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
For a time-dependent trajectory 𝗿(t) we then compute 𝗿’’ = ??
𝗿’’ = r’‘𝗲ᵣ + r’𝗲’ᵣ + r’θ’𝗲θ + rθ’𝗲’θ
= (r’’ - rθ’²)𝗲ᵣ + (2r’θ’ + rθ’‘)𝗲θ
What is the kinetic energy of a particle?
T = 1/2 m|𝗿’|²
Motion in higher dimensions:
E = T + V = 1/2 m|𝗿’|² + V(𝗿) is conserved if the force 𝗙 = 𝗙(r) takes the form….
𝗙 = -∇V
= (−∂ₓV, −∂ᵧV, −∂𝓏V )
Higher dimensions:
The work done a force 𝗙 in moving a particle from 𝗿₁ to 𝗿₂ along a curve C is W =
W = ∫𝒸 𝗙 . d𝗿
Higher dimensions:
Show that the work done by a force is the change in kinetic energy?
W = ᵗ²∫ₜ₁ 𝗙 . 𝗿’ dt = m ᵗ²∫ₜ₁ 𝗿’’ . 𝗿’ dt = 1/2 m ᵗ²∫ₜ₁ d/dt |𝗿’|² dt = T(t₂) - T(t₁)
Higher dimensions:
Work done by a force = change in KE.
Now suppose the total energy E is conserved.
W = ??
E = T(t₁) + V(𝗿₁) = T(t₂) + V(𝗿₂) W = ∫𝒸 𝗙 . d𝗿 = V(𝗿₁) - V(𝗿₂)
Higher dimensions:
A force 𝗙 = 𝗙(r) is said to be conservative if….
if there exists a potential energy function
V = V (𝗿) such that
𝗙 = - ∇V
Note that as in one dimension the potential V is only defined up to an additive constant
Let 𝗙 : S → R
3 be a vector field, where the domain S ⊂ R³ is open and path connected. Then the following three statements are equivalent:
1) 𝗙 is conservative
2) Given any two points 𝗿₁, 𝗿₂ in S, and any curve C in S starting at 𝗿₁ and ending at 𝗿₂, then the integral [ ] is independent of the choice of C
3) For a simple closed curve C in S we have ∫𝒸 𝗙 . d𝗿 = ??
1) 𝗙 is conservative
2) Given any two points 𝗿₁, 𝗿₂ in S, and any curve C in S starting at 𝗿₁ and ending at 𝗿₂, then the integral ∫𝒸 𝗙 . d𝗿 is independent of the choice of C
3) For a simple closed curve C in S we have ∫𝒸 𝗙 . d𝗿 = 0
What is a central force?
A force that is always directed along the line joining a particle to a fixed position in an inertial frame is called a central force
It is usually convenient to choose this point as the origin
of the frame, meaning that 𝗙 ∝ 𝗿 where 𝗿 is the position vector of the particle, measured from the origin O
What is the Angular Momentum of a particle about a point P in an inertial frame?
The angular momentum 𝗟 = 𝗟ₚ of a particle about a point P in an inertial frame is
the moment of linear momentum 𝗽 = m𝗿’ about P. That is,
𝗟ₚ ≡ (𝗿 − 𝘅) ∧ m𝗿’ = (𝗿 − 𝘅) ∧ 𝗽
Here 𝘅 is the position vector of the point P, while 𝗿 is the position of the particle (both measured from the origin O). Notice that 𝗿’ is the velocity of the particle in the inertial frame, not the velocity relative to P (which could in principle be moving, 𝘅 = 𝘅(t))
If a particle is acted on by a central force with centre O then the angular momentum 𝗟 = 𝗟ₒ is [ ]. Describe the motion
Conserved, and the path of the particle lies entirely in a fixed plane through O. That
is, the motion is planar
If the angular momentum 𝗟 is conserved, then the quantity h ≡ r²θ’ θ specific angular momentum is [ ]
What does h represent?
Conserved
|𝗟| = m|h|, so that h is also the angular momentum per unit mass
The torque τ = τₚ of a force 𝗙, about a point P with position vector 𝘅, acting on a particle with position vector 𝗿 is….
τₚ ≡ (𝗿 − 𝘅) ∧ 𝗙
In other words, the torque is the moment of the force about P.
What direction is the torque in?
