MPH-006 Solved Assignment
PART A
Question:-01
a) Obtain the Lagrangian, the Hamiltonian and the equations of motion for a projectile near the surface of the earth. Take the mass of the projectile to be
m
m
m m m
z
z
z z z
Answer:
To solve for the Lagrangian, Hamiltonian, and equations of motion for a projectile near the surface of the Earth, we follow these steps:
1. Lagrangian
The Lagrangian
L
L
L L L is defined as the difference between the kinetic energy
T
T
T T T and the potential energy
V
V
V V V :
L
=
T
−
V
L
=
T
−
V
L=T-V L = T – V L = T − V Kinetic Energy (
T
T
T T T
The kinetic energy of a projectile of mass
m
m
m m m moving with velocity
v
=
(
v
x
,
v
y
,
v
z
)
v
=
(
v
x
,
v
y
,
v
z
)
v=(v_(x),v_(y),v_(z)) \mathbf{v} = (v_x, v_y, v_z) v = ( v x , v y , v z ) is given by:
T
=
1
2
m
(
v
x
2
+
v
y
2
+
v
z
2
)
T
=
1
2
m
(
v
x
2
+
v
y
2
+
v
z
2
)
T=(1)/(2)m(v_(x)^(2)+v_(y)^(2)+v_(z)^(2)) T = \frac{1}{2}m(v_x^2 + v_y^2 + v_z^2) T = 1 2 m ( v x 2 + v y 2 + v z 2 ) Expressing velocity components in terms of the time derivatives of the coordinates:
T
=
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
T
=
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
T=(1)/(2)m(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2)) T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) T = 1 2 m ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) Potential Energy (
V
V
V V V
The potential energy of a projectile in a uniform gravitational field near the surface of the Earth is given by:
V
=
m
g
z
V
=
m
g
z
V=mgz V = mgz V = m g z where
g
g
g g g is the acceleration due to gravity, and
z
z
z z z is the height above the reference point.
Lagrangian (
L
L
L L L
Combining the kinetic and potential energies, we get the Lagrangian:
L
=
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
L
=
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
L=(1)/(2)m(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))-mgz L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) – mgz L = 1 2 m ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) − m g z 2. Hamiltonian
The Hamiltonian
H
H
H H H is the total energy of the system and is given by the Legendre transform of the Lagrangian. It is expressed in terms of the generalized coordinates
q
i
q
i
q_(i) q_i q i and the conjugate momenta
p
i
p
i
p_(i) p_i p i .
Conjugate Momenta (
p
i
p
i
p_(i) p_i p i
The conjugate momenta are defined as:
p
x
=
∂
L
∂
x
˙
=
m
x
˙
p
x
=
∂
L
∂
x
˙
=
m
x
˙
p_(x)=(del L)/(del(x^(˙)))=mx^(˙) p_x = \frac{\partial L}{\partial \dot{x}} = m\dot{x} p x = ∂ L ∂ x ˙ = m x ˙
p
y
=
∂
L
∂
y
˙
=
m
y
˙
p
y
=
∂
L
∂
y
˙
=
m
y
˙
p_(y)=(del L)/(del(y^(˙)))=my^(˙) p_y = \frac{\partial L}{\partial \dot{y}} = m\dot{y} p y = ∂ L ∂ y ˙ = m y ˙
p
z
=
