Dynamics Flashcards
What is a reference frame?
A reference frame S is specified by a choice of origin O, together with a set of perpendicular (right handed) Cartesian coordinate axes at O
Thus if r = (x, y, z) denotes the position of a point P
in the frame S, and r’ = (x’, y’, z’) is the position of the same point in the frame S’, we have
r’ = [ ]
r’ = R(r - x)
where R is a 3 × 3 orthogonal matrix. Recall these are characterised by Rᵀ = R⁻¹
What is a point particle?
A point particle is an idealized object that at a given instant of time t is located at a point r(t), as measured in some reference frame S.
What is Newton’s first law?
N1:
N1: In an inertial frame a particle moves with constant momentum, unless acted on
by an external force
How do you know if a reference frame is inertial?
According to N1 it is inertial
if a particle with no identifiable forces acting on it travels in a straight line with constant speed
v = |v|
What is Newton’s second law? (In an inertial frame)
N2:
In an inertial frame, the dynamics of a point particle is governed by
N2: The rate of change of linear momentum is equal to the net force acting on the
particle: F = p’
Assuming the mass m is constant the right hand side of Newton’s second law is p’ = mr’’
What is Newton’s third law?
N3:
If particle 1 exerts a force F = F₂₁ on particle 2, then particle 2 also exerts a force
F₁₂ = −F on particle 1.
In other words, F₁₂ = −F₂₁. This is often paraphrased by saying that every action has an equal and opposite reaction
Describe linear drag (force)
A linear drag holds when viscous forces predominate, i.e. this is due to the “stickiness” of
the fluid. The force is
F = −b r’, where b > 0 is a constant (the friction coefficient), and r’ is the particle velocity.
Describe quadratic drag (force)
A quadratic drag holds when the resistance is due to the body having to push fluid to the side as it moves, for example a rowing boat moving through water. The force is
F = −D |r’| r’, where the constant D > 0
What is Hooke’s law for springs?
F = −k(x − l)i l = natural length Spring is lying along the x axis x = length of the spring k>0 spring constant This is a restoring force; that is, the force opposes any motion away from the equilibrium position x = l
What is the ODE that represents simple harmonic oscillators?
x₀’’ + ω²x₀ = 0
where ω ≡√k/m
What is the general solution to the ODE representing simple harmonic oscillators?
x₀(t) = C cos ωt + D sin ωt = A cos (ωt + φ)
Without loss of generality we may take the
integration constant A > 0, which is called the amplitude, while the constant φ is called the
phase. The motion is periodic, with period T = 2π/ω in t. The parameter ω is called the (angular) frequency of the oscillator
In general, a particle of charge q moving in an electromagnetic field experiences a force given by the Lorentz force law
F =
F = q E + q r’ ∧ B
Here r’ is the velocity of the particle, E is the electric field, and B is the magnetic field.
(All 3 vectors)
E = E(r, t) and B = B(r, t) depend on both position and time, making them time-dependent
vector fields.
What is the kinetic energy of a particle? And in terms of momentum?
The kinetic energy of the particle is T =1/2mx’². We may also write this in terms of momentum p = mx’ as T = p²/2m
T’ =
When T = kinetic energy
T’ = mx’x’’ = F(x) x’
Suppose the particle starts at position x₁ at time t₁, and
finishes at x₂ at time t₂. Integrating T’ with respect to time t gives
T(t₂) − T(t₁) = ᵗ²∫ₜ₁T’ dt = ᵗ²∫ₜ₁ F(x(t))x’ dt = ˣ²∫ₓ₁ F(x) dx
What is the work done W by the force in moving the particle from x₁ to x₂?
W = ˣ²∫ₓ₁ F(x) dx
What is the work-energy theorem?
W = T(t₂) − T(t₁)
What is the potential energy of a particle?
V (x) = − ˣ∫ₓ₀ F(y) dy where x₀ is arbitrary
potential energy is understood to be defined only up to an [ ]
overall additive constant
Using the Fundamental Theorem of Calculus we may write the force as
F(x) =
F(x) = − dV/dx = -V’(x)
What is the potential if F=-mg?
V (x) = mgx
What is the potential if F=-k(x-l)?
V (x) = 1/2k(x − l)²
What is the Conservation of Energy Theorem?