The direction of τₚ is normal to the plane containing (𝗿 − 𝘅) and 𝗙, and may be regarded as the axis about which the force tends to rotate the particle about P
If P is a fixed point in the inertial frame, so that 𝘅 = constant, the using 𝗟ₚ ≡ (𝗿 − 𝘅) ∧ m𝗿’ = (𝗿 − 𝘅) ∧ 𝗽 and Newton’s second Law we have 𝗟’ₚ = ….
𝗟’ₚ = (𝗿 − 𝘅) ∧ m𝗿’’ = (𝗿 − 𝘅) ∧ 𝗙 = τₚ
and the torque is the rate of change of angular momentum
A force is a central force about P if and only if the torque about P is [ ] or equivalently 𝗟ₚ is [ ]
zero
conserved
What is the assumption made about the constraint force N?
It is always perpendicular to the constraint space
𝗡 · 𝗿’ =
N is a constraint force
0
What is the work done by the force N (constraint force) when a particle moves along a curve C in the constraint space?
What does this tell us about the constraint space?
W = ∫𝒸 𝗡 · d𝗿 = ∫ 𝗡 · 𝗿’ dt = 0
Hence the constrained space is smooth/frictionless
If we consider a particle of mass m, acted on by a force 𝗙₀, that is then further constrained to
move on a smooth constraint space, Newton’s second law simply reads…
m𝗿’’ = 𝗙 = 𝗙₀ + 𝗡
where 𝗡 is the normal reaction/constraint force
What is the conservation of energy theorem for constrained motion?
Suppose that the force 𝗙₀ = -∇V is conservative, with potential V = V(𝗿). Then the total energy E = T + V is conserved in the constrained motion of the particle.
5.2
Haven’t included anything
Constrained Surfaces:
Gravitational force =
Potential =
𝗙₀ = -mg𝗸 V(𝗿) = mgz
Constrained Surfaces:
The surface of revolution maybe be specified as ….
f(r, θ, z) ≡ z − H(r) = 0
where (r, θ, z) are cylindrical polar coordinates. Recall this means that the Cartesian (x, y) coordinates are given by x = r cos θ, y = r sin θ
Constrained surface
𝗡 · 𝗲θ = ??
𝗡 · 𝗲θ = 0
What is the position vector of a particle moving on a surface of revolution?
Hence, what is Newton’s second law?
𝗿 = r𝗲ᵣ + z𝗲𝓏
where 𝗲𝓏 = 𝗸
m𝗿’’ = 𝗙 = -mg𝗲𝓏 + 𝗡
For a particle constrained to a surface of revolution, write an equation for 𝗟 · 𝗲𝓏
𝗟 · 𝗲𝓏 = mr²θ’ = mh
Hence the component of angular momentum in the direction of the axis of symmetry 𝗲𝓏 is conserved.
Constrained motion on a surface of revolution:
What is an equation for a normal vector? What is this normal vector proportional to?
𝗻 (normal) = ∇f = 𝗲𝓏 - H’(r)𝗲ᵣ
Proportional to the constraint force 𝗡
Constrained motion on a surface of revolution:
We know that 𝗲θ is tangent to the surface. What other independent vector is tangent?
𝘁 = H’(r)𝗲𝓏 + 𝗲ᵣ
Constrained motion on a surface of revolution:
If 𝘁 = H’(r)𝗲𝓏 + 𝗲ᵣ and 𝗻 = ∇f = 𝗲𝓏 - H’(r)𝗲ᵣ, what does 𝘁 · 𝗻 = ??
𝘁 · 𝗻 = 0
Constrained motion on a surface of revolution:
Having calculated tangent, and normal vectors to your surface, what do you do?
Hint use your eqn for N2
Take the dot product of m𝗿’’ = 𝗙 = -mg𝗲𝓏 + 𝗡 with 𝘁 = H’(r)𝗲𝓏 + 𝗲ᵣ (cancelling an overall factor of the mass m)
Answer:
(r’’ - rθ’²) + H’(r)z’’ = -gH’(r)
Constrained motion on a surface of revolution:
After dotting your N2 eqn with a tangent vector, you get (r’’ - rθ’²) + H’(r)z’’ = -gH’(r)
What is the next step?