∂
L
∂
z
˙
=
m
z
˙
p
z
=
∂
L
∂
z
˙
=
m
z
˙
p_(z)=(del L)/(del(z^(˙)))=mz^(˙) p_z = \frac{\partial L}{\partial \dot{z}} = m\dot{z} p z = ∂ L ∂ z ˙ = m z ˙ Hamiltonian (
H
H
H H H
The Hamiltonian is given by:
H
=
∑
i
p
i
q
˙
i
−
L
H
=
∑
i
p
i
q
˙
i
−
L
H=sum _(i)p_(i)q^(˙)_(i)-L H = \sum_i p_i \dot{q}_i – L H = ∑ i p i q ˙ i − L Substitute the momenta and velocities into the Hamiltonian:
H
=
p
x
x
˙
+
p
y
y
˙
+
p
z
z
˙
−
(
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
)
H
=
p
x
x
˙
+
p
y
y
˙
+
p
z
z
˙
−
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
H=p_(x)x^(˙)+p_(y)y^(˙)+p_(z)z^(˙)-((1)/(2)m(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))-mgz) H = p_x \dot{x} + p_y \dot{y} + p_z \dot{z} – \left( \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) – mgz \right) H = p x x ˙ + p y y ˙ + p z z ˙ − ( 1 2 m ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) − m g z ) Since
x
˙
=
p
x
m
x
˙
=
p
x
m
x^(˙)=(p_(x))/(m) \dot{x} = \frac{p_x}{m} x ˙ = p x m ,
y
˙
=
p
y
m
y
˙
=
p
y
m
y^(˙)=(p_(y))/(m) \dot{y} = \frac{p_y}{m} y ˙ = p y m , and
z
˙
=
p
z
m
z
˙
=
p
z
m
z^(˙)=(p_(z))/(m) \dot{z} = \frac{p_z}{m} z ˙ = p z m , we have:
H
=
p
x
2
m
+
p
y
2
m
+
p
z
2
m
−
(
1
2
m
(
p
x
2
m
2
+
p
y
2
m
2
+
p
z
2
m
2
)
−
m
g
z
)
H
=
p
x
2
m
+
p
y
2
m
+
p
z
2
m
−
1
2
m
p
x
2
m
2
+
p
y
2
m
2
+
p
z
2
m
2
−
m
g
z
H=(p_(x)^(2))/(m)+(p_(y)^(2))/(m)+(p_(z)^(2))/(m)-((1)/(2)m((p_(x)^(2))/(m^(2))+(p_(y)^(2))/(m^(2))+(p_(z)^(2))/(m^(2)))-mgz) H = \frac{p_x^2}{m} + \frac{p_y^2}{m} + \frac{p_z^2}{m} – \left( \frac{1}{2}m\left(\frac{p_x^2}{m^2} + \frac{p_y^2}{m^2} + \frac{p_z^2}{m^2}\right) – mgz \right) H = p x 2 m + p y 2 m + p z 2 m − ( 1 2 m ( p x 2 m 2 + p y 2 m 2 + p z 2 m 2 ) − m g z ) Simplifying:
H
=
p
x
2
2
m
+
p
y
2
2
m
+
p
z
2
2
m
+
m
g
z
H
=
p
x
2
2
m
+
p
y
2
2
m
+
p
z
2
2
m
+
m
g
z
H=(p_(x)^(2))/(2m)+(p_(y)^(2))/(2m)+(p_(z)^(2))/(2m)+mgz H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m} + mgz H = p x 2 2 m + p y 2 2 m + p z 2 2 m + m g z 3. Equations of Motion
The equations of motion can be derived using the Euler-Lagrange equations for the Lagrangian formulation or Hamilton’s equations for the Hamiltonian formulation.
Euler-Lagrange Equations:
The Euler-Lagrange equations are given by:
d
d
t
(
∂
L
∂
q
˙
i
)
−
∂
L
∂
q
i
=
0
d
d
t
∂
L
∂
q
˙
i
−
∂
L
∂
q
i
=
0
(d)/(dt)((del L)/(delq^(˙)_(i)))-(del L)/(delq_(i))=0 \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) – \frac{\partial L}{\partial q_i} = 0 d d t ( ∂ L ∂ q ˙ i ) − ∂ L ∂ q i = 0 For the coordinates
x
,
y
,
z
x
,
y
,
z
x,y,z x, y, z x , y , z :
d
d
t
(
m
x
˙
)
=
0
⟹
m
x
¨
=
0
⟹
x
¨
=
0
d
d
t
(
m
x
˙
)
=
0
⟹
m
x
¨
=
0
⟹
x
¨
=
0
(d)/(dt)(mx^(˙))=0Longrightarrowmx^(¨)=0Longrightarrowx^(¨)=0 \frac{d}{dt} (m\dot{x}) = 0 \implies m\ddot{x} = 0 \implies \ddot{x} = 0 d d t ( m x ˙ ) = 0 ⟹ m x ¨ = 0 ⟹ x ¨ = 0
