Instructions:
This is a closed note/book exam. Work on the problem by following the order of the questions. Question 5 can be solved independently of question 4.
A spinning top 1 \((O,\boldsymbol{\hat{x}}_1, \boldsymbol{\hat{y}}_1,\boldsymbol{\hat{z}}_1)\) of mass center \(G\) and mass \(m\) is in motion in a referential 0 \((O,\boldsymbol{\hat{x}}_0, \boldsymbol{\hat{y}}_0,\boldsymbol{\hat{z}}_0)\) in such a way that its tip \(O\) remains stationary relative to 0. The connection between 0 and 1 is equivalent to a frictionless spherical joint. The upward vertical is directed along \(\boldsymbol{\hat{z}}_0\). The orientation of body 1 relative to 0 is defined in terms of the three angles \((\psi,\theta, \phi)\) according the following sequence of rotations: \[ (\boldsymbol{\hat{x}}_0, \boldsymbol{\hat{y}}_0,\boldsymbol{\hat{z}}_0) \xrightarrow{{\cal R}_{\psi , \boldsymbol{\hat{z}}_0}} (\boldsymbol{\hat{u}}, \boldsymbol{\hat{v}},\boldsymbol{\hat{z}}_0) \xrightarrow{{\cal R}_{\theta, \boldsymbol{\hat{u}}}} (\boldsymbol{\hat{u}}, \boldsymbol{\hat{w}},\boldsymbol{\hat{z}}_1) \xrightarrow{{\cal R}_{\phi , \boldsymbol{\hat{z}}_1}} (\boldsymbol{\hat{x}}_1, \boldsymbol{\hat{y}}_1,\boldsymbol{\hat{z}}_1) \] The top is axisymmetric about axis \((O,\boldsymbol{\hat{z}}_1)\). The position of its mass center is defined by \(\boldsymbol{r}_{OG}= l \boldsymbol{\hat{z}}_1\). Its inertia operator about point \(O\) is entirely defined by its axial moment of inertia \[ C= \boldsymbol{\hat{z}}_1 \cdot {\cal I}_O (\boldsymbol{\hat{z}}_1) \] and its transverse moment of inertia \[ A= \boldsymbol{\hat{x}}_1 \cdot {\cal I}_O (\boldsymbol{\hat{x}}_1)= \boldsymbol{\hat{y}}_1 \cdot {\cal I}_O (\boldsymbol{\hat{y}}_1) \] The top is under the sole effect of gravity and the contact force at \(O\).
Note: Since a single body is at play, you may use the notation \(\boldsymbol{\omega}= \boldsymbol{\omega}_{1/0}\), \(\boldsymbol{H}_O = \boldsymbol{H}_{O, 1/0}\), etc.
Question 1 (25 pts). Sketch the rotational diagrams which map basis of 0 to basis of body 1. Find the derivatives of unit vectors \(\boldsymbol{\hat{w}}\) and \(\boldsymbol{\hat{z}}_1\). Find the kinematic screw \(\{{\cal V}_{1/0}\}\) resolved at \(G\). Then find acceleration \(\boldsymbol{a}_{G/0}\).
Question 2 (10 pts). Find the expression of kinetic screw \(\{{\cal H}_{1/0}\}\).
Question 3 (10 pts). Find the expression of the action screw \(\{{\cal A}_{\bar{1}\to 1}\}\).
Question 4 (30 pts). Apply the FTD to find the 3 equations of motion and the contact force at \(O\). Can you identify one or more first integrals of motion?
Question 5 (25 pts). Assume that the initial rate of spin angle \(\dot{\phi}_0\) is very large compared to the other contributions to angular velocity \(\boldsymbol{\omega}\), so that the angular momentum \(\boldsymbol{H}_O\) can be approximated as \(\boldsymbol{H}_O \approx C\dot{\phi}\boldsymbol{\hat{z}}_1\). Show that \(\dot{\phi}= Cst = \dot{\phi}_0\), \(\theta=\theta_0\) and that \(\dot{\psi}\) is given by \[ \dot{\psi}= \frac{mgl}{C\dot{\phi}_0} \] where \(\theta_0 \neq 0\) or \(\pi\) is the initial value given to angle \(\theta\). Comment (is the assumed approximation acceptable?).