(Substitution)
Substitute:
θ’ = h/r², and z = H(r)
and use the chain rule to get
[1 + (H’(r))²]r’’ + H’(r)H’‘(r)r’² - h²/r³ = -gH’(r)
Kepler problem: Let r = |𝗿| and let 𝗲ᵣ = 𝗿/r = ˆ𝗿 be a unit vector in the direction of 𝗿, where the latter is the position vector of the particle. The forces in the form 𝗙 = F(r)𝗲ᵣ what type of force is this? Describe the potential
Conservative central forces
V = V(r) depends only on the distance r to the origin
What is The inverse square law force is a conservative central force with V(r) = − κ/r ??
V(r) = − κ/r =⇒ 𝗙 = − κ/r² 𝗲ᵣ
where κ is constant
According to Newton, the gravitational force on a point particle at position 𝗿₁ due to a point particle at position 𝗿₂ is given by…
𝗙 = 𝗙₁₂ = -Gₙ m₁m₂/|𝗿₁ - 𝗿₂|² . (𝗿₁ - 𝗿₂)/|𝗿₁ - 𝗿₂| = -Gₙ m₁m₂/|𝗿₁ - 𝗿₂|² 𝗿ˆ₁₂
Here m₁,m₂ are the (gravitational) masses of the two particles, we have defined the unit vector 𝗿ˆ₁₂ = (𝗿₁ - 𝗿₂)/|𝗿₁ - 𝗿₂|,
and Gₙ ≈ 6.67 x 10⁻¹¹ Nm²kg⁻² is Newton’s gravitational constant
Newton’s law of universal gravitation
The force is proportional to the product of the masses; given the overall minus sign and the
fact that masses are positive, the gravitational force is always [ ]
The force acts in the direction of the vector joining the two masses, and is inversely proportional to the [ ]
The force is proportional to the product of the masses; given the overall minus sign and the
fact that masses are positive, the gravitational force is always attractive.
The force acts in the direction of the vector joining the two masses, and is inversely proportional to the square of the d
Newton’s law of universal gravitation:
Let us now put the second mass at the origin O (that is, we put r₂ = 0), and relabel m₂ = M.
We also write 𝗿₁ = 𝗿 and m₁ = m. Then we may restate Newton’s law of universal gravitation in this set up as:
What is the potential of the mass at the original?
A point mass M at the origin O exerts a gravitational force 𝗙 on a point mass m at
position 𝗿, where κ = GₙMm > 0
V(r) = -GₙMm/r
What is the Newtonian gravitational potential, or Newtonian gravitational field, generated by
the mass M?
Φ(𝗿) = - GₙM/r
Bold phi
Write potential in terms of Newtonian gravitational potential
V = mΦ
Write total conserved energy in terms of Newtonian gravitational potential
E = 1/2 m |𝗿’|² + mΦ(𝗿)
What is the Newtonian Gravitational potential generated by point masses M₁, ….., Mₙ at positions 𝗿₁, …, 𝗿ₙ?
Φ(𝗿) = - Gₙ ᴺΣᵢ₌₁ Mᵢ/|𝗿 - 𝗿ᵢ|
Newton’s law of universal gravitation:
What is the total force acting on a mass m at position 𝗿? (In terms of Φ, then in terms of a summation)
𝗙 = −∇(mΦ) = −Gₙm ᴺΣᵢ₌₁ Mᵢ/|𝗿 - 𝗿ᵢ|³ (𝗿 - 𝗿ᵢ)
What is Newton’s Shell Theorem?
The Newtonian gravitational potential external to a spherically symmetric body of total mass M is the same as that generated by a point mass M at the
centre of mass. That is, the gravitational potential is, where the origin is at the centre of mass.
What is the electrostatic force experienced by the first charge due to the second charge where q₁, q₂ are two charges, at positions 𝗿₁, 𝗿₂ respectively.
𝗙₁₂ = 1/4πε₀ q₁q₂/|𝗿₁ - 𝗿₂|𝗿ˆ₁₂
The constant ε₀ ≈ 8.85 x 10⁻¹² C²N⁻¹m⁻² is called the permittivity of free space
Is the Coulomb force attractive or repulsive?
Can be either, depending on the sign of the charges
same sign = repulsive, different = attractive
Move the second charge to the origin (𝗿₂ = 0), and relabel q₂ = Q. We also write 𝗿₁ = 𝗿 and q₁ = q. Restate Coulomb’s law
A point charge Q at the origin O exerts an electrostatic force 𝗙 on a point charge q at position 𝗿 (given by inverse square law), where κ = -Qq/4πε₀
In electrostatics (the study of charges at rest), define the electric field?
