Section 12.1 Power, Work, and Kinetic Energy
Recall that the power \(\Pow\) generated by force \(\bF\) acting on particle \(P\) relative to referential \(\cE (O, \be_x, \be_y, \be_z)\) is defined as the scalar product \(\bF \cdot \vel_{P/\cE}\text{.}\) The integral defined by
\begin{equation}
\cW_{t_1}^{t_2} = \int_{t_1}^{t_2} \Pow dt =
\int_{P_1}^{P_2} ( F_x dx + F_y dy + F_z dz)\tag{12.1.1}
\end{equation}
is the work done by force \(\bF = F_x \be_x + F_y \be_y + F_z \be_z\) during the interval of time \(t_1 \leq t \leq t_2\text{.}\)
The work done by a time-independent force \(\bF\) is only a function of the initial and final positions of a particle \(P\) in \(\cE\) whenever there exists a scalar field \(\pot (P)\text{,}\) called potential energy such that
\begin{equation}
\cW_{t_1}^{t_2} = \int_{P_1}^{P_2}\bF_P \cdot d\br_{OP} = \pot(P_1) - \pot(P_2)\tag{12.1.2}
\end{equation}
In this case, force \(\bF\) is said to be conservative, and that it derives from potential \(\pot\text{.}\) If it exists, potential \(\pot\) is defined by
\begin{equation}
\bF = - \grad \pot = - {\partial \pot\over \partial x} \be_x -
{\partial \pot\over \partial y} \be_y -
{\partial \pot\over \partial z} \be_z\tag{12.1.3}
\end{equation}
Now consider a particle \(P\) of mass \(m\) moving in a Newtonian referential \(\cE\) under the effect of a resultant force \(\bF\text{.}\) Then according to Newton’s second law we have
\begin{equation*}
m \ba_{P/\cE} = \bF
\end{equation*}
Upon taking the scalar product of this equation with velocity \(\vel_{P/\cE}\) and integrating from instant \(t=t_1\) to \(t=t_2\) we obtain
\begin{equation*}
\int_{t_1}^{t_2} m \ba_{P/\cE}\cdot \vel_{P/\cE} dt
=
\int_{t_1}^{t_2} \bF \cdot \vel_{P/\cE} dt
\end{equation*}
We recognize in this last equation that:
the integral \(\int_{t_1}^{t_2} m \ba_{P/\cE}\cdot \vel_{P/\cE} dt =
{1\over 2}m \vel_{P/\cE}^2 (t_2) - {1\over 2}m \vel_{P/\cE}^2 (t_1)\) is the variation of the kinetic energy of \(P\) between the two instants.
the integral \(\int_{t_1}^{t_2} \bF \cdot \vel_{P/\cE} dt\) is the work done by force \(\bF\) along the motion of \(P\) from \(t=t_1\) to \(t=t_2\) in \(\cE\text{.}\)
We can then write the following equation
\begin{equation}
\frac{1}{2} m \vel_{P/\cE}^2 (t_2) -\frac{1}{2} m \vel_{P/\cE}^2 (t_1)
= \cW_{t_1}^{t_2}\tag{12.1.4}
\end{equation}
known as the Work-Energy Theorem: The work done by the resultant force \(\bF\) during the motion of \(P\) from \(t=t_1\) to \(t=t_2\) in a Newtonian referential \(\cE\) is equal to the change of the particle’s kinetic energy.