d
d
t
(
m
y
˙
)
=
0
⟹
m
y
¨
=
0
⟹
y
¨
=
0
d
d
t
(
m
y
˙
)
=
0
⟹
m
y
¨
=
0
⟹
y
¨
=
0
(d)/(dt)(my^(˙))=0Longrightarrowmy^(¨)=0Longrightarrowy^(¨)=0 \frac{d}{dt} (m\dot{y}) = 0 \implies m\ddot{y} = 0 \implies \ddot{y} = 0 d d t ( m y ˙ ) = 0 ⟹ m y ¨ = 0 ⟹ y ¨ = 0
d
d
t
(
m
z
˙
)
=
−
m
g
⟹
m
z
¨
=
−
m
g
⟹
z
¨
=
−
g
d
d
t
(
m
z
˙
)
=
−
m
g
⟹
m
z
¨
=
−
m
g
⟹
z
¨
=
−
g
(d)/(dt)(mz^(˙))=-mgLongrightarrowmz^(¨)=-mgLongrightarrowz^(¨)=-g \frac{d}{dt} (m\dot{z}) = -mg \implies m\ddot{z} = -mg \implies \ddot{z} = -g d d t ( m z ˙ ) = − m g ⟹ m z ¨ = − m g ⟹ z ¨ = − g Thus, the equations of motion are:
x
¨
=
0
x
¨
=
0
x^(¨)=0 \ddot{x} = 0 x ¨ = 0
y
¨
=
0
y
¨
=
0
y^(¨)=0 \ddot{y} = 0 y ¨ = 0
z
¨
=
−
g
z
¨
=
−
g
z^(¨)=-g \ddot{z} = -g z ¨ = − g Hamilton’s Equations:
Hamilton’s equations are given by:
q
˙
i
=
∂
H
∂
p
i
q
˙
i
=
∂
H
∂
p
i
q^(˙)_(i)=(del H)/(delp_(i)) \dot{q}_i = \frac{\partial H}{\partial p_i} q ˙ i = ∂ H ∂ p i
p
˙
i
=
−
∂
H
∂
q
i
p
˙
i
=
−
∂
H
∂
q
i
p^(˙)_(i)=-(del H)/(delq_(i)) \dot{p}_i = -\frac{\partial H}{\partial q_i} p ˙ i = − ∂ H ∂ q i For the coordinates
x
,
y
,
z
x
,
y
,
z
x,y,z x, y, z x , y , z :
x
˙
=
∂
H
∂
p
x
=
p
x
m
x
˙
=
∂
H
∂
p
x
=
p
x
m
x^(˙)=(del H)/(delp_(x))=(p_(x))/(m) \dot{x} = \frac{\partial H}{\partial p_x} = \frac{p_x}{m} x ˙ = ∂ H ∂ p x = p x m
p
˙
x
=
−
∂
H
∂
x
=
0
p
˙
x
=
−
∂
H
∂
x
=
0
p^(˙)_(x)=-(del H)/(del x)=0 \dot{p}_x = -\frac{\partial H}{\partial x} = 0 p ˙ x = − ∂ H ∂ x = 0
y
˙
=
∂
H
∂
p
y
=
p
y
m
y
˙
=
∂
H
∂
p
y
=
p
y
m
y^(˙)=(del H)/(delp_(y))=(p_(y))/(m) \dot{y} = \frac{\partial H}{\partial p_y} = \frac{p_y}{m} y ˙ = ∂ H ∂ p y = p y m
p
˙
y
=
−
∂
H
∂
y
=
0
p
˙
y
=
−
∂
H
∂
y
=
0
p^(˙)_(y)=-(del H)/(del y)=0 \dot{p}_y = -\frac{\partial H}{\partial y} = 0 p ˙ y = − ∂ H ∂ y = 0
z
˙
=
∂
H
∂
p
z
=
p
z
m
z
˙
=
∂
H
∂
p
z
=
p
z
m
z^(˙)=(del H)/(delp_(z))=(p_(z))/(m) \dot{z} = \frac{\partial H}{\partial p_z} = \frac{p_z}{m} z ˙ = ∂ H ∂ p z = p z m
p
˙
z
=
−
∂
H
∂
z
=
−
m
g
p
˙
z
=
−
∂
H
∂
z
=
−
m
g
p^(˙)_(z)=-(del H)/(del z)=-mg \dot{p}_z = -\frac{\partial H}{\partial z} = -mg p ˙ z = − ∂ H ∂ z = − m g Thus, we have:
x
˙
=
p
x
m
,
y
˙
=
p
y
m
,
z
˙
=
p
z
m
x
˙
=
p
x
m
,
y
˙
=
p
y
m
,
z
˙
=
p
z
m
x^(˙)=(p_(x))/(m),quady^(˙)=(p_(y))/(m),quadz^(˙)=(p_(z))/(m) \dot{x} = \frac{p_x}{m}, \quad \dot{y} = \frac{p_y}{m}, \quad \dot{z} = \frac{p_z}{m} x ˙ = p x m , y ˙ = p y m , z ˙ = p z m
p
˙
x
=
0
,
p
˙
y
=
0
,
p
˙
z
=
−
m
g
p
˙
x
=
0
,
p
˙
y
=
0
,
p
˙
z
=
−
m
g
p^(˙)_(x)=0,quadp^(˙)_(y)=0,quadp^(˙)_(z)=-mg \dot{p}_x = 0, \quad \dot{p}_y = 0, \quad \dot{p}_z = -mg p ˙ x = 0 , p ˙ y = 0 , p ˙ z = − m g These equations confirm the same equations of motion derived from the Euler-Lagrange equations.
