Principles of Dynamics

Tom Charnock

Contents

1 Coordinates, Vectors and Matrices 4

1.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Cylindrical Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.2 Spherical Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Identity Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3.2 Levi Civita Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Products Of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4.1 Scalar (Dot) Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4.2 Vector (Cross) Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4.3 Product of Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Position, Velocity and Acceleration 6

2.1 Position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Operators 7

3.1 Gradient Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.2 Gradient (Grad) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.3 Divergence (Div) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.4 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Classical Mechanics 8

4.1 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4.1.1 Kepler’s Law of Planetary Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4.2 Galileo’s Principle of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4.3 Isaac Newton’s Principia Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4.3.1 Newton’s First Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.3.2 Newton’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.3.3 Newton’s Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.3.4 Law of Universal Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.4 From Newton to Kepler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

5 Action 13

5.1 Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5.2 Calculus of Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5.3 Euler-Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

6 Many Particle Systems 15

7 Hamilton’s Principle 16

7.1 Physical Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

7.2 Euler-Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2

8 Rigid Bodies 178.1 Euler’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178.2 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178.3 Moment of Inertia Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

8.3.1 Principal Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188.3.2 Perpendicular Axis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188.3.3 Parallel Axis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8.4 Kinetic Energy of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198.5 Gravitational Potential of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208.6 Compound Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208.7 Chasles’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

9 Double Pendulum 22

10 Constraints 2410.1 Newton’s Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2410.2 Holonomic Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2410.3 Non-Holonomic Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

11 Symmetries 2611.1 Constant Generalised Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2611.2 Constant Pseudo Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2611.3 Hamilton’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

11.3.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

12 Gyroscope 2812.1 Lagrangian Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

12.1.1 Euler’s Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2812.1.2 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

12.2 Hamiltonian Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3012.2.1 General Motion of the Gyroscope . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3012.2.2 Stability of a Vertical Top . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3

Chapter 1

Coordinates, Vectors and Matrices

1.1 Cartesian Coordinates

Cartesian coordinates are given by a set of vectors which can be written as:

(x, y, z) = v˜ =

v1

v2

v3

= v1i+ v2j + v3k

In index notation this can be written as:

v˜ =

3∑i=1

e˜ivix, y, z can be replaced with x1, x2, x3 so that a generalisation can be made to arbitrary dimensions ofindex notation.

1.2 Polar Coordinates

The symmetry of a system can often be solved more easily using a different coordinate system suchfocuses on the space around an object.

1.2.1 Cylindrical Polar Coordinates

This coordinate system is useful for calculating fields around a line such as a wire and is denoted by(%, θ, z) which gives the vector:

v˜ = v%%+ vθ θ + vz z

1.2.2 Spherical Polar Coordinates

This coordinate system is useful for calculating fields around a point, such as an atom and us denotedby (r, θ, ϕ), so that the vector is:

v˜ = vr r + vθ θ + vϕϕ

1.3 Matrices

A matrix can be written in terms of index notation as Mij so that a matrix looks like:

M =

M11 M12 M13

M21 M22 M23

M31 M32 M33

A vector can be multiplied by a matrix to obtain another vector where the sum is written in indexnotation as:

ui =

3∑j=1

Mijvj

4

1.3.1 Identity Matrix

The identity matrix is the matrix which when multiplied gives by a vector gives the same vector.

I =

1 0 00 1 00 0 1

This can be given by the Kronecker Delta δij where δij = 0 when i 6= j and δij = 1 when i = j.

1.3.2 Levi Civita Tensor

The Levi Civita Tensor can be used to indicate whether a matrix has cyclic permutations, anti-cyclicpermutations or non-cyclic permutations. It is indicated by:

ε123 = ε231 = ε312 = 1

ε132 = ε213 = ε321 = −1

Any other permutation of the matrix is given by εijk = 0.

1.4 Products Of Vectors

1.4.1 Scalar (Dot) Product

u˜ · v˜ = |u˜||v˜| cos θ

Where |u˜||v˜| is the norm of the vectors and θ is the angle between the vector. Using this it can be shownthat orthogonality is defined by two unit vectors pointing in the same direction:

u˜ · v˜ = 1

This is because cos θ = 1 when θ is 0. In index notation this can be shown by:

ei · ej = δij

So that when i = j then δij = 1 and when i 6= j then δij = 0.

1.4.2 Vector (Cross) Product

u˜× v˜ = |u˜||v˜| sin θnWhere n is the unit vector orthogonal to both u˜ and v˜. The vector product only works in three dimensions.If u˜ and v˜ are parallel then sin θ = 0 so that u˜× v˜ = 0. This can be written as:

ei × ej =

3∑k=1

εijkek

If ω˜ = u˜× v˜ then the components are given by:

ωi =∑ijk

εijkujvk

ω˜ =

u2v3 − v2u3

u3v1 − v1u3

u1v2 − v1u1

1.4.3 Product of Products

Some identities of the products of products are such that:

u˜ · (u˜× u˜) = 0

a˜ · (b˜× c˜) = c˜ · (a˜× b˜) = b˜ · (c˜× a˜)a˜× (b˜× c˜) = (a˜ · c˜)b˜− (a˜ · b˜)c˜

5

Chapter 2

Position, Velocity and Acceleration

2.1 Position

Position can be given by a vector:

r˜(t) =

x1(t)x2(t)x3(t)

=

3∑i=1

xi(t)ei

Distance is not the same as position because distance is a scalar quantity given by r(t) = |r˜(t)| =√x1(t)2 + x2(t)2 + x3(t)2.

