# What are the expressions for rotational and translational kinetic energies of a system of point particles?

Consider a system of point particles , where the mass of particle $i$ is $\mu_i$ and its position vector is $r_i$. What are the expressions for translational kinetic energy and rotational kinetic energy of the system of particles?

The expression for the total kinetic energy for the system is given as: $$\frac{1}{2}\sum_i \mu_i(\dot{r}_i\cdot\dot{r}_i).$$ The sum of the rotational and translational kinetic energies should be equal to the total kinetic energy of the system.

My guess is that the translational kinetic energy of the system is given as: $$\frac{1}{2}(\sum_i \mu_i)(\dot{r}_{cm}\cdot\dot{r}_{cm}),$$ where $r_{cm}$ is the position vector of the center of mass of the system.

However, I am not able to get the expression for the rotational kinetic energy of the system and appreciate some help.

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The definition of rotational kinetic energy is $$E_\text{rot}{}_{(i)} = \frac{1}{2} J_i \omega_i^2 = \frac{\;\; L_i^2}{2J_i}$$ where $J_i$ is moment of inertia, $\omega_i$ is angular velocity and $L_i = J_i\omega_i$ is angular momentum of the particle.

If you select $\vec{r}_\text{cm}$ as the center of rotation these values can be calculated for each particle as follows: $$J_i = \mu_i (\vec{r}_i - \vec{r}_\text{cm})^2$$ $$\vec{L}_i = \mu_i \Bigl[(\vec{r}_i - \vec{r}_\text{cm}) \times \dot{\vec{r}}_i\Bigr]$$ The rotational energy of the system is the sum of rotational energies of the particles: $$E_\text{rot} = \sum_i \frac{\;\; L_i^2}{2J_i}$$

There are two translational energies:

• translational energy of the whole system $$E_\text{cm}=\frac{1}{2}\sum_i \mu_i \dot{\vec{r}}_\text{cm}^2$$
• internal translational energy (broadening and shrinking without rotation) $$E_\text{int} = \frac{1}{2}\sum \mu_i \left(\frac{d}{dt}\left|\vec{r}_i - \vec{r}_\text{cm}\right|\right)^2$$

Total energy is the sum: $$E_\text{tot} = \frac{1}{2}\sum_i \mu_i \dot{\vec{r}}_i^2 = E_\text{cm} + E_\text{int} + E_\text{rot}$$

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thank you for the answer. Here by $J$ you mean instataneous moment of inertia i guess as it changes with time, please clarify – Rajesh Dachiraju Dec 19 '11 at 7:51
the sigma's (summations) are missing for rotational kinetic enerhy, please include them to make it complete and to clear ambiguity of where to place sigma's – Rajesh Dachiraju Dec 19 '11 at 8:04
@RajeshD Yes $J$ changes with time so the total internal energy is redistributed between $E_\text{int}$ and $E_\text{rot}$. – Maksim Zholudev Dec 19 '11 at 8:19

The total kinetic energy $E$ can be represented as a kinetic energy of the Center of Mass $E_{CM}=\frac{M_{tot}\vec{V_{CM}}^2}{2}$ and the relative motion kinetic energy (internal energy) $E_{int}=E-E_{CM}$. The latter can be represented as kinetic energy of the radial motion (relative to CM) plus a kinetic energy of angular motion. Normally there may be sudden exchange of radial and angular energies if particles collide. Only their sum is conserved.

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Collision is not necessary for the exchange of radial and angular energies. The energy of radial motion is zero at the point closest to the center and approaches total kinetic energy when the particle goes to infinity. The only exception is a particle going directly through the center. – Maksim Zholudev Dec 19 '11 at 14:44
@MaksimZholudev: You are right, of course. In case of only two particles it is clear that at the closest distance between them the radial kinetic energy $\propto \dot{r}$ passes its minimum (zero). – Vladimir Kalitvianski Dec 19 '11 at 15:04