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Advanced Engineering Dynamics

R. Valéry Roy

Section 5.5 Acceleration Field of a Rigid Body

We now seek to relate the accelerations of points of rigid body \(\cB\) in motion relative to referential \(\cA\) as we did for velocities. For any two points \(P\) and \(Q\) fixed in \(\cB\) at all times, we have according to the kinematic screw formula
\begin{equation*} \vel_{Q /\cA}= \vel_{P /\cA} + \bomBA \times \br_{PQ}. \end{equation*}
Then, by time-differentiation relative to \(\cA\text{,}\) we obtain
\begin{equation*} \ba_{Q /\cA}= \left( {d\over dt} {\vel_{Q /\cA}} \right)_{\cA}= \ba_{P /\cA} + \bal _{\cB / \cA }\times \br_{PQ} + \bomBA\times \left( {d \over dt} \br_{PQ} \right)_{\cA} . \end{equation*}
But \(( d \br_{PQ} / dt)_{\cA} = \vel_{Q /\cA}- \vel_{P /\cA} = \bomBA \times \br_{PQ}\text{.}\) Finally we obtain the formula governing the accelerations of any two points attached to \(\cB\text{:}\)
\(\danger\)The presence of the last term \(\bomBA\times(\bomBA\times \br_{PQ})\) in equation (5.5.1) shows that the acceleration field \(P\in\cB \mapsto \ba_{P /\cA}\) of \(\cB\) relative to \(\cA\) does not define a screw.

Remark 5.5.2.

We can find the time-derivative of kinematic screw \(\{\cV_{\cB/\cA} \}\) according to the definition of Section 4.9:
\begin{equation*} \Big\{ \frac{d}{dt}\cV_{\cB/\cA} \Big\}_\cA = \left\{ \begin{array}{c} \bal _{\cB / \cA } \\ \ba_{P /\cA} + \vel_{P/\cA}\times\bomBA \end{array} \right\}_P \end{equation*}
This shows that the field \(P\in\cB \mapsto \ba_{P /\cA} +\vel_{P/\cA}\times\bomBA\) defines a screw. This fact can be used to formulate interesting properties of the acceleration field.

Remark 5.5.3.

To emphasize that \(P\) is a point attached to \(\cB\text{,}\) its acceleration may be denoted as \(\ba_{P\in\cB/\cA}\text{.}\) We may also want to find the acceleration of a point \(P\) attached to \(\cB\) only instantaneously: in that case, acceleration \(\ba_{P\in\cB /\cA}\) cannot be determined as the time-derivative \(( d \vel_{P \in \cB / \cA} / dt)_{\cA}\text{,}\) but rather must be found by using formula (5.5.1) (by relating it to the known acceleration of a point, say \(B\text{,}\) attached to \(\cB\)):
\begin{equation*} \ba_{P\in\cB /\cA}=\ba_{B /\cA} + \bal _{\cB / \cA }\times \br_{BP} +\bomBA\times(\bomBA\times \br_{BP}) \end{equation*}

Remark 5.5.4.

The notation \(\ba_{P \in \cB / \cA}\) is vital whenever \(P\) is considered a point of \(\cB\) only in an instantaneous way: in this case \(\ba_{P \in \cB / \cA}\) is neither equal to \(\ba_{P/\cA} = (d \vel_{P / \cA} / dt)_{\cA}\) nor equal to \((d \vel_{P \in \cB / \cA} / dt )_{\cA}\text{.}\) We shall learn in Section 6.3 of Chapter 6 how the three vectors \(\ba_{P \in \cB / \cA}\text{,}\) \(( d \vel_{P \in \cB / \cA} / dt )_{\cA}\text{,}\) and \(( d \vel_{P / \cA} / dt )_{\cA}\) are related to each other.

Example 5.5.5.

Find the acceleration \(\ba_{I \in \cB / \cA}\) for the point \(I\) of Example 5.4.3 by relating it to the acceleration \(\ba_{B / \cA}\) using equation (5.5.1) assuming \(\om_A\) and \(\om_B\) constant. Compare the expression found with \((d \vel_{I \in \cB / \cA} / dt )_{\cA}\text{.}\)
Solution.
Point \(I\) is not attached to \(\cB\text{.}\) We cannot find \(\ba_{I \in \cB / \cA}\) as the time-derivative of \(\vel_{I \in \cB / \cA} = \bze\) relative to \(\cA\text{.}\) Instead, we relate point \(I\) to point \(B\text{:}\)
\begin{equation*} \ba_{I \in \cB / \cA} = \ba_{B/\cA} + \bal_{\cB/\cA}\times \br_{BI} +\bom_{\cB/\cA}\times(\bom_{\cB/\cA}\times\br_{BI} ) \end{equation*}
with acceleration \(\ba_{B/\cA} = (d\vel_{B/\cA} /dt)_\cA\) given by
\begin{equation*} \ba_{B/\cA} = -\om_A^2 (a\be_3\times\be_2 + b\be_3\times\be_1)= \om_A^2 (a\be_1 - b\be_2) \end{equation*}
Taking into account \(\bal_{\cB/\cA} = \bze\text{,}\) we find
\begin{equation*} \ba_{I \in \cB / \cA} =\om_A^2 (a\be_1 - b\be_2) - (\om_B-\om_A)^2 \br_{BI} = \om_A \om_B (a\be_1 -b\be_2) \end{equation*}
As expected, \(\ba_{I \in \cB / \cA} \neq \bze\text{.}\)
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