2.2 Velocity

Velocity is the rate of change of position and can be denoted:

v˜(t) =δr˜(t)δt

= r˜(t) =

x1(t)x2(t)x3(t)

=

3∑i=1

xi(t)ei

Where e is a basis vector fixed with time. The speed is a scalar quantity like the distance, which is relatedto the velocity by v(t) = |v˜(t)| = |r˜| where |r˜| 6= r in general.

|r˜| =√x2

1 + x22 + x2

3

r =x1x1 + x2x2 + x3x3√

x21 + x2

2 + x23

In circular motion the radius is given by the scalar value r, which remains constant and r = 0. This doesnot imply that the speed has vanished because |r˜| 6= 0.

2.3 Acceleration

The acceleration is the rate of change of velocity or the rate of change of the rate of change of position.It is denoted by:

a(t) =δv˜(t)δt

=δ2r˜(t)δt2

= r˜ =

x1(t)x2(t)x3(t)

=

3∑i=1

xi(t)ei

The scalar acceleration is:

|r˜| =√x2

1 + x22 + x2

3

6

Chapter 3

Operators

Vector Calculus allows the rate of change in space to be calculated.

3.1 Gradient Operator

The gradient operator is given by ∇ which in index notation can be written as ∂∂xi

or even just ∂xi,where the latter is shortened terminology.

3.2 Gradient (Grad)

For scalar valued functions of space f(x1, x2, x3) the gradient (grad) is defined as:

∇f =

∂f∂x1∂f∂x2∂f∂x3

=

3∑i=1

∂f

∂xiei

The operator maps the scalar value function into a vector function.

3.3 Divergence (Div)

∇ can also act on vector functions. This is called the divergence or div.

∇ · F˜ =∂F1

∂x1+∂F2

∂x2+∂F3

∂x3=

3∑i=1

∂Fi∂xi

Divergence maps the vector quantity into a scalar function.

3.4 Curl

The curl maps a vector to a vector such that:

∇× F˜ =

3∑i=1

(εijk

∂Fk∂xj

)ei

For any scalar function ∇×∇f = 0.

7

Chapter 4

Classical Mechanics

4.1 Orbits

4.1.1 Kepler’s Law of Planetary Motion

Kepler’s First Law

Planets move in elliptical orbits with the sun at one of the foci.

Kepler’s Second Law

A line drawn between the sun and planet sweeps out equal areas in equal times.

Figure 4.1: Orbit Around a Foci with Equal Areas in Equal Times

Kepler’s Third Law

The square of the period is proportional to the cube of the radius of the semi-major axis:

T 2 ∝ r3

This comes from:

v =2πr

T=

√Gm

r

So that:

T 2 =

(4π2

Gm

)r3

4.2 Galileo’s Principle of Inertia

Galileo stated that the masses of falling bodies were independent from the uniform acceleration of thebody when resistance was negligible. This means that a body moving on a level surface would remaintravelling in the same direction with the same constant speed unless it were disturbed. Although thiswas discovered there was no real theory or mathematics to prove it.

4.3 Isaac Newton’s Principia Mathematica

Isaac Newton developed three laws of motion and a law of universal gravitation which proposed thetheory from which Kepler’s laws can be derived.

8

4.3.1 Newton’s First Law

An object at rest will remain at rest or an object moving at constant speed will remain at constant speedif the net force is zero. This is the same as Galileo’s principle of inertia.

4.3.2 Newton’s Second Law

The force applied on a body is equal to the rate of change of momentum.

F˜ =δp˜δt

= p˜When the mass remains constant then the velocity is equal to p˜ = mv˜ and so the second law can be

written as:

F˜ = mδv˜δt

= ma˜4.3.3 Newton’s Third Law

Every force occurs as one member of an action/reaction pair of forces.

4.3.4 Law of Universal Gravitation

Newton’s law of universal gravitation states that two masses can exert a force on each other at a distanceof r2.

F˜ =GMm

r2(−r˜)

Where the direction of the force is along the line joining the two masses.

4.4 From Newton to Kepler

Newton’s laws can be uses to find Kepler’s laws. Doing this uncovers the conservation of energy and theconservation of angular momentum. As r˜ =

r

r then the equation of motion from Newton’s second lawand law of universal gravitation can be written as:

r˜ = −GMr3

r˜By using scalar multiplication by r˜ on both sides of the equation then this becomes:

1

2r2 = −GM

r

This is because r˜ · r˜ = 12δr2

δt and r˜ · r˜ = 12δr2

δt . The equation found can be rearranged to obtain:

1

2r2 − GM

r= E

Where E is the total energy of the system and is constant due to δEδt = 0. This is the conservation of

energy where the two parts are the kinetic and potential energy of the system.Now by using vector multiplication on the original equation of motion by r˜ on both sides then it is found

that because r˜× r˜ =δ(r˜×r˜)δt due to r˜× r˜ = 0 and because r˜× r˜ = 0 then it can be seen that:

δ(r˜× r˜)δt

=δh˜δt

= 0

This is the conservation of angular momentum. Without loss of generality, h˜ can point along the z axisdue it being constant so that:

h˜ =

00h

9

With this claim the motion of the planet’s position is now confined to the x-y plane.

r˜ =

xy0

θ

r

rθ

x

y

Figure 4.2: Planar Polar Coordinates

Now converting this into planar polar coordinates then x = r cos θ and y = r sin θ This makes:

r =

cos θsin θ

0

, θ =

− sin θcos θ

0

The unit vectors r and θ are perpendicular so that r · r = 0. The velocity can be calculated so that:

r˜ =

xy0

=

r cos θ − rθ sin θ

r sin θ + rθ cos θ0

= rr + rθθ

Writing the angular momentum h˜ = r˜× r˜ in polar coordinates gives:

h˜ = (rr)× (rr + rθθ)

