Screws (Part 7): An Energy Theorem

1. Introduction

Recall that the Principle of Virtual Power states that the sum of the virtual power of body and contact forces acting on a material system $$\Sigma$$ and of the virtual power of inertial forces relative to a Newtonian referential $${\cal E}$$ adds to $$0$$: $\int_\Sigma{\bf f}^g _{\bar{\Sigma}\to \Sigma} (P) \cdot \, {\bf v}_P ^* \, dV(P) + \int_{\partial\Sigma}{\bf f}^c _{\bar{\Sigma}\to \Sigma} (Q) \cdot \, {\bf v}_Q ^* \, dA(Q) = \int_\Sigma{\bf a}_{P/{\cal E}} \, \cdot \, {\bf v}_P^* \, dm(P) \qquad\qquad (1)$ for all rigidifying virtual velocity fields $$P\mapsto {\bf v}_P^*$$, that is, vector fields satisfying $${\bf v}_Q^*={\bf v}_P^*+{\bf V}\times{\bf r}_{PQ}$$ (screws).

We are now assuming that system $$\Sigma$$ is a rigid body $${\cal B}$$ of mass $$m$$, mass center $$G$$ and inertia operator $${\cal I}_B$$ (about a particular point $$B$$ of the body).

We will choose a particular virtual field as the actual velocity field $$P\in{\cal B}\mapsto {\bf v}_{P/{\cal E}}$$. This is possible since this field defines the kinematic screw $$\{{\cal V}_{{\cal B}/{\cal B}} \}$$ of the rigid body.

2. Power Generated by Body/Contact Forces

Recall that the power generated by a force $${\bf F}$$ acting on a particle $$P$$ of velocity $${\bf v}_{P/{\cal E}}$$ is the instantaneous scalar quantity defined as $${\bf v}_{P/{\cal E}} \cdot {\bf F}$$.

The total power generated by the external (gravitational) body forces acting on body $${\cal B}$$ is then naturally defined as the quantity $\mathbb{P}^g_{\bar{{\cal B}}\to {\cal B}/{\cal E}} = \int_{{\cal B}} {\bf f}^g _{\bar{{\cal B}}\to {\cal B}} (P) \cdot \, {\bf v}_{P/{\cal E}} \, dV(P)$ Similarly, the total power generated by the contact forces acting on body $${\cal B}$$ is defined as the quantity $\mathbb{P}^c_{\bar{{\cal B}}\to {\cal B}/{\cal E}} = \int_{\partial{\cal B}} {\bf f}^c _{\bar{{\cal B}}\to {\cal B}} (Q) \cdot \, {\bf v}_{Q/{\cal E}} \, dA(Q)$ Recall that the body and contact forces define the action screws $$\{{\cal A}^g_{\bar{{\cal B}}\to {\cal B}}\}$$ and $$\{{\cal A}^c_{\bar{{\cal B}}\to {\cal B}}\}$$, of resultant $${\bf F}^g_{\bar{{\cal B}}\to{\cal B}}$$ and $${\bf F}^c_{\bar{{\cal B}}\to {\cal B}}$$, respectively. Hence, in practice, the integrals which define the powers generated by the body and contact forces can be expressed in the form of the dot product of the kinematic screw with the corresponding action screw: $\mathbb{P}^g_{\bar{{\cal B}}\to {\cal B}/{\cal E}} = \{ {\cal V}_{\bar{{\cal B}}/{\cal E}} \} \cdot \{{\cal A}^g_{\bar{{\cal B}}\to {\cal B}}\} = {\bf v}_{A \in {\cal B}/{\cal E}}\cdot {\bf F}^g_{\bar{{\cal B}}\to {\cal B}} + \boldsymbol{\omega}_{{\cal B}/{\cal E}}\cdot {\bf M}^g_{A,\bar{{\cal B}}\to{\cal B}}$ $\mathbb{P}^c_{\bar{{\cal B}}\to {\cal B}/{\cal E}} = \{ {\cal V}_{\bar{{\cal B}}/{\cal E}} \} \cdot \{{\cal A}^c_{\bar{{\cal B}}\to {\cal B}}\} = {\bf v}_{A\in {\cal B}/{\cal E}} \cdot {\bf F}^c_{\bar{{\cal B}}\to {\cal B}} + \boldsymbol{\omega}_{{\cal B}/{\cal E}}\cdot {\bf M}^c_{A,\bar{{\cal B}}\to{\cal B}}$