The instantaneous version of this theorem, known as the Kinetic Energy Theorem can be stated as follows:
\begin{equation}
\frac{d\kin}{dt} = \Pow\tag{12.1.5}
\end{equation}
Hence, the time rate of change of kinetic energy \(\kin\) of particle \(P\) is equal to the power \(\Pow = \bF \cdot \vel_{P/\cE}\) generated by resultant force \(\bF\) acting on \(P\) in Newtonian referential \(\cE\text{.}\)
Consider now a system \(\Si\) of \(N\) particles \(P_1\text{,}\) \(P_2\text{,}\) \(\ldots\text{,}\) \(P_N\) of mass \(m_1\text{,}\) \(m_2\text{,}\) \(\ldots\text{,}\) \(m_N\text{,}\) respectively, in motion in a Newtonian referential \(\cE\text{.}\) The forces acting on each particle \(P_i\) are comprised of the external force \(\bF_{\bSi \to i}\) and a sum of internal forces \(\sum_{j=1}^N
\bF_{j\to i}\) due to the other particles of \(\Si\text{.}\) The power generated by external force \(\bF_{\bSi \to i}\) acting on particle \(P_i\) relative to referential \(\cE\) is given by \(\Pow_{\bSi \to i /\cE} = \vel_{P_i /\cE}\cdot\bF_{\bSi \to i}\text{.}\) The power generated by the resultant external forces \(\bF_{\bSi \to \Si}= \sum_{i=1}^N
\bF_{\bSi\to i}\) acting on \(\Si\) relative to \(\cE\) can be defined as the sum
\begin{equation}
\Pow_{\bSi \to \Si / \cE} = \sum_{i=1}^N \vel_{P_i /\cE}\cdot\bF_{\bSi \to i}
= \sum_{i=1}^N \Pow_{\bSi \to i / \cE}\tag{12.1.6}
\end{equation}
Similarly the power generated by the sum of all internal forces \(\bF_{i \to j}\) can be determined according to
\begin{equation*}
\sum_{i=1}^N \sum_{j=1}^N \vel_{P_i /\cE}\cdot\bF_{j \to i}
\end{equation*}
Despite the law of action and reaction, this power does not, in general, vanish. Consider the terms due to a pair of interacting forces \(\bF_{i \to j}=F_{ij} \, \be_{ij}= -\bF_{j \to i}\) between particles \(P_i\) and \(P_j\) directed along \(\br_{P_i P_j} = r_{ij} \, \be_{ij}\text{:}\)
\begin{equation*}
\bF_{j \to i}\cdot\vel_{P_i /\cE} + \bF_{i \to j}\cdot\vel_{P_j /\cE} =
\bF_{i \to j}\cdot(\vel_{P_j /\cE}-\vel_{P_i /\cE})
=F_{ij}\be_{ij}\cdot {d\over dt} r_{ij} \be_{ij}
=F_{ij} {d r_{ij} \over dt}
\end{equation*}
where we have used the fact that unit vector \(\be_{ij}\) satisfies \(\be_{ij}\cdot {d\over dt}\be_{ij} = 0\text{.}\)
Hence, when viewed pairwise, the internal power is only a function of the time rate of change of the relative distance between the particles of \(\Si\text{,}\) and, consequently, is not a function of the chosen referential \(\cE\text{:}\) we denote this internal power between particles \(P_i\) and \(P_j\) as \(\Pow_{i \leftrightarrow j}\text{.}\) Then the power generated by all internal forces can be written as
\begin{equation}
\sum_{1\leq i \lt j \leq N} \Pow_{i \leftrightarrow j}\tag{12.1.7}
\end{equation}
Next we sum the \(N\) equations derived by the Kinetic Energy Theorem applied to each particle (relative to referential \(\cE\)) to find
\begin{equation*}
\frac{d}{dt} \kin_{\Si /\cE} = \sum_{i=1}^N \bF_{i} \cdot \vel_{P_i/\cE}
\end{equation*}
We then decompose \(\bF_i\) in terms of external and internal forces. The power generated by all (internal and external) forces can be written as the sum:
\begin{equation*}
\sum_{i=1}^N \bF_{i} \cdot \vel_{P_i/\cE} = \Pow_{\bSi \to \Si / \cE}+
\sum_{1\leq i \lt j \leq N} \Pow_{i \leftrightarrow j}
\end{equation*}
We conclude with the following theorem:
Theorem 12.1.3. Kinetic Energy Theorem (Systems of Particles).
The time rate of change of the kinetic energy of a system \(\Si\) of \(N\) particles in motion in a Newtonian referential \(\cE\) is equal to the power generated by the external and internal forces acting on \(\Si\)
\begin{equation}
\frac{d}{dt} \kin_{\Si /\cE} = \Pow_{\bSi \to \Si / \cE} +
\sum_{1\leq i \lt j \leq N} \Pow_{i \leftrightarrow j}\tag{12.1.8}
\end{equation}
Next, we generalize these results for systems of rigid bodies.
