# Screws (Part 4): The Kinetic & Dynamic Screws of a Material System

### 1. The Kinetic Screw

Consider a material system \(\Sigma\) (not necessarily a rigid body) of mass \(m\) and mass center \(G\) in motion
relative to a referential \({\cal E}\). Each particle \(P\) of mass \(dm\) of \(\Sigma\) has a linear momentum
\(dm{\bf v}_{P/{\cal E}}\). The angular momentum about a (arbitrary) point \(A\) of particle \(P\) is
\({\bf r}_{AP}\times dm {\bf v}_{P/{\cal E}}\).
It is straightforward to show that the vector field \(A\mapsto {\bf H}_{A, \Sigma/{\cal E}} = \int_\Sigma{\bf r}_{AP}\times {\bf v}_{P/{\cal E}} \, dm\) defines a screw. This quantity, which we can simply denote as \({\bf H}_A\), is the *angular momentum of system \(\Sigma\) about A*. It satisfies
the property: \({\bf H}_B = {\bf H}_A + {\bf r}_{BA}\times m{\bf v}_{G/{\cal E}}\). The corresponding screw, called *kinetic screw* of system \(\Sigma\) relative to referential \({\cal E}\) is denoted as follows
\[
\{ {\cal H}_{\Sigma/{\cal E}} \}
=
\begin{Bmatrix}
m {\bf v}_{G/{\cal E}}
\\
{\bf H}_{A, \Sigma/{\cal E}}
\end{Bmatrix}
\]

In practice, the angular momentum of a rigid body \({\cal B}\) is found by determining its

*inertia operator*about a point \(B\) defined as the mapping \({\bf u}\mapsto {\cal I}_B ({\bf u}) = \int_{\cal B}{\bf r}_{BP}\times ({\bf u}\times {\bf r}_{BP}) dm\). After choosing a basis \((\boldsymbol{\hat{b}}_1,\boldsymbol{\hat{b}}_2,\boldsymbol{\hat{b}}_3)\) attached to \({\cal B}\), the inertia operator is characterized by the body’s moments \((I_{Bxx}, I_{Byy}, I_{Bzz} )\) and products of inertia \((I_{Bxy}, I_{Byz}, I_{Bxz})\) about the axes \((B,\boldsymbol{\hat{b}}_1,\boldsymbol{\hat{b}}_2,\boldsymbol{\hat{b}}_3)\). For instance, the angular momentum about mass center \(G\) can be found as \[ {\bf H}_G = {\cal I}_G (\boldsymbol{\omega}) = \begin{pmatrix} I_{Bxx} & I_{Bxy} & I_{Bxz}\\ I_{Bxy} & I_{Byy} & I_{Byz}\\ I_{Bxz} & I_{Byz} & I_{Bzz} \end{pmatrix} \begin{pmatrix} \omega_1\\ \omega_2\\ \omega_3 \end{pmatrix} \] where \(\boldsymbol{\omega}= \omega_1 \boldsymbol{\hat{b}}_1 +\omega_2 \boldsymbol{\hat{b}}_2 +\omega_3 \boldsymbol{\hat{b}}_3\) is the angular velocity of body \({\cal B}\) relative to \({\cal E}\). It is also possible to derive a more general expression of angular momentum \({\bf H}_A\) about an arbitrary point \(A\) in terms of \({\cal I}_B\).The kinetic screw of a system \(\Sigma\) of \(N\) rigid bodies \({\cal B}_1\), \({\cal B}_2\), \(\ldots\), \({\cal B}_N\) is found to be given by \[ \{ {\cal H}_{\Sigma/{\cal E}} \} =\{ {\cal H}_{{\cal B}_1 /{\cal E}} \} + \{ {\cal H}_{{\cal B}_2 /{\cal E}} \} + \cdots + \{ {\cal H}_{{\cal B}_N /{\cal E}} \} \]