3. The Kinetic Energy Theorem

Let us now apply the Principle of Virtual Power (1) by using $${\bf v}^*_p = {\bf v}_{P/{\cal E}}$$, that is, by substituting the true velocity field in the integrals of equation (1) in place of the virtual velocity field. Then, we recognize the integrals of the left-hand-side of equation (1) (virtual powers) as the powers $$\mathbb{P}^g_{\bar{{\cal B}}\to {\cal B}/{\cal E}}$$ and $$\mathbb{P}^c_{\bar{{\cal B}}\to {\cal B}/{\cal E}}$$ generated by the body and contact forces.

The integral of the right-hand-side of (1) can be expressed in terms of the kinetic energy of the body since $${\bf a}_{P/{\cal E}}\cdot {\bf v}_{P/{\cal E}}= \frac{d}{dt}(\frac{1}{2} {\bf v}^2 _{P/{\cal E}})$$: $\mathbb{K}_{{\cal B}/{\cal E}} = \int_{\cal B}\frac{1}{2} {\bf v}_{P/{\cal E}}^2 \, dm(P)$

This yields the expression of the Kinetic Energy Theorem: $\frac{d}{dt}\mathbb{K}_{{\cal B}/{\cal E}} \, = \, \mathbb{P}^g_{\bar{{\cal B}}\to {\cal B}/{\cal E}} \, + \, \mathbb{P}^c_{\bar{{\cal B}}\to {\cal B}/{\cal E}} \qquad\qquad (2)$ which shows that the time rate of change of the kinetic energy of a rigid body is equal to the total power of the external body and contact forces acting on the body (relative to a Newtonian referential).

• In practice, the kinetic energy of the body is found from the knowledge of $$\{{\cal V}_{{\cal B}/{\cal E}}\}$$ and the inertia operator $${\cal I}_B$$: $\mathbb{K}_{{\cal B}/{\cal E}} = \frac{1}{2} m {\bf v}^{2}_{B\in{\cal B}/{\cal E}} + m {\bf v}_{B\in{\cal B}/{\cal E}} \cdot (\boldsymbol{\omega}_{{\cal B}/{\cal E}} \times {\bf r}_{BG})+ \frac{1}{2} \boldsymbol{\omega}_{{\cal B}/{\cal E}} \cdot {\cal I}_B (\boldsymbol{\omega}_{{\cal B}/{\cal E}})$

• If we choose the mass center $$G$$ for point $$B$$, we find the expression $\mathbb{K}_{{\cal B}/{\cal E}} = \frac{1}{2} m {\bf v}^{2}_{G/{\cal E}} + \frac{1}{2} \boldsymbol{\omega}_{{\cal B}/{\cal E}} \cdot {\cal I}_G (\boldsymbol{\omega}_{{\cal B}/{\cal E}})$

4. Example

Consider the disk $$1$$ of center $$G$$, radius $$R$$ and mass $$m$$ first treated in Part 4. It rolls on a horizontal plane $$(O,\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}})$$ of a 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 apply the Kinetic Energy Theorem to body $$1$$ (assuming referential $$0$$ Newtonian).

