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Section 3.5 Angular Acceleration
Definition 3.5.1 . Angular acceleration of a rigid body.
The angular acceleration of rigid body \(\cB\) relative to referential (or rigid body) \(\cA\text{,}\) denoted by \(\bal_{\cB/\cA}\text{,}\) is defined as
\begin{equation}
\bal_{\cB/\cA} = \left( {d\over dt} \bom _{\cB / \cA} \right)_{\cA}\tag{3.5.1}
\end{equation}
\(\ddanger\) Note that the loop identity for angular velocities does not extend to angular accelerations, that is
\begin{equation*}
\bal_{\cA/\cB}+ \bal_{\cB/\cC} +\bal_{\cC/\cA} \neq \bze
\end{equation*}
for general motions between
\(\cA\text{,}\) \(\cB\text{,}\) and
\(\cC\text{.}\) See
Exercise 3.7.1 .
It is also possible to express angular acceleration
\(\bal_{\cB/\cA}\) in terms of unit quaternion
\(Q_{\cB/\cA}\) and its second derivative: by differentiating
(3.4.1) we find that
\(\bal_{\cB/\cA}\) is given by
\begin{equation}
\bal_{\cB/\cA} = 2 \, \text{Vect}( \frac{d^2 Q_{\cB/\cA} }{dt^2}\, \conjQ_{\cB/\cA} )\tag{3.5.2}
\end{equation}
where \(\text{Vect}(Q)\) denotes the vector part of quaternion \(Q\text{.}\)
Proof.
We take the time-derivative of
(3.4.1) to obtain
\begin{equation*}
\bal_{\cB/ \cA} = 2 \,\frac{d^2Q}{dt^2} \conjQ + 2\,\frac{dQ}{dt}\frac{d\conjQ}{dt}
\end{equation*}
We then need to show that quaternion \(R= \frac{dQ}{dt}\frac{d\conjQ}{dt}\) is a scalar: we find \(R= - \frac{dQ}{dt}\conjQ\frac{dQ}{dt}\conjQ\text{.}\) It is easily shown that \(\overline{R}= R\text{.}\)
Example 3.5.3 .
Find
\(\bal_{\cC / \cE}\) for the gyroscopic system of
Example 3.2.15 . Compare
\(\bal_{\cC / \cE}\) with
\(\bal_{\cC / \cB}+\bal_{\cB / \cA} +\bal_{\cA / \cE}\text{.}\)
Solution .
With \(\bom_{\cC /\cE} = \dot{\al} \be_{3}
+\dot{\beta} \bha_{1}+\dot{\ga} \bhb_{2}\text{,}\) we find
\begin{align*}
\bal_{\cC / \cE} \amp = \ddot{\al} \be_{3}
+\ddot{\beta} \bha_{1} + \dot{\beta} \bom_{\cA/ \cE}\times \bha_{1}
+\ddot{\ga} \bhb_{2} + \dot{\ga} \bom_{\cB / \cE} \times \bhb_{2} \\
\amp = \ddot{\al} \be_{3} +\ddot{\beta} \bha_{1} +
\dot{\beta} \dot{\al}\bha_{2} +\ddot{\ga} \bhb_{2} + \dot{\ga}(\dot{\beta}
\bhb_{3}
- \dot{\al}\cos\beta \bha_{1})
\end{align*}
One can easily verify that \(\bal_{\cC /\cE}\) is not equal to the sum \(\bal_{\cC / \cB}+\bal_{\cB / \cA} +\bal_{\cA / \cE}\text{,}\) since \(\bal_{\cC / \cB} = \ddot{\ga} \,\bhb_{2} ,\quad
\bal_{\cB / \cA} =\ddot{\beta} \,\bha_{1}\text{,}\) and \(\quad \bal_{\cA / \cE} = \ddot{\al} \,\be_{3}\text{.}\)