As rrr× r = 0 and r2θr× θ = 1 in the z direction then it can be seen that h = r2θ, which is the angularmomentum around z.The conservation of energy can also be written in polar coordinates in the form:

r˜2 = (rr + rθθ) · (rr + rθθ)

This expands to:r˜2 = r2r · r + 2rrθr · θ + r2θ2θ · θ

And because the middle of this is orthogonal then it is equal to zero. This leaves r˜2 = r2 + r2θ2 whichmeans the conservation of energy in polar form is:

E =1

2(r2 + r2θ2)− GM

r

The polar forms of the conservation of momentum and energy can be used to derive Kepler’s Laws ofPlanetary Motion. As r

θ= δr

δθ and θ = hr2 then:

E =1

2

(h

r2

)2[(

δr

δθ

)2

+ r2

]− GM

r

Substituting in u = 1r then:

E =1

2(hu2)62

[(− 1

u2

δu

δθ

)2

+1

u2

]−GMu

10

This can be rearranged to get:(δu

δθ

)2

= −(u− GM

h2

)2

+2E

h62+

(GM

h2

)2

Now taking another substitution of ω = u− GMh2 then it is seen that:

δω

δθ=

√2E

h2+GM

h2− ω2

In general, when integrating a kth variable then a trigonometric function should be used, so making

ω = αθ cosβ gives where α =

√2Eh2 +

(GMh2

)2:

δω

δθ= −α sinβ

δβ

δθ

This can be placed into the conservation of energy equation in polar coordinates to give:

α2 sin2 β

(δβ

δθ

)2

= α2 sin2 β

This means that β = ±(θ − θo) where θo is the constant of integration. When p = h2

GM and ε = αp thenthe position can be written as:

r =p

1 + ε cos(θ − θo)For a planet to reach ∞ there must be an angle θ∞ where 1 + ε cos(θ − θo) = 0. For ε < 1 there is nosolution and so the system is bound and the planet is in orbit, so the energy is E < 0. If ε > 1 then thereis a value of θ∞ and the planet is unbounded, which is the case when the energy is E > 0.

Kepler’s First Law

By rearranging the equation for position and placing in the values of x and y from polar coordinatesgives:

x2 + y2 = p2 − 2εpx+ ε2x2

Which can be rearranged to obtain:

(1− ε2)2

p2

(x+

εp

1− ε2

)2

+p2

1− ε2y2 = 1

This has the general form of an ellipse:

(x+ c)2

a2+y2

b2= 1

This therefore has shown Kepler’s First Law.

Kepler’s Second Law

The second law can been found from using the rate of change of area. A small change in area is given byδA = 1

2r2δθ for small values of θ. The rate of change of the area is given by A = 1

2r2θ and using θ = h

r2

the change in area in a given time is shown to be a constant of A = h2 .

Kepler’s Third Law

Finally Kepler’s Third Law can be found by using the total area of an ellipse:

Atot =

∫ T

0

δA

δtδt =

h

2T

The equation for the area of an ellipse is Atot = πab where b = a√

1− ε2. This leaves:

T 2

a3=

4π2a(1− ε2)

h2

11

And using the values of ε, a and p stated earlier this leaves:

T 2 =4π2

GMa3

Which is Kepler’s Third Law.

12

Chapter 5

Action

If a path is denoted by x˜(t) then after a short time the path may change from x˜(t) to x˜(t+ δt).

5.1 Lagrangian

A Lagrangian is generally a function of L(t, x˜(t), x˜(t)) and it can be used to calculated the action using:

S =

∫ t2

t1

Lδt

Where the boundary conditions come from x˜(t1) = x˜1 and x˜(t2) = x˜2. Nature always selects the pathwhere the functional S[x] is minimised. This can be found using calculus of variation.

5.2 Calculus of Variation

For F (t) the minima or maxima occurs at a time t = t∗ and is given by F ′(t∗) = 0. This definition of aderivative at a maxima or minima is:

limε→0

F (t∗ + ε)− F (t∗)ε

= 0

When functionals, such as the action, are involved then this becomes:

limε→0

S[x˜∗ + εf˜(t)]− S[x˜∗]ε

= 0

The boundary conditions on this are that x˜∗(t1) = x˜1, x˜∗(t2) = x˜2 and f˜(t1) = f˜(t2) = 0. This can then

be used to calculate the Euler-Lagrange equations of a system.

5.3 Euler-Lagrange Equations

If the general Lagrangian is given by L(t, xi(t), xi(t)) then the action is given by:

S[xi] =

∫ t2

t1

L (t, xi, xi) δt

If an arbitrary vector valued function fi(t) is introduced where the function can be minimised by fi(t1) =fi(t2) = 0 then the Lagrangian can be minimised.

S[xi + εfi] =

∫ t2

t1

L(t, xi + εfi, xi + εfi)δt

Using Taylor Expansion on the Lagrangian gives:

L(t, xi + εf1, xi + εfi) = L(t, xi, xi) + εfi∂L∂xi

+ εfi∂L∂xi

+O(ε2)

13

Now ignoring the orders of ε2 and using integration by parts on fi∂L∂xi

then the action becomes:

S[xi + εfi] = S[xi] +

[εfi

∂L∂xi

]t2t1

+ ε

∫ t2

t1

D∑i=1

fi∂L∂xi− fi

δ

δt

∂L∂xi

δt

And because[εfi

∂L∂xi

]t2t1

= 0 then finally minimising this gives:

0 =

∫ t2

t1

D∑i=1

fi

(∂L∂xi− δ

δt

∂L∂xi

)δt

And since fi(t) is arbitrary up to the boundary then for all i:

