Kinetic energy in center of mass reference frame

[zenterix]

Homework Statement

In MIT OCW's 8.01 "Classical Mechanics" textbook, there is a derivation to show that change in kinetic energy is independent of the choice of relatively inertial reference frames.

Relevant Equations

$K=\frac{1}{2}mv^2$

Here is the relevant chapter.

Suppose we have two masses $m_1$ and $m_2$ interacting via some force, and two reference frames, $S$ and $CM$. The $CM$ frame is the center of mass reference frame. The origin of this reference frame is at the location of the center of mass of the system.

The position of particle $j$ in frame $S$ is $\vec{r}_j$ and in frame $CM$ is $\vec{r}_{j}'$.

We have

$$\vec{r}_j=\vec{r}_j'+\vec{r}_{cm}$$

$$\vec{v}_j=\vec{v}_j'+\vec{v}_{cm}$$

Kinetic energy in the CM frame can be shown to be

$$K_{cm}=\frac{1}{2}(m_1v_1'^2+m_2v_2'^2)=\frac{1}{2}\mu\vec{v}_{1,2}^2$$

where $\mu=\frac{m_1m_2}{m_1+m_2}$ and $\vec{v}_{1,2}=\vec{v}_1-\vec{v}_2$.

My question is about kinetic energy in frame $S$.

We start with

$$K_S=\frac{1}{2}(m_1v_1^2+m_2v_2^2)$$

And after subbing in $v_j^2=\vec{v}_j\cdot\vec{v_j}=(\vec{v}_j'+\vec{v}_cm)\cdot(\vec{v}_j'+\vec{v}_{cm})$ we end up with

$$K_S=K_{cm}+\frac{1}{2}(m_1+m_2)v_{cm}^2$$

Now, if $v_{cm}$ is constant then $K_S$ differs from $K_{cm}$ by a constant.

But how do we know that $v_{cm}$ is constant?

After writing this up I think I figured out the answer.

There are no external forces on the system of two particles. Thus, the acceleration of the center of mass is zero. Thus the velocity of the center of mass is constant.

 

[zenterix]

Since I already solved the question, I think this question can be deleted.

 

[kuruman]

Since I already solved the question, I think this question can be deleted.

Why? This is not only about you. It's a good illustration of how one grapples with a question and reasons out the answer. Others might profit from it by seeing that, more often than not, getting to the answer is not as smooth a process as textbooks want us to believe.

 

[PeroK]

Since I already solved the question, I think this question can be deleted.

You can also get the result without considering the CoM frame. If two inertial frames are related by some relative velocity $\vec v$, then the KE of mass $m_1$ is related by:
$$KE_1' = \frac 1 2 m_1v_1'^2 = \frac 1 2 m_1|\vec v_1 +\vec v|^2$$$$= \frac 1 2 m_1(v_1^2 + 2\vec v_1\cdot \vec v + v^2)$$$$= KE_1 + \vec p_1 \cdot \vec v + \frac 1 2 m_1v^2$$Hence, the change in KE is:
$$\Delta KE_1' = \Delta KE_1 + \Delta \vec p_1 \cdot \vec v$$Hence, for a system of (any number of) particles:
$$\Delta KE' = \Delta KE + \Delta \vec P \cdot \vec v$$So, if the total momentum of the system is conserved ($\Delta \vec P = 0$), then the change in KE is independent of the inertial reference frame.