ADEOBA OLAMIDE JOSEPH

(1808008014)

FACULTY OF SCIENCE

DEPARTMENT OF PHYSICS

EKITI STATE UNIVERSITY (EKSU), ADO EKITI, EKITI

STATE

PHY 413 ASSIGNMENT

ANALYTICAL MECHANICS

PROF. HAMEED
1. Rigid body of a homogenous body is define as the quantity expressed by the body
resisting angular acceleration which is the sum of the product of the mass of every
particle with its square of a distance from the axis of rotation.

2. (i) At one end

I = ∫ ρ R dv
2

Density(𝛒) = Volume
Mass

Volume = Area× Length

V= A × L
V= A × xdv
dv =Adx

I = ∫ ρ R2 Adx
I = ∫ ρ x 2 Adx
I = ρA ∫ x 2 dx

I = ρA ∫ x dx
L
2

I = ρA [ ]
L
x3
3 0

I = 3 ρA L
1 3

I = 3 ( ρAL)L
1 2

Mass = Density × Volume

I = 3ML
1 2

(ii) At Centre
 Recall: I = ∫ ρ R dv
2

dv =Adx

R= x

I=
+L
2

∫ ρ x 2 Adx
−L
2

I = ρA [ ]
+L
x3 2

3 −L
2

[][ ]
I = ρA 2 − 2
3 3
L −L

3 3

I = 12 M L
1 2

3. A rigid body is a body in which the distance between all its

components and particles remain fixed under the application of a
force.

Types motion rigid body can undergo:

a. Translational Motion: All the particles in a rigid body are parallel
path so that the line pointing any two points in the body always
remains parallel to the initial position. All of the body’s particles
have the same velocity at any one time. We treat the entire
system as a single point like particles with mass M at the centre
of mass, travelling at the centre of the mass’s velocity V cm. At the
centre of the mass, the external force acting on the system.
Where
F= dt = dt (MT VCM)
dp d

b. Rotational Motion: The motion of a rigid body is rotating around

an axis when all the particles describe circular path around a
line called the axis of rotation. The Newton Law of motion is
applied. That is, unless an external torque is given to a body to
remain in a state of rest or uniform rotational motion around a
 fixed axis. If we calculate acceleration for a frame that is pure
translating with an acceleration, we get
∑F=m∑a+m ∑a

M Y
Z r
4.

Z’
Let angular momentum = ι
in z – direction
ι z =Pr=L Z

Then LZ = MVr
LZ = M Z V Z r Z

But V =ωr
LZ = M Z ω r Z ∙r Z
2
LZ = M Z ω r Z

For the whole body

L=∑ L Z , L=l 1 +l 2 +… l N
2 2 2
L=M Z1 r Z 1 ω+ M Z2 r Z 2 ω+… + M Z N r Z N ω
LZ =¿

Recall that moment of Inertial

I =∑ M i r i
2

LZ =Iω

5. Hamilton’s principle states that the motion of a physical system

from time t 0 to t 1 is such that the integral I =∫ L d t when L=T −V
has a stationary value for the correct path of the motion.

6. The Hamiltonian function is given as:

H=∑ q̇i p i−L(q i , q̇ i)

H= q̇ p−L(q q̇) ……….. (eqn 1)

i

where P is the generalized momentum defined as

…………………..
dL
P= ( eqn 2 a )
d q̇
 dq ………………………
dL
P= (eqn 2 b)

taking the differential at (0)

∂ p …………….
∂H ∂H
dH = dq+ dP (eqn 3)
∂q

taking the ∂differentiation of eqn(1)

d q̇ …………….. (eqn 4)
dH =q̇ dp+ pd q̇−
[ ∂L
∂q
dq+
∂L
∂ q̇ ]
put eqn ( 2 a ) ∧eqn ( 2 b ) ∈eqn(4)
dH =q̇ dp+ Pd q̇− Ṗ dq−Pd q̇

dH =q̇ dp− Ṗ dq ………………… (eqn 5)

compare ( eqn 3 )∧¿(eqn 5) we obtain:

∂H
− Ṗ=
∂q
∂H
−q̇=
∂p
Hamiltonian canonical equations
7. From the equations obtained P and q are said to be canonical conjugates.
8. Recall that Moment of Inertia
I =∑ M i r i
2

mass=12 kg

Frequency = 400rpm
radius=16 cm≃ 0.16 m
2
I =( 12 ) ×( 0.16)
I =12 ×0.0256
I =0.3072
−1 3
I =3.072 ×10 kg m
¿ get the Kinetic energy
1 2
K . E= M V
2
1 2
K . E= I ω
2
where ω=2 πf
2 π (400)
ω=
60
−1
ω=41.89rad s
 1
K . E= ( 3.072 ×10 ) (41.89)
−1 2
2

K . E=269.5 Joules

9. F= f(x, y) and I= ∫ f ( y , ẏ )dx . Derive Euler Lagrange’s Equation

∂ I =0 …………….(eqn 1)
x0

If F =f(x, y)

dy …………….(eqn 2)
∂F ∂F
dF= dx+
∂x dy

and I =∫ f ( y , ẏ )dx …………….(eqn 3)