Recall that the expression of the kinematic screw of the body is given by $\{ {\cal V}_{1/0} \} = \begin{Bmatrix} \boldsymbol{\omega}_{1/0} \\ {\bf v}_{I\in 1/0} \end{Bmatrix} = \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{k}}+ \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{w}} \\ {\bf 0} \end{Bmatrix}_I$ leading to $${\bf v}_{G/0}= \boldsymbol{\omega}_{1/0}\times {\bf r}_{IG}= (\dot{\psi}\boldsymbol{\hat{k}}+ \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{w}}) \times R \boldsymbol{\hat{v}}= -R(\dot{\psi}\cos\theta +\dot{\phi}) \boldsymbol{\hat{u}}+ R\dot{\theta}\boldsymbol{\hat{w}}$$. Since $${\bf v}_{G/0}$$ is also given by $$\frac{d}{dt}(x\boldsymbol{\hat{\imath}}+y\boldsymbol{\hat{\jmath}}+ R\boldsymbol{\hat{v}}) = \dot{x}\boldsymbol{\hat{\imath}}+\dot{y}\boldsymbol{\hat{\jmath}}+R (\dot{\theta}\boldsymbol{\hat{w}}-\dot{\psi}\cos\theta\boldsymbol{\hat{u}})$$, we find the non-holonomic equations $\dot{x}= -R \dot{\phi}\cos\psi, \qquad \dot{y}= - R\dot{\phi}\sin\psi$ The kinetic energy of the body is obtained by using inertia operator $${\cal I}_G$$ on basis $$(\boldsymbol{\hat{u}},\boldsymbol{\hat{v}},\boldsymbol{\hat{w}})$$: $\mathbb{K}_{0/1} = \frac{1}{2} m {\bf v}^{2}_{G/0} + \frac{1}{2} \boldsymbol{\omega}_{1/0} \cdot {\bf H}_G = \frac{1}{2} m R^2 \big( (\dot{\phi}+\dot{\psi}\cos\theta)^2 + \dot{\theta}^2 \big) +\frac{1}{4}mR^2 \big( \dot{\theta}^2 + \dot{\psi}^2 \sin^2\theta +2 (\dot{\phi}+\dot{\psi}\cos\theta)^2 \big)$ which gives $\mathbb{K}_{0/1} = \frac{1}{4} m R^2 \big( 3\dot{\theta}^2 + \dot{\psi}^2 \sin^2\theta +4 (\dot{\phi}+\dot{\psi}\cos\theta)^2 \big)$ The total action screw on the body can be expressed as the sum of two contributions $\{ {\cal A}_{\bar{1}\to 1} \} = \begin{Bmatrix} -mg \boldsymbol{\hat{k}} \\ {\bf 0} \end{Bmatrix}_G + \begin{Bmatrix} N\boldsymbol{\hat{k}}+ {\bf F} \\ {\bf 0} \end{Bmatrix}_I$ with $${\bf F}\cdot \boldsymbol{\hat{k}}=0$$. We can easily find the power generated by the action screw: $\mathbb{P}_{\bar{1}\to 1/0} = \begin{Bmatrix} -mg \boldsymbol{\hat{k}} \\ {\bf 0} \end{Bmatrix}_G \cdot \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{k}}+ \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{w}} \\ {\bf v}_{G/0} \end{Bmatrix}_G + \begin{Bmatrix} N\boldsymbol{\hat{k}}+ {\bf F} \\ {\bf 0} \end{Bmatrix}_I \cdot \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{k}}+ \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{w}} \\ {\bf 0} \end{Bmatrix}_I = -mg \boldsymbol{\hat{k}}\cdot {\bf v}_{G/0} = -mg R \dot{\theta}\cos\theta$ Finally, application of the Kinetic Energy Theorem gives $\frac{d}{dt} \mathbb{K}_{1/0} = -mgR \dot{\theta}\cos\theta = -\frac{d}{dt} (mg R \sin\theta)$ which gives the first integral $\frac{1}{4} m R^2 \big( 3\dot{\theta}^2 + \dot{\psi}^2 \sin^2\theta +4 (\dot{\phi}+\dot{\psi}\cos\theta)^2 \big) + mgR \sin\theta = \text{Constant}$

5. Relationship between the Kinetic Energy Theorem and the Fundamental Theorem of Dynamics

Let us write the Fundamental Theorem of Dynamics for rigid body $${\cal B}$$: $\{{\cal D}_{{\cal B}/{\cal E}} \} = \{ {\cal A}_{\bar{{\cal B}}\to{\cal B}} \} \qquad\qquad (3)$ We can take the dot product of (3) with the kinematic screw of body $${\cal B}$$ to obtain $\{{\cal D}_{{\cal B}/{\cal E}} \} \cdot \{{\cal V}_{{\cal B}/{\cal E}} \} = \{ {\cal A}_{\bar{{\cal B}}\to{\cal B}} \} \cdot \{{\cal V}_{{\cal B}/{\cal E}} \}$ The right-hand side is nothing but the power $$\mathbb{P}_{\bar{{\cal B}}\to {\cal B}/{\cal E}}$$.

