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:



$$=\frac{1}{2}(m_1+m_2)v^2-(m_1+m_2)g(b\sin(\alpha)) \quad, $$ where $b$ is how far up the plane the COM is.

But my questions are this:

  1. 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).

  2. 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.


$$=\frac{1}{2}I\omega ^2 - \frac{1}{2}kd^2 \quad ,$$ where $d$ is the distance past the equilibrium point.

My question for this part:

  1. 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.
  • $\begingroup$ For one thing, you've forgotten about the kinetic energy in the relative coordinate. $\endgroup$ – garyp Dec 9 '16 at 20:38
  • $\begingroup$ @garyp do you mean an additional KE term? Because I wrote that the KE in the relative motion is the rotational energy. $\endgroup$ – whatwhatwhat Dec 9 '16 at 20:40
  • $\begingroup$ Two related answers that I hope don't waste your time, maybe just a quick skim, if you have not read them already physics.stackexchange.com/questions/32609/… and math.stackexchange.com/questions/1093749/… but they don't cover the sliding part, sorry. $\endgroup$ – user108787 Dec 9 '16 at 21:17
  • $\begingroup$ Your $T_{rel}$ assumes the moment of inertia is constant. And note that $b(t)$ is a dynamic variable. I don't want to tell you how to proceed but I think that if you sit down and think for a while you can find a better approach. Look at several worked-out examples. $\endgroup$ – garyp Dec 10 '16 at 2:30
  • $\begingroup$ @CountTo10 I read both of the links. Could you help me understand something about the first link? In Emilio Pisanty's answer, is his first term rotational KE and the 2nd term is oscillatory KE? $\endgroup$ – whatwhatwhat Dec 10 '16 at 5:25

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