# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, ACuriousMind, Martin, Jim, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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

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 $\omega$.