The left-hand side is given by $$\int_{\cal B}{\bf v}_{P/{\cal E}} \cdot {\bf a}_{P/{\cal E}} dm$$: it is recognized as the time rate of change of the kinetic energy of the body.

We have recovered the Kinetic Energy Theorem.

6. Power of Interaction

The Kinetic Energy Theorem can also be derived for a system $$\Sigma$$ of $$N$$ rigid bodies. Since the bodies are interconnected, there are inevitably internal actions between the bodies. The question is then whether the contributions of these internal actions cancel each other out in the context of power. To answer this question, consider two bodies $$i$$ and $$j$$ of $$\Sigma$$ and the corresponding powers $$\mathbb{P}_{i\to j/0}$$ and $$\mathbb{P}_{j\to i/0}$$ generated by the action $$\{{\cal A}_{i\to j}\}$$ and the reaction $$\{{\cal A}_{j\to i}\}$$ relative to referential $$0$$. These powers are given by $\mathbb{P}_{i\to j/0}= \{{\cal V}_{j/0} \} \cdot \{{\cal A}_{i\to j}\} , \qquad \mathbb{P}_{j\to i/0}= \{{\cal V}_{i/0} \} \cdot \{{\cal A}_{j\to i}\}$ Addition of these two terms gives, upon using $$\{{\cal A}_{j\to i}\}= -\{{\cal A}_{i\to j}\}$$: $\mathbb{P}_{i\to j/0} \, + \, \mathbb{P}_{j\to i/0}= ( \{{\cal V}_{j/0} \} - \{{\cal V}_{i/0} \}) \cdot \{{\cal A}_{i\to j}\} = \{{\cal V}_{j/i} \} \cdot \{{\cal A}_{i\to j}\} \qquad\qquad (4)$ This result shows that the sum of the powers generated by the action and the reaction, which we call power of interaction between the two bodies and denote $$\mathbb{P}_{i\leftrightarrow j}$$, does not vanish. In fact, it is independent of Newtonian referential $$0$$, and is only a function of the relative motion between the two bodies. In practice, it is found by using (4): $\mathbb{P}_{i\leftrightarrow j} \, = \, {\bf v}_{A\in j/i}\cdot {\bf F}_{i\to j} \, + \, \boldsymbol{\omega}_{j/i}\cdot {\bf M}_{A, i\to j}$

7. The Kinetic Energy Theorem for a System of Rigid Bodies

To derive the Kinetic Energy Theorem for a system $$\Sigma$$ of $$N$$ rigid bodies in motion relative to a Newtonian referential $$0$$, we apply (1) for bodies $$1$$, $$2$$, $$\ldots$$, $$N$$ and sum these $$N$$ equations to obtain: $\sum_{i=1}^N \frac{d}{dt}\mathbb{K}_{i/0} = \sum_{i=1}^N \mathbb{P}_{\bar{i} \to i /0} = \underbrace{\sum_{i=1}^N \mathbb{P}_{\bar{\Sigma}\to i/0}}_{\text{external}} + \underbrace{\sum_{i,j=1}^N \mathbb{P}_{i \to j/0}}_{\text{internal}}$ To obtain the last terms of this expression, we have split the actions on body $$i$$ into contributions external to the system and contributions internal to the system.

The sum of the left-hand side is the time rate of change kinetic energy of the system $$\mathbb{K}_{\Sigma/0}$$.

The first sum of the left hand side is the contribution to the power of all actions external to the system $$\mathbb{P}_{\bar{\Sigma}\to \Sigma/0}$$.

The second sum of the left hand side is the contribution to the power of all actions internal to the system: we have learned in the previous section that this power term does not vanish: it amounts to all pairwise powers of interaction $$\mathbb{P}_{i \leftrightarrow j}$$.

In conclusion, the Kinetic Energy Theorem applied to the system takes the following form $\frac{d}{dt}\mathbb{K}_{\Sigma/0} = \mathbb{P}_{\bar{\Sigma}\to \Sigma/0}+ \sum_{1\leq i<j \leq N} \mathbb{P}_{i \leftrightarrow j}$