Instructions:
Read the questions carefully. Solve the questions in the indicated order.
Use the formulae sheet provided at the end of this document.
Give brief explanations as to how you will carry out your solution.
Problem 1
The goal of this problem is to study the effect of the rotating airplane propeller on the motion of an airplane during a flat turn maneuver, that is, a turn of the airplane without banking its wings. Figure 1 shows the airplane 1\((G_1, \boldsymbol{\hat{x}}_1,\boldsymbol{\hat{y}}_1,\boldsymbol{\hat{z}}_0)\) in motion in a horizontal plane \((O,\boldsymbol{\hat{x}}_0,\boldsymbol{\hat{y}}_0)\) of a Newtonian referential 0\((O,\boldsymbol{\hat{x}}_0,\boldsymbol{\hat{y}}_0,\boldsymbol{\hat{z}}_0)\). The airplane’s orientation is defined by angle \(\psi\) about unit vector \(\boldsymbol{\hat{z}}_0\). The propeller 2\((G_2, \boldsymbol{\hat{x}}_1,\boldsymbol{\hat{y}}_2,\boldsymbol{\hat{z}}_2)\) of mass center \(G_2\) is in rotation about axis \((G_2, \boldsymbol{\hat{x}}_1)\) at angular velocity \(\boldsymbol{\omega}_{2/1} = \dot{\phi}\boldsymbol{\hat{x}}_1\). The orientation of basis \(b_2 (\boldsymbol{\hat{x}}_1,\boldsymbol{\hat{y}}_2,\boldsymbol{\hat{z}}_2)\) is defined by: \[ b_0 (\boldsymbol{\hat{x}}_0,\boldsymbol{\hat{y}}_0,\boldsymbol{\hat{z}}_0) \xrightarrow{{\cal R}_{\psi , \, \boldsymbol{\hat{z}}_0}} b_1(\boldsymbol{\hat{x}}_1,\boldsymbol{\hat{y}}_1,\boldsymbol{\hat{z}}_0) \xrightarrow{{\cal R}_{\phi , \, \boldsymbol{\hat{x}}_1}} b_2(\boldsymbol{\hat{x}}_1,\boldsymbol{\hat{y}}_2,\boldsymbol{\hat{z}}_2) \] The inertia matrix of the propeller about \(G_2\) takes the expression \[ [I_{G_2, 2}]_{b_2} = \begin{pmatrix} A_2 & 0 & 0 \\ 0 & B_2 & 0 \\ 0 & 0 & C_2 \end{pmatrix}_{b_2} \] Assume that \(\dot{\psi}\) and \(\dot{\phi}\) are constant, with \(|\dot{\psi}|\ll \dot{\phi}\). For simplicity we assume that the action screw \(\{ {\cal A}_{\bar{2}\to 2}\}\) on body 2 can be approximated as the sole action of body 1 \(\{ {\cal A}_{1 \to 2}\}\) and we denote \(\boldsymbol{M}_{G_2, 1\to 2} = L_1 \boldsymbol{\hat{x}}_1 + M_1 \boldsymbol{\hat{y}}_1 + N_1 \boldsymbol{\hat{z}}_0\).
1. Find the expression of \(\boldsymbol{\omega}_{2/0}\). Then deduce the expression of angular momentum \(\boldsymbol{H}_{G_2, 2/0}\).
2. Use the results of 1 to find the expression of dynamic moment \(\boldsymbol{D}_{G_2, 2/0} = \frac{d}{dt} \boldsymbol{H}_{G_2, 2/0}\).
3. Now assume that \(B_2 = C_2\). By applying the FTD to body 2, find the components \(L_1, M_1, N_1\) of moment \(\boldsymbol{M}_{G_2, 1\to 2}\) as a function of \(\dot{\psi}\), \(\dot{\phi}\) and the moment of inertia \((A_2, B_2)\). Deduce the qualitative effects of the propeller’s rotation on the motion of the airplane.