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 \[ \boldsymbol{D}_A = \frac{d \boldsymbol{H}_A}{dt} + \boldsymbol{v}_A \times m \boldsymbol{v}_G \]
Solution
Question 1. First we sketch the rotational diagrams which map basis of 0 to basis of body 1.
The time-derivatives of \(\boldsymbol{\hat{w}}\) and \(\boldsymbol{\hat{z}}_1\) are easily found by using the angular velocity \(\dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\theta}\boldsymbol{\hat{u}}\) of the referential in which these two vectors are attached: \[ \frac{d\boldsymbol{\hat{w}}}{dt} = (\dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\theta}\boldsymbol{\hat{u}})\times \boldsymbol{\hat{w}}=-\dot{\psi}\cos\theta\boldsymbol{\hat{u}}+ \dot{\theta}\boldsymbol{\hat{z}}_1, \qquad \frac{d\boldsymbol{\hat{z}}_1}{dt} =(\dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\theta}\boldsymbol{\hat{u}})\times \boldsymbol{\hat{z}}_1=\dot{\psi}\sin\theta\boldsymbol{\hat{u}}- \dot{\theta}\boldsymbol{\hat{w}} \] With \(\boldsymbol{\omega}= \dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{z}}_1\) and \(\boldsymbol{v}_{O\in 1/0} =\boldsymbol{0}\), we find the expression of the kinematic screw of the body \[ \{ {\cal V} _{1/0} \} = \begin{Bmatrix} \dot{\psi}\boldsymbol{\hat{z}}_0 + \dot{\theta}\boldsymbol{\hat{u}}+ \dot{\phi}\boldsymbol{\hat{z}}_1 \\ \boldsymbol{0} \end{Bmatrix} _O \] We can find velocity \(\boldsymbol{v}_G= \boldsymbol{\omega}\times \boldsymbol{r}_{OG}\) or simply as \(l \, d\boldsymbol{\hat{z}}_1/dt\): \[ \boxed{ \boldsymbol{v}_G = l(\dot{\psi}\sin\theta\boldsymbol{\hat{u}}- \dot{\theta}\boldsymbol{\hat{w}}) } \] and acceleration (using \(d\boldsymbol{\hat{w}}/dt\)) \[ \boldsymbol{a}_G = l(\ddot{\psi}\sin\theta+ \dot{\psi}\dot{\theta}\cos\theta) \boldsymbol{\hat{u}} + l\dot{\psi}^2 \sin\theta\boldsymbol{\hat{v}}- l\ddot{\theta}\boldsymbol{\hat{w}} - l\dot{\theta}(-\dot{\psi}\cos\theta\boldsymbol{\hat{u}}+\dot{\theta}\boldsymbol{\hat{z}}_1) \] or \[ \boxed{ \boldsymbol{a}_G = l(\ddot{\psi}\sin\theta+ 2\dot{\psi}\dot{\theta}\cos\theta) \boldsymbol{\hat{u}} + l\dot{\psi}^2 \sin\theta\boldsymbol{\hat{v}}- l\ddot{\theta}\boldsymbol{\hat{w}}- l\dot{\theta}^2 \boldsymbol{\hat{z}}_1 } \]
Question 2. To find the kinetic screw \(\{{\cal H}_{1/0}\}\), we need to find the angular momentum \(\boldsymbol{H}_O\) of the top as a body rotating about fixed point \(O\). We are also given the moments of inertia about axes passing through \(O\). We can represent inertia operator \({\cal I}_O\) on basis \(b (\boldsymbol{\hat{u}},\boldsymbol{\hat{w}},\boldsymbol{\hat{z}}_1)\) by taking advantage of the axisymmetry of the body.