∂L∂xi− δ

δt

∂L∂xi

= 0

14

Chapter 6

Many Particle Systems

In a system with N particles in D dimensions the kth particle has the position x(k)i (t). This system is

described by the Lagrangian L(t, x

(k)i (t), x

(k)i (t)

).

x(1)(t)

x(2)(t)

x(3)(t)

XA(t)

Figure 6.1: Systems with Many Particles, or One Particle in Many Dimensions

Many paths can be described as a single path in DN dimensions:

X˜ (t) =

x(1)1...

x(1)D

x(2)1...

x(2)D...

x(N)D

In index notation this is XA(t) where A goes from 1→ DN . The Lagrangian in terms of DN dimensions

is L(t,XA(t), XA(t)

). From this the Euler-Lagrange equations are:

∂L∂x

(k)i

− δ

δt

∂L∂x

(k)i

= 0

15

Chapter 7

Hamilton’s Principle

”For a system described by a set of generalised coordinates qA, the correct path of motion qA(t) betweenan initial state qA(to) at time to and the final state qA(t1) at time t1, corresponds to a stationary path

of the ”action” S =∫ t1toLδt where L = L(t, qA, qA) is the Lagrangian describing the system.”

7.1 Physical Lagrangian

The physical Lagrangian is in the form of L = T −V where T is the kinetic energy and V is the potentialenergy.

7.2 Euler-Lagrange Equations

The stationary path is given by the solution of the Euler-Lagrange equations:

∂L∂qA

− δ

δt

∂L∂qA

= 0

16

Chapter 8

Rigid Bodies

Rigid bodies are many particle systems which are held together by rigid bonds. The sum over the particlesbecomes an integral over the volume and the mass becomes the mass density at a position xi, which isgiven by

∫volume

%(xi)δV .

8.1 Euler’s Theorem

r

r e(t)

r e(t

+t

)

O

ne

Figure 8.1: Rigid Body

The general displacement of a rigid body with one fixed point, O, is a rotation about some axis, throughO. The point is displaced by δr˜ = r˜(t+ δt)− r˜(t) where δr˜ is perpendicular to n˜. δr˜ is also perpendicularto r˜(t) when the movement is infinitesimal rotation. This means that δr˜ is parallel to n˜ × r˜(t). Thismeans that:

|δr˜| = |n˜||r˜| sin θδϕ = (n˜ × r˜)δϕWhen the rate of change of an infinitesimally small amount of time is taken into account then:

r˜ = (n˜ × r˜)ϕ = (ϕn˜)× r˜This leads to the angular velocity ω˜ = ϕn˜ so that r˜ = ω˜ × r˜.8.2 Angular Momentum

A rigid body as a system of particles of mass mk and position r(k)i relative to the fixed point. Considering

the kth particle then the linear momentum is p(k)i = mkr

(k)i (t) and the angular momentum is h

(k)i =

17

r(k)i × p

(k)i = mk

[r

(k)i ×

(ω

(k)i × r

(k)i

)]. Using vector product identities this can be shown to be:

h(k)i = mk

[|r(k)i |

2ω(k)i − (r

(k)i · ω

(k)i )r

(k)i

]The total angular momentum of the system can be found by summing over all the particles, and so fora rigid body the total angular momentum is given by:

hi =

∫body

%(ri)[|ri|2ωi − (ri · ωi)ri

]δV

This leads to hi = Iωi where I is the moment of inertia tensor. The total angular momentum of thesystem is:

hi =∑j

∫body

%(ri)[|r˜|2δij − rirj]ωjδV

So the moment of inertia tensor is:

Iij =

∫body

%(ri)[|r˜|2δij − rirj] δV

8.3 Moment of Inertia Tensor

In Cartesian coordinates the moment of inertia tensor is given by:

I =

∫ % [y2 − z2]δV −

∫% [xy] δV −

∫% [xz] δV

−∫% [xy] δV

∫%[x2 + z2

]δV −

∫% [yz] δV

−∫% [xz] δV −

∫% [yz] δV

∫%[x2 + y2

]δV

The moment of inertia about an axis along the n˜ direction is given by:

Inn =∑ij

Iij ninj

Using the moment of inertia tensor the moment of inertia bout the x axis is selected from the matrixIxx =

∫%(y2 + z2)δV . The moment of inertia tensor is both real and symmetric, which means that the

axis can be chosen so that I is diagonal.

8.3.1 Principal Axis

To find the principal axes the eigenvectors of the moment of inertia tensor must be computed. Thecorresponding eigenvalues are the values of the principal moment of inertia.

8.3.2 Perpendicular Axis Theorem

The moment of inertia of a lamina about an axis perpendicular to the lamina is the sum of the momentof inertia about two perpendicular axes in the plane of the lamina.

Figure 8.2: Lamina in the x-y plane

18

I =

∫%y2δV −

∫%xyδV 0

−∫%xyδV

∫%x2δV 0

0 0∫%(x2 + y2)δV

So the moments of inertia in the different axes are Ixx =

∫%y2δV and Iyy =

∫%x2δV and Izz is given by:

Izz = Ixx + Iyy

8.3.3 Parallel Axis Theorem

The moment of inertia about a parallel axis is I = ICM +Md2 where d is the perpendicular distance tothe parallel axis.

