence of gravity along a general curve y = h(x).
1. Write down the Lagrangian for the problem and the constraint.
2. Write down the equations of motion.
3. Obtain the force that constraints the motion of the skier.
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4. The skier flies off the curve when the vertical force of constraint Fy starts to become
negative, because the curve can only supply a positive normal force. Suppose the skier
starts from rest at a height y0 and determine the point x at which the skier detaches
from the curve.
Solution:
1. The position of the skier is given by x, y and the constraint y = h(x). So the Lagrangian
with the constraint is
1
L = m ẋ2 + ẏ 2 − mgy + λ (y − h(x))
2
2. We use the Euler-Lagrange equation to receive the equations of motion.
d ∂L
d ∂L
= mẍ
= mÿ
dt ∂ ẋ
dt ∂ ẏ
∂L
∂L
= −λh0 (x)
= −mg + λ
∂x
∂y
∂L
∂L
= 0
= y − h(x)
∂λ
∂ λ̇
Where
dh(x)
dx
(1)
(2)
(3)
(4)
= h0 (x), this gives us
mẍ = −λh0 (x)
and
mÿ + mg = λ
(5)
We need to find the equations of motion meaning we need some ẍ(essentially there is
only one parameter x because of the constraint) which we could possibly solve. We use
the constraint equation to find ÿ which we can then use to find ẍ.
y = h(x)
ẏ = h0 (x)ẋ
ÿ = h00 (x)ẋ2 + h0 (x)ẍ
−mẍ = mh0 (x) h00 (x)ẋ2 + h0 (x)ẍ + mh0 (x)g
h0 (x)h00 (x)ẋ2 + h0 (x)g
ẍ = −
1 + h02 (x)
2
(6)
(7)
(8)
3. The force is in face λ so we need to isolate λ. Using our perviously found ẍ we find
that
00
h (x)ẋ2 + g
0
(9)
mẍ = −λh (x)
⇒
λ=m
1 + h02 (x)
our problem is that we are now stuck with ẋ, luckily this we can find using the energy
conservation.
mgy0 =
1
1 + h02 (x) ẋ2 + mgh(x)
2
⇒
ẋ2 =
2g (y0 − h(x))
1 + h02 (x)
(10)
This we can now plug into equation (9) and get the constraining force
λ=
mg
1 + h02 (x) + 2 (y0 − h(x)) h00 (x)
2
(1 + h02 (x))
(11)
4. As mentioned in the item we want the point where λ stops begin positive, namely λ = 0.
This means that the equation which determines the detachment point is
1 + h02 (x) + 2 (y0 − h(x)) h00 (x) = 0
(12)
if h(x) was given we would have been able (hopefully) to solve this to obtain a specific
x.
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Exercise 2.2: Falling Ladder
A uniform ladder of length l and mass m has one end on a smooth horizontal floor and
the other end against a smooth vertical wall. The ladder is initially at rest and makes an
angle θ0 with respect to the horizon.
1. Write the two constrains reflecting the fact that the ladder is touching the wall and
the floor.
2. Write down the Lagrangian in terms of the center of mass coordinates and the angle
θ.
3. Write down the equations of motion.
4. Write the constraining forces as a function of θ.
There is guidance at the end of the exercise, but try solving it on your own before
C1 ẍẋ = C2 f (x) ẋ
C1 2
ẋ = C2 (f (x1 ) − f (x2 ))
2
⇒
5. Find the critical height at which the ladder will detach from the wall.
6. Express the time of detachment as a well defined integral (including borders).
Solution:
4
(13)
1. The two constraints reflecting the fact that the ladder touches the wall and the floor
are
1
G1 (x, y, θ) = x − l cos θ = 0
(14)
2
1
(15)
G2 (x, y, θ) = y − l sin θ = 0
2
2. Using the center of mass coordinates and the angle θ our Lagrangian is
1 2
1
1
1
2
2
L = m ẋ + ẏ + I θ̇ − mgy + λ1 x − l cos θ + λ2 y − l sin θ
2
2
2
2
(16)
3. We have one proper coordinate and so we know that essentially we will only need one
EOM and it will be for θ. Using E-L we will get
l
I θ̈ (λ1 sin θ − λ2 cos θ)
2
we now use the constraint equations to find ẍ and ÿ to find
l
l
2
2
;
ÿ =
θ̈ cos θ − θ̇ sin θ
ẍ = − θ̈ sin θ + θ̇ cos θ
2
2
this we plug into the θ EOM to get
mẍ = λ1
mÿ + mg = λ2
θ̈ = −
where α =
I
,
I0
I0 =
ml2
4
and ω0 =
pg
l
2ω02 cos θ
1+α
(17)
(18)
(19)
.
