Solution

Question 1. Motion 1/0: we need \(\boldsymbol{v}_{A/0} = \frac{d}{dt} (l sin\alpha\boldsymbol{\hat{u}})\) and \(\boldsymbol{\omega}_{1/0}=\dot{\psi}\boldsymbol{\hat{z}}_0\).

Motion 2/1: we can use \(\boldsymbol{v}_{A/2}= \boldsymbol{0}\). \[ \{ {\cal V}_{1/0} \} = \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{z}}_0 \\ l \dot{\psi}\sin\alpha\boldsymbol{\hat{v}} \end{Bmatrix}_A \qquad \{ {\cal V}_{2/1} \} = \begin{Bmatrix} \dot{\phi}\boldsymbol{\hat{x}}_1 \\ \boldsymbol{0} \end{Bmatrix}_A \qquad \{ {\cal V}_{2/0} \} = \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{z}}_0+ \dot{\phi}\boldsymbol{\hat{x}}_1 \\ \dot{\psi}\boldsymbol{\hat{z}}_0 \end{Bmatrix}_A = \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\phi}\boldsymbol{\hat{x}}_1 \\ (l \sin\alpha+ e \cos\beta ) \dot{\psi}\boldsymbol{\hat{v}} \end{Bmatrix}_G \]

Question 2. Here we use the gyroscopic approximation: we approximate angular momentum \(\boldsymbol{H}_G\) of body 2 as \(C\dot{\phi}\boldsymbol{\hat{x}}_1\) and assume \(\dot{\phi}= Cst\): \[ \{{\cal H}_{2/0}\} = \begin{Bmatrix} m (l \sin\alpha+ e \cos\beta ) \dot{\psi}\boldsymbol{\hat{v}} \\ C \dot{\phi}\boldsymbol{\hat{x}}_1 \end{Bmatrix}_G \qquad \{{\cal D}_{2/0}\} = \begin{Bmatrix} -m (l \sin\alpha+ e \cos\beta ) \dot{\psi}^2 \boldsymbol{\hat{u}} \\ C \dot{\phi}\dot{\psi}\cos\beta \boldsymbol{\hat{v}} \end{Bmatrix}_G \] where we used \(\frac{d}{dt}\boldsymbol{\hat{u}}= \dot{\psi}\boldsymbol{\hat{v}}\) and \(\frac{d}{dt}\boldsymbol{\hat{x}}_1 = \dot{\psi}\cos\beta \boldsymbol{\hat{v}}\).

Question 3. In the absence of axial moment on body 2, \(\boldsymbol{\hat{x}}_1 \cdot\boldsymbol{M}_{G, \bar{2}\to 2} = 0\), we obtain \(\ddot{\phi}= 0\): \(\dot{\phi}\) remains constant. Gyroscopic stiffness follows from a very large value of \(\dot{\phi}\): the value of angle \(\alpha\) set at \(t=0\) (which gives the orientation of the gyroscope’s axis in the plane \((O,\boldsymbol{\hat{u}},\boldsymbol{\hat{z}}_0)\)) remains sensibly constant. However the axis \((G,\boldsymbol{\hat{x}}_1)\) will precess about vertical axis (at precessional angular speed \(\dot{\psi}\boldsymbol{\hat{z}}_0\)) as a reaction to the action of the cable. The angle \(\beta\) taken by the cable most likely adjusts its value in response to the gyroscope’s motion (we still treat it as a constant angle).

Question 4. The total external action screw on system \(\Sigma\) is due to the cable and gravity \[ \{{\cal A}_{\bar{\Sigma}\to \Sigma} \} = \begin{Bmatrix} - m g\boldsymbol{\hat{z}}_0 \\ 0 \end{Bmatrix}_G + \begin{Bmatrix} T (\cos\alpha\boldsymbol{\hat{z}}_0 -\sin\alpha\boldsymbol{\hat{u}}) \\ 0 \end{Bmatrix}_A = \begin{Bmatrix} (T\cos\alpha- m g) \boldsymbol{\hat{z}}_0 - T \sin\alpha\boldsymbol{\hat{u}} \\ e T \cos(\beta-\alpha) \boldsymbol{\hat{v}} \end{Bmatrix}_G \] Application of the FTD to system \(\Sigma\) \[ \{{\cal A}_{\bar{\Sigma}\to \Sigma} \}= \{{\cal D}_{2/0}\} \] yields 3 equations \[ T \cos\alpha= mg, \qquad T \sin\alpha= m (l \sin\alpha+ e \cos\beta ) \dot{\psi}^2 \qquad \text{(1-2)} \] \[ C \dot{\phi}\dot{\psi}\cos\beta = e T \cos(\beta-\alpha) \qquad \text{(3)} \] These equations solve for \((\beta, T, \dot{\psi})\) as a function of \(\dot{\phi}\) and the parameters \((m, C, e, l, \alpha)\). With \(T= mg/\cos\alpha\), we find the precession rate and a relationship between angles \(\alpha\) and \(\beta\): \[ \boxed{\dot{\psi}= \frac{m g e \cos (\beta-\alpha)}{C \dot{\phi}\cos\alpha\cos\beta}}, \quad \boxed{ \tan\alpha= g \Big( \frac{m e \cos (\beta-\alpha)}{C \dot{\phi}\cos\alpha\cos\beta} \Big)^2 ( e \cos\beta + l \sin\alpha)} \quad \text{(4-5)} \] As expected \(\dot{\psi}\) is inversely proportional to \(C\dot{\phi}\). The second relationship shows that the two angles \(\alpha\) and \(\beta\) cannot be preset independently. In practice, it is quite likely that they both adjust to steady values to match equation (5). See the final note.

