3.5 Energy in a System of Multiple Particles

3.5.1 Motion Relative to the Centre of Mass

Let’s look at motion relative to a centre of mass. Consider a macroscopic body composed of many particles with a centre of mass at $\boldsymbol{r}_{\text{COM}}$ .

We can label the position of particle $i$ inside the body as $\boldsymbol{r}_{i}$ , and its position relative to the centre of mass at $\boldsymbol{r}_{i}^{\prime}$ . Then

\boldsymbol{r}_{i}=\boldsymbol{r}_{\text{COM}}+\boldsymbol{r}_{i}^{\prime}.

(3.69)

If we differentiate this equation, we can find the velocity of particle $i$ :

	$\displaystyle\boldsymbol{v}_{i}=\dot{\boldsymbol{r}}_{i}$	$\displaystyle=\dot{\boldsymbol{r}}_{\text{COM}}+\dot{\boldsymbol{r}}_{i}^{\prime}$		(3.70)
		$\displaystyle=\boldsymbol{v}_{\text{COM}}+\boldsymbol{v}_{i}^{\prime},$		(3.71)

where $\boldsymbol{v}_{i}^{\prime}$ is the velocity of particle $i$ relative to the centre of mass.

3.5.2 Total Kinetic Energy

Let’s look at the total kinetic energy of the system. It is the sum of the kinetic energy of all the constituent particles.

$\displaystyle K$	$\displaystyle=\frac{1}{2}\sum_{i=1}^{N}m_{i}v_{i}^{2}$	(3.72)
	$\displaystyle=\frac{1}{2}\sum_{i=1}^{N}m_{i}(\boldsymbol{v}_{\text{COM}}+% \boldsymbol{v}_{i}^{\prime})^{2}$	(3.73)
	$\displaystyle=\frac{1}{2}\sum_{i=1}^{N}m_{i}v_{\text{COM}}^{2}+\frac{1}{2}\sum% _{i=1}^{N}m_{i}v_{i}^{\prime 2}+\frac{1}{2}\sum_{i=1}^{N}m_{i}(2\boldsymbol{v}% _{\text{COM}}\cdot\boldsymbol{v}_{i}^{\prime}).$	(3.74)

In the first term, $v_{\text{COM}}^{2}$ is constant and can be pulled out of sum, so the sum is simply the total mass $M$ . The second term is the total kinetic energy in the centre of mass frame of reference, we call this the kinetic energy in the centre of mass. Looking at the last term, we can once again take $\boldsymbol{v}_{\text{COM}}$ out of the sum, so the term becomes $\boldsymbol{v}_{\text{COM}}\cdot\sum_{i=1}^{N}m_{i}\boldsymbol{v}_{i}^{\prime}$ . The sum is the total momentum in the centre of mass frame, which is zero. This can be justified by noting that in the centre of mass frame we have $\sum_{i=1}^{N}m_{i}\boldsymbol{r}_{i}^{\prime}=0$ , so

\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t}\left(\sum_{i=% 1}^{N}m_{i}\boldsymbol{r}_{i}^{\prime}\right)=\sum_{i=1}^{N}m_{i}\boldsymbol{v% }_{i}^{\prime}=0.

(3.75)

Therefore the total expression for the kinetic energy of the system is

K=\frac{1}{2}Mv_{\text{COM}}^{2}+\frac{1}{2}\sum_{i=1}^{N}m_{i}v_{i}^{\prime 2}.

(3.76)

This is the kinetic energy associated with the total mass moving with speed $v_{\text{COM}}$ and the internal kinetic energy from motion of particles relative to the centre of mass (the kinetic energy in the centre of mass). Depending on the problem, we can ignore the second term as it may not be relevant. For example, if we are examining the acceleration of a train, for the most part we don’t care about the kinetic energy of objects within the vehicle. However, if we are looking at something that is rotating, like a flywheel for instance, we do care about the second term because there is a lot of motion relative to the centre of mass.

3.5.3 Reduced Mass

Consider a system of two particles.

In the centre of mass frame we have that $\sum_{i=1}^{N}m_{1}\boldsymbol{r}_{i}=0$ , which in this case implies that

m_{1}\boldsymbol{r}_{1}=-m_{2}\boldsymbol{r}_{2}.

(3.77)

We can define a vector $\boldsymbol{r}$ that points from $m_{1}$ to $m_{2}$ , given by either

	$\displaystyle\boldsymbol{r}=\boldsymbol{r}_{2}-\boldsymbol{r}_{1},$	(3.78)
or equivalently,
	$\displaystyle\boldsymbol{r}=\boldsymbol{r}_{2}^{\prime}-\boldsymbol{r}_{1}^{% \prime}.$	(3.79)

Substituting these relations into equation 3.77 above and rearranging, we get

\boldsymbol{r}_{1}^{\prime}=\frac{-m_{2}}{m_{1}+m_{2}}\boldsymbol{r},\quad% \boldsymbol{r}_{2}^{\prime}=\frac{m_{1}}{m_{1}+m_{2}}\boldsymbol{r}.

(3.80)

Now let’s look at the kinetic energy. In the centre of mass frame, it is

K^{\prime}=\frac{1}{2}m_{1}v_{1}^{\prime 2}+\frac{1}{2}m_{2}v_{2}^{\prime 2},

(3.81)

were $\boldsymbol{v}_{1}^{\prime}=\dot{\boldsymbol{r}}_{1}^{\prime}$ and $\boldsymbol{v}_{2}^{\prime}=\dot{\boldsymbol{r}}_{2}^{\prime}$ . Using the relations for $\boldsymbol{r}_{1}^{\prime}$ and $\boldsymbol{r}_{2}^{\prime}$ above, we get

\boldsymbol{v}_{1}^{\prime}=\frac{-m_{2}}{m_{1}+m_{2}}\boldsymbol{v},\quad% \boldsymbol{v}_{2}^{\prime}=\frac{m_{1}}{m_{1}+m_{2}}\boldsymbol{v},

(3.82)

where $\boldsymbol{v}=\dot{\boldsymbol{r}}$ is the relative velocity between the two particles. Therefore, the kinetic energy in the centre of mass frame becomes

$\displaystyle K^{\prime}$	$\displaystyle=\frac{1}{2}m_{1}\left(\frac{-m_{2}}{m_{1}+m_{2}}\boldsymbol{v}% \right)^{2}+\frac{1}{2}m_{2}\left(\frac{m_{1}}{m_{1}+m_{2}}\boldsymbol{v}% \right)^{2}$	(3.83)
	$\displaystyle=\frac{1}{2}\frac{m_{1}m_{2}^{2}+m_{1}^{2}m_{2}}{(m_{1}+m_{2})^{2% }}v^{2}$	(3.84)
	$\displaystyle=\frac{1}{2}\frac{m_{1}m_{2}(m_{1}+m_{2})}{(m_{1}+m_{2})^{2}}v^{2}$	(3.85)
	$\displaystyle=\frac{1}{2}\mu v^{2},$	(3.86)

where we have defined the reduced mass $\mu$ as

\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}.

(3.87)

In a general inertial reference frame, the total energy (under the action of a conservative force) is given by

E=\frac{1}{2}Mv_{\text{COM}}^{2}+\frac{1}{2}\mu v^{2}+V(\boldsymbol{r}).

(3.88)