4. We take a second look at the θ EOM, if we multiply it by θ̇ we have a full derivative
which we integrate over to get
1
(1 + α) θ̇2 + 2ω02 sin θ = 2ω02 sin θ0
2
which is nothing but energy conservation. From here it follows that
θ̇2 =
4ω02 (sin θ0 − sin θ)
1+α
Since we found both θ̈ and θ̇ we can now plug them into λ1 and λ2 to get
mg
λ1 (θ) = −
(3 sin θ − 2 sin θ0 ) cos θ
1+α
mg
λ2 (θ) =
((3 sin θ − 2 sin θ0 ) sin θ + α)
1+α
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(20)
(21)
(22)
(23)
5. To find the critical height we need the detachment angle, which is given when λ1 = 0
which will give us
sin θd =
2
sin θ0
3
(24)
hence the critical height of the upper end of the ladder when it leaves the wall is yd =
2
l sin θ0
3
6. The time of detachment is
√
Z
Z
Z
dθ
1 + α θ0
dθ
√
Td (θ0 ) = dt =
=
2ω0
sin θ0 − sin θ
θ̇
θd
6
(25)
Exercise 2.3: Heavy bead sliding on a rotating wire
A bead slides without friction on a thin wire which is rotated about a vertical axis by
a motor at a constant angular frequency ω as shown in the figure. The wire is tilted away
from the z axis by a fixed angle α. The bead is constrained to move on the wire. Make sure
to consider gravity.
1. How many degrees of freedom are required to describe the motion of the bead?
2. Choose an appropriate set of generalized coordinates and use them to express the time
dependent Cartesian coordinates.
3. Write down the Lagrangian in terms of these.
4. Write down the constraints of the problem.
5. Write down the equations of motion.
6. Find the constraining forces on the bead.
7. Solve the equations of motion for the bead.
Solution:
1. There is only one dynamical variable needed to describe this system, so it has only one
degree of freedom.
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2. The best set of coordinates will be spherical coordinates
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
3. Using spherical coordinates the Lagrangian will be
1 2
2
2
2
2 2
L = m ṙ + r θ̇ + r sin θφ̇
2
4. We now look at the constraints of the problem
and
G1 φ̇ = φ̇ − ω 2 = 0
G2 (θ) = θ − α = 0
5. We can plug these straight into our Lagrangian to receive
1 L = m ṙ2 + r2 φ2 sin2 θ + r2 θ̇2 − mgr cos θ + λ1 φ̇ − ω 2 + λ2 (θ − α)
2
(26)
(27)
(28)
(29)
(30)
(31)
and using E-L we get
mr̈
mr2 φ̈ sin2 θ
mr2 θ̈
φ̇
θ
=
=
=
=
=
mrφ̇2 sin2 θ + mrθ̇2 − mg cos θ
λ1 − 2mṙφ̇ sin2 θ − +mr2 φ̇2θ̇ sin θ cos θ
mgr sin θ + λ2 − 2mrṙθ̇
ω2
α
(32)
(33)
(34)
(35)
(36)
Plugin the constraints we get the constraining forces
λ1 (r) = 2mrṙω 2 sin2 α
λ2 (r) = mgr sin α
(37)
(38)
6. The only EOM we have to solve is for r which is
r̈ − rω 2 sin2 α = −g cos α
(39)
You can solve this by first solving the homogeneous equation first and then the private
equation. The solution for this is
g cos α
r = 2 2 + Aeω sin αt + Be−ω sin αt
(40)
ω sin α
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Guidance for question 2.2: Solve the equation of motion for θ (to do this you will first
need to express the constraining forces as a function of θ and it’s derivatives) using the given
equation (Can you identify this equation?) and plug it back into the constraints.
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