The total energy of the particle
E = T + V
is conserved, i.e. is constant when evaluated on a solution to Newton’s second law
Prove the conservation of energy theorem (proof 1)
From the Work-Energy Theorem we already have
T(t₂) − T(t₁) = W = ˣ²∫ₓ₁ F(x) dx = V(x₁) - V(x₂)
earranging thus gives
E = T(t₁) + V (x₁) = T(t₂) + V (x₂) .Since the initial and final positions and times here are arbitrary, this proves E is conserved
Prove the conservation of energy theorem (proof 2)
More precisely we first write the right hand side of E = T + V as T(t) + V (x(t)). Using the
chain rule we then have
E’ = T’ + V’ = mx’x’’ + dV/dx dx/dt = x’(mx’’ - F)
It follows that E’ = 0 is implied by Newton’s second law
Write W(t) (Work done) in terms of an integral and separately in terms of potential
W(t) = ˣ⁽ᵗ⁾∫ₓ₀ F(y) dy = V(x₀) - V(x(t))
Define power
Power P = rate of work done, so that
P = dW/dt = Fx’
2/m (E - V(x)) =
2/m (E - V(x)) = x’’
Solve this ODE
2/m (E - V(x)) = x’’
t = ± ∫1/(√2/m (E - V(x))) dx
What is an equilibrium configuration?
An equilibrium configuration is a solution to Newton’s second law with x = xₑ =
constant.
Since this implies x’’ = 0 for all time t, Newton’s second law implies that F(xₑ) = 0, and there is no net force acting on the particle.
Describe equilibrium configuration when the force acting is conservative
For a conservative force 0 = F(xₑ) = −V’(xₑ) implies that xₑ is a critical point of the potential V (x)
Consider motion near an equilibrium point x = xₑ. Expand Newton’s second law around this point.
Make a change of variable o ξ ≡ x − xₑ
Now write a linear differential equation for ξ, stating what assumptions you’ve made
mx’’ = F(x) = F(xₑ) + (x - xₑ)F’(xₑ) + O((x - xₑ)²)
F(xₑ) = 0
We change variables to ξ ≡ x − xₑ, so that the equilibrium point is now at ξ = 0. Assuming we are sufficiently close to the latter, so that the quadratic and higher
order terms in the eqn are small, we may write down the following approximate linear differential equation for ξ:
mξ’’ = F’(xₑ)ξ
What is the linearized equation of motion, and what are linearized solutions?
(Equilibrium positions)
mξ’’ = F’(xₑ)ξ is called the linearized equation of motion. Solutions to this linear
homogeneous equation are called linearized solutions
Linearized equation of motion: mξ’’ = F’(xₑ)ξ
Let K ≡ −F’(xₑ)
There are 3 cases depending on the sign of K.
Describe the qualitative difference between the cases.
K > 0 : point of stable equilibrium
K = 0 : Unknown
K < 0 : point of unstable equilibrium
Linearized equation of motion: mξ’’ = F’(xₑ)ξ
Let K ≡ −F’(xₑ)
K > 0
How do we know that this is a point of stable equilibrium?
ω = √K/m > 0
ξ’’ + ω²ξ = 0
The general solution is ξ(t) = A cos (ωt + φ)
In this case ξ = 0 is called a point of stable
equilibrium – for amplitude A small enough so that it is consistent to ignore the higher order
terms in the expansion of the force, the system executes small oscillations around the equilibrium point. The frequency of these oscillations is ω. Crucially, this analysis applies to any point of stable equilibrium, and it is for this reason that the harmonic oscillator is so important.
Linearized equation of motion: mξ’’ = F’(xₑ)ξ
Let K ≡ −F’(xₑ)
K < 0
How do we know that this is a point of unstable equilibrium?
In this case we may define p =√−K/m > 0
ξ’’ - p²ξ = 0 which has general solution
ξ(t) = Aeᵖᵗ + Be⁻ᵖᵗ
A generic small displacement of the system at time t = 0 will have both A and B non-zero, and the solution grows exponentially with t, for both t > 0 and t < 0. The higher order terms in the Taylor expansion, that we ignored, quickly become relevant. Such equilibria are hence termed unstable.
Linearized equation of motion: mξ’’ = F’(xₑ)ξ
Let K ≡ −F’(xₑ)
K = 0
How do we know that this is unknown stable-ness?