X

x0

F=F(y, ẏ )
∂F ∂F
∂F = ∂ y+ ∂ ẏ
∂y ∂y
d
Where ∂ y= (∂ y)
dx
∂y ∂y
= lim
∂ x ∂ x →0 ∂ x
From equation 3

∂ I =∫ ∂ F ¿ ¿)dx …………….(eqn 4)
X

x0

∂ ẏ ¿ …………….(eqn 5)
X
∂F ∂F
∂ I =∫ ∂ y+ ¿
x0 dy dy

(∂ y)]dx ¿ …………….(eqn 6)
X
∂F ∂F d
∂ I =∫ [ ∂ y +¿
x0 dy d ẏ dx

[ ] [ ( ∂ y ) dx …………….(eqn 7)
]
X X
∂F ∂F d
∂ I =∫ ∂ y dx +∫
x0 dy x 0 d ẏ dx

u dv
Recall: ∫ udv =uv−¿∫ vdu ¿

[ ] [ ] )∂ ydx ¿ ¿…………….(eqn 8)
X X
∂F d ∂F d ∂F
∂ I =∫ ∂ y dx= ∂ y −∫ (¿ ¿
x0 d ẏ dx d ẏ x 0 dx d ẏ

Put equation 8 into 7

 [ ] [ ] [ ] ∂ ydx …………….(eqn 9)
X X
∂F ∂F d ∂F
∂ I =∫ ∂ y dx + ∂ y −∫
x0 dy d ẏ x0 dx dy

∫[ ] [ ] ∂ y x1−x 0 …………….(eqn 10)

X
∂F d ∂F ∂F
∂ I= − ( ) ∂ ydx +
x0 dy dx dy d ẏ

∂ y ¿0) …………….(eqn 11)

∂F ∂F
¿ ∂ y ¿1) −¿
dy d ẏ
¿ 0−0
Since ∂ y ¿0) =0 and ∂ y ¿1) =0
To determine the minimum value at x and y ∂ I =0
If ∂ I −0 , thenthe integral must be equal ¿ zero .

[ ∂ ydx …………….(eqn 12)

]
X
∂F d ∂F
∂ I =∫ −
x0 dy dx dy
∂ I =0
∂F d ∂F
− =0 Euler Lagrange’s Equation
dy dx d ẏ

10.i. They are used in the advance treatment of mechanics and quantum
mechanics.
ii. they are used in writing the equation of motion in any system of
coordinates.
iii. It can be use to predict how a classical system behaves
11.i. in generalized coordinates:
Lagrange L=T −V
whereT is Kinetic Energy
V is Potential Energy of the system
L can be expressed as:
1
L= M ( ẋ + ẏ + ż )−V ( x , y , z )
2 2 2
2
1 2
T = M ẋ
2
V =V (x)
1 2
L= M ẋ −V ( x )
2
dL
=m ẋ
d ẋ
m ẋ=Px(Momentum)
 dL −d v( x) = F (Conservative force)
= x
dx dx
Momentum= mass X velocity
Force X time = momentum
momentum
Force = time
dP
F=
dt
dL d dL
=
dx dt d ẋ( )
Therefore,
dL d dL
( )
d qi = dt d q̇i

ii. in generalized momenta:

dL
Pi=
d qi

iii. . in generalized force:

−dL
F i=
d qi

12. Recall that: L=T −V …………. eqn (1)

V =mgh

h¿ l−l cos θ
l=r

V =mg ( l−l cos θ )

V =mgl−mgl cos θ ……………eqn(2)

1 2
T= mv
2
V =ωr=l θ̇
1 2 2
T = ml θ̇ …………………….eqn(3)
2
Substitute eqn (2) and eqn (3) into eqn (1)
 1 2 2
L= m l θ̇ −(mgl−mgl cos θ)
2

1 2 2
L= m l θ̇ −mgl+mgl cos θ ………………. *
2
Recall from Lagrange equation
dP
F=
dt
d dL
=( )
dL
dt d q̇ dq

q=θ andq̇=θ̇

d dL
=
( )
dL
dt d θ̇ d θ

Differentiate * wrtθ
dL
=−mgl sin θ
dθ
Differentiate * wrtθ̇
dL 2
=ml θ̇
d θ̇
d
( ml 2 θ̇ ) =−mgl sin θ
dt
dθ
Let θ̇= dt

d
dt
ml 2 (
dθ
dt )
=−mgl sinθ

2
2 d θ
ml 2
=−mgl sin θ
dt
2
d θ
l 2
=−g sin θ
dt
2
d θ −g sin θ
2
=
dt l
2
d θ g
2
+ sin θ=0 (pendulum equation)
dt l

13. The curve is represented by an equation y =y(x) having 2 close point

∂ y ¿0) =0 …………. eqn (1)
∂ y ¿1) =0 …………. eqn (2)

dl 2=dy2+dx2

dl = (dy2+dx2)1/2 (dx2)1/2
dl
=¿2+dx2)1/2/dx = ¿2+dx2)1/2/(dx2)1/2
dx
=[(dy /dx)2+ (dx /dx )2]1/2
=dL/dx = (1+ ẏ 2)1/2
dL = (1+ ẏ 2)1/2 dx
X X

L= ∫ ∂ L=∫ (1+ ẏ 2)1/2 dx

x0 x0

L = ∫ (1+ ẏ 2)1/2 dx: Lagrange’s in Differential Form