Formulas:
1. Time-Derivative Formula. Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), and an arbitrary vector function \(\boldsymbol{U}(t)\), \[ \Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal E}=\Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal F}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{U} \]
2. Triple product formula: \[ \boldsymbol{A}\times (\boldsymbol{B}\times \boldsymbol{C})= (\boldsymbol{A}\cdot \boldsymbol{C}) \boldsymbol{B}- (\boldsymbol{A}\cdot \boldsymbol{B}) \boldsymbol{C}$ \]
3. Rigid Body Kinematics: Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), for any two points \(P\) and \(Q\), and at any given time, \[ \boldsymbol{v}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{v}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} \] \[ \boldsymbol{a}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{a}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\alpha}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} + \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times (\boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ}) \]
4. Change of Referential Formulas. Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), for any particle \(P\), and at any given time, \[ \boldsymbol{v}_{P/{\cal E}} = \boldsymbol{v}_{P/{\cal F}} + \boldsymbol{v}_{P\in {\cal F}/ {\cal E}} \] \[ \boldsymbol{a}_{P/{\cal E}} = \boldsymbol{a}_{P/{\cal F}} + \boldsymbol{a}_{P\in {\cal F}/ {\cal E}} + 2 \boldsymbol{\omega}_{{\cal F}/{\cal E}} \times \boldsymbol{v}_{P/{\cal F}} \]
5. Loop formulas: Given 3 rigid bodies \({\cal A}\), \({\cal B}\), \({\cal C}\), and an arbitrary point \(P\), \[ \boldsymbol{\omega}_{{\cal A}/{\cal B}}+ \boldsymbol{\omega}_{{\cal B}/{\cal C}}+ \boldsymbol{\omega}_{{\cal C}/{\cal A}} = \boldsymbol{0} \] \[ \boldsymbol{v}_{P\in {\cal A}/ {\cal B}}+ \boldsymbol{v}_{P\in {\cal B}/ {\cal C}}+ \boldsymbol{v}_{P\in {\cal C}/ {\cal A}}=\boldsymbol{0} \]
6. Angular Momentum: The angular momentum of a material system \(\Sigma\) of constant mass \(m\) and mass center \(G\) satisfies \[ \boldsymbol{H}_{A} = \boldsymbol{H}_B + \boldsymbol{r}_{AB} \times m\boldsymbol{v}_G \] for any two points \(A\) and \(B\). The angular momentum of a rigid body \({\cal B}\) of mass \(m\) and mass center \(G\) can be found according to \[ \boldsymbol{H}_{B} = {\cal I}_B (\boldsymbol{\omega}) + \boldsymbol{r}_{BG} \times m\boldsymbol{v}_{B\in {\cal B}} \] where \({\cal I}_B\) is the inertia operator of \({\cal B}\) about point \(B\).
7. The Dynamic Moment of a system \(\Sigma\) of mass \(m\) and mass center \(G\) satisfies \[ \boldsymbol{D}_{A} = \boldsymbol{D}_B + \boldsymbol{r}_{AB} \times m\boldsymbol{a}_G \] for any two points \(A\) and \(B\). The dynamic moment \(\boldsymbol{D}_A\) of a system \(\Sigma\) can be found from its angular momentum \(\boldsymbol{H}_A\) according to \[ \boldsymbol{D}_A = \frac{d \boldsymbol{H}_A}{dt} + \boldsymbol{v}_A \times m \boldsymbol{v}_G \]
Solution
Note: prior to working out your solution, you should sketch the rotation diagrams. This will facilitate your solution.
