📓
Lecture Notes/ Week 1 Lecture 1 & 2
Contents
Introduction and Review
Constraints
Example: A Simple Pendulum
Example: A Parabolic Bowl
Degrees of Freedom
Example - a rigid body with centre of mass fixed
Coordinates and Notation
Generalised Coordinates
Example A system with 2 particles and 0 constraints ∴ 6 degrees of freedom
Introduction and Review
📌 Newton → Lagrange→Hamilton
Newton: F = ma
Lagrange: Lagrangian L =T −V
T = kinetic energy and V = potential energy
Hamilton: Hamiltonian H =T +V
Lagrange and Hamilton deal with scalars not vectors
Constraints
Exist when a system is not able to move freely in 3D
Example: A Simple Pendulum
∣L∣ = L = length of inextensible c
⨀ = out of page
⨂ = into page
Lecture Notes/ Week 1 Lecture 1 & 2 1
, a = fixed vector r = vector describing the position of M
Holonomic: Equations of Constraint
1. Motion is in the x-y plane, z =0
2. Have L = r − a or ∣L∣ = L = ∣r − a∣
Non-Holonomic: Inequalities of Constraint
3. θ is limited to −90° → +90°⇒ −90° < θ < 90°
Example: A Parabolic Bowl
1. A mass m slides on the surface of the bowl (no rotation or friction)
m described by 3 coordinates (x, y, z ) and 1 holonomic constraint
z = ax2 + ay2
2 degrees of freedom
2. A mass moves in the bowl
m described by 3 coordinates (x, y, z ) and 1 non-holonomic constraint
z ≥ ax2 + ay2
Lecture Notes/ Week 1 Lecture 1 & 2 2
, 3 degrees of freedom
In Case 1, system only requires 2 coordinates to describe motion of m
In Case 2, system requires 3 coordinates to describe motion of m
Degrees of Freedom
a particle in 3D with no constraints has 3 degrees of freedom
If we have N particles, then
m is Holonomic Constraints
n = no.DOF = 3N − m
Example - a rigid body with centre of mass fixed
x′ , y′ and z ′ axes are fixed with respect to the object, origin at the centre of mass position
There are 3 rotational degrees of freedom associated with rotation about (x, y, z). Coordinates could
be 3 angles (αx , αy , αz ) of axes x′ , y′ , z ′ with respect to (x, y, z)
Coordinates and Notation
Lecture Notes/ Week 1 Lecture 1 & 2 3
, N = 1 → (1 particle)
m = 2 → (2 constraints)
Generalised Coordinates
q! , q2 , q3 , ..., qn where n is degrees of freedom
Example A system with 2 particles and 0 constraints ∴ 6 degrees of freedom
⎧x1 q1 ⎫
Particle 1: ⎨ y1 q2 ⎬
⎩ ⎭
z1 q3
⎧x2 q4 ⎫
Particle 1: ⎨ y2 q5 ⎬
⎩ ⎭
z2 q6
d
For rate of change dt use
Velocities
dx dθ
= ẋ, = θ̇
dt dt
Accelerations
dẋ dθ̇
= ẍ, = θ̈
dt dt
∴ also write qi , q˙i , q¨i
Lecture Notes/ Week 1 Lecture 1 & 2 4
Lecture Notes/ Week 1 Lecture 1 & 2
Contents
Introduction and Review
Constraints
Example: A Simple Pendulum
Example: A Parabolic Bowl
Degrees of Freedom
Example - a rigid body with centre of mass fixed
Coordinates and Notation
Generalised Coordinates
Example A system with 2 particles and 0 constraints ∴ 6 degrees of freedom
Introduction and Review
📌 Newton → Lagrange→Hamilton
Newton: F = ma
Lagrange: Lagrangian L =T −V
T = kinetic energy and V = potential energy
Hamilton: Hamiltonian H =T +V
Lagrange and Hamilton deal with scalars not vectors
Constraints
Exist when a system is not able to move freely in 3D
Example: A Simple Pendulum
∣L∣ = L = length of inextensible c
⨀ = out of page
⨂ = into page
Lecture Notes/ Week 1 Lecture 1 & 2 1
, a = fixed vector r = vector describing the position of M
Holonomic: Equations of Constraint
1. Motion is in the x-y plane, z =0
2. Have L = r − a or ∣L∣ = L = ∣r − a∣
Non-Holonomic: Inequalities of Constraint
3. θ is limited to −90° → +90°⇒ −90° < θ < 90°
Example: A Parabolic Bowl
1. A mass m slides on the surface of the bowl (no rotation or friction)
m described by 3 coordinates (x, y, z ) and 1 holonomic constraint
z = ax2 + ay2
2 degrees of freedom
2. A mass moves in the bowl
m described by 3 coordinates (x, y, z ) and 1 non-holonomic constraint
z ≥ ax2 + ay2
Lecture Notes/ Week 1 Lecture 1 & 2 2
, 3 degrees of freedom
In Case 1, system only requires 2 coordinates to describe motion of m
In Case 2, system requires 3 coordinates to describe motion of m
Degrees of Freedom
a particle in 3D with no constraints has 3 degrees of freedom
If we have N particles, then
m is Holonomic Constraints
n = no.DOF = 3N − m
Example - a rigid body with centre of mass fixed
x′ , y′ and z ′ axes are fixed with respect to the object, origin at the centre of mass position
There are 3 rotational degrees of freedom associated with rotation about (x, y, z). Coordinates could
be 3 angles (αx , αy , αz ) of axes x′ , y′ , z ′ with respect to (x, y, z)
Coordinates and Notation
Lecture Notes/ Week 1 Lecture 1 & 2 3
, N = 1 → (1 particle)
m = 2 → (2 constraints)
Generalised Coordinates
q! , q2 , q3 , ..., qn where n is degrees of freedom
Example A system with 2 particles and 0 constraints ∴ 6 degrees of freedom
⎧x1 q1 ⎫
Particle 1: ⎨ y1 q2 ⎬
⎩ ⎭
z1 q3
⎧x2 q4 ⎫
Particle 1: ⎨ y2 q5 ⎬
⎩ ⎭
z2 q6
d
For rate of change dt use
Velocities
dx dθ
= ẋ, = θ̇
dt dt
Accelerations
dẋ dθ̇
= ẍ, = θ̈
dt dt
∴ also write qi , q˙i , q¨i
Lecture Notes/ Week 1 Lecture 1 & 2 4
