# 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: $\{ {\cal H}_{1/0} \} = \begin{Bmatrix} -mR(\dot{\psi}\cos\theta +\dot{\phi}) \boldsymbol{\hat{u}}+ mR\dot{\theta}\boldsymbol{\hat{w}} \\ \frac{1}{4}mR^2 \Big( \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\psi}\sin\theta \boldsymbol{\hat{v}}+(\dot{\phi}+\dot{\psi}\cos\theta)\boldsymbol{\hat{w}}\Big) \end{Bmatrix}_G$