Sliding down a Sliding Plane

Consider the case of a particle of mass $m$ sliding down a smooth inclined plane of mass $M$ which is, itself, free to slide on a smooth horizontal surface, as shown in Fig. 48. This is a two degree of freedom system, so we need two coordinates to specify the configuration. Let us choose $x$, the horizontal distance of the plane from some reference point, and $x'$, the parallel displacement of the particle from some reference point on the plane.

Figure 48: A sliding plane.

Defining $x$- and $y$-axes, as shown in the diagram, the $x$- and $y$-components of the particle's velocity are clearly given by

$$v_x = \dot{x} + \dot{x}'\,\cos\theta,$$ (707)

$$v_y = -\dot{x}'\,\sin\theta,$$ (708)

respectively, where $\theta$ is the angle of inclination of the plane with respect to the horizontal. Thus,

$$v^2 = v_x^{\,2} + v_y^{\,2} = \dot{x}^{\,2} + 2\,\dot{x}\,\dot{x}'\,\cos\theta+ \dot{x}'^{\,2}.$$ (709)

Hence, the kinetic energy of the system takes the form

$$K = \frac{1}{2}\,M\,\dot{x}^{\,2} + \frac{1}{2}\,m\,(\dot{x}^{\,2} + 2\,\dot{x}\,\dot{x}'\,\cos\theta+ \dot{x}'^{\,2}),$$ (710)

whereas the potential energy is given by

$$U = - m\,g\,x'\,\sin\theta + {\rm constant}.$$ (711)

It follows that the Lagrangian is written

$$L = \frac{1}{2}\,M\,\dot{x}^{\,2} + \frac{1}{2}\,m\,(\dot{x}... ...os\theta+ \dot{x}'^{\,2})+ m\,g\,x'\,\sin\theta + {\rm const}.$$ (712)

The equations of motion,

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}}\right) - \frac{\partial L}{\partial x} = 0,$$ (713)

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{x}'}\right) - \frac{\partial L}{\partial x'} = 0,$$ (714)

thus yield

$$M\,\ddot{x}+ m\,(\ddot{x} + \ddot{x}'\,\cos\theta) = 0,$$ (715)

$$m\,(\ddot{x}'+ \ddot{x}\,\cos\theta) - m\,g\,\sin\theta = 0.$$ (716)

Finally, solving for $\ddot{x}$ and $\ddot{x}'$, we obtain

$$\ddot{x} = - \frac{g\,\sin\theta\,\cos\theta}{(m+M)/m-\cos^2\theta},$$ (717)

$$\ddot{x}' = \frac{g\,\sin\theta}{1 - m\,\cos^2\theta/(m+M)}.$$ (718)