If we set \(\alpha=\beta=0\), we find \(\dot{\psi}=0\) from (2) and \(\dot{\psi}\neq 0\) from (3): No solution is possible.

Question 5. Here we need to find the velocity of \(G\) to account for the variation of \(\alpha\) and \(\beta\): \[\begin{align*} \boldsymbol{v}_{G/0} & = \frac{d}{dt} \left\{ (l \sin\alpha+ e \cos\beta) \boldsymbol{\hat{u}}+ (-l \cos\alpha+ e \sin\beta) \boldsymbol{\hat{z}}_0\right\} \\ & = (l \dot{\alpha}\cos\alpha- e \dot{\beta}\sin\beta) \boldsymbol{\hat{u}}+ (l \sin\alpha+ e \cos\beta) \dot{\psi}\boldsymbol{\hat{v}} + (l \dot{\alpha}\sin\alpha+ e \dot{\beta}\cos\beta) \boldsymbol{\hat{z}}_0 \end{align*}\] This gives the KE of body 2: \[ 2\mathbb{K}_{2/0} = m (l^2 \dot{\alpha}^2 + e^2 \dot{\beta}^2 + 2 el \dot{\alpha}\dot{\beta}\sin(\alpha-\beta) ) + m (l \sin\alpha+ e \cos\beta)^2 \dot{\psi}^2 + C \dot{\phi}^2 \] The gravitational potential energy is given by \[ \mathbb{U}^g = m g (-l \cos\alpha+e \sin\beta) \] The internal power is given by \[ \mathbb{P}_{1\leftrightarrow 2} = \{{\cal V}_{2/1}\} \cdot \{{\cal A}_{1\to 2}\} = \begin{Bmatrix} \dot{\phi}\boldsymbol{\hat{x}}_1 \\ \boldsymbol{0} \end{Bmatrix}_G \cdot \{{\cal A}_{1\to 2}\} = -\mu \dot{\phi}^2 + {\cal C}\dot{\phi} \] The external power is given by \[ \mathbb{P}_{cable \to 1/0} = \{{\cal V}_{0/1}\} \cdot \{{\cal A}_{cable\to 1}\} = \boldsymbol{v}_{A/0} \cdot \boldsymbol{T} =0 \] since \(\boldsymbol{v}_{A/0} = l d\boldsymbol{\tau}/dt\) and \(\boldsymbol{T}= -T \boldsymbol{\tau}\) (where \(\boldsymbol{\tau}\) is the unit vector along line \(OA\)). Finally the TKE applied to system \(\Sigma\) gives: \[ \boxed{ \frac{d}{dt} (\mathbb{K}_{2/0}+ \mathbb{U}^g ) = -\mu \dot{\phi}^2 + {\cal C}\dot{\phi} } \]

Final Note: you should watch this video for a demonstration of this system.

https://www.youtube.com/watch?v=GEKtnlZfksI

Formula Sheet

1. Rodrigues formula. The mapping of (arbitrary) vector \(\boldsymbol{V}\) by rotation \({\cal R}_{\alpha, \boldsymbol{\hat{u}}}\) of angle \(\alpha\) about unit vector \(\boldsymbol{\hat{u}}\) can be obtained as \[ {\cal R}_{\alpha, \boldsymbol{\hat{u}}} (\boldsymbol{V}) = \boldsymbol{V}+ \sin\alpha \boldsymbol{\hat{u}}\times\boldsymbol{V}+ (1- \cos\alpha) \boldsymbol{\hat{u}}\times(\boldsymbol{\hat{u}}\times\boldsymbol{V}) \]

2. Time-Derivative Formula. Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), and an arbitrary vector function \(\boldsymbol{U}(t)\), \[ \Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal E}=\Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal F}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{U} \]

3. Triple product formula: \[ \boldsymbol{A}\times (\boldsymbol{B}\times \boldsymbol{C})= (\boldsymbol{A}\cdot \boldsymbol{C}) \boldsymbol{B}- (\boldsymbol{A}\cdot \boldsymbol{B}) \boldsymbol{C} \]

4. 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, \[ \boldsymbol{v}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{v}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} \] \[ \boldsymbol{a}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{a}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\alpha}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} + \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times (\boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ}) \]

5. Change of Referential Formulas. Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), for any particle \(P\), and at any given time, \[ \boldsymbol{v}_{P/{\cal E}} = \boldsymbol{v}_{P/{\cal F}} + \boldsymbol{v}_{P\in {\cal F}/ {\cal E}} \] \[ \boldsymbol{a}_{P/{\cal E}} = \boldsymbol{a}_{P/{\cal F}} + \boldsymbol{a}_{P\in {\cal F}/ {\cal E}} + 2 \boldsymbol{\omega}_{{\cal F}/{\cal E}} \times \boldsymbol{v}_{P/{\cal F}} \] 6. Loop formulas: Given 3 rigid bodies \({\cal A}\), \({\cal B}\), \({\cal C}\), and an arbitrary point \(P\), \[ \boldsymbol{\omega}_{{\cal A}/{\cal B}}+ \boldsymbol{\omega}_{{\cal B}/{\cal C}}+ \boldsymbol{\omega}_{{\cal C}/{\cal A}} = \boldsymbol{0} \] \[ \boldsymbol{v}_{P\in {\cal A}/ {\cal B}}+ \boldsymbol{v}_{P\in {\cal B}/ {\cal C}}+ \boldsymbol{v}_{P\in {\cal C}/ {\cal A}}=\boldsymbol{0} \]

7. 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 \(B\).

8. 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 \[ \boldsymbol{D}_A =\frac{d \boldsymbol{H}_A}{dt} + \boldsymbol{v}_A \times m \boldsymbol{v}_G \]