Particle Kinematics

Position, velocity, acceleration

The two basic geometric objects we are using are positions and vectors. Positions describe locations in space, while vectors describe length and direction (no position information). To describe the kinematics (motion) of bodies we need to relate positions and vectors to each other.

Add the table summary shown in Fig \ref fig:PosVelAcce

Fig: PosVelAcce

Graphical understanding

Complete in reference page "Position, velocity, and acceleration"

Rotating frames

Angular velocity

Complete in reference page "Rotations and angular velocity"

A rotation of a vector is a change which only alters the direction, not the length, of a vector. A rotation consists of a rotation axis and a rotation rate. By taking the rotation axis as a direction and the rotation rate as a length, we can write the rotation as a vector, known as the angular velocity vector$ \vec{\omega} $. We use the right-hand rule to describe the direction of rotation. The units of $ \vec{\omega} $ are $\rm rad/s$ or $ {}^\circ/s $.

Rotation axis: $ \hat\imath $$ \hat\jmath $$ \hat{k} $$ \hat\imath + \hat\jmath $$ \hat\imath + \hat\jmath + \hat{k} $

Angular velocity vector $\vec\omega$. The direction of $\vec\omega$ is the axis of rotation, while the magnitude is the speed of rotation (positive direction given by the right-hand rule).

Did you know?

The Greek letter ω (lowercase omega) is the last letter of the Greek alphabet, leading to expressions such as “from alpha to omega” meaning “from start to end”. Omega literally means O-mega, meaning O-large, as capital Omega (written Ω) developed from capital Omicron (written Ο) by breaking the circle and turning up the edges. Omicron is literally O-micron, meaning O-small, and it is the ancestor of the Latin letter O that we use today in English.

Α	α	alpha	Ι	ι	iota	Ρ	ρ	rho
Β	β	beta	Κ	κ	kappa	Σ	σ	sigma
Γ	γ	gamma	Λ	λ	lambda	Τ	τ	tau
Δ	δ	delta	Μ	μ	mu	Υ	υ	upsilon
Ε	ε	epsilon	Ν	ν	nu	Φ	φ	phi
Ζ	ζ	zeta	Ξ	ξ	xi	Χ	χ	chi
Η	η	eta	Ο	ο	omicron	Ψ	ψ	psi
Θ	θ	theta	Π	π	pi	Ω	ω	omega

The Greek alphabet, shown above, was the first true alphabet, meaning that it has letters representing phonemes (basic significant sounds) and includes vowels as well as consonants. The Greek alphabet was was derived from the earlier Phoenician alphabet, which was probably the original parent of all alphabets. This shows that the idea of an alphabet is so non-obvious that it has only ever been invented once, and then always copied after that.

Vector derivatives and rotations

If a unit vector $ \hat{a} $ is rotating, then the angular velocity vector $ \vec{\omega} $ is defined so that:

Derivative of unit vectors. #rkr-ew

$$ \dot{\hat{a}} = \vec{\omega} \times \hat{a} $$

Derivative of general vectors. #rkr-ed

$$ \begin{aligned}\dot{\vec{a}} = \underbrace{\dot{a}\hat{a}}_{\operatorname{Proj}(\dot{\vec{a}}, \vec{a})} +\underbrace{\vec{\omega} \times\vec{a}}_{\operatorname{Comp}(\dot{\vec{a}}, \vec{a})}\end{aligned} $$

Using the same approach as #rvc-em we write $ \vec{a} =a\hat{a} $ and differentiate this and use rkr-ew to find:

$$ \begin{aligned}\dot{\vec{a}} &= \frac{d}{dt} \big( a \hat{a} \big) \\&= \dot{a} \hat{a} + a \dot{\hat{a}} \\&= \dot{a} \hat{a} + a (\vec\omega \times \hat{a}) \\&= \dot{a} \hat{a} + \vec\omega \times (a \hat{a}) \\&= \dot{a} \hat{a} + \vec\omega \times \vec{a}.\end{aligned} $$

Comparing this to #rvc-em shows that the two components are the projection and the complementary projection, respectively.

Derivative of constant-length vectors. #rkr-el

$$ \dot{\vec{a}} = \vec{\omega} \times \vec{a}\qquad\text{if $\vec{a}$ is constant length}\ $$

This can be seen from the fact that $ \dot{a} = 0 $ if $a$ is constant (a fixed length vector), substituted into #rkr-ed.