Summary
Lagrangian :
L
=
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
L
=
1
2
m
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
L=(1)/(2)m(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))-mgz L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) – mgz L = 1 2 m ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) − m g z Hamiltonian :
H
=
p
x
2
2
m
+
p
y
2
2
m
+
p
z
2
2
m
+
m
g
z
H
=
p
x
2
2
m
+
p
y
2
2
m
+
p
z
2
2
m
+
m
g
z
H=(p_(x)^(2))/(2m)+(p_(y)^(2))/(2m)+(p_(z)^(2))/(2m)+mgz H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m} + mgz H = p x 2 2 m + p y 2 2 m + p z 2 2 m + m g z Equations of Motion :
x
¨
=
0
,
y
¨
=
0
,
z
¨
=
−
g
x
¨
=
0
,
y
¨
=
0
,
z
¨
=
−
g
x^(¨)=0,quady^(¨)=0,quadz^(¨)=-g \ddot{x} = 0, \quad \ddot{y} = 0, \quad \ddot{z} = -g x ¨ = 0 , y ¨ = 0 , z ¨ = − g
b) A particle of mass
m
m
m m m has the Lagrangian:
L
=
1
2
μ
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
L
=
1
2
μ
x
˙
2
+
y
˙
2
+
z
˙
2
−
m
g
z
L=(1)/(2)mu(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))-mgz L=\frac{1}{2} \mu\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)-m g z L = 1 2 μ ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) − m g z where
μ
=
m
M
M
+
m
μ
=
m
M
M
+
m
mu=(mM)/(M+m) \mu=\frac{m M}{M+m} μ = m M M + m .
Determine the Routhian and the equation of motion.
Answer:
To determine the Routhian and the equations of motion for the given Lagrangian, we need to follow these steps:
1. Routhian
The Routhian
R
R
R R R is a function similar to the Hamiltonian but is used in problems with cyclic (ignorable) coordinates. It’s defined for a system where some coordinates are cyclic, meaning their corresponding conjugate momenta are constants of motion.
Given the Lagrangian:
L
=
1
2
μ
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
L
=
1
2
μ
x
˙
2
+
y
˙
2
+
z
˙
2
−
m
g
z
L=(1)/(2)mu(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))-mgz L = \frac{1}{2} \mu \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) – mgz L = 1 2 μ ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) − m g z where
μ
=
m
M
M
+
m
μ
=
m
M
M
+
m
mu=(mM)/(M+m) \mu = \frac{mM}{M + m} μ = m M M + m , we note that
x
x
x x x and
y
y
y y y are cyclic coordinates because they do not appear explicitly in the Lagrangian.