Figure 8.3: A Rigid Body Rotating about a Parallel Axis

If the inertia of the centre of mass is about Ixx and is shifted from y → y − d then:∫%(y2 + z2)δV →

∫%((y − d)2 + z2)δV

This gives the moment inertia as:

I = ICM − 2d

∫%yδV +Md2

As∫%yδV is the y component of the centre of mass then is goes to zero if at the origin. It is also trivial

that the integral of the density of the volume is the same as the total mass of the rigid body. Thistherefore leaves:

I = ICM +Md2

8.4 Kinetic Energy of a Rigid Body

Treating a rigid body as a system of N particles with the kth particle having mass mk and position x˜(k)

it is seen that:

Tl =1

2mk

∣∣∣x˜(k)∣∣∣2 =

1

2mkx˜(k) ·

(ω˜ × x˜(k)

)Where x˜(k) = ω˜ × x˜(k). Using scalar identities it can be seen that:

Tk =1

2ω˜ ·[x˜(k) ×mkx˜(k)

]=

1

2ω˜ · h˜k

19

And summing over these gives the total kinetic energy.

T =1

1ω˜h˜

Because h˜ = Iω˜ This shows that the total kinetic energy is the total angular momentum of the body.

T =1

2

∑ij

Iijωiωj

8.5 Gravitational Potential of a Rigid Body

For a single particle then the Gravitation Potential Energy of the particle (in the field of Earth) isV = −mg˜ · x˜, where g˜ = −gz. Considering a system of N particles, then the Gravitational Potential

Energy of the particle is Vk = −mkg˜ ·x˜(k). Using Newton’s Third Law then it is known hat if one particle

exerts a force F on another particle then the other particle exerts a force of −F on the first particle. Thismeans that for a rigid body internal forces cancel out leaving the total Gravitational Potential Energyas:

V =∑k

Vk = −g˜ ·∑k

m+ kx˜(k)

The definition of the centre of mass is states that:

Mx˜CM =∑k

mkx˜(k)

This means that for a rigid body the potential energy due to gravity of the body can be idealised as aparticle with its entire mass located at the centre of mass x˜CM .

V = −Mg˜ · x˜CM8.6 Compound Pendulum

Figure 8.4: Compound Pendulum

A compound pendulum is a shaped lamina which can swing freely about a point O. The moment ofinertia about the axis perpendicular to O is given by I and the centre of mass of the pendulum is a fixeddistance l from the origin. θ(t) is the angle made to the vertical at a time t. To find the Lagrangian ofthe system then the Kinetic Energy and the Potential Energy need to be found.The Kinetic Energy of a compound pendulum comes from the angular velocity, θ about the pivot point.Therefore in this case the Kinetic Energy is T = 1

2Iθ2.

The Potential Energy of a system with mass m is given by V = −g˜ ·x˜CM = −mgl cos θ, This means that

the Lagrangian is:

L = T − V =1

2Iθ2 +mgl cos θ

20

And using the Euler-Lagrange equations it can be seen that:

θ = −mgl sin θI

This states that there are equilibrium points at both θ(t) = 0 and θ(t) = π. These points correspondto when the pendulum is below and above the vertical axis. Slightly perturbing the system will show ifthe points are stable or not. Using the small angle approximation of sin θ = θ the equation of motionbecomes:

θ = −mglθI

This is the equation of motion for a simple harmonic oscillator with a frequency of ω =√

mglI and the

solution is θ(t) = A cosωt+ B sinωt. The fact that this is oscillating about the point θ(t) = 0 indicatesthat the system is stable.For θ(t) = π then small fluctuations give θ = π + δθ so:

δθ = −mglI

sin (π + δθ)

This gives a phase change to δθ = mglI sin δθ. This means that the solution can be written as:

δθ = Aeωt +Be−ωt

This is an exponential growth and shows that the system is unstable.

8.7 Chasles’ Theorem

So far it has been assumed that the pivot has been fixed, which is Euler’s Theorem. The corollary of thisis Chasles’ Theorem. It states that the most general displacement of a rigid body is a translation plus arotation.

Figure 8.5: Translation and Rotation of a Rigid Body

Since the centre of mass is always fixed in the rigid body the motion of the body relative to the centreof mass can only be a rotation. A system of N particles with the kth particle having a mass of mk andposition x˜(k)(t). The centre of mass position is x˜CM = 1

M

∑kmkx˜(k)(t). As each particle has a position

of S(k) = x˜(k) − x˜CM then relative to the centre of mass∑kmkS

(k) = 0. This can be used to calculatethe total kinetic energy of the system.

T =1

2

∑k

mk

∣∣∣x˜(k)∣∣∣2

This gives a equation with two parts. this first part is the Kinetic Energy of the total mass at the centreof mass and the second is the total Kinetic Energy of the particles relative to the centre of mass.

T =∑k

mk

∣∣∣S(k)∣∣∣2 +

1

2

∑k

mk

∣∣x˜CM ∣∣2

21

Chapter 9

Double Pendulum

(t)

(t)

l1, m1

l2, m2

Figure 9.1: Double Pendulum

Two rods are uniform and can swing freely at a fixed point O. The Kinetic Energy is made up ofT = T1 + T2. The first rod has a moment of inertia of I = 1

3m1l1 and an angular velocity of θ and using

T = 12Iθ

2 then the Kinetic Energy of the first rod is:

T1 =1

6m1l

21θ

2

The second rod is not attached to a fixed point and so its kinetic energy is given by T2 = TCM2 + T rel2 .TCM2 has a velocity of v and is given by TCM2 = 1

2m2v2. The components of v can be found from the

values of xCM and yCM . xCM = l1 sin θ + 12 l2 sinϕ and yCM = l1 cos θ + 1

2 l2 cosϕ. This therefore gives:

TCM2 =1

2m2

(l21θ

2 +1

4l22ϕ

2 + l1l2θϕ cos(θ − ϕ)

)T rel2 has an angular velocity ϕ with a moment of inertia I = 1

12m2l22.