Rotations in 2D

In 2D the angular velocity can be thought of as a scalar (positive for counter-clockwise, negative for clockwise). This scalar is just the out-of-plane component of the full angular velocity vector. We can draw the angular velocity as either a vector pointing out of the plane, or as a circle-arrow in the plane, which is simpler for 2D diagrams.

Show:

Comparison of the vector and scalar representations of $\vec\omega$ for 2D rotations.

In 2D the angular velocity scalar $\omega$ is simply the derivative of the rotation angle $\theta$ in the plane:

Magnitude $\omega$ is derivative of angle $\theta$ in 2D. #rkr-e2

$$ \omega = \dot\theta $$

Take $ \hat{a} $ to be a unit vector rotating in the 2D $ \hat\imath–\hat\jmath $ plane, making an angle of $\theta$ with the $x$-axis, as in Figure #rkr-f2. Then:

$$ \hat{a} = \cos\theta \,\hat\imath + \sin\theta \,\hat\jmath. $$

Differentiating this expression gives:

$$ \dot{\hat{a}} = -\sin\theta \, \dot\theta \,\hat\imath+ \cos\theta \,\dot\theta \,\hat\jmath. $$

We now consider an angular velocity vector $\vec\omega$. Because the rotation is in the $\hat\imath$–$\hat\jmath$ plane, the angular velocity vector must be in the $ \hat{k} $ direction. Thus $ \vec\omega = \omega \hat{k} $. Now we can compute the derivative of $ \hat{a} $ using #rkr-ew, giving:

$$ \begin{aligned}\dot{\hat{a}} &= \vec\omega \times \hat{a} \\&= \omega\hat{k} \times \big( \cos\theta \,\hat\imath+ \sin\theta \,\hat\jmath \big) \\&= \omega \cos\theta \,(\hat{k} \times \hat\imath)+ \omega \sin\theta \,(\hat{k} \times \hat\jmath) \\&= \omega \cos\theta \,\hat\jmath+ \omega \sin\theta \,(-\hat\imath) \\&= - \omega \sin\theta \,\hat\imath+ \omega \cos\theta \,\hat\jmath.\end{aligned} $$

Comparing this expression to the earlier one for $ \dot{\hat{a}} $we see that $\omega = \dot\theta$.

The right-hand rule convention for angular velocities means that counter-clockwise rotations are positive, just like the usual angle direction convention.

Did you know?

Angular directions have long been considered to have magical or spiritual significance. In Britain the counterclockwise direction was once known as widdershins, and it was considered unlucky to travel around a church in a widdershins direction.

Interestingly, right-handed people tend to naturally draw circles in a counterclockwise direction, and clockwise drawing in right-handed children is an early warning sign for the later development of schizophrenia [Blau, 1977].

References

T. H. Blau. Torque and schizophrenic vulnerability. American Psychologist, 32(12):997–1005, 1977. DOI: 10.1037/0003-066X.32.12.997.

Rotations and vector “positions”

The fact that vectors don't have positions means that vector rotations are independent of where vectors are drawn, just like for derivatives.

Show:

Rotational motion of vectors which are drawn moving about. Note that the drawn position does not affect the angular velocity $\omega$ or the derivative vectors.

Properties of rotations

Rotations are rigid transformations, meaning that they keep constant all vector lengths and all relative vector angles. These facts are reflected in the following results, which all consider two vectors $ \vec{a} $ and $ \vec{b} $ that are rotating with angular velocity $ \vec\omega $.

Derivative of rotating vector is orthogonal. #rkr-e2a

$$ \dot{\vec{a}} \cdot \vec{a} = 0 $$

Using #rkr-el and the scalar triple product formula #rvi-es gives:

$$ \begin{aligned}\vec{a} \cdot \dot{\vec{a}}&= \vec{a} \cdot \big( \vec{\omega} \times \vec{a} \big) \\&= \vec{\omega} \cdot \big( \vec{a} \times \vec{a} \big) \\&= 0.\end{aligned} $$

Angle $\theta$ between rotating vectors is constant. #rkr-e2b

$$ \theta = \cos^{-1}\left(\frac{\vec{b} \cdot\vec{a}}{b a}\right) = \text{constant} $$