𝗘 = 𝗘(𝗿) is by definition the force a unit test charge (i.e q = 1) at rest experiences at point 𝗿.
Eg, for Coulomb’s law implies that a point charge Q at the origin generates an electric field
𝗘(𝗿) = 1/4πε₀ Q/|𝗿|² 𝗿ˆ
What is the force on another charge q at position 𝗿 is….. (in terms of electric fields)
𝗙 = q𝗘(𝗿)
In electromagnetism, there is an electric potential φ = φ(𝗿) for which 𝗘 =
𝗘 = -∇φ
What is the electric potential generated by a point charge Q at the origin?
φ(𝗿) = Q/4πε₀r
What is the potential energy of a charge q at position 𝗿? And the total energy?
V = qφ E = 1/2 m|𝗿'|² + qφ(𝗿) this is conserved!
Let F be a conservative central force.
r²˙θ’ = h =
Constant
H is the specific angular momentum
F(r) = -dV/dr
Wrote two equations using the effective potential
mr'' = -dVₑ𝒻𝒻/dr Vₑ𝒻𝒻(r) = V(r) + mh²/2r²
Write the energy of a particle in terms of the effective potential
E = 1/2 m r’² + Vₑ𝒻𝒻(r)
What is Kepler’s first law?`
K1: The path of each planet is an ellipse with the sun at the focus
What is Kepler’s second law?
K2: A straight line joining the Sun and a planet sweeps out equal areas in equal times
What is Kepler’s third law?
K3: The square of each planet’s period is proportional to the cube of the semi-major
axis of its elliptical orbit.
Describe the difference between coulomb scattering and the normal Kepler problem
Coulomb: κ < 0
So the force F(r) = -κ/r² s repulsive rather than attractive
Galilean transformations:
Describe a spatial translation
r’ = r − x, where x is a constant vector
all vectors
Galilean transformations:
Describe constant rotations
r’ = R r, where R is a constant 3 × 3 orthogonal matrix
Galilean transformations:
Describe a Galilean boosts
r’ = r − u t, where u is a constant velocity
rs and us vectors
Consider a system of N point particles. With respect to an inertial frame Sˆ, we denote the position
vector of the Ith particle from Oˆ by rᵢ , which has mass mᵢ
Write down the linear momentum.
pᵢ = mᵢr˙ᵢ , I = 1, . . . , N
Consider a system of N point particles. With respect to an inertial frame Sˆ, we denote the position
vector of the Ith particle from Oˆ by rᵢ
We suppose that particle J exerts a force Fᵢⱼ on particle I, for I ≠ J
Write Newton’s 3rd law
Fᵢⱼ = - Fⱼᵢ
for each I ≠ J
Consider a system of N point particles. With respect to an inertial frame Sˆ, we denote the position
vector of the Ith particle from Oˆ by rᵢ
We suppose that particle J exerts a force Fᵢⱼ on particle I, for I ≠ J
Write Newton’s second law
mᵢ¨rᵢ = p˙ ᵢ = Fᵢ = Fᵉˣᵗᵢ +
I≠J Σ Fᵢⱼ
The centre of mass of the system of particles is the point G, with position vector ???
R𝓰 = 1/M ᴺΣᵢ₌₁ mᵢrᵢ
where M = ᴺΣᵢ₌₁ mᵢ is the total mass
What is the total momentum of system of particles?
P = ᴺΣᵢ₌₁ pᵢ = MR’𝓰
The centre of mass of the system behaves like a point particle of mass M acted on by the [ ]
total external force
What is a closed system?
A closed system is one in which all forces are internal, acting between the constituents
of the system. That is,
Fᵉˣᵗᵢ = 0, I = 1, . . . , N
For a closed system, the total momentum is [ ]
Conserved, P˙ = 0
What is the centre of mass frame?
For a system with Fᵉˣᵗ = 0, the inertial frame in which the centre of mass R𝓰 = 0 is called the centre of mass frame
What is the total angular momentum of a system of particles about a point P
Lₚ = ᴺΣᵢ₌₁ (rᵢ - x) ∧ pᵢ
where P has position vector x from the origin Oˆ
That is, L is the vector sum of the angular momenta Lᵢ = (rᵢ − x) ∧ pᵢ for each particle I about P
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