T rel2 =1

24m2l

22θ

2

This gives the total kinetic energy of the system as:

T =1

6m1l

21θ +

1

2

(l21θ

2 +1

3l22ϕ

2 + l1l2θϕ cos(θ − ϕ)

)The potential energy of the system due to gravity for the two systems is V = V1 + V2 and as each rodhas a potential energy of V1 = − 1

2m1gl1 cos θ and V2 = −m2g(l1 cos θ+ 12 cosϕ) so the Lagrangian of the

system is:

L =1

6(m1 + 3m2)l21θ

2 +1

6m2l

22ϕ

2 +1

2m2l1l2θϕ cos(θ − ϕ) +

1

2(m1 + 2m2)gl1 cos θ +

1

2m2gl2 cosϕ

22

This can be solved by the Euler-Lagrange equations:

∂L∂θ− δ

δt

∂L∂θ

= 0

∂L∂ϕ− δ

δt

∂L∂ϕ

= 0

This gives the equation for θ:

0 =− 1

2m2l1l2θϕ sin(θ − ϕ)− 1

2(m1 + 2m2)gl1 sin θ +

1

3(m1 + 3m2)l21θ

− 1

2m2l1l2

(ϕ cos(θ − ϕ)− (θϕ− ϕ2) sin(θ − ϕ)

)And the equation for ϕ:

0 = −1

2m2l1l2θϕ sin(θ − ϕ)− 1

2m2gl2ϕ+

1

3m2l

22ϕ−

1

2m2l1l2

(θ cos(θ − ϕ)− (θ2 − θϕ) sin(θ − ϕ)

)At the equilibrium points where θ(t) = 0 and ϕ(t) = 0, small fluctuations can be used to see if the system

is stable or not. In this case θ and ϕ are small so that cos θ ≈ 1 − θ2

2 + O(θ4), cosϕ ≈ 1 − ϕ2

2 + O(ϕ4)

and cos(θ − ϕ) ≈ 1− (θ−ϕ)2

2 +O((θ − ϕ)4

). By neglecting any terms beyond the quadratic order makes

the Lagrangian:

L ≈ 1

6(m1 +3m2)l21θ

2 +1

6m2l

22ϕ

2 +1

2m2l2θϕ−

1

4(m1 +2m2)gl1θ

2− 1

4m2gl2ϕ

2 +1

2(m1 +2m2)gl1 +

1

2m2gl2

The last two parts of this equation are constant and so fall out of the Euler-Lagrange equations. TheEuler-Lagrange equations can then be written in matrix form as:(

13 (m1 + 3m2)l21

12m2l1l2

12m2l1l2

13m2l

22

)(θϕ

)+

(12 (m1 + 2m2)gl1 0

0 12m2gl2

)(θϕ

)= 0

This is in the form kψ +mpsi = 0. As the equations are coupled they are difficult to solve. The normalmodes are combinations of θ and ϕ that oscillate with a well defined frequency. these frequencies are thenormal frequencies which can be found if the above equation is written in the form ψ˜ + Aψ˜ = 0 where

A = K−1M . This can be done because detK 6= 0 and the inverse is well defined. Two vectors can bemade u˜1 and u˜2 which are linearly independent eigenvectors of A with eigenvalues λ1 and λ2.

Au˜1 = λ1u˜1, Au˜2 = λ1u˜2

If ψ = Θ(t)u˜1 + Φ(t)u˜2 where Θ and Φ are the linear combinations of θ and Φ then:

Θ(t) =

(∣∣u˜2

∣∣2 u˜1 −(u˜1 · u˜2

)u˜2

)· ψ∣∣u˜1

∣∣2 ∣∣u˜2

∣∣2 − (u˜1 · u˜2

)2Φ(t) =

(∣∣u˜1

∣∣2 u˜2 −(u˜1 · u˜2

)u˜1

)· ψ∣∣u˜1

∣∣2 ∣∣u˜2

∣∣2 − (u˜1 · u˜2

)2This can be found by dotting both sides of ψ˜ = Θu˜1 + Φu˜2 with u˜1 and u˜2 and solving the two equations

simultaneously. As these are linearly independent then:

Θ = −λ1Θ, Φ = −λ2Φ

These are the normal modes of oscillation with corresponding normal frequencies ωΘ and ωΦ. Θ and Φsatisfy simple harmonic motion with a frequency of ωΘ =

√λ1 and ωΦ =

√λ2 respectively. Finding the

eigenvalues and eigenvectors of A gives the values ωΘ and ωΦ.

Mu˜ = λKu˜23

Chapter 10

Constraints

Some dynamical systems involve constraints. There are two ways of deal with constraints. This first usesNewton’s Laws and the second uses the Lagrangian.

10.1 Newton’s Constraints

l

m

Figure 10.1: Mass on a Fixed Rope

For a simple pendulum with mass m at the end of a rope with a fixed length l the constraint is definedby the length of the rope. It can be seen that the constraint for this system is x2 + y2 = l2. By solvingin the x and y directions the equations of motion can be found. Horizontally mx = −Txl and vertically

my = mg − Tyl . By setting x = l sin θ and y = l cos θ then this constraint is immediately imposed such

that x = lθ cos θ, y = −lθ sin θ, x = lθ cos θ− lθ2 sin θ and y = −lθ sin θ− lθ2 cos θ. The constraint on thetension is then T = mlθ2 +mg cos θ and θ = − gl sin θ. Using vector multiplication with this and θ it can

be seen that θ2 = 2gl cos θ + c and so the tension is given by T = 3mg cos θ +mlc.

