7. Dynamics — Classical Mechanics

Solution.

Kinematics. Using \(x\), \(\theta\) as generalized coordinates,

\[\begin{split}\begin{cases} \vec{r}_A = x \hat{x} \\ \vec{r}_B = ( x + \ell \sin \theta ) \hat{x} + \ell \cos \theta \hat{y} \\ \end{cases}\end{split}\]

\[\begin{split}\begin{cases} \vec{v}_A = \dot{x} \hat{x} \\ \vec{v}_B = ( \dot{x} + \ell \dot{\theta} \cos \theta ) \hat{x} - \ell \dot{\theta} \sin \theta \hat{y} \\ \end{cases}\end{split}\]

Lagrangian function.

\[\mathscr{L} = K + U\]

\[\begin{split}\begin{aligned} K & = \frac{1}{2} m_A |\vec{v}_A|^2 + \frac{1}{2} m_B |\vec{v}_B|^2 = \\ & = \frac{1}{2} m_A \dot{x}^2 + \frac{1}{2} m_B \left( \dot{x}^2 + \ell^2 \dot{\theta}^2 + 2 \ell \dot{x} \dot{\theta} \cos \theta \right) \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} U & = - \frac{1}{2} k x_A^2 + m_B g y_B = \\ & = - \frac{1}{2} k x^2 + m_B g \ell \cos \theta \end{aligned}\end{split}\]

Lagrange equations (II type) RHS of Lagrange equations read

\[\begin{split}\begin{aligned} \dfrac{d}{dt}\left( \frac{\partial \mathscr{L}}{\partial \dot{x}} \right) & = \dfrac{d}{dt} \left( m_A \dot{x} + m_B \dot{x} + m_B \ell \dot{\theta} \cos \theta \right) = \left( m_A + m_B \right) \ddot{x} + m_B \ell \ddot{\theta} \cos \theta - m_B \ell \dot{\theta}^2 \sin \theta \\ \frac{\partial \mathscr{L}}{\partial x } & = -k x \\ \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \dfrac{d}{dt}\left( \frac{\partial \mathscr{L}}{\partial \dot{\theta}} \right) & = \dfrac{d}{dt} \left( m_B \ell^2 \dot{\theta} + m_B \ell \dot{x} \cos \theta \right) = m_B \ell^2 \ddot{\theta} + m_B \ell \ddot{x} \cos \theta - m_B \ell \dot{x} \dot{\theta} \sin \theta \\ \frac{\partial \mathscr{L}}{\partial \theta } & = - m_B \ell \dot{x} \dot{\theta} \sin \theta - m_B g \ell \sin \theta \\ \end{aligned}\end{split}\]

Generalized forces read

\[Q_x = F\]

\[Q_\theta = C\]

so that the pure equations of motion follows from Lagrange equations \(\frac{d}{dt} \left( \frac{\partial \mathscr{L}}{\partial \dot{q}} \right) - \frac{\partial \mathscr{L}}{\partial q} = Q_q\)

\[\begin{split}\begin{cases} \left( m_A + m_B \right) \ddot{x} + m_B \ell \ddot{\theta} \cos \theta - m_B \ell \dot{\theta}^2 \sin \theta + k x = F \\ m_B \ell^2 \ddot{\theta} + m_B \ell \ddot{x} \cos \theta - m_B \ell \dot{x} \dot{\theta} \sin \theta + m_B \ell \dot{x} \dot{\theta} \sin \theta + m_B g \ell \sin \theta = C \\ \end{cases}\end{split}\]

and after the simplifications

\[\begin{split}\begin{cases} \left( m_A + m_B \right) \ddot{x} + m_B \ell \ddot{\theta} \cos \theta - m_B \ell \dot{\theta}^2 \sin \theta + k x = F \\ m_B \ell \ddot{x} \cos \theta + m_B \ell^2 \ddot{\theta} + m_B g \ell \sin \theta = C \\ \end{cases}\end{split}\]

Obs. The first equation is the \(x\)-component of the momentum equation of the whole system. The second equation is the angular momentum equation of the rod around the hinge in \(A\).

