Homework 14 Solutions. HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS ......
5
Homework 14 Solutions. There is a typo in eqn (2) in the solution to 7-34. 7-34. x R r m M θ The coordinates of the wedge and the particle are cos 0 s M m M m x x x r y y r in x θ θ = = = =− + (1) The Lagrangian is then ( ) 2 2 2 2 2 cos 2 sin sin 2 M m m L x r r xr xr mgr r θ θ θ θ + = + + + − + θ (2) Note that we do not take r to be constant since we want the reaction of the wedge on the particle. The constraint equation is ( ) , , 0 f x r r R θ = − = . a) Right now, however, we may take r = R and 0 r r = = to get the equations of motion for x and θ. Using Lagrange’s equations, ( ) 2 sin cos x aR θ θ θ θ = + (3) sin cos x g R θ θ θ + = (4) where ( ) a m M m ≡ + . b) We can get the reaction of the wedge from the Lagrange equation for r 2 cos sin mx mR mg λ θ θ = − − θ (5) We can use equations (3) and (4) to express in terms of θ and x θ , and substitute the resulting expression into (5) to obtain ( ) 2 2 1 sin 1 sin a R g a λ θ θ ª º − = + « » − ¬ ¼ θ (6) To get an expression for θ , let us use the conservation of energy ( ) 2 2 2 0 2 sin sin sin 2 2 M m m H x R xR mgR mgR θ θ θ θ θ + = + − − =− (7) where 0 θ is defined by the initial position of the particle, and 0 sin mgR θ − is the total energy of the system (assuming we start at rest). We may integrate the expression (3) to obtain si R n x a θ θ = , and substitute this into the energy equation to obtain an expression for θ ( ) ( ) 0 2 2 2 sin sin 1 sin g R a θ θ θ θ − = − (8) Finally, we can solve for the reaction in terms of only θ and 0 θ ( ) ( ) ( ) 3 0 2 2 3 sin sin 2 sin 1 sin mMg a M m a θ θ θ θ − − =− + − λ (9)
Transcript of Homework 14 Solutions. HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS ......
Homework14Solutions.Thereisatypoineqn(2)inthesolutionto7-34.
HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 225
7-34.
x
Rr
m
M
θ
The coordinates of the wedge and the particle are
cos
0 s
M m
M m
x x x r
y y r in
xθ
θ
= =
= = −
+ (1)
The Lagrangian is then
( )2 2 2 2 2 cos 2 sin sin2
M m mL x r r xr xr mgr
rθ θ θ θ
+= + + + − +� �� � � � � θ (2)
Note that we do not take r to be constant since we want the reaction of the wedge on the particle. The constraint equation is ( ), , 0f x r r Rθ = − = .
a) Right now, however, we may take r = R and 0r r= =� �� to get the equations of motion for x and θ. Using Lagrange’s equations,
( )2sin cosx aR θ θ θ θ= +�� ��� (3)
sin cosx g
Rθ θ
θ+
=���� (4)
where ( )a m M m≡ + .
b) We can get the reaction of the wedge from the Lagrange equation for r
2cos sinmx mR mgλ θ θ= − −��� θ (5)
We can use equations (3) and (4) to express in terms of θ and x�� θ� , and substitute the resulting expression into (5) to obtain
( )22
1sin
1 sina
R ga
λ θθ
−= +
−� θ (6)
To get an expression for θ� , let us use the conservation of energy
( )2 2 202 sin sin sin
2 2M m m
H x R xR mgR mgRθ θ θ θ θ+
= + − − = −� �� � (7)
where 0θ is defined by the initial position of the particle, and 0sinmgR θ− is the total energy of the system (assuming we start at rest). We may integrate the expression (3) to obtain
siR nx a θ θ=� � , and substitute this into the energy equation to obtain an expression for θ�
( )( )
022
2 sin sin
1 sin
g
R a
θ θθ
θ−
=−
� (8)
226 CHAPTER 7
Finally, we can solve for the reaction in terms of only θ and 0θ
( )( )( )
30
22
3 sin sin 2 sin
1 sin
mMg a
M m a
θ θ θ
θ
− −= −
+ −λ (9)
7-35. We use iz and as our generalized coordinates, the subscript i corresponding to the ith particle. For a uniform field in the z direction the trajectories z = z(t) and momenta p = p(t) are given by
ip
2
0 0
0
12i i i
i i
z z v t gt
p p mgt
= + −
= −
(1)
where 0iz , , and 0ip 0 0i iv p= m are the initial displacement, momentum, and velocity of the ith particle.