10.2 Holonomic Constraints

For a dynamical system where the position is at xA(t) then the holonomic constraint can be writtenF (xA, t) = 0. If there are N − n constraints then Fα(xA, t) = 0 where α = n + 1, ..., N . This canbe incorporated into a Lagrangian by introducing N − n new variables, λn. These are the Lagrangemultipliers. If an unconstrained Lagrangian is L(t, xA, xA) then the constrained Lagrangian is:

L′ = L(t, xA, xA) +

N∑α=n+1

λαFα(xA, t)

The Euler-Lagrange equations for λβ are:

∂L′

∂λβ− δ

δt

∂L′

∂λβ= 0

As the second part goes to zero then it can be seen that:

∂λαλβ

= δαβ

24

And so the constraint is recovered for λβ . The Euler-Lagrange equations become:

∂L∂xA

− δ

δt

∂L∂xA

+

N∑α=n+1

λα∂Fα∂xA

= 0

In the case of the simple pendulum then the Lagrange multiplier is λ = − T2l . This shows that with the

Lagrange multiplier approach the nature of the constraint forces do not matter.

10.3 Non-Holonomic Constraints

If the constraint is velocity dependent so that F (xa, xa, t) then this can be absorbed into the Lagrangianformalism, and so can be solved. An example of a velocity dependent constraint is particles travellinginto a magnetic field.

25

Chapter 11

Symmetries

Symmetries are often connected to the action principle. The theories connected the symmetries of asystem with physically conserved quantities such as conserved momentum and conserved pseudo-energy.

11.1 Constant Generalised Momentum

If L(t, qA, qA) is such that a particular generalised coordinate qj the Lagrangian does not depend uponit then:

∂L∂qj

= 0

Now if the generalised momentum is pj = ∂L∂qA

then the Euler-Lagrange equations imply that:

δ

δt

∂L∂qA

= 0

This means that pj is constant. The coordinate, qj , is cyclic. If qj is a linear coordinate then it followsthat pj is the linear momentum and so for an angular coordinate qj then pj is the angular momentum.

11.2 Constant Pseudo Energy

The Lagrangian does not have to explicitly depend upon time so that L = L(qi, qi). This means that∂L∂t = 0. In this case time is cyclic. This leads to:

E =

N∑i=1

q∂L∂qi− L(qiqi) = const

For a general Lagrangian:

δLδt

=∂L∂t

+

N∑i=1

∂L∂qi

qi +∂L∂qi

qi

Now considering a function H =∑Ni=1 qi

∂L∂qi− L then the rate of change of this is:

H =

N∑i=1

∂L∂qi

qi + qiδ

δt

∂L∂qi− δLδt

This leaves:∂H∂t

= −∂L∂t

When L is independent of time then H is constant and this defines the Hamiltonian. The Hamiltonianuses a Lagrangian with no explicit time dependence. This is related to the conservation of energy. Itgives E = T + V which is the total energy of the system. This may not always be the case depending onthe initial conditions.

26

11.3 Hamilton’s Equations

Hamilton’s equations are an extension of the Lagrangian method. They use generalised momenta, p1, p2,..., pN , instead of generalised velocities q1, q2, ..., qN , which is a useful technique when the generalisedmomenta are constants of motion. This makes the principle well suited to finding conserved quantities.The Euler-Lagrange equations are used to find the generalised momenta:

pα =∂L∂qα

Then using 2N variables, every part of the system can be uniquely specified. Hamilton’s approach is tosolve pα = ∂L

∂qαfor qα(q, p). With the 2N new variables q and p serve equally well to specify the position

and velocity of every particle.

11.3.1 The Hamiltonian

H(q, p) =

N∑β=1

pβ qβ(q, p)− L(q, q(q, p))

Hamilton’s First Equation

∂H∂pα

= qα +N∑β=1

pβ∂qβ∂pα

− ∂L∂qβ

∂qβ∂pα

As pβ = ∂L∂qβ

then the last two terms cancel out and leave:

∂H∂pα

= qα

Hamilton’s Second Equation

∂H∂qα

= − ∂L∂qα

+

N∑β=1

pβ∂qβ∂qα− ∂L∂qβ

∂qβ∂qα

Again the last two terms cancel and using ∂L∂qα

= pα then the second equation is:

∂L∂qα

= −pα

The Euler-Lagrange equations are a set of N 2nd order differential equations whereas Hamilton’s equationsare a set of 2N 1st order differential equations.

27

Chapter 12

Gyroscope

R cos

R

! m

R

mω

θ

Ω

The potential energy of the system is given by mgR cos θ and the kinetic energy can be calculated byfinding the moment of inertia tensor.

12.1 Lagrangian Approach

The principle axes are e1, e2 and e3, which are orthonormal such that:

Iij =

I1 0 00 I2 00 0 I3

These axes are fixed to the rigid body and therefore must rotate with it. From this the angular velocitycan be obtained so that ω˜ = ω1e1 + ω2e2 + ω3e3. The angular momentum is then J˜ = Iijω˜ = I1ω1e1 +I2ω2e2 + I3ω3e3. This gives the kinetic energy as:

T =1

2

(I1ω

21 + I2ω

22 + I3ω

23

)The coordinate system for this has not been set, and so although a generalised kinetic energy has beenfound, this cannot yet be used to find the dynamics of the system.

12.1.1 Euler’s Angles

Three angles are needed to specify the orientation of an object. Two of the angles are used to fix the e3

axis and then one is used to fix the rotation about e3.

28

ke(e3e0)

ie(e1e000)

je(e2e00)

Figure 12.1: The axes are orientated along i, j and k.ke(e3e0)

e2e0

e1e00

'

'

Figure 12.2: The axes are rotated by ϕ about k to give a still orthonormal set.

e2e0

e1e0

e3e

Figure 12.3: The axis are rotated by θ about the e′2 axis.e3e e2

ee1e

Figure 12.4: Finally rotate about the e3 axis by angle ψ.

