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