We first consider the dot product $ \vec{a} \cdot \vec{b} $ and show that this is not changing with time. We do this by using the scalar triple product formula #rvi-es to find:

$$ \begin{aligned}\frac{d}{dt} \big( \vec{a} \cdot \vec{b} \big)&= \dot{\vec{a}} \cdot \vec{b} + \vec{a} \cdot \dot{\vec{b}} \\&= (\vec{\omega} \times \vec{a}) \cdot \vec{b} + \vec{a} \cdot (\vec{\omega} \times \vec{b}) \\&= \vec{b} \cdot (\vec{\omega} \times \vec{a}) + \vec{b} \cdot (\vec{a} \times \vec{\omega}) \\&= \vec{b} \cdot (\vec{\omega} \times \vec{a}) - \vec{b} \cdot (\vec{\omega} \times \vec{a}) \\&= 0.\end{aligned} $$

Now $ \vec{a} \cdot \vec{b} $ is constant and the lengths $a$ and $b$ are constant, so the angle $\theta$ between the vectors must be constant.

Rotating vectors parallel to $\vec\omega$ are constant. #rkr-e2c

$$ \dot{\vec{a}} = 0 \qquad \text{if $\vec{a}$ is rotating and parallel to $\vec\omega$} $$

From #rkr-el we know that the derivative is

$$ \dot{\vec{a}} = \vec\omega \times \vec{a}, $$

but the cross product is zero for parallel vectors, so this the derivative is zero.

Rodrigues’ rotation formula

Rodrigues’ rotation formula gives an explicit formula for a vector rotated by an angle about a given axis.

Rodrigues’ rotation formula for $\vec{a}$ rotated by $\theta$ about $\hat{b}$. #rkr-er

$$ \operatorname{Rot}(\vec{a}; \theta, \hat{b}) =\vec{a} \cos\theta + (\hat{b} \times\vec{a}) \sin\theta + \hat{b} (\hat{b} \cdot\vec{a}) (1 - \cos\theta) $$

Assume $ \vec{a} $ is not parallel to $ \hat{b} $. Then let $ \vec{v} = \hat{b} \times \vec{a} $ and $ \vec{u} = \vec{v} \times \vec{b} $, so $ \hat{u}, \hat{v}, \hat{b} $ is a right-handed orthonormal basis. Take $\phi$ to be the angle between $ \vec{a} $ and $ \hat{b} $. Then we do a rotation by $\theta$ in the $ \hat{u}-\hat{v} $ plane:

$$ \begin{aligned} \vec{a} &= a \sin\phi\,\hat{u} + a \cos\phi \,\hat{b} \\\operatorname{Rot}(\vec{a};\theta,\hat{b}) &= a\cos\theta \sin\phi \,\hat{u} + a \sin\theta \sin\phi\,\hat{v} + a \cos\phi \,\hat{b}.\end{aligned} $$

Now we want to convert from the $ \hat{u},\hat{v},\hat{b} $ basis to write the rotated result in terms of $ \vec{a}, \hat{b} \times \vec{a}), \hat{b} $. To do this, we need to work out what $ \hat{u},\hat{v},\hat{b} $are in terms of these other vectors.

$$ \begin{aligned} \hat{v} &= \frac{\hat{b} \times\vec{a}}{\|\hat{b} \times \vec{a}\|} =\frac{\hat{b} \times \vec{a}}{a \sin\phi} \\\hat{u} &= \frac{\vec{v} \times\hat{b}}{\|\vec{v} \times \vec{b}\|} =\frac{\vec{v} \times \hat{b}}{a \sin\phi} =\frac{(\hat{b} \times \vec{a}) \times \hat{b}}{a\sin\phi} = \frac{\hat{b} \times (\vec{a} \times\hat{b})}{a \sin\phi} \\&= \frac{\vec{a} - (\hat{b}\cdot \vec{a}) \hat{b}}{a \sin\phi} = \frac{1}{a\sin\phi} \vec{a} - \frac{\hat{b} \cdot\vec{a}}{a \sin\phi} \hat{b}.\end{aligned} $$