Using the initial conditions given, we have
201 0
12
p tz z
m= + − gt
t
(2a)
1 0p p mg= − (2b)
202 0 0
12
p tz z z g
m= + ∆ + − t
t
(2c)
2 0p p mg= − (2d)
( )0 0 2
3 012
p p tz z
m+ ∆
= + − gt
t
(2e)
3 0 0p p p mg= + ∆ − (2f)
( )0 0 2
4 0 012
p p tz z Z g
m+ ∆
= + ∆ + − t
t
(2g)
4 0 0p p p mg= + ∆ − (2h)
The Hamiltonian function corresponding to the ith particle is
22
const.2 2
iii i i i i
pmzV mgz mgz
m= + = + = + =
�H T (3)
From (3) the phase space diagram for any of the four particles is a parabola as shown below.
226 CHAPTER 7
Finally, we can solve for the reaction in terms of only θ and 0θ
( )( )( )
30
22
3 sin sin 2 sin
1 sin
mMg a
M m a
θ θ θ
θ
− −= −
+ −λ (9)
7-35. We use iz and as our generalized coordinates, the subscript i corresponding to the ith particle. For a uniform field in the z direction the trajectories z = z(t) and momenta p = p(t) are given by
ip
2
0 0
0
12i i i
i i
z z v t gt
p p mgt
= + −
= −
(1)
where 0iz , , and 0ip 0 0i iv p= m are the initial displacement, momentum, and velocity of the ith particle.
Using the initial conditions given, we have
201 0
12
p tz z
m= + − gt
t
(2a)
1 0p p mg= − (2b)
202 0 0
12
p tz z z g
m= + ∆ + − t
t
(2c)
2 0p p mg= − (2d)
( )0 0 2
3 012
p p tz z
m+ ∆
= + − gt
t
(2e)
3 0 0p p p mg= + ∆ − (2f)
( )0 0 2
4 0 012
p p tz z Z g
m+ ∆
= + ∆ + − t
t
(2g)
4 0 0p p p mg= + ∆ − (2h)
The Hamiltonian function corresponding to the ith particle is
22
const.2 2
iii i i i i
pmzV mgz mgz
m= + = + = + =
�H T (3)
From (3) the phase space diagram for any of the four particles is a parabola as shown below.
HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 227
p0 + ∆p0
z0 + ∆z0
p0
Area at t = 0
Area at t = t1
∆p0
z0
∆z0
p
z
12
3 4
From this diagram (as well as from 2b, 2d, 2f, and 2h) it can be seen that for any time t,
1p p2= (4)
3p p4= (5)
Then for a certain time t the shape of the area described by the representative points will be of the general form
p
z
1 2
3 4
(p1,z1) (p2,z2)
(p3,z3) (p4,z4)
where the base 1 must parallel to the top 2 3 4 . At time t = 0 the area is given by 0 0z p∆ ∆ , since it corresponds to a rectangle of base 0z∆ and height 0p∆ . At any other time the area will be given by
( ){( ) }( ){( ) }
1 1
1
1 1
1
2 1
4 3 0
3 1
4 2 0
0 0
base of parallelogram
height of parallelogram
=
t t t t
t t
t t t t
t t
A z z
z z z
x p p
p p p
p z
= =
=
= =
=
= = −
= − = ∆
= −
= − = ∆
∆ ∆ (6)
Thus, the area occupied in the phase plane is constant in time.