Generalized forces on rigid bodies

Following the derivation of the Lagrange equations for rigid bodies, generalized forces are

\[Q_q = \frac{\partial \vec{r}_G}{\partial q} \cdot \vec{R}^e + \frac{\partial \vec{\theta}}{\partial q} \cdot \vec{M}_G^e\]

With the definition of \(\theta_{\delta}\) in the increment of the rigid body motion

\[d \vec{r}_A = d \vec{r}_B + \theta_{\delta} \times (A - B) \ ,\]

and writing the resultant of forces and moments as

\[\begin{split}\begin{aligned} \vec{R}^e & = \sum_i \vec{F}_i \\ \vec{M}^e_G & = \sum_i \left( \vec{r}_G - \vec{r}_i \right) \times \vec{F}_i + \sum_i \vec{C}_i \\ \end{aligned}\end{split}\]

the generalized force can be recast as

\[\begin{split}\begin{aligned} Q_q & = \frac{\partial \vec{r}_G}{\partial q} \cdot \vec{R}^e + \frac{\partial \vec{\theta}}{\partial q} \cdot \vec{M}_G^e = \\ & = \frac{\partial \vec{r}_G}{\partial q} \cdot \sum_i \vec{F}_i + \frac{\partial \vec{\theta}}{\partial q} \cdot \left[ \sum_i \left( \vec{r}_G - \vec{r}_i \right) \times \vec{F}_i + \sum_i \vec{C}_i \right] = \\ & = \sum_i \left[ \frac{\partial \vec{r}_G}{\partial q} + \frac{\partial \vec{\theta}}{\partial q} \times \left( \vec{r}_G - \vec{r}_i \right) \right] \cdot \vec{F}_i + \sum_i \frac{\partial \vec{\theta}}{\partial q} \cdot \vec{C}_i = \\ & = \sum_i \frac{\partial \vec{r}_i}{\partial q} \cdot \vec{F}_i + \sum_i \frac{\partial \vec{\theta}}{\partial q} \cdot \vec{C}_i \ , \end{aligned}\end{split}\]

i.e. as the contribution of the single forces \(\vec{F}_i\) acting on different points \(\vec{r}_i\) the rigid body and the overall contribution of the couples of forces \(\vec{C}_i\).

With \(\vec{r}_C\left(q(t),t \right)\) and \(\mathbb{R}\left(q(t), t \right)\),

\[\vec{r}_i(q(t), t) - \vec{r}_C(q(t), t) = \mathbb{R}(q(t), t) \cdot \left( \vec{r}_i^0 - \vec{r}_C^0 \right)\]

Time derivative becomes

\[\begin{split}\begin{aligned} \vec{v}_i - \vec{v}_C & = \left[ \dot{q}(t) \, \frac{\partial \mathbb{R}}{\partial q} + \frac{\partial \mathbb{R}}{\partial t} \right] \cdot \left( \vec{r}_i^0 - \vec{r}_C^0 \right) = \\ & = \left[ \dot{q}(t) \, \frac{\partial \mathbb{R}}{\partial q} + \frac{\partial \mathbb{R}}{\partial t} \right] \cdot \mathbb{R}^T \cdot \underbrace{\mathbb{R} \cdot \left( \vec{r}_i^0 - \vec{r}_C^0 \right)}_{\vec{r}_i - \vec{r}_C} \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \vec{\omega}_\times & = \dot{\mathbb{R}} \cdot \mathbb{R}^T \\ \delta \vec{\theta}_{\times} & = \delta \mathbb{R} \cdot \mathbb{R}^T \\ \frac{\partial \vec{\theta}_{\times}}{\partial q} & = \dfrac{\partial \mathbb{R}}{\partial q} \cdot \mathbb{R}^T \end{aligned}\end{split}\]

Solution

Geometry.

\[\begin{split}\begin{aligned} R & = d \sin \alpha \\ b & = d \cos^2 \alpha \\ \end{aligned}\end{split}\]

so that \(\frac{b}{R} = \frac{\cos^2 \alpha}{\sin \alpha}\).

Kinematics.

Position of the center of mass, \(C\)

\[\begin{split}\begin{aligned} \vec{r}_C & = b \cos \theta \hat{x} + b \sin \theta \hat{y} - h \hat{z} \\ \vec{v}_C & = -b \dot{\theta} \sin \theta \hat{x} + b \dot{\theta} \cos \theta \hat{y} = b \dot{\theta} \hat{y}_1 \\ \end{aligned}\end{split}\]

Angular velocity \(\vec{\omega}\) of the rigid body

\[\vec{\omega} = \dot{\theta} \hat{z} + \dot{\varphi} \hat{x}_1\]