Substituting these into the rotated vector expression above gives

$$ \begin{aligned}\operatorname{Rot}(\vec{a};\theta,\hat{b}) &= a\cos\theta \sin\phi \, \left( \frac{1}{a \sin\phi}\vec{a} - \frac{\hat{b} \cdot \vec{a}}{a\sin\phi} \hat{b} \right) \\& \qquad + a \sin\theta \sin\phi \, \left(\frac{\hat{b} \times \vec{a}}{a \sin\phi} \right)+ a \cos\phi \,\hat{b} \\ &= \cos\theta\,\vec{a} - \cos\theta \,(\hat{b} \cdot\vec{a}) \,\hat{b} + \sin\theta \,(\hat{b} \times\vec{a}) + a\cos\theta \,\hat{b} \\ &=\cos\theta \,\vec{a} + (1 - \cos\theta) (\hat{b}\cdot \vec{a}) \,\hat{b} + \sin\theta \,(\hat{b}\times \vec{a}).\end{aligned} $$

Angular acceleration

Complete in reference page "Position, velocity, and acceleration"

This equation is displayed under the title "Velocity and acceleration in polar basis". Add the description of each term as shown in Fig \ref fig:AngularAccelerationEq

Fig: AngularAcceleration

Tangential

Normal basis

Complete in reference page "Tangential/normal basis"

Consider a particle moving with position vector $ \vec{r} $ and corresponding velocity $ \vec{v} $ and acceleration $ \vec{a} $. The tangential/normal basis $ \hat{e}_t,\hat{e}_n,\hat{e}_b $ is:

Tangential/normal basis vectors. #rkt-eb

$$ \begin{aligned}\hat{e}_t &= \hat{v}& &\text{tangential basis vector} \\\hat{e}_n &= \frac{\dot{\hat{e}}_t}{\|\dot{\hat{e}}_t\|}= \frac{\operatorname{Comp}(\vec{a},\vec{v})}{\|\operatorname{Comp}(\vec{a},\vec{v})\|}& &\text{normal basis vector} \\\hat{e}_b &= \hat{e}_t \times \hat{e}_n& &\text{binormal basis vector} \\\end{aligned} $$

These equations are definitions of the basis vectors, so the only thing to derive is the alternative formula for $ \hat{e}_n $. Using the definition of $ \hat{e}_t $ above and #rvc-eu, we see that

$$ \dot{\hat{e}}_t = \dot{\hat{v}} = \frac{1}{v} \operatorname{Comp}(\dot{\vec{v}}, \vec{v}) = \frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v}). $$

Normalizing both sides gives the desired expression:

$$ \frac{\dot{\hat{e}}_t}{\|\dot{\hat{e}}_t\|} = \frac{\frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v})}{ \left\|\frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v})\right\|} = \frac{\frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v})}{ \frac{1}{v} \|\operatorname{Comp}(\vec{a}, \vec{v})\|} = \frac{\operatorname{Comp}(\vec{a}, \vec{v})}{ \|\operatorname{Comp}(\vec{a}, \vec{v})\|}. $$

The tangential basis vector $ \hat{e}_t $ points tangential to the path, the normal basis vector $ \hat{e}_n $ points perpendicular (normal) to the path towards the instantaneous center of curvature, and the binormal basis vector $ \hat{e}_b $ completes the right-handed basis.

Show:

Tangential/normal basis associated with movement around a curve in 3D. Observe that the velocity $ \vec{v} $ is always in the $ \hat{e}_t $ direction and that the acceleration $ \vec{a} $ always lies in the $ \hat{e}_t,\hat{e}_n $ plane (the osculating plane). The center of curvature and osculating circle are defined below.

Basis derivatives and angular velocity

As the point $P$ moves along its path, the associated tangential/normal basis rotates with an angular velocity vector $\omega$ given by:

Angular velocity of the tangential/normal basis. #rkt-ew

$$ \begin{aligned}\vec{\omega} &= v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b\end{aligned} $$

We start by writing the angular velocity in the tangential/normal basis, giving:

$$ \vec{\omega} = \omega_t\,\hat{e}_t + \omega_n\,\hat{e}_n+ \omega_b\,\hat{e}_b. $$