Then, using \(\boldsymbol{\omega}\cdot\boldsymbol{\hat{z}}_1 = \dot{\psi}\cos\theta+\dot{\phi}\), \(\boldsymbol{\omega}\cdot \boldsymbol{\hat{w}}= \dot{\psi}\sin\theta\), and \(\boldsymbol{\omega}\cdot \boldsymbol{\hat{u}}= \dot{\theta}\), we find: \[ \boldsymbol{H}_O = {\cal I}_O (\boldsymbol{\omega})= \begin{bmatrix} A & 0 & 0 \\ 0 & A & 0\\ 0 & 0 & C \end{bmatrix}_{b} \begin{bmatrix} \dot{\theta}\\ \dot{\psi}\sin\theta\\ \dot{\psi}\cos\theta+\dot{\phi} \end{bmatrix}_{b} = A\dot{\theta}\boldsymbol{\hat{u}}+ A\dot{\psi}\sin\theta\boldsymbol{\hat{w}}+ C(\dot{\psi}\cos\theta+\dot{\phi})\boldsymbol{\hat{z}}_1 \] This gives the expression of the kinetic screw of the body resolved at \(O\): \[ \{{\cal H}_{1/0}\} = \begin{Bmatrix} m\boldsymbol{v}_G \\ A\dot{\theta}\boldsymbol{\hat{u}}+ A\dot{\psi}\sin\theta\boldsymbol{\hat{w}}+ C(\dot{\psi}\cos\theta+\dot{\phi})\boldsymbol{\hat{z}}_1 \end{Bmatrix}_O \]
Question 3. We find the gravitational and contact contributions to action screw \(\{{\cal A}_{\bar{1}\to 1}\}\) resolved at point \(O\) \[ \{{\cal A}_{\bar{1}\to 1}\}= \begin{Bmatrix} -m g \boldsymbol{\hat{z}}_0 \\\boldsymbol{0} \end{Bmatrix}_G + \begin{Bmatrix} N\boldsymbol{\hat{z}}_0 + F_u \boldsymbol{\hat{u}}+ F_v \boldsymbol{\hat{v}}\\ \boldsymbol{0} \end{Bmatrix}_O = \begin{Bmatrix} (N-m g) \boldsymbol{\hat{z}}_0 + F_u \boldsymbol{\hat{u}}+ F_v \boldsymbol{\hat{v}} \\ mg l \sin\theta\boldsymbol{\hat{u}} \end{Bmatrix}_O \] We used the fact that the joint at \(O\) is equivalent to a frictionless spherical joint, thus implying \(\boldsymbol{M}_O = \boldsymbol{0}\).
Question 4. We now apply the FTD \(\{{\cal D}_{1/0} \}= \{{\cal A}_{\bar{1} \to 1} \}\) by resolving all screws at point \(O\): this will decouple the 3 unknowns \((\psi,\theta,\phi)\) from the 3 components \((N,F_u,F_v)\) of the contact force. This will give us the 3 equations of motion and the contact force at \(O\) separately: \[ \begin{Bmatrix} m \boldsymbol{a}_G \\ \frac{d}{dt} \left( A\dot{\theta}\boldsymbol{\hat{u}}+ A\dot{\psi}\sin\theta\boldsymbol{\hat{w}}+ C(\dot{\psi}\cos\theta+\dot{\phi})\boldsymbol{\hat{z}}_1 \right) \end{Bmatrix}_O = \begin{Bmatrix} (N-m g) \boldsymbol{\hat{z}}_0 + F_u \boldsymbol{\hat{u}}+ F_v \boldsymbol{\hat{v}} \\ mg l \sin\theta\boldsymbol{\hat{u}} \end{Bmatrix}_O \] We can write the moment equation as \[ \frac{d}{dt} (\boldsymbol{H}_O\cdot \boldsymbol{X}) - \frac{d\boldsymbol{X}}{dt} \cdot \boldsymbol{H}_O = mg l \sin\theta\boldsymbol{\hat{u}}\cdot\boldsymbol{X} \] with \(\boldsymbol{X}= \boldsymbol{\hat{z}}_0, \boldsymbol{\hat{u}}, \boldsymbol{\hat{z}}_1\). This will give the 3 equations of motion governing angles \(\psi\), \(\theta\) and \(\phi\).