12.1.2 Angular Velocity

The first rotation in ϕ gives the angular velocity ϕk, the second rotation in θ gives the angular velocity˙thetae′2 and the final rotation in ψ gives the angular velocity ψe3. This means that the angular velocity

is:

ω˜ = ϕk + θe′2 + ψe3

This involves three different base axes and so an assumption can be made that, for a symmetrical gy-roscope, I1 = I2. his means that rotating about e3 will remain in the principle axes. Any pair oforthonormal vectors can be used to define the e1-e2 plane. There is also a relationship between hatkand θ that can be seen from the rotation diagram. This is that k = − sin θe′1 + cos θe3. This can besubstituted into the angular velocity to give ω˜ = −ϕ sin θe′1 + θe′2 + (ψ+ ϕ cos θ)e3. This means that the

kinetic energy is T = 12I1

[ϕ2 sin2 θ + θ2

]+ 1

2I3

(ψ + ϕ cos θ

)2

and so the Lagrangian is:

L =1

2I1

[ψ2 sin2 θ + θ2

]+

1

2I3

(ψ + ϕ cos θ

)2

−mgR cos θ

Both ϕ and ψ are cyclic coordinates and so the momenta are given by:

∂L∂ψ

= ω3 = ψ + ϕ cos θ

∂L∂ϕ

= ω2 = I1ϕ sin2 θ + I3ω3

In the special case when θ is constant then ϕ is constant and becomes Ω which is the precession frequency.

29

12.2 Hamiltonian Approach

Using the Lagrangian for the gyroscope and the calculated values of the generalised moment gives:

pϕ = I1ϕ sin2 θ + I3(ψ + ϕ cos θ) cos θ

pθ = I1θ

pψ = I3(ψ + ϕ cos θ)

By rearranging the generalised momenta equation and substituting in the velocities into the Hamiltonianfunction then it is found to be:

H =(pϕ − pψ cos θ)2

2I1 sin2 θ+

p2θ

2I1+p2ψ

2I3+MgR cos θ

From this it can be seen that ψ and ϕ are cyclic such that pψ and pϕ are constant. This means that thesystem only has one degree of freedom, which is in the θ component. This gives the Hamiltonian finallyin the form:

H =p2θ

2I1+ U(θ)

Where U(θ) =(pϕ−pψ cos θ)2

2I1 sin2 θ+

p2ψ2I3

+MgR cos θ.

12.2.1 General Motion of the Gyroscope

Using Hamilton’s equations then:

θ =∂H∂pθ

=pθI1

−pθ =∂H∂θ

=∂U

∂θ

The Hamiltonian is independent of time and therefore ∂H∂t = 0 and H = E is a constant. In the case

where θ = 0 then pθ = 0 which means that U(θ) = E. When the motion is confined to the regionU(θ) ≤ E the potential energy is a quadratic shape with a minimum at θo.

U()

E

1 2 o

Figure 12.5: Confined to the region U(θ) ≤ E

It is clear that as U(θ) → ∞, θ → 0 or θ → π. The minima occurs when δUδθ = 0. This minima

corresponds to the steady precession. This is the equilibrium point, and so shows that the gyroscopewill oscillate between θ = θ1 and θ = θ2. For the angular velocity about the vertical axis to vanish thencos θvanish =

pϕpψ

.

30

1

2

Figure 12.6: θ1 < θvanish > θ2 so ϕ will never vanish and precession will continue to oscillate

1

2

Figure 12.7: θ1 < cos−1 pϕpψ

< θ2 so ϕ→ 0 when θ → θvanish. This means that the precession is positive

at the top of the loop, but negative near the bottom.

1

2

Figure 12.8: θ1 = cos−1 pϕpψ

= θvanish which means the precession comes to rest at the top of each loop.

12.2.2 Stability of a Vertical Top

If the gyroscope is spinning with an angular velocity ω3 with its axis vertical, so that θ = 0 then thepotential energy U(0) must be finite. The only way this is possible is in the case that pϕ = pψ = I3ω3.

This is true because ψ = ω3. The motion when θ = π also has a similar relation, but is always stable,and so, it is not as interesting a problem. Now using pϕ = pψ = I3ω3, the potential energy is now:

U(θ) =I23ω

23

2I1tan2 θ

2+

1

2I3ω

23 +MgR cos θ

Where tan2 θ2 = 1−cos θ

sin θ . For small values of θ, tan2 θ2 and cos θ can be expanded to give:

U(θ) ≈(

1

2I3ω

23 +MgR

)+

1

2

(I23ω

23

4I1−MgR

)θ2

This can be seen to stable when θ = 0. Also if the value of θ is positive then there is be a stable position,but for negative θ the equilibrium point is unstable. There is also a minimum value of ω3 where thevertical top will be stable, and this is given by:

ω23 =

4MgRI1I23

= ω2o

This shows that as long as ω3 > ωo then the top will remain upright, but if it drops below ωo then the

top will begin to wobble. If the vertical top has an energy given by E =p2θ2I1

+ U(θ) and θ = θ = 0 then:

E =1

2I3ω

23 +MgR

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Using this then the condition the U(θ) = E is:

I23ω

23

2I1tan2 θ

2+

1

2I3ω

23 +MgR cos =

1

2I3ω

23 +MgR

Which gives θ = θ1 = 0 and cos2 θ22 =

ω23

ω20. If the top is set spinning slowly with ω3 < ωo and with its axis

vertical then the gyroscope will oscillate between the vertical and θ2. As θ2 increases, ω4 decreases suchthat θ2 → π when ω3 → 0. When ω3 = 0 then the top behaves like a compound pendulum and swings ina circle through the upward and downward verticals.

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