Now the basis vector derivatives are given by the cross product by $ \vec{\omega} $ from #rkr-ew, so we can evaluate the expressions #rkt-ek for curvature and torsion to give:

$$ \begin{aligned}\kappa &= \frac{1}{v} \dot{\hat{e}}_t \cdot \hat{e}_n \\&= \frac{1}{v} (\vec{\omega} \times \hat{e}_t) \cdot \hat{e}_n \\&= \frac{1}{v} (\omega_b\,\hat{e}_n - \omega_n\,\hat{e}_b) \cdot \hat{e}_n \\&= \frac{1}{v} \omega_b\end{aligned} $$

and

$$ \begin{aligned}\tau &= -\frac{1}{v} \dot{\hat{e}}_b \cdot \hat{e}_n \\&= -\frac{1}{v} (\vec{\omega} \times \hat{e}_b) \cdot \hat{e}_n \\&= -\frac{1}{v} (\omega_n\,\hat{e}_t - \omega_t\,\hat{e}_n) \cdot \hat{e}_n \\&= \frac{1}{v} \omega_t.\end{aligned} $$

Rearranging the final expressions in each case gives $ \omega_b = v\kappa $ and $ \omega_t = v\tau $. From the definition #rkt-eb of $ \hat{e}_n $, we see that

$$ \begin{aligned}\hat{e}_n &= \frac{\dot{\hat{e}}_t}{\|\dot{\hat{e}}_t\|} \\\hat{e}_n \cdot \hat{e}_b &= \frac{1}{\|\dot{\hat{e}}_t\|} (\vec{\omega} \times \hat{e}_t) \cdot \hat{e}_b \\0 &= \frac{1}{\|\dot{\hat{e}}_t\|} (\omega_b\,\hat{e}_n - \omega_n\,\hat{e}_b) \cdot \hat{e}_b \\&= - \frac{1}{\|\dot{\hat{e}}_t\|} \omega_n.\end{aligned} $$

from which we conclude that $ \omega_n = 0 $, giving us all three components of $ \vec{\omega} $.

Knowing the angular velocity vector of the tangential/normal basis allows us to easily compute the time derivatives of each tangential/normal basis vector, as follows:

Tangential/normal basis vector derivatives. #rkt-ed

$$ \begin{aligned}\dot{\hat{e}}_t &= \phantom{-v\kappa\,\hat{e}_t - } v\kappa\,\hat{e}_n \\\dot{\hat{e}}_n &= -v\kappa\,\hat{e}_t \phantom{ - v\kappa\,\hat{e}_n + } + v\tau\,\hat{e}_b \\\dot{\hat{e}}_b &=\phantom{-v\kappa\,\hat{e}_t } - v\tau\,\hat{e}_n\end{aligned} $$

We can use the expression #rkt-ew for $ \vec{\omega} $ together with #rkr-ew to find the basis vector derivatives:

$$ \begin{aligned}\dot{\hat{e}}_t &= \vec{\omega} \times \hat{e}_t= (v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b) \times \hat{e}_t= v \kappa \,\hat{e}_n \\\dot{\hat{e}}_n &= \vec{\omega} \times \hat{e}_n= (v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b) \times \hat{e}_n= - v \kappa \,\hat{e}_t + v \tau \,\hat{e}_b \\\dot{\hat{e}}_b &= \vec{\omega} \times \hat{e}_b= (v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b) \times \hat{e}_b= -v \tau \,\hat{e}_n.\end{aligned} $$

Notation note

The tangential/normal basis is also called the Frenet–Serret frame after Jean Frédéric Frenet and Joseph Alfred Serret, who discovered it independently around 1850. The equations #rkt-ed for the basis derivatives are often called the Frenet-Serret formulas, typically written in terms of $s$ derivatives:

$$ \begin{aligned}\frac{d\hat{e}_t}{ds} &= \phantom{-\kappa\,\hat{e}_t - } \kappa\,\hat{e}_n \\\frac{d\hat{e}_n}{ds} &= -\kappa\,\hat{e}_t \phantom{ - \kappa\,\hat{e}_n + } + \tau\,\hat{e}_b \\\frac{d\hat{e}_b}{ds} &=\phantom{-\kappa\,\hat{e}_t } - \tau\,\hat{e}_n.\end{aligned} $$

If we divide the angular velocity vector #rkt-ew by $v$ then we obtain the vector

$$ \frac{1}{v} \vec{\omega} = \tau\,\hat{e}_t + \kappa\,\hat{e}_b, $$

which is known as the Darboux vector after its discoverer, Jean Gaston Darboux.