The first choice \(\boldsymbol{X}=\boldsymbol{\hat{z}}_0\) gives a first integral \(\boldsymbol{H}_O \cdot \boldsymbol{\hat{z}}_0 = Cst\) since \(\boldsymbol{\hat{z}}_0\) is constant: \[ \boxed{(A\sin^2\theta+C \cos^2 \theta)\dot{\psi}+ C\dot{\phi}\cos\theta= Cst } \quad (1) \] The second choice \(\boldsymbol{X}= \boldsymbol{\hat{u}}\) (using \(d\boldsymbol{\hat{u}}/dt = \dot{\psi}\boldsymbol{\hat{v}}\)) gives \[ \boxed{ A\ddot{\theta}+ (C-A)\dot{\psi}^2 \sin\theta\cos\theta+ C\dot{\psi}\dot{\phi}\sin\theta=mg l \sin\theta }\quad (2) \] Finally, using \(d\boldsymbol{\hat{z}}_1/dt\) of question 1, we obtain (with \(\boldsymbol{X}=\boldsymbol{\hat{z}}_1\)) \[ C\frac{d}{dt}(\dot{\psi}\cos\theta+\dot{\phi}) - (A\dot{\theta}\dot{\psi}\sin\theta-A \dot{\psi}\dot{\theta}\sin\theta) =0 \] which gives another first integral of motion: \[ \boxed{ \dot{\psi}\cos\theta+\dot{\phi}= Cst}\quad (3) \] Finally, using the expression of \(\boldsymbol{a}_G\) found in question 1, we can find the components \((F_u,F_v,N)\) of the contact force: \[\begin{align*} F_u & = m\boldsymbol{a}_G \cdot \boldsymbol{\hat{u}}= l(\ddot{\psi}\sin\theta+ 2\dot{\psi}\dot{\theta}\cos\theta) & \quad(4) \\ F_v & = m\boldsymbol{a}_G \cdot \boldsymbol{\hat{v}}= l\dot{\psi}^2 \sin\theta- l\ddot{\theta}\cos\theta+ l\dot{\theta}^2\sin\theta & \quad (5) \\ N-mg & = m\boldsymbol{a}_G \cdot \boldsymbol{\hat{z}}_0 = - l\ddot{\theta}\sin\theta- l\dot{\theta}^2 \cos\theta & \quad (6) \end{align*}\]
Question 5. Now we assume that \(\boldsymbol{H}_O \approx C\dot{\phi}\boldsymbol{\hat{z}}_1\). It is then straightforward to find \(d\boldsymbol{H}_O/dt\) (using the expression of \(d\boldsymbol{\hat{z}}_1/dt\)): \[ \frac{d\boldsymbol{H}_O}{dt} = C \ddot{\phi}\boldsymbol{\hat{z}}_1 + C\dot{\phi}(\dot{\psi}\sin\theta\boldsymbol{\hat{u}}-\dot{\theta}\boldsymbol{\hat{w}}) \] Now equations (1-3) are replaced by much simpler equations \[\begin{align*} C\dot{\phi}\dot{\psi}\sin\theta& = mgl \sin\theta \\ -C\dot{\phi}\dot{\theta}& = 0 \\ C\ddot{\phi}& =0 \end{align*}\] which give \[ \boxed{ \dot{\phi}= \dot{\phi}_0, \quad \theta= \theta_0 , \quad \dot{\psi}= \frac{mgl}{C\dot{\phi}_0} } \] where we have assumed \(\theta_0 \neq 0\) or \(\pi\). This seems, at first, a strange result: the gravitational moment acting on the top gives rise to a precessional motion (i.e. a rotation about vertical axis \((O,\boldsymbol{\hat{z}}_0)\)) instead of the expected rotation about axis \((O,\boldsymbol{\hat{u}})\). We also note that, if \(C\dot{\phi}_0\gg mgl\), this precessional motion is very slow. This is in fact the expected motion of the spinning top. The approximation seems reasonable.
© 2022 R. Valéry Roy.