Coupled wheel and rod (analytical mechanics) [closed]

I am struggling with formulating the equations of motion.

Consider a coordinate system with origin in $O$ ($y$ upwards and $x$ to the right), label the center of mass of rod $AB$ with $G$ then:

$$\begin{cases} x_A = ( R - r) \sin \theta \\ y_A = -(R - r) \cos \theta \end{cases} \quad \text{&} \quad \begin{cases} x_G = (R - r) \sin \theta + \frac l2 \sin \varphi \\ y_G = - (R - r) \cos \theta - \frac l2 \cos \varphi \end{cases}$$

The plan is that we want to use $L = T - V$ together with $$\frac {\partial L}{\partial q_k} = \frac { d }{dt} \left( \frac {\partial L}{\partial \dot{q}_k } \right)$$

where $T$ is the sum of the kinetic energy of the rod and wheel (given individually by $\frac 12 m v^2 + \frac 12 I \omega^2$) and $V$ is the potential energy. We can differentiate the position equations in order to obatin $v_G^2$ and $v_A^2$ respectively. Ultimately I have arrived at

$$\begin{cases} T_G = \frac 12 m \bigg( (R-r)^2 \dot{\theta}^2 + \frac {l^2}4 \dot{\varphi}^2 + (R - r) \dot{\theta} l \dot{\varphi} \cos (\theta - \varphi) \bigg) \\ T_A = \frac {\dot{\theta}^2}4 \bigg( 2m (R - r)^2 + mr^2 \bigg) \end{cases}$$

and

$$\begin{cases} V_A = -mg (R - r)\cos \theta + \text{const.} \\ V_G = -mg \left( (R - r)\cos \theta + \frac l2 \cos \varphi \right) \end{cases}$$

I worry though that there must first be some relationship established between the angular velocities of the two bodies but I am unable to pinpoint it. Also I am not entirely confident of my hitherto progress, is it correct thus far?

nevermind, solved the problem eventually

closed as off-topic by Kyle Kanos, ACuriousMind♦, Martin, Jim, Qmechanic♦May 11 '15 at 21:31

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• There is no necessary relationship between the two angular velocities since there is a hinge - however there will be a torque on one due to acceleration of the other, so their accelerations will be linked. – Floris Dec 21 '14 at 20:30
• Could you post the solution you found? It would be interesting for the rest of us. – Floris Dec 24 '14 at 1:43

1 Answer

The angular velocity for the disk is given by $\dot{x}_G^2 + \dot{y}_G^2 = v_G^{\, 2} = \omega_{disk}^2 r^2$. Insert $\omega_{\text{disk}}^2$ into $$T_A = \frac 12 m v_G^2 + \frac 12 I_A \omega_{\text{disk}}^2$$

We can then add up $T_A$ and $T_G$ to form $T$, likewise for $V = V_A + V_G$. The Lagrange function is then given be $L = T - V$ and we need to study $$\begin{gather*} \frac {\partial L}{\partial \theta} = \frac {d}{dt} \left( \frac {\partial L}{\partial \dot{\theta}} \right) \end{gather*} \\ \frac {\partial L}{\partial \varphi} = \frac d{dt} \left( \frac { \partial L }{\partial \dot{\varphi}} \right)$$

These equations will give rise to two differential which upon linearization can be written as $$\begin{gather*} 5(R -r) \ddot{\theta} + l \ddot{ \varphi } + 4g\theta = 0 \\ 3(R-r)\ddot{\theta} + 2l\ddot{ \varphi } + 3g \varphi = 0\end{gather*}$$

Let $\theta = A \cos {\omega t + \alpha}$ and $\varphi = B \cos {\omega t + \alpha}$, and insert into the differential equations. A linear system will arise with $A$ and $B$ as unknowns. Upon computing the determinant one can determine the frequencies $\omega$.