Curvature

Complete in reference page "Tangential/normal basis"

Curvature and torsion. #rkt-ek

To better understand the geometry of the tangential/normal basis, we can use the curvature $\kappa$ to describe the curving of the path, and the torsion $\tau$ to describe the rotation of the basis about the path. These quantities are defined by:

$$ \begin{aligned}\kappa &= \frac{d\hat{e}_t}{ds} \cdot \hat{e}_n= \frac{1}{v} \dot{\hat{e}}_t \cdot \hat{e}_n & &\text{curvature} \\\rho &= \frac{1}{\kappa} & &\text{radius of curvature} \\\tau &= -\frac{d\hat{e}_b}{ds} \cdot \hat{e}_n= -\frac{1}{v} \dot{\hat{e}}_b \cdot \hat{e}_n & &\text{torsion} \\\sigma &= \frac{1}{\tau} & &\text{radius of torsion}\end{aligned} $$

These equations are definitions, so we need only check that the two expressions for each of $\kappa$ and $\tau$ are equivalent. This follows immediately, however, from the chain-rule conversions #rkt-ea between $d/ds$ and $d/dt$.

The radius of curvature $\rho$ is the radius of equivalent circular motion, and the torsion determines the rate of rotation of the osculating plane, as described below in Section #rkt-so.

Given a parametric curve, its curvature can be directly evaluated with:

Curvature of parametric curve $\vec{r}(u)$ in 3D. #rkt-ec

$$ \kappa = \frac{\|\vec{r}' \times \vec{r}''\|}{\|\vec{r}'\|^3} $$

Because the tangential/normal basis vectors #rkt-eb are all normalized, it does not matter how fast the point moves along the line. We can thus evaluate the parametric curve as a function of time, giving $ \vec{r}(t) $. Now:

$$ \begin{aligned}\|\vec{r}'' \times \vec{r}'\| &= \|\vec{a} \times \vec{v}\| \\&= a v \sin\theta \\&= \frac{v^3}{\rho},\end{aligned} $$

where we used the cross product length formula #rvv-el and equation #rkt-er for the radius of curvature $ \rho $. By definition #rkt-ek the curvature is $ \kappa = 1/\rho $, so

$$ \begin{aligned}\|\vec{r}'' \times \vec{r}'\|&= v^3 \kappa \\\kappa &= \frac{\|\vec{r}'' \times \vec{r}'\|}{\|\vec{r}'\|^3}.\end{aligned} $$

While the above formula can be used in 2D by taking the third component to be zero, it can also be written in an explicitly 2D form:

Curvature of parametric curve $x = x(u)$, $y = y(u)$ in 2D. #rkt-e2

$$ \kappa = \frac{|x''y' - y''x'|}{(x'^2 + y'^2)^{3/2}} $$

This equation is just #rkt-ec written in explicit 2D coordinates. To see this, we first take the position vector $ \vec{r}(u) $ to be

$$ \vec{r}(u) = x(u)\,\hat\imath + y(u)\,\hat\jmath. $$

Now $ r'(u) = \sqrt{(x'(u))^2 + (y'(u))^2} $ and

$$ \begin{aligned}\vec{r}'' \times \vec{r}'&= (x''\,\hat\imath + y''\,\hat\jmath)\times (x'\,\hat\imath + y'\,\hat\jmath) \\&= (x''y' - y''x')\,\hat{k},\end{aligned} $$

so evaluating #rkt-ec gives the desired expression.

We can take this a step further, and obtain an expression for an explicitly defined function.

Curvature of an explicitly defined function $y = f(x).$ #rkt-e3

$$ \kappa = \frac{|y''(x)|}{(1 + y'(x)^2)^{3/2}} $$

Use equation #rkt‑e2 and parametrize the curve using $x$ as the parametrization variable. Namely:

$$ \begin{aligned}x = u, \quad \quad y = y(u) = y(x)\end{aligned} $$

This yields a very elegant expression, as $ x'(u) = 1 $ and $ x''(u) = 0 $, which lets us arrive at the desired expression #rkt‑e3.

System comparison: Cartesian/Polar/Tangent-Normal

Add summary table as in Fig \ref fig:SystemsComparison

Fig: SystemsComparison

Applications

Train wheels

This was mentioned in lecture but it was not expanded. It would be a great idea to include a picture or example.

Refers to "angular velocity".

Variable inertia flywheel

This is mentioned in L07-Notes, Slide 6. This paper is cited https://doi.org/10.1016/j.egyr.2020.01.001. This picture is included \ref fig:AppFlywheel .

Refers to "Angular accelerations".

Fig: AppFlywheel

Celestial velocities

Complete in "Celestial Velocities".

