Given: 2 masses are connected by a spring and are sliding down a frictionless inclined plane. They also have a nonzero angular momentum about their COM.
My attempt:
$$L_{com}=T_{com}-U_{com}$$
$$=\frac{1}{2}mv^2-mgh$$
$$=\frac{1}{2}(m_1+m_2)v^2-(m_1+m_2)g(b\sin(\alpha)) \quad, $$ where $b$ is how far up the plane the COM is.
But my questions are this:
Do I simply add up the masses for the motion of the center of mass? Is it really that simple? Note - this is classical mechanics (if it wasn't already obvious).
The problem says that the masses are in motion on the plane because of the gravitational field. So if they are already in motion, then $b$ will change with time. Therefore, I need to express $b$ as a function of time, right?
Now for the second part, I also need the Lagrangian for the relative motion.
$$L_{rel}=T_{rel}-U_{rel}$$
$$=\frac{1}{2}I\omega ^2 - \frac{1}{2}kd^2 \quad ,$$ where $d$ is the distance past the equilibrium point.
My question for this part:
- It seems strange that the relative kinetic energy deals with rotation and the relative potential energy deals with distance between the masses. Shouldn't they be more closely related? For example, when we talk about the KE and PE of a pendulum, the relationship is that the PE that builds up will eventually be "given" to the KE when it starts to fall. In this case, the potential energy of the spring's stretching or compressing does not seem to augment the rotational KE.