Respuesta :
Newton-Euler Approach
The equations of motion for the system can be derived by summing the forces and moments on the system.
Derivation for newton-Euler approach :
For the block:
F_x = ma = me \ddot x \\force calculation
M_y = 0 = me \ddot x r \sin \theta
For the sphere:
F_x = 0 = m_y \ddot x + m_y \dot \theta^2 r \cos \theta \\
M_y = 0 = m_y \ddot x r \sin \theta - m_y r^2 \dot \theta
Combining the equations:
me \ddot x = m_y \ddot x + m_y \dot \theta^2 r \cos \theta \\
0 = m_y \ddot x r \sin \theta - m_y r^2 \dot \theta
Combining the two equations gives the following equation of motion:
\ddot x = \frac{m_y}{me} \dot \theta^2 r \cos \theta + \frac{m_y}{me} \frac{r^2}{r \sin \theta} \dot \theta
Lagrange's Equations
The Lagrangian for the system is given by:
L = \frac{1}{2} m_e \dot x^2 + \frac{1}{2} m_y \dot x^2 + \frac{1}{2} m_y \dot \theta^2 r^2
The equations of motion can be derived from the Lagrangian by taking the partial derivatives with respect to the generalized coordinates.
\frac{\partial L}{\partial x} = m_e \ddot x + m_y \ddot x = 0 \\
\frac{\partial L}{\partial \theta} = m_y \ddot \theta r^2 = 0
Combining the two equations gives the following equation of motion:
\ddot x = -\frac{m_y}{m_e} \dot \theta^2 r \cos \theta - \frac{m_y}{m_e} \frac{r^2}{r \sin \theta} \dot \theta
Comparing the two solutions, we see that they are the same.
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