Conjugate Momenta:
For the cyclic coordinates
x
x
x x x and
y
y
y y y :
p
x
=
∂
L
∂
x
˙
=
μ
x
˙
p
x
=
∂
L
∂
x
˙
=
μ
x
˙
p_(x)=(del L)/(del(x^(˙)))=mux^(˙) p_x = \frac{\partial L}{\partial \dot{x}} = \mu \dot{x} p x = ∂ L ∂ x ˙ = μ x ˙
p
y
=
∂
L
∂
y
˙
=
μ
y
˙
p
y
=
∂
L
∂
y
˙
=
μ
y
˙
p_(y)=(del L)/(del(y^(˙)))=muy^(˙) p_y = \frac{\partial L}{\partial \dot{y}} = \mu \dot{y} p y = ∂ L ∂ y ˙ = μ y ˙ For the non-cyclic coordinate
z
z
z z z :
p
z
=
∂
L
∂
z
˙
=
μ
z
˙
p
z
=
∂
L
∂
z
˙
=
μ
z
˙
p_(z)=(del L)/(del(z^(˙)))=muz^(˙) p_z = \frac{\partial L}{\partial \dot{z}} = \mu \dot{z} p z = ∂ L ∂ z ˙ = μ z ˙ Velocity in terms of momenta:
x
˙
=
p
x
μ
x
˙
=
p
x
μ
x^(˙)=(p_(x))/(mu) \dot{x} = \frac{p_x}{\mu} x ˙ = p x μ
y
˙
=
p
y
μ
y
˙
=
p
y
μ
y^(˙)=(p_(y))/(mu) \dot{y} = \frac{p_y}{\mu} y ˙ = p y μ Routhian
R
R
R R R
The Routhian is defined by treating the cyclic coordinates as constants and is given by:
R
=
∑
i
∈
cyclic
p
i
q
˙
i
−
L
R
=
∑
i
∈
cyclic
p
i
q
˙
i
−
L
R=sum_(i in”cyclic”)p_(i)q^(˙)_(i)-L R = \sum_{i \in \text{cyclic}} p_i \dot{q}_i – L R = ∑ i ∈ cyclic p i q ˙ i − L For our system:
R
=
p
x
x
˙
+
p
y
y
˙
−
(
1
2
μ
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
)
R
=
p
x
x
˙
+
p
y
y
˙
−
1
2
μ
(
x
˙
2
+
y
˙
2
+
z
˙
2
)
−
m
g
z
R=p_(x)x^(˙)+p_(y)y^(˙)-((1)/(2)mu(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))-mgz) R = p_x \dot{x} + p_y \dot{y} – \left( \frac{1}{2} \mu (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) – mgz \right) R = p x x ˙ + p y y ˙ − ( 1 2 μ ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) − m g z ) Substitute
x
˙
x
˙
x^(˙) \dot{x} x ˙ and
y
˙
y
˙
y^(˙) \dot{y} y ˙ in terms of the momenta:
R
=
p
x
p
x
μ
+
p
y
p
y
μ
−
(
1
2
μ
(
p
x
2
μ
2
+
p
y
2
μ
2
+
z
˙
2
)
−
m
g
z
)
R
=
p
x
p
x
μ
+
p
y
p
y
μ
−
1
2
μ
p
x
2
μ
2
+
p
y
2
μ
2
+
z
˙
2
−
m
g
z
R=p_(x)(p_(x))/(mu)+p_(y)(p_(y))/(mu)-((1)/(2)mu((p_(x)^(2))/(mu^(2))+(p_(y)^(2))/(mu^(2))+z^(˙)^(2))-mgz) R = p_x \frac{p_x}{\mu} + p_y \frac{p_y}{\mu} – \left( \frac{1}{2} \mu \left( \frac{p_x^2}{\mu^2} + \frac{p_y^2}{\mu^2} + \dot{z}^2 \right) – mgz \right) R = p x p x μ + p y p y μ − ( 1 2 μ ( p x 2 μ 2 + p y 2 μ 2 + z ˙ 2 ) − m g z ) Simplifying:
R
=
p
x
2
μ
+
p
y
2
μ
−
(
1
2
(
p
x
2
μ
+
p
y
2
μ
+
μ
z
˙
2
)
−
m
g
z
)
R
=
p
x
2
μ
+
p
y
2
μ
−
1
2
p
x
2
μ
+
p
y
2
μ
+
μ
z
˙
2
−
m
g
z
R=(p_(x)^(2))/(mu)+(p_(y)^(2))/(mu)-((1)/(2)((p_(x)^(2))/(mu)+(p_(y)^(2))/(mu)+muz^(˙)^(2))-mgz) R = \frac{p_x^2}{\mu} + \frac{p_y^2}{\mu} – \left( \frac{1}{2} \left( \frac{p_x^2}{\mu} + \frac{p_y^2}{\mu} + \mu \dot{z}^2 \right) – mgz \right) R = p x 2 μ + p y 2 μ − ( 1 2 ( p x 2 μ + p y 2 μ + μ z ˙ 2 ) − m g z )
R
=
1
2
(
p
x
2
μ
+
p
y
2
μ
)
+
1
2
μ
z
˙
2
+
m
g
z
R
=
1
2
p
x
2
μ
+
p
y
2
μ
+
1
2
μ
z
˙
2
+
m
g
z
R=(1)/(2)((p_(x)^(2))/(mu)+(p_(y)^(2))/(mu))+(1)/(2)muz^(˙)^(2)+mgz R = \frac{1}{2} \left( \frac{p_x^2}{\mu} + \frac{p_y^2}{\mu} \right) + \frac{1}{2} \mu \dot{z}^2 + mgz R = 1 2 ( p x 2 μ + p y 2 μ ) + 1 2 μ z ˙ 2 + m g z 2. Equations of Motion
The equations of motion for
z
z
z z z can be derived from the Lagrangian.