We normally think of our classroom or laboratory as being stationary when we are doing dynamics. But how valid is this assumption?

We will consider motion due to:

Spinning of the Earth about the North-South axis.
Orbit of the Earth about the Sun.
Rotation of the Sun around center the Milky Way.
Motion of the Milky Way through the universe.

As we will see below, some of these velocities are not small. Why is normally valid to assume that we are in an inertial reference frame?

Reference material

Elementary motions

Concepts applied

Angular velocity
Angular acceleration

Periods of rotation

The different types of motion have magnitudes roughly given in the following table. These motions are not all in the same direction, and may add to each other or act in opposite or even orthogonal directions.

	Earth spin	Earth orbit	Milky Way	Through CMB
period $T$	$24\rm\ h$	$365\rm\ d$	$200\rm\ My$
	$8.64 \times 10^4\rm\ s$	$3.16 \times 10^7\rm\ s$	$ 6.31 \times 10^{15}\rm\ s $
radius $r$	$6370\rm\ km$	$1\rm\ AU$	$27.2\rm\ kly$
	$6.37 \times 10^6\rm\ m$	$ 1.50 \times 10^{11}\rm\ s $	$ 2.57 \times 10^{20}\rm\ s $
ang. vel.	$15.0^\circ/\rm h$	$0.986^\circ/\rm d$	$1.8^\circ/\rm My$
$\omega = 2\pi/T$	$ 7.27 \times 10^{-5}\rm\ s $	$ 1.99 \times 10^{-7}\rm\ s $	$ 9.97 \times 10^{-16}\rm\ s $
velocity	$1670\rm\ km/h$	$107\,000\rm\ km/h$	$922\,000\rm\ km/h$	$1\,990\,000\rm\ km/h$
$v = r\omega$	$4.63 \times 10^2\rm\ m/s$	$2.98 \times 10^4\rm\ m/s$	$2.56 \times 10^5\rm\ m/s$	$5.52 \times 10^5\rm\ m/s$
acceleration	$0.343\%\ g$	$0.0605\%\ g$	$2.60 \times 10^-9 \%\ g$
$a = r\omega^2$	$ 3.37 \times 10^{-2}\rm\ m/s^2 $	$ 5.93 \times 10^{-3}\rm\ m/s^2 $	$ 2.55 \times 10^{-10}\rm\ m/s^2 $

Did you know?

The first people to have a rough idea of the radius of the Earth and the distance to the Sun were the ancient Greeks. Eratosthenes (276–195 BCE) computed the radius of the Earth to be 40 000 stadia (6800 km) by measuring the difference in Sun angle between Aswan and Alexandria.

Aristarchus of Samos (310–230 BCE) obtained the first estimates of the distance to the Sun by using observations of lunar eclipses and solar parallax. While his method was in principle correct, poor observational data meant that his computed Earth-Sun distance was quite inaccurate.

Solar and sidereal time

We all know that one day is 24 hours long. But the period of the Earth's rotation is not 24 hours! This is because of the difference between solar time and sidereal time. Solar time is the time measured against the Sun, as we normally do. Sidereal time is measured against the stars, which is slightly different.

Schematic of the Earth's rotation about its own axis and about the sun, counting solar days and sidereal days. Here the Earth has just 8 solar days per year for better visualization.

As we can see above, the Earth rotates one more sidereal day each year than solar days. This means:

$$ \text{1 solar year} = \text{365 solar days}= \text{366 sidereal days} $$

and so:

$$ \begin{aligned}\text{sidereal day} &= \frac{365}{366}\ \text{solar day} \\&= \frac{365}{366} \times 24\ {\rm h} \\&= 23.93{\rm\ h} \\&= 23{\rm\ h}\ 56{\rm\ min}\end{aligned} $$

The Earth's orbital angular velocity is thus actually $ \omega = 360^\circ / (23.93{\rm\ h}) = 15.04^\circ/\rm h $.

The relationship between solar and sidereal days can also be computed by considering just a single day, as shown below.

Diagram of one solar day and one sidereal day for the Earth, not drawn to scale. The fact that $ \omega_{\rm E} $ and $ \omega_{\rm S} $ are in the same direction means that sidereal days are shorter than solar days (this is not a coincidence).