The first two terms in the Taylor expansion in mx’’ = F(x) = F(xₑ) + (x - xₑ)F’(xₑ) + O((x - xₑ)²) are zero, and we need
to expand to higher order to determine what happens
Linearized equation of motion: mξ’’ = F’(xₑ)ξ
Let K ≡ −F’(xₑ)
K > 0
Describe the potential
A stable equilibruim point with K > 0 is then a local
minimum of the potential (for example xₑ = xₘᵢₙ)
Linearized equation of motion: mξ’’ = F’(xₑ)ξ
Let K ≡ −F’(xₑ)
K < 0
Describe the potential
An unstable equilibruim point with K < 0 is instead a local maximum (for example xₑ = xₘₐₓ)
Explain why the potential has minima/maxima at equilibria
We similarly expand
V (x) = V (xₑ) + (x − xₑ)V’(xₑ) + 1/2(x − xₑ)²V’‘(xₑ) + O((x − xₑ)³) .
Wlog choose the arbitrary additive constant in V so that V (xₑ) = 0.
Moreover, V’(xₑ) = −F(xₑ) = 0.
This means that near equilibrium the potential is approximately quadratic: Vᵩᵤₐ𝒹(x) = 1/2 K(x - xₑ)²
Where K = V’‘(xₑ) = -F’(xₑ)
3.4 Damping - Not explicity on syllabus
Will leave until have revised everything else
Suppose we have a dynamical system described by the coupled ODEs
x’’ = F(x, y)
y’’ = G(x,y)
An equilibrium configuration is a solution with x = xₑ, y = yₑ both constant
Linearize the equations of motion
x = xₑ + ξ , y = yₑ + η where ξ and η are small, and then Taylor expand x’’ and y’’ leading to:
ξ’’ = F(xₑ + ξ, yₑ + η) = F(xₑ, yₑ) + ξ ∂F/∂x (xₑ, yₑ) + η ∂F/∂x (xₑ, yₑ) + …
η’’ = G(xₑ + ξ, yₑ + η) = G(xₑ, yₑ) + ξ ∂G/∂x (xₑ, yₑ) + η ∂G/∂x (xₑ, yₑ) + …
where · · · denote terms of quadratic and higher order in ξ, η. The linearized equations of motion are hence:
ξ'' = a ξ + b η , η'' = c ξ + d η a = ∂F/∂x (xₑ, yₑ) b = ∂F/∂y (xₑ, yₑ) c = ∂G/∂x (xₑ, yₑ) d = ∂G/∂y (xₑ, yₑ)
Suppose we have a dynamical system described by the coupled ODEs x'' = F(x, y) y'' = G(x,y) Linearized equations ξ'' = a ξ + b η , η'' = c ξ + d η
How do we solve these?
Write as a matrix equation
(ξ'') = (a b) (ξ)
(η'') (c d) (η)
Then we seek solutions in the form
( ξ(t) ) = ( α ) exp(λt)
( η(t) ) ( β )
Subbing in and cancelling the exp.
λ² (α) = (a b) (α)
(β) (c d) (β)
This says that λ² of the a,b,c,d matrix, with eigenvector (α,β).
The characteristic eqn is
λ⁴ - (a+d) λ² + (ad - bc) = 0
which gives eigenvalues
λ² = 1/2 (a + d ± √((a+ d)² - 4(ad-bc)) ≡ λ²± (± is subscript)What makes a stable equilibrium for a coupled dynamical system?
What are the normal frequencies of the system?
If all solutions for λ = ±λ± (susbcript(±)) are purely imaginary (equivalently both λ²± < 0), we say the equilibrium point is stable
We write λ = ±iω±, where ω± > 0 are called the
normal frequencies of the system
Coupled dynamical system. Linearized solution
Writing exp(λt) = exp(±iω±t) (Second ± is subscript) in terms of trigonometric functions, the linearized solution is ??
What is a normal mode?
( ξ(t) ) = ( α₊ ) cos(ω₊t + φ₊) + (α₋) cos(ω₋t + φ₋)
( η(t) ) ( β₊ ) (β₋)
where (α±)
(β±) are the eigenvectors corresponding to the eigenvalues λ²±, respectively, and φ± are
constants. The solution for a given eigenvector is called a normal mode
Polar Coordinates:
What is 𝗲ᵣ and𝗲θ?
𝗲ᵣ = cosθ i + sinθ j 𝗲θ = - sinθ i + cosθ j
For a time-dependent trajectory 𝗿(t) we then compute 𝗿’ = ????
𝗿’ = r’𝗲ᵣ + r𝗲’ᵣ = r’𝗲ᵣ + rθ’𝗲θ