### 2. The Dynamic Screw

The dynamic moment about \(A\) of system \(\Sigma\) is the vector \({\bf D}_{A, \Sigma/{\cal E}} = \int_\Sigma{\bf r}_{AP}\times {\bf a}_{P/{\cal E}} \, dm\), where \({\bf a}_{P/{\cal E}}\) is the acceleration vector of point \(P\).
It is straightforward to show that the vector field \(A\mapsto {\bf D}_{A, \Sigma/{\cal E}} = \int_\Sigma{\bf r}_{AP}\times {\bf v}_{P/{\cal E}} \, dm\) defines a screw: it satisfies the property \({\bf D}_B = {\bf D}_A + {\bf r}_{BA}\times m{\bf a}_{G/{\cal E}}\). This screw, called *dynamic screw* of system \(\Sigma\) relative to referential \({\cal E}\) is formally the time-derivative of the kinetic screw. It is denoted as follows
\[
\{ {\cal D}_{\Sigma/{\cal E}} \}
=
\frac{d}{dt} \{ {\cal H}_{\Sigma/{\cal E}} \}
=
\begin{Bmatrix}
m {\bf a}_{G/{\cal E}}
\\
{\bf D}_G = \frac{d}{dt} {\bf H}_G
\end{Bmatrix}
=
\begin{Bmatrix}
m {\bf a}_{G/{\cal E}}
\\
{\bf D}_A = \frac{d}{dt} {\bf H}_A + {\bf v}_A \times m {\bf v}_G
\end{Bmatrix}
\]
The dynamic screw of a system \(\Sigma\) of \(N\) rigid bodies \({\cal B}_1\), \({\cal B}_2\), \(\ldots\), \({\cal B}_N\) is found to be given by
\[
\{ {\cal D}_{\Sigma/{\cal E}} \}
=\{ {\cal D}_{{\cal B}_1 /{\cal E}} \} + \{ {\cal D}_{{\cal B}_2 /{\cal E}} \} + \cdots + \{ {\cal D}_{{\cal B}_N /{\cal E}} \}
\]

### 3. Example

Consider the disk \(1\) of mass center \(G\), radius \(R\), uniform mass \(m\), in contact with plane \((O,\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}})\) of referential \(0(O,\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\). Assume that the contact at \(I\) is without slip, so that \({\bf v}_{I\in 1/0}= {\bf 0}\). The position of \(1\) relative to \(0\) is defined by the Cartesian coordinates \((x,y)\) of \(I\) and the (Euler) angles \((\psi,\theta,\phi)\) as shown below.

We want to find the expression of the kinetic screw of body \(1\).

We first derive the velocity of center \(G\) by determining the expression of the kinematic screw of the body: \[ \{ {\cal V}_{1/0} \} = \begin{Bmatrix} \boldsymbol{\omega}_{1/0} \\ {\bf v}_{I\in 1/0} \end{Bmatrix} = \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{k}}+ \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{w}} \\ {\bf 0} \end{Bmatrix}_I \] which gives \({\bf v}_{G/0}= \boldsymbol{\omega}_{1/0}\times {\bf r}_{IG}= (\dot{\psi}\boldsymbol{\hat{k}}+ \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{w}}) \times R \boldsymbol{\hat{v}}= -R(\dot{\psi}\cos\theta +\dot{\phi}) \boldsymbol{\hat{u}}+ R\dot{\theta}\boldsymbol{\hat{w}}\). Given the expression of inertia operator \({\cal I}_G\) on basis \((\boldsymbol{\hat{u}},\boldsymbol{\hat{v}},\boldsymbol{\hat{w}})\) we obtain the angular momentum of the body about \(G\): \[ {\bf H}_G = {\cal I}_G (\boldsymbol{\omega}) = mR^2 \begin{pmatrix} \frac{1}{4} & 0 & 0\\ 0 & \frac{1}{4} & 0\\ 0 & 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} \dot{\theta}\\ \dot{\psi}\sin\theta\\ \dot{\phi}+\dot{\psi}\cos\theta \end{pmatrix} =\frac{1}{4}mR^2 \Big( \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\psi}\sin\theta \boldsymbol{\hat{v}}+2 (\dot{\phi}+\dot{\psi}\cos\theta)\boldsymbol{\hat{w}}\Big) \]

We can now give the expression of the kinetic screw: \[ \{ {\cal H}_{1/0} \} = \begin{Bmatrix} -mR(\dot{\psi}\cos\theta +\dot{\phi}) \boldsymbol{\hat{u}}+ mR\dot{\theta}\boldsymbol{\hat{w}} \\ \frac{1}{4}mR^2 \Big( \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\psi}\sin\theta \boldsymbol{\hat{v}}+(\dot{\phi}+\dot{\psi}\cos\theta)\boldsymbol{\hat{w}}\Big) \end{Bmatrix}_G \]