2 Home exercise sheet 2.1 Equations of motion and constraint systems

2
Home exercise sheet
2.1
Equations of motion and constraint systems
Exercise 2.1: Rotating marching band stick
Two points of mass m are joined by a rigid weightless rod of length l, the center of which
is constrained to move on a circle of radius a.
1. Set up the kinetic energy in generalized coordinates.
2. Find the equations of motion and solve them.
Solution:
1. We simply need to define the correct coordinates for each mass.
l
cos φ ; y1
2
l
= a cos θ − cos φ ; y2
2
l
= −aθ̇ sin θ − φ̇ sin φ ;
2
l
= −aθ̇ sin θ + φ̇ sin φ ;
2
x1 = a cos θ +
x2
ẋ1
ẋ2
l
sin φ
2
l
= a sin θ − sin φ
2
l
ẏ1 = aθ̇ cos θ + φ̇ cos φ
2
l
ẏ2 = aθ̇ cos θ − φ̇ cos φ
2
= a sin θ +
All that is left now is to plug these into the Lagrangian and do some algebra
1
l2 2
1
2
2
2
2
2 2
L = m ẋ1 + ẏ1 + m ẋ2 + ẏ2 = ... = m a θ̇ + φ̇
2
2
4
1
(1)
(2)
(3)
(4)
(5)
2. Notice how the Lagrangian has no dependence on the coordinates, only on the velocities,
when this happens the coordinates are called cyclic coordinates and they usually point
to some conservation law. So after using E-L our EOM are
1 2
ml φ̇ = c2
2ma2 θ̇ = c1 ;
2
2c2
φ = ml
2 t + c3
⇒
c1
θ = 2ma
2 t + c4
(6)
(7)
Look at the two EOM we received do they look familiar? They should! This is angular
momentum conservation for both angles.
Exercise 2.2: Linear pendulum using Lagrange multipliers
Solve the problem of a linear pendulum confined to the (x, y) plane using Lagrange
multipliers:
1. Choose x and y as the coordinates of the mass and write down the Lagrangian in terms
of these.
2. Are there any constraints in the system? Write down their equation.
3. Using Lagrange multipliers derive the equation of motion for the pendulum using the
x and y as variables.
4. Write down the Lagrangian in spherical coordinates, find the EOM and obtain the
strength of the force that constraints the system.
5. Suppose the mass, initially at rest at it’s equilibrium position, is given an impulse
S in the positive x direction. Determine the range of impulses for which the string
crumples and the angle θc at which it crumples. Guidance: The impulse S is just the
momentum of the pendulum at the start of the movement. Use the fact that the energy
is conserved and don’t forget to use 4.
2
Solution:
1. For the pendulum the Lagrangian in the Cartesian coordinates system is simple
1
L = m ẋ2 + ẏ 2 − mgy
2
2. The constraint for the problem is the non elastic string, meaning that
p
x2 + y 2 = l 2 ⇒
x2 + y 2 = l
(8)
(9)
3. We now write the full Lagrangian (including the Lagrange multipliers)
p
1
2
2
2
2
L = m ẋ + ẏ − mgy − λ
x +y −l
(10)
2
notice that for λ to be the force eventually, we need that the constraint will be in units
of distance.
Using E-L we get the EOm
y
x
; mÿ = −mg −
(11)
mẍ = −λ
l
l
where we have already inserted the constraints.
4. We now move to polar coordinates, this time the constraint is r = l
1 L = m ṙ2 + r2 θ̇ + mgr cos θ − λ (r − l)
2
and the EOM are
mr̈ − mrθ̇2 − λ = 0
mr2 θ̈ + 2mrṙθ̇ + mgr sin θ = 0
r−l = 0
so we can now plug the constraints into the equations to receive
g
θ̈ = − sin θ
l
T = λ = mlθ̇2 + mg cos θ
(12)
(13)
(14)
(15)
(16)
(17)
λ is the string tension where the first term is the centrifugal force and the second one
is the gravitational force.
3
5. Since the Lagrangian is time independent we know that the energy is conserved so
E=
1
s2
− mgl = ml2 θ̇2 − mgl cos θ
2m
2
(18)
looking carefully we see that the left hand side is similar to our new found force
1
1
lT = ml2 θ̇2 − mgl cos θ
2
2
(19)
plugin this into the energy conservation we get
T =
s2
+ 3mg cos θ − 2mg
ml
(20)
We now look for the point where T stops ”pulling”, meaning that T = 0
s2
1
s2
0=
+ 3mg cos θc − 2mg ⇒
20 2
= cos θc
ml
3
m gl
Now thinking about it , the string we can only actually crumble for
us the following range of impulses
5
m2 gl < s2 < m2 gl
2
π
2
(21)
< θc < π giving
(22)
Exercise 2.3: Two masses, one swinging
Two equal masses m, connected by a massless string, hang over two pulleys (of negligible size). The left one moves in a vertical line, but the right one is free to swing back and
forth in the plane of the masses and pulleys.
1. Find the equations of motion for r and θ, as shown.(Try both with Lagrange multiplier
and without - check to see if there is a difference)
4
2. Assume that the left mass starts at rest, and the right mass undergoes small oscillations
with angular amplitude (with << 1). What is the initial average acceleration
(averaged over a few periods) of the left mass? In which direction does it move?
Solution:
1. This can be solved by either using Lagrange multipliers or by plugin the constraint
straight into the Lagrangian.
The constraint of the problem is
⇒
r1 + r2 = R
r1 = R − r2
We now plug the constraint into the Lagrangian
1 2 1 2
mṙ1 + m ṙ2 + r22 θ̇2 + mgr1 + mgr2 cos θ =
L =
2
2
1
= mṙ2 + mr2 θ̇2 − mgr (1 − cos θ)
2
where we have renamed r2
EOM are
→
(23)
(24)
(25)
r and have removed the constant term mgR. The
2r̈ = rθ̇2 − g (1 − cos θ)
r2 θ̈ = −2rṙθ̇ − gr sin θ
Doing the same exercise with the following Lagrangian
1
1 L = mṙ12 + m ṙ22 + r22 θ̇2 + mgr1 + mgr2 cos θ − λ (r1 + r2 − R)
2
2
(26)
(27)
(28)
will give us the exact same EOM.
2. We now want to solve small oscillations with the initial condition ṙ = 0 with sin θ = θ
and cos θ = 1 and we get
r̈ = 0
(29)
θ̈ = − gr θ
5
where we neglected θ̇2 because << 1.This is harmonic oscillator for θ and no motion
for r.
So we got that the left mass doesn’t move at all, but if we want the leading term for r
we need a bit less coarse approximation.
So we now leave the leading terms giving us
θ2
2r̈ = rθ̇ − g
2
g
θ̈ = − θ
r
2
The solution to the second equation given the known amplitude is
r
g
θ = cos (ωt + φ) ; ω =
r
we can plug this into the first equation now and get
1
2
2
2
r̈ = g sin (ωt + φ) − cos (ωt + φ)
2
since we want the average over a full (or a few) periods we get
2
1
2
2
hr̈i = g sin (ωt + φ) −
cos (ωt + φ)
2
2 g
=
8
This a second order effect, since it is positive the left mass starts to slowly climb.
6
(30)
(31)
(32)
(33)
(34)
(35)