Velocity of point contact point \(A\) is zero, \(\vec{v}_A = \vec{0}\) for pure rolling constraint. Being \((A-C) = R \hat{z}_1\), the general expression of \(\vec{v}_A\) as function of \(\theta\) and \(\varphi\) reads

\[\begin{split}\begin{aligned} \vec{v}_A & = \vec{v}_C + \vec{\omega} \times \left( A - C \right) = \\ & = \hat{y}_1 b \dot{\theta} + \left( \dot{\theta} \hat{z} + \dot{\varphi} \hat{x}_1 \right) \times R \hat{z}_1 = \\ & = \hat{y}_1 \left( b \dot{\theta} + R \dot{\theta} \sin \alpha - R \dot{\varphi} \right) = \\ \end{aligned}\end{split}\]

so that the kinetatic constraint (integrable, with arbitrary initial condition) between \(\theta\) and \(\varphi\) is

\[R \varphi = \left( R \sin \alpha + b \right) \, \theta \ .\]

Lagrangian function.

With

\[\mathbb{I}_C = I_{x} \hat{x}_1 \hat{x}_1 + I_{y} \hat{y}_1 \hat{y}_1 + I_{z} \hat{z}_1 \hat{z}_1 \ ,\]

\[\vec{\omega} = \dot{\theta} \hat{z} + \dot{\varphi} \hat{x}_1 = \left( \dot{\varphi} - \dot{\theta} \sin \alpha \right) \hat{x}_1 + \dot{\theta} \cos \alpha \hat{z}_1 \]

and using

\[\dot{\varphi} - \dot{\theta} \sin \alpha = \frac{b}{R} \dot{\theta} \ ,\]

the Lagrangian function becomes

\[\begin{split}\begin{aligned} \mathscr{L} & = K + U = \\ & =\frac{1}{2} m |\vec{v}_C|^2 + \frac{1}{2} \vec{\omega} \cdot \mathbb{I}_C \cdot \vec{\omega} + m g x_C = \\ & =\frac{1}{2} m b^2 \dot{\theta}^2 + \frac{1}{2} \left[ I_x \left( \dot{\varphi} - \dot{\theta} \sin \alpha \right)^2 + I_z \dot{\theta}^2 \cos^2 \alpha \right] + m g b \cos \theta = \\ & =\frac{1}{2} m b^2 \dot{\theta}^2 + \frac{1}{2} \left[ I_x \left( \frac{b}{R} \right)^2 + I_z \cos^2 \alpha \right] \dot{\theta}^2 + m g b \cos \theta = \\ & =\frac{1}{2} \widetilde{I} \dot{\theta}^2 + m g b \cos \theta \ , \end{aligned}\end{split}\]

with the equivalent inertia

\[\widetilde{I} = m b^2 + I_x \left( \frac{b}{R} \right)^2 + I_z \cos^2 \alpha \ .\]

Method 1. Lagrange equation (II). Legrange equation gives a pure equation of motion

\[\begin{split}\begin{aligned} 0 & = \dfrac{d}{dt}\left( \frac{\partial \mathscr{L}}{\partial \dot{\theta}} \right) - \frac{\partial \mathscr{L}}{\partial \theta} = \widetilde{I} \ddot{\theta} + m g b \sin \theta \ . \\ \end{aligned}\end{split}\]

Method 2. Kinetic energy theorem - or energy conservation

Solution

Kinematics. todo check kinematic constraints. No influence of \(\theta\)?

\[\varphi_2 = \frac{R_1}{R_2} \varphi_1 = r \varphi_1\]

Lagrangian function.

\[\begin{split}\begin{aligned} K & = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 + \frac{1}{2} m_1 |\vec{v}_1|^2 = \\ & = \frac{1}{2} I_1 \dot{\varphi}_1^2 + \frac{1}{2} I_2 \dot{\varphi}_2^2 + \frac{1}{2} m_1 \ell^2 \dot{\theta}^2 = \\ & = \frac{1}{2} \left( I_1 + I_2 r^2 \right) \dot{\varphi}_1^2 + \frac{1}{2} m_1 \ell^2 \dot{\theta}^2 = \\ \end{aligned}\end{split}\]

\[\begin{aligned} U = - \frac{1}{2} k \theta^2 - m g \ell \sin \theta \end{aligned}\]

Generalized forces read

\[\begin{split}\begin{aligned} Q_{\varphi_1} & = C - R_1 F \\ Q_{\theta} & = C \end{aligned}\end{split}\]

Lagrange functions (II).

\[\begin{split}\begin{cases} \left( I_1 + I_2 r^2 \right) \ddot{\varphi} = C - R_1 F \\ m_1 \ell^2 \ddot{\theta} + k \theta + m g \ell \cos \theta = C \\ \end{cases}\end{split}\]

Solution

Dynamics

7. Dynamics#