1. The angular velocity \(\boldsymbol{\omega}_{2/0}= \boldsymbol{\omega}_{2/1}+\boldsymbol{\omega}_{1/0}\) is easily found: \[ \boldsymbol{\omega}_{2/0}= \dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\phi}\boldsymbol{\hat{x}}_1 \] Then we can find the expression of angular momentum \(\boldsymbol{H}_{G_2, 2/0}\) of body 2 about mass center \(G_2\) using the inertia matrix about \(G_2\): \[ \boldsymbol{H}_{G_2, 2/0}= {\cal I}_{G_2, 2}(\boldsymbol{\omega}_{2/0}) = \begin{pmatrix} A_2 & 0 & 0 \\ 0 & B_2 & 0 \\ 0 & 0 & C_2 \end{pmatrix}_{b_2} \begin{pmatrix} \dot{\phi}\\ \dot{\psi}\sin\phi \\ \dot{\psi}\cos\phi\end{pmatrix}_{b_2} = \boxed{ A_2 \dot{\phi}\boldsymbol{\hat{x}}_1 + B_2 \dot{\psi}\sin\phi \boldsymbol{\hat{y}}_2 + C_2 \dot{\psi}\cos\phi \boldsymbol{\hat{z}}_2} \]
2. Assuming \(\dot{\psi}\) and \(\dot{\phi}\) constant, we then obtain the dynamic moment about \(G_2\) by taking the time-derivative of \(\boldsymbol{H}_{G_2, 2/0}\): \[\begin{align*} \boldsymbol{D}_{G_2, 2/0} & = \frac{d}{dt} \boldsymbol{H}_{G_2, 2/0} = A_2 \dot{\phi}\dot{\psi}\boldsymbol{\hat{y}}_1 +B_2 \dot{\psi}\dot{\phi}\cos\phi \boldsymbol{\hat{y}}_2 + B_2 \dot{\psi}\sin\phi (\dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\phi}\boldsymbol{\hat{x}}_1)\times\boldsymbol{\hat{y}}_2 -C_2 \dot{\psi}\dot{\phi}\sin\phi \boldsymbol{\hat{z}}_2\\ &\qquad\qquad\qquad+ C_2 \dot{\psi}\cos\phi (\dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\phi}\boldsymbol{\hat{x}}_1) \times\boldsymbol{\hat{z}}_2\\ & = \boxed{ (C_2-B_2)\dot{\psi}^2 \sin\phi\cos\phi \boldsymbol{\hat{x}}_1 + A_2 \dot{\phi}\dot{\psi}\boldsymbol{\hat{y}}_1 + (B_2 - C_2) \dot{\psi}\cos\phi \dot{\phi}\boldsymbol{\hat{y}}_2 + (B_2 -C_2) \dot{\psi}\dot{\phi}\sin\phi \boldsymbol{\hat{z}}_2 } \end{align*}\] where we made sure to take the time-derivative of \(\boldsymbol{\hat{y}}_2\) and \(\boldsymbol{\hat{z}}_2\).
3. Assuming that \(B_2 = C_2\), to find the components \(L_1, M_1, N_1\) of \(\boldsymbol{M}_{G_2, 1\to 2}\), we apply the moment equation of the FTD applied to body 2 (Euler’s second principle) and take into account the approximation \(\{ {\cal A}_{\bar{2}\to 2}\} \approx \{ {\cal A}_{1 \to 2}\}\): this gives \[ \boldsymbol{M}_{G_2, 1\to 2} \approx \boldsymbol{M}_{G_2,\bar{2}\to 2} = \boldsymbol{D}_{G_2, 2/0} \] In this case the expression of \(\boldsymbol{D}_{G_2, 2/0}\) found in question 2 simplifies to: \[ \boldsymbol{D}_{G_2, 2/0}=A_2 \dot{\phi}\dot{\psi}\boldsymbol{\hat{y}}_1 \] This gives on basis \((\boldsymbol{\hat{x}}_1 , \boldsymbol{\hat{y}}_1 , \boldsymbol{\hat{z}}_0)\): \[\begin{align*} \boxed{ L_1 = 0} \\ \boxed{M_1 = A_2 \dot{\phi}\dot{\psi}} \\ \boxed{N_1 =0} \end{align*}\] Since \(\boldsymbol{M}_{G_2, 1\to 2} = -\boldsymbol{M}_{G_2, 2\to 1}= -A_2 \dot{\phi}\dot{\psi}\boldsymbol{\hat{y}}_1\) the rotation of the propeller is inducing a moment on the airplane directed along \(\boldsymbol{\hat{y}}_1\): if \(\dot{\phi}\dot{\psi}>0\), a turn of the airplane is followed by a tendency of the airplane to rise. Conversely, if \(\dot{\phi}\dot{\psi}>0\), a turn of the airplane is followed by a tendency of the airplane to dive. This effect is quite noticeable to pilots of light aircrafts.
© 2022 R. Valéry Roy.