Euler-Lagrange Equation for
z
z
z z z
d
d
t
(
∂
L
∂
z
˙
)
−
∂
L
∂
z
=
0
d
d
t
∂
L
∂
z
˙
−
∂
L
∂
z
=
0
(d)/(dt)((del L)/(del(z^(˙))))-(del L)/(del z)=0 \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{z}} \right) – \frac{\partial L}{\partial z} = 0 d d t ( ∂ L ∂ z ˙ ) − ∂ L ∂ z = 0 Since:
∂
L
∂
z
˙
=
μ
z
˙
∂
L
∂
z
˙
=
μ
z
˙
(del L)/(del(z^(˙)))=muz^(˙) \frac{\partial L}{\partial \dot{z}} = \mu \dot{z} ∂ L ∂ z ˙ = μ z ˙
∂
L
∂
z
=
−
m
g
∂
L
∂
z
=
−
m
g
(del L)/(del z)=-mg \frac{\partial L}{\partial z} = -mg ∂ L ∂ z = − m g The equation of motion is:
d
d
t
(
μ
z
˙
)
+
m
g
=
0
d
d
t
(
μ
z
˙
)
+
m
g
=
0
(d)/(dt)(muz^(˙))+mg=0 \frac{d}{dt} (\mu \dot{z}) + mg = 0 d d t ( μ z ˙ ) + m g = 0
μ
z
¨
=
−
m
g
μ
z
¨
=
−
m
g
muz^(¨)=-mg \mu \ddot{z} = -mg μ z ¨ = − m g
z
¨
=
−
m
g
μ
z
¨
=
−
m
g
μ
z^(¨)=-(mg)/(mu) \ddot{z} = -\frac{mg}{\mu} z ¨ = − m g μ Substitute
μ
=
m
M
M
+
m
μ
=
m
M
M
+
m
mu=(mM)/(M+m) \mu = \frac{mM}{M + m} μ = m M M + m :
z
¨
=
−
g
M
+
m
M
z
¨
=
−
g
M
+
m
M
z^(¨)=-g((M+m)/(M)) \ddot{z} = -g \frac{M + m}{M} z ¨ = − g M + m M So, the equations of motion are:
z
¨
=
−
g
M
+
m
M
z
¨
=
−
g
M
+
m
M
z^(¨)=-g((M+m)/(M)) \ddot{z} = -g \frac{M + m}{M} z ¨ = − g M + m M Summary
Routhian :
R
=
1
2
(
p
x
2
μ
+
p
y
2
μ
)
+
1
2
μ
z
˙
2
+
m
g
z
R
=
1
2
p
x
2
μ
+
p
y
2
μ
+
1
2
μ
z
˙
2
+
m
g
z
R=(1)/(2)((p_(x)^(2))/(mu)+(p_(y)^(2))/(mu))+(1)/(2)muz^(˙)^(2)+mgz R = \frac{1}{2} \left( \frac{p_x^2}{\mu} + \frac{p_y^2}{\mu} \right) + \frac{1}{2} \mu \dot{z}^2 + mgz R = 1 2 ( p x 2 μ + p y 2 μ ) + 1 2 μ z ˙ 2 + m g z
Equations of Motion :
z
¨
=
−
g
M
+
m
M
z
¨
=
−
g
M
+
m
M
z^(¨)=-g((M+m)/(M)) \ddot{z} = -g \frac{M + m}{M} z ¨ = − g M + m M