# How to solve this exercise in Mechanics?

Jan 4, 2017

See below.

#### Explanation:

Calling $v = \dot{x}$ and $\omega = \dot{\theta}$ and

${v}_{c} = r \omega \left(\cos \left(\theta\right) , - \sin \left(\theta\right)\right) + \left(v , 0\right)$ we have

${E}_{K} = \frac{1}{2} \left(J {\omega}^{2} + {m}_{2} \left\langle{v}_{c} , {v}_{c}\right\rangle + {m}_{1} {v}^{2}\right)$
=1/2 ((m_1 + m_2)dot x^2 + 2 m_2 r cos(theta) dot x dot theta+ (J + m_2 r^2)dot theta^2)
${E}_{P} = \left(R - r\right) \left(1 - \cos \left(\theta\right)\right) {m}_{2} g + \frac{1}{2} {x}^{2}$

so

$L = {E}_{K} - {E}_{P}$

The movement equations are obtained solving for $\dot{v} , \dot{\omega}$

$\frac{d}{\mathrm{dt}} \left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = {F}_{q}$

where $q = \left(x , \theta\right)$ and ${F}_{q} = \left(F , 0\right)$ obtaining

$\ddot{x} = \frac{\left(J + {m}_{2} {r}^{2}\right) \left(F - k x + {m}_{2} r \sin \left(\theta\right) {\dot{\theta}}^{2}\right) + {m}_{2} g r \left(R - r\right) \cos \left(\theta\right) \sin \left(\theta\right)}{\left({m}_{1} + {m}_{2}\right) \left(J + {m}_{2} {r}^{2}\right) - {m}_{2}^{2} {r}^{2} {\cos}^{2} \left(\theta\right)}$
$\ddot{\theta} = - \frac{{m}_{2} \cos \left(\theta\right) \left(F r + g \left({m}_{1} + {m}_{2}\right) \left(R - r\right) \tan \left(\theta\right) - k r x + {m}_{2} {r}^{2} \sin \left(\theta\right) {\dot{\theta}}^{2}\right)}{\left({m}_{1} + {m}_{2}\right) \left(J + {m}_{2} {r}^{2}\right) - {m}_{2}^{2} {r}^{2} {\cos}^{2} \left(\theta\right)}$