Section 9.7 Kinetic Energy of a Rigid Body
With the assumptions of
Section 9.6, we seek to evaluate the kinetic energy
\(\kin_{\cB}= \half \int_{\cB} \vel^{2}_{P} dm\) of a rigid body
\(\cB\) relative to referential
\(\cE\text{.}\)
We can find the velocity of any point \(P\) of the body from that of point \(B\in\cB\text{:}\)
\begin{equation*}
\vel_{P} = \vel_{B\in\cB} + \bom_{\cB} \times
\br_{BP}
\end{equation*}
Then, we find
\begin{equation*}
\kin_{\cB }= \half m \vel^{2}_{B\in\cB} + \vel_{B\in\cB} \cdot
\int_{\cB} \bom_{\cB } \times \br_{BP} \, dm +
\half
\int_{\cB} \, ( \bom_{\cB} \times \br_{BP} )^{2} \, dm
\end{equation*}
The second term \(\int_{\cB} \bom_{\cB} \times \br_{BP} \, dm\) is simply \(\bom_{\cB} \times m \br_{BG}\text{.}\) The third term \(\int_{\cB} \, ( \bom_{\cB } \times \br_{BP})^{2} \,dm\) can be simplified by recalling the identity
\begin{equation*}
(\bu\times\bv)^{2}= \bu \cdot (\bv\times(\bu\times\bv)) .
\end{equation*}
Hence
\begin{equation*}
\int_{\cB} \, ( \bom_{\cB} \times \br_{BP} )^{2} \, dm
=
\int_{\cB} \bom_{\cB}\cdot \left( \br_{BP}\times(\bom_{\cB}
\times \br_{BP}) \right) \, dm
= \bom_{\cB} \cdot \iner (\bom_{\cB})
\end{equation*}
We have derived the following result:
Theorem 9.7.1. Kinetic Energy of a Rigid Body (I).
The kinetic energy of rigid body \(\cB\) of mass \(m\text{,}\) mass center \(G\text{,}\) and inertia operator \(\iner\) about point \(B\) is given by
\begin{equation}
\kin_{\cB}= \half m \vel^{2}_{B\in\cB} +
m \vel_{B\in\cB} \cdot (\bom_{\cB} \times \br_{BG})+
\half \bom_{\cB} \cdot \iner (\bom_{\cB}) \tag{9.7.1}
\end{equation}
Simplifications can be obtained in formula
(9.7.1) in two special cases:
Case 1: If point
\(B\) is fixed in
\(\cE\) at all time, or if
\(B\) is an instantaneous center of rotation, then
\(\vel_{B\in\cB} = \bze\) and
(9.7.1) simplifies to
\begin{equation}
\kin_{\cB}= \half \bom_{\cB} \cdot \iner (\bom_{\cB})\tag{9.7.2}
\end{equation}
Case 2: If point
\(B\) is chosen to be mass center
\(G\text{,}\) then
(9.7.1) simplifies to
\begin{equation}
\kin_{\cB}= \half m \vel^{2}_{G}+ \half \bom_{\cB} \cdot \inerG (\bom_{\cB})\tag{9.7.3}
\end{equation}
Finally, we give below a theorem which is useful for the practical determination of the kinetic energy of a rigid body and its time-rate of change:
Theorem 9.7.2. Kinetic Energy of a Rigid Body: Screw Formula.
The kinetic energy of a rigid body \(\cB\) can be obtained from the product (or comoment) between the kinematic screw and the kinetic screw of \(\cB\) as follows:
\begin{equation}
\kin_{\cB / \cE} = \half \{ \cV_{\cB /\cE} \} \cdot \{ \cH_{\cB /\cE} \}\tag{9.7.4}
\end{equation}
Similarly, the time-rate of change of \(\kin_{\cB / \cE}\) can be found as the product between the kinematic screw and the dynamic screw of \(\cB\)
\begin{equation}
{d\over dt}\kin_{\cB / \cE} = \{ \cV_{\cB /\cE} \} \cdot \{ \cD_{\cB /\cE} \} \tag{9.7.5}
\end{equation}
Proof.
To prove
(9.7.4), we first find the product
\(\{ \cV_{\cB /\cE} \} \cdot \{ \cH_{\cB /\cE} \}\) by resolving both screws about mass center
\(G\) of body
\(\cB\text{:}\)
\begin{equation*}
\{ \cV_{\cB /\cE} \} \cdot \{ \cH_{\cB /\cE} \}= \bom_\cB\cdot \bH_G + \vel_G \cdot m\vel_G
\end{equation*}
With
\(\bH_G = \cI_G (\bom)\text{,}\) we recognize
\(2\kin_{\cB}\) given by
(9.7.3). As for
(9.7.5), we take the time-derivative of
\(\kin_{\cB}\) to find
\begin{align*}
{d\over dt}\kin_{\cB / \cE} \amp = \int_{\cB} \vel_P\cdot \ba_P dm = \int_\cB (\vel_G+ \bom_\cB \times \br_{GP}) \cdot \ba_P dm\\
\amp = \vel_G \cdot \int_\cB \ba_P dm + \bom_\cB \cdot \int_\cB \br_{GP}\times \ba_P dm\\
\amp = \vel_G \cdot m \ba_G + \bom_\cB \cdot \bD_G
\end{align*}
This last expression is recognized as \(\{ \cV_{\cB /\cE} \} \cdot \{ \cD_{\cB /\cE} \}\text{.}\)
