Relative kinematics is discussed here using two Cartesian reference frames.
\[\begin{split}\begin{aligned}
\hat{e}^1_i
= \hat{e}^{1}_i \cdot \hat{e}^0_k \, \hat{e}^0_k
& = \hat{e}^{1}_j \cdot \hat{e}^0_k \, \hat{e}^0_k \otimes \hat{e}^0_j \cdot \hat{e}^0_i = \\
& = R^{0\rightarrow 1}_{kj} \hat{e}^0_k \otimes \hat{e}^0_j \cdot \hat{e}^0_i
= \mathbb{R}^{0\rightarrow 1} \cdot \hat{e}^0_i \ .
\end{aligned}\end{split}\]
1.6.1. Points
Position.
Given two reference frames \(Ox^i\), \(O' x^{i'}\), for the position of a point \(P\) reads
(1.7)\[(P - O_0) = ( O_1 - O_0 ) + ( P - O_1) \ ,\]
\[x^0_{P/O_0,i} \hat{e}^0_i = x^0_{O_1/O_0,i} \hat{e}^0_i + x^1_{P/O_1,k} \hat{e}^1_k \ ,\]
i.e. the position vector \(P-O\) of the point \(P\) w.r.t. point \(O\) - origin of the reference frame \(O x^i\) - is the sum of the position vector \(P-O'\) of the point \(P\) w.r.t. to the point \(O'\) - origin of the reference frame \(O' x^{'i}\) - and the position vector \(O' - O\), of the origin \(O'\) w.r.t. to \(O\).
Velocity. Time derivative of relative position relation (1.7) w.r.t. to reference frame \(0\) is performed keeping \(\hat{e}^0_i\) constant.
\[\begin{split}\begin{aligned}
\dfrac{{}^0 d}{dt} (P-O_0)
& = \dfrac{{}^0 d}{dt} \left[ (O_1 - O_0) + (P - O_1) \right] = \\
& = \dfrac{{}^0 d}{dt} \left( x^0_{O_1/O_0,i} \hat{e}^0_i \right) + \dfrac{{}^0 d}{dt} \left( x^1_{P/O_1,k} \hat{e}^1_k \right) = \\
& = \dfrac{{}^0 d}{dt} x^0_{O_1/O_0,i} \, \hat{e}^0_i + \dfrac{{}^0 d}{dt} x^1_{P/O_1,k} \, \hat{e}^1_k + x^1_{P/O_1,k} \, \dfrac{{}^0 d}{dt} \hat{e}^1_k = \\
& = \vec{v}^0_{O_1/O_0} + \vec{v}^1_{P/O_1} + \vec{\omega}_{1/0} \times \hat{e}^1_k x^1_{P/O_1,k} = \\
& = \vec{v}^0_{O_1/O_0} + \vec{v}^1_{P/O_1} + \vec{\omega}_{1/0} \times ( P - O_1 ) \ ,
\end{aligned}\end{split}\]
so that
(1.8)\[ \vec{v}^0_{P/O} = \vec{v}^0_{O_1/O_0} + \vec{v}^1_{P/O_1} + \vec{\omega}_{1/0} \times ( P - O_1 ) \ .\]
Acceleration. Time derivative of relative velocity relation (1.8) w.r.t. reference frame \(0\) reads
\[\begin{split}\begin{aligned}
\dfrac{{}^0 d}{dt} \vec{v}^0_{P/O_0}
& = \dfrac{{}^0 d}{dt} \left[ \vec{v}^0_{O_1/O_0} + \vec{v}^1_{P/O_1} + \vec{\omega}_{1/0} \times ( P - O_1 ) \right] = \\
& = \dots \\
& = \vec{a}^0_{O_1/O_0} + \vec{a}^{1}_{P/O_1} + 2 \vec{\omega}_{1/0} \times \vec{v}^1_{P/O_1} + \vec{\alpha}_{1/0} \times (P - O_1) + \vec{\omega}_{1/0} \times [ \, \vec{\omega}_{1/0} \times (P - O_1) \, ]
\end{aligned}\end{split}\]
so that
(1.9)\[
\vec{a}^0_{P/O_0} = \vec{a}^0_{O_1/O_0} + \vec{a}^{1}_{P/O_1} + \underbrace{\vec{\alpha}_{1/0} \times (P - O_1)}_{\text{tangential}} + \underbrace{2 \vec{\omega}_{1/0} \times \vec{v}^1_{P/O_1}}_{\text{Coriolis}} + \underbrace{\vec{\omega}_{1/0} \times [ \, \vec{\omega}_{1/0} \times (P - O_1) \, ]}_{\text{centripetal}} \ .
\]
where:
the “tangential component” is orthogonal to the instantaneous angular acceleration and radius,
the “centripetal component” is orthogonal w.r.t. the instantaneous angular velocity
todo tangent to what, centripetal w.r.t. what? state it clearly, otherwise delete this
