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Vector Magnitudes

Magnitudes in mechanics are vectors in a 3-dimensional space: Position, Velocity, Acceleration, Force, Torque, etc. We will denote such quantities by $\vec{u}, \vec{v}, \vec{w}$ Addition:

\begin{aligned} \vec{w} & =\vec{u}+\vec{v} \\ \vec{v} & =a \vec{u} \end{aligned}

Basis

A basis is a maximal set of linearly independent vectors. In the Euclidean three-dimensional space, every basis has 3 vectors:

\{a\} \equiv\left\{\vec{a}_1, \vec{a}_2, \vec{a}_3\right\}

Vectors can then be expressed as a linear combination of the elements of a basis:

\vec{u}=u_1^a \vec{a}_1+u_2^a \vec{a}_2+u_3^a \vec{a}_3

Basis & Coordinates

Once we express a vector magnitude using a basis, we can represent it as a column matrix:

\vec{u}=u_1^a \vec{a}_1+u_2^a \vec{a}_2+u_3^a \vec{a}_3 \equiv \mathbf{u}^a=\left[\begin{array}{c} u_1^a \\ u_2^a \\ u_3^a \end{array}\right]

Note:

The matrix representation is always related to a basis. This will be denoted by a right superscript.
Vector magnitude equations hold independent of the basis used to represent the vectors.
It is convenient to derive vector equations independent of any basis.
The matrix representation then allows us to make calculations and write computer programs, but they depend on a basis.

Examples

Inner product
$\vec{u} \cdot \vec{v} \quad \equiv \quad\left(\mathbf{u}^a\right)^{\top} \mathbf{v}^a=\left[\begin{array}{lll} u_1^a & u_2^a & u_3^a \end{array}\right]\left[\begin{array}{c} v_1^a \\ v_2^a \\ v_3^a \end{array}\right]$
Vector Product
$\vec{w}=\vec{u} \times \vec{v} \quad \equiv \quad \mathbf{w}^a=\mathbf{S}\left(\mathbf{u}^a\right) \mathbf{v}^a$
Skew-symmetric operator
$\quad \mathbf{S}(\mathbf{u})=-\mathbf{S}^{\boldsymbol{\top}}(\mathbf{u})=\left[\begin{array}{ccc}0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0\end{array}\right]$

Coordinate transformations

Consider two bases $\{a\}$ and $\{b\}$ . Then

\begin{gathered} \vec{u}=u_1^a \vec{a}_1+u_2^a \vec{a}_2+u_3^a \vec{a}_3=u_1^b \vec{b}_1+u_2^b \vec{b}_2+u_3^b \vec{b}_3 \\ u_1^b=\vec{u} \cdot \vec{b}_1, \quad u_2^b=\vec{u} \cdot \vec{b}_2, \quad u_3^b=\vec{u} \cdot \vec{b}_3 \\ u_1^b=\left(u_1^a \vec{a}_1+u_2^a \vec{a}_2+u_3^a \vec{a}_3\right) \cdot \vec{b}_1, \\ u_2^b=\left(u_1^a \vec{a}_1+u_2^a \vec{a}_2+u_3^a \vec{a}_3\right) \cdot \vec{b}_2, \\ u_3^b=\left(u_1^a \vec{a}_1+u_2^a \vec{a}_2+u_3^a \vec{a}_3\right) \cdot \vec{b}_3 \end{gathered}

\left[\begin{array}{l} u_1^b \\ u_2^b \\ u_3^b \end{array}\right]=\left[\begin{array}{lll} \left(\vec{a}_1 \cdot \vec{b}_1\right) & \left(\vec{a}_2 \cdot \vec{b}_1\right) & \left(\vec{a}_3 \cdot \vec{b}_1\right) \\ \left(\vec{a}_1 \cdot \vec{b}_2\right) & \left(\vec{a}_2 \cdot \vec{b}_2\right) & \left(\vec{a}_3 \cdot \vec{b}_2\right) \\ \left(\vec{a}_1 \cdot \vec{b}_3\right) & \left(\vec{a}_2 \cdot \vec{b}_3\right) & \left(\vec{a}_3 \cdot \vec{b}_3\right) \end{array}\right]\left[\begin{array}{l} u_1^a \\ u_2^a \\ u_3^a \end{array}\right]

Coordinate transformation matrix:

\mathbf{u}^b=\mathbf{C}_a^b \mathbf{u}^a

Coordinate transformation matrices are orthogonal:

\mathbf{C}_a^b\left(\mathbf{C}_a^b\right)^{\top}=\mathbf{I} \quad \mathbf{C}_b^a=\left(\mathbf{C}_a^b\right)^{-1}=\left(\mathbf{C}_a^b\right)^{\top}

1st order Tensors

A magnitude whose coordinates in different bases are linearly related according to

u_i^b=\sum_j C_{i j} u_j^a

where $C_{i j}$ are the elements of the coordinate transformation matrix, is called a vector or first order tensor. Hence we call $\vec{u}, \vec{v} \quad$ the tensor forms

2nd order Tensors

A second-order tensor allows transforming one vector into another. Examples Cross-product tensor, $\vec{S}_u: \quad \vec{w}=\vec{S}_u \cdot \vec{v} \quad(\equiv \vec{u} \times \vec{v})$ Inertia tensor, $\quad \vec{I}_c: \quad \vec{L}_c=\vec{I}_c \cdot \vec{\omega}$ Think of the tensor "." as an operation that maps a vector into another. When a $2^{\text {nd }}$ order tensor is expressed in a basis, it becomes a $3 \times 3$ matrix.