7-36. The initial volume of phase space accessible to the beam is
20 0V R p2
0π π= (1)
After focusing, the volume in phase space is
2 21 1V R p1π π= (2)
HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 227
p0 + ∆p0
z0 + ∆z0
p0
Area at t = 0
Area at t = t1
∆p0
z0
∆z0
p
z
12
3 4
From this diagram (as well as from 2b, 2d, 2f, and 2h) it can be seen that for any time t,
1p p2= (4)
3p p4= (5)
Then for a certain time t the shape of the area described by the representative points will be of the general form
p
z
1 2
3 4
(p1,z1) (p2,z2)
(p3,z3) (p4,z4)
where the base 1 must parallel to the top 2 3 4 . At time t = 0 the area is given by 0 0z p∆ ∆ , since it corresponds to a rectangle of base 0z∆ and height 0p∆ . At any other time the area will be given by
( ){( ) }( ){( ) }
1 1
1
1 1
1
2 1
4 3 0
3 1
4 2 0
0 0
base of parallelogram
height of parallelogram
=
t t t t
t t
t t t t
t t
A z z
z z z
x p p
p p p
p z
= =
=
= =
=
= = −
= − = ∆
= −
= − = ∆
∆ ∆ (6)
Thus, the area occupied in the phase plane is constant in time.
7-36. The initial volume of phase space accessible to the beam is
20 0V R p2
0π π= (1)
After focusing, the volume in phase space is
2 21 1V R p1π π= (2)
HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 229
2
11 1 2 2
d xm g m
dtλ 0− + = (2)
2
22 2 2 0
d xm g m
dtλ− + = (3)
2
33 3 2 0
d xm g m
dtλ− + = (4)
Combining (1)–(4) we find
1 2
44 1 1
g
m m m
λ
3
−=
+ +
Finally, the string tension that acts on m is (see Eq. (2)) 1
2
11 1 1 2
1 2
82 4 1 1
gd xg m
dtm m m
λ= − = − =+ +
3
T m
7-38. The Hamiltonian of the system is
2 22 4 21
2 2 4 2 2pdx kx bx kx bx
U mdt m
= + = + + = + +4
4H T
The Hamiltonian motion equations that follow this Hamiltonian are
pdx H
dt p m∂
= =∂
3( )dp H
kx bxdt x
∂= − = − +
∂
7-39.
z
The Lagrangian of the rope is
2 2 21 1
2 2 2mgzdz mz z dz
U m g mdt b dt b
= − = − − = +2
L T
230 CHAPTER 7
from which follows the equation of motion
2
2
mgzL d L d zm
z dt z b dt∂ ∂
= ⇒ =∂ ∂�
7-40.
m
m
2 mx
θ2
θ1
We choose the coordinates for the system as shown in the figure.
The kinetic energy is
22 22 1 1
1
2 21 2 1 2
1 2 1
1 12 2 cos
2 2
1cos cos sin sin
2
d ddx dx dxT m m b b
dt dt dt dt dt
d d d ddxm b b b b
dt dt dt dt dt
= + + +
+ + + + +
θ θθ
θ θ θ θθ θ θ 2θ
2
The potential energy is
U m 1 1cos ( cos cos )gb mg b bθ θ θ= − − +
And the Lagrangian is
2 222 21 1 2
1
22 1 22 1 2 1
12 2 cos
2
cos cos ( ) 2 cos cos
d d ddx dxL T U m mb mb mb
dt dt dt dt dt
d d ddxmb mb mgb mgb
dt dt dt dt
= − = + + +
+ + − + +
θ θ θθ
θ θ θ2θ θ θ θ θ
From this follow 3 equations of motion
2 221 2
1 22 2 2
2 21 2
1 2
0 4 2 cos cos
2 sin sin
d dL d L d xb
x dt x dt dt dt
d db
dt dt
∂ ∂= ⇒ = + +
∂ ∂
− +
�θ θ
θ θ
θ θθ θ