Just as for the definition of a day, there is a similar distinction between the synodic lunar month, which is the time between passes of the Moon between the Earth and the Sun, and the sidereal lunar month, which is the orbital period of the Moon in an inertial frame. In common usage, the phrase lunar month refers to the synodic month.

Did you know?

Just as sidereal days and solar days are different, the exact length of one year depends on how we define it. The sidereal year is the time for the Earth to complete one orbit relative to the fixed stars, and has length 365.256363 solar days. The tropical year is the time for the Earth to return to the same point in the seasons, which varies around a value of about 365.242189 solar days (about 20 minutes shorter than the sidereal year). These years are different because of the axial precession of the Earth.

In common usage the word “year” refers to the tropical year, as the seasons have historically been more important for people than the motion of the stars. Observe that:

$$ 365.242 \approx 365 + \frac{1}{4} - \frac{1}{100} + \frac{1}{400} $$

This is why leap years in our Gregorian calendar add an extra day every 4 years, unless the year is a whole century, except every 400 years. If our calendar used sidereal years instead of tropical, the fact that 365.256363 is slightly larger than 365.25 would mean that about every 200 years we would need to have a double leap year, when there would be two extra days (February 30?).

While leap years occur because the year is not an exact integer number of days, a similar problem is caused by the fact that the solar day is not exactly 24 × 60 × 60 seconds. The length of the day actually varies somewhat unpredictably due to tidal friction as well as climactic and geologic events such as glacier formation and mass redistribution in the mantle, both of which change the Earth's moment of inertia. To correct for these variations a leap second is occasionally added, in which the last minute of the day has 61 seconds.

Track transition curves

Complete in "Track transition curves".

Refers to "position, velocity, and acceleration" and "Tangential".

Roller coaster

This topic needs to be created, it was mentioned in lecture with no information or figure

Movement:	circle	var-circle	ellipse	arc
	trefoil	eight	comet	pendulum
Show:

Origin:	\(O_1\)	\(O_2\)

	Earth spin	Earth orbit	Milky Way	Through CMB
period \(T\)	\(24\rm\ h\)	\(365\rm\ d\)	\(200\rm\ My\)
	\(8.64 \times 10^4\rm\ s\)	\(3.16 \times 10^7\rm\ s\)	\( 6.31 \times 10^{15}\rm\ s \)
radius \(r\)	\(6370\rm\ km\)	\(1\rm\ AU\)	\(27.2\rm\ kly\)
	\(6.37 \times 10^6\rm\ m\)	\( 1.50 \times 10^{11}\rm\ s \)	\( 2.57 \times 10^{20}\rm\ s \)
ang. vel.	\(15.0^\circ/\rm h\)	\(0.986^\circ/\rm d\)	\(1.8^\circ/\rm My\)
\(\omega = 2\pi/T\)	\( 7.27 \times 10^{-5}\rm\ s \)	\( 1.99 \times 10^{-7}\rm\ s \)	\( 9.97 \times 10^{-16}\rm\ s \)
velocity	\(1670\rm\ km/h\)	\(107\,000\rm\ km/h\)	\(922\,000\rm\ km/h\)	\(1\,990\,000\rm\ km/h\)
\(v = r\omega\)	\(4.63 \times 10^2\rm\ m/s\)	\(2.98 \times 10^4\rm\ m/s\)	\(2.56 \times 10^5\rm\ m/s\)	\(5.52 \times 10^5\rm\ m/s\)
acceleration	\(0.343\%\ g\)	\(0.0605\%\ g\)	\(2.60 \times 10^-9 \%\ g\)
\(a = r\omega^2\)	\( 3.37 \times 10^{-2}\rm\ m/s^2 \)	\( 5.93 \times 10^{-3}\rm\ m/s^2 \)	\( 2.55 \times 10^{-10}\rm\ m/s^2 \)

TAM 2xx References

Particle Kinematics

Position, velocity, acceleration

Graphical understanding

Rotating frames

Angular velocity

Vector derivatives and rotations

Rotations in 2D

References

Rotations and vector “positions”

Properties of rotations

Rodrigues’ rotation formula

Angular acceleration

Tangential

Normal basis

Basis derivatives and angular velocity

Normal and Tangential acceleration

Curvature

System comparison: Cartesian/Polar/Tangent-Normal

Applications

Train wheels

Variable inertia flywheel

Celestial velocities

Periods of rotation

Solar and sidereal time

Track transition curves

Roller coaster