When a $2^{\text {nd }}$ order tensor is expressed in a basis, it becomes a matrix:

\begin{gathered} \vec{S}_u: \quad \vec{w}=\vec{S}_u \cdot \vec{v} \quad(\equiv \vec{u} \times \vec{v}) \\ \mathbf{w}^a=\mathbf{S}\left(\mathbf{u}^a\right) \mathbf{v}^a \\ \mathbf{S}\left(\mathbf{u}^a\right)=\left[\begin{array}{ccc} 0 & -u_3^a & u_2^a \\ u_3^a & 0 & -u_1^a \\ -u_2^a & u_1^a & 0 \end{array}\right] \end{gathered}

Tensor operations are independent of the basis in which the vectors are expressed. Matrix forms are always related to a basis.

\begin{array}{rll} \vec{u} \cdot \vec{v} & \equiv & \left(\mathbf{u}^a\right)^T \mathbf{v}^a \\ \vec{w}=\vec{u} \times \vec{v} & \equiv & \mathbf{w}^a=\mathbf{S}\left(\mathbf{u}^a\right) \mathbf{v}^a \end{array}

Reference Frame & Coordinate Systems

A reference frame is perspective from which the motion of a body or an object is described by an observer. A reference frame can be defined by a set of at least 3 non-co-linear points in space that are rigidly connected.

A coordinate system is a mathematical entity that allows us to establish a one-to-one correspondence between vector magnitudes and scalars called coordinates. A basis defines a coordinate system. Therefore, a reference frame is not the same as a coordinate system, one is physical and the other is mathematical.

We will denote reference frames by $\mathcal{A}, \mathcal{B}$ and bases associated with reference frames by $\{a\},\{b\}$

Time-derivative of a vector

The time derivative of a scalar magnitude is independent of the reference frame in which the magnitude is observed. The time derivative of a vector magnitude depends, in general, on the reference frame in which the magnitude is observed. We will use a notation that indicates this explicitly by using a left superscript:

\frac{\mathcal{A} d \vec{r}}{d t}=\lim _{\Delta t \rightarrow 0}^{\mathcal{A}} \frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}

In general,

\frac{\mathcal{A}_{\mathrm{d}} \vec{r}}{\mathrm{~d} t} \not \equiv \frac{\mathcal{B}^{\mathcal{B}} \vec{r}}{\mathrm{~d} t}

(Rate of) Transport Theorem (TT)

Theorem 1 (Rate of Change Transport Theorem): Consider the scenario depicted in the figure, and assume that $\mathcal{B}$ only rotates with respect of $\mathcal{A}$ . Then there exist a unique vector $\vec{\omega}_{\mathcal{B} / \mathcal{A}}$ called the angular velocity of $\mathcal{B}$ with respect of $\mathcal{A}$ such that This formula is key to derive kinematic models. We will be using it a lot!

Addition of angular velocities

\begin{gathered} \frac{\mathcal{A}_{\mathrm{d} u}}{\mathrm{~d} t}=\frac{\mathcal{B}_{\mathrm{d} u}}{\mathrm{~d} t}+\vec{\omega}_{\mathcal{B} / \mathcal{A}} \times \vec{u}, \\ \frac{\mathcal{B}_{\mathrm{d}} \vec{u}}{\mathrm{~d} t}=\frac{\mathcal{C}_{\mathrm{d} \vec{u}}}{\mathrm{~d} t}+\vec{\omega}_{\mathcal{C} / \mathcal{B}} \times \vec{u}, \\ \frac{\mathcal{C}_{\mathrm{d} u}}{\mathrm{~d} t}=\frac{\mathcal{D}_{\mathrm{d} u}}{\mathrm{~d} t}+\vec{\omega}_{\mathcal{D} / \mathcal{C}} \times \vec{u} . \\ \frac{\mathcal{A}_{\mathrm{d}} \vec{u}}{\mathrm{~d} t}=\frac{\mathcal{D}_{\mathrm{d} \vec{u}}}{\mathrm{~d} t}+\left(\vec{\omega}_{\mathcal{D} / \mathcal{C}}+\vec{\omega}_{\mathcal{C} / \mathcal{B}}+\vec{\omega}_{\mathcal{B} / \mathcal{A}}\right) \times \vec{u} \\ \vec{\omega}_{\mathcal{D} / \mathcal{A}}=\vec{\omega}_{\mathcal{D} / \mathcal{C}}+\vec{\omega}_{\mathcal{C} / \mathcal{B}}+\vec{\omega}_{\mathcal{B} / \mathcal{A}} \end{gathered}

Note how the subscript notation works like fraction multiplications.

Position

The position vector denoted by $r_{P/A}$ indicates “the position of the point P with respect to the point A.”