# Kinetic Energy of a Body in Combined Rotation and Translation

## Kinetic Energy of a Body in Combined Rotation and Translation

Consider a body in combined translational and rotational motion in the lab frame. Suppose in the frame of the centre of mass, the body is making a pure rotation with an angular velocity ω. The centre of mass itself is moving in the lab frame at a velocity $$\overrightarrow{{{v}_{0}}}$$.

The velocity of a particle of mass mi is $$\overrightarrow{{{v}_{i,CM}}}$$ with respect to the centre of mass frame and $$\overrightarrow{{{v}_{i}}}$$ with respect to the lab frame. We have:

$$\overrightarrow{{{v}_{i}}}=\overrightarrow{{{v}_{i,CM}}}+\overrightarrow{{{v}_{o}}}$$.

The kinetic energy of the particle in the lab frame is:

$$\frac{1}{2}{{m}_{i}}{{v}_{i}}^{2}=\frac{1}{2}{{m}_{i}}\left( \overrightarrow{{{v}_{i,cm}}}+\overrightarrow{{{v}_{0}}} \right)\left( \overrightarrow{{{v}_{i,cm}}}+\overrightarrow{{{v}_{0}}} \right)$$.

$$\frac{1}{2}{{m}_{i}}{{v}_{i}}^{2}=\frac{1}{2}{{m}_{i}}{{v}^{2}}_{i,cm}+\frac{1}{2}{{m}_{i}}{{v}_{o}}^{2}+\frac{1}{2}{{m}_{i}}\left( 2\overrightarrow{{{v}_{i,cm}}}.\overrightarrow{{{v}_{0}}} \right)$$.

Summing over all the particles, the total kinetic energy of a body in the ab frame is:

$$K=\sum\limits_{i}{\frac{1}{2}{{m}_{i}}{{v}_{i}}^{2}}=\sum\limits_{i}{\frac{1}{2}}{{m}_{i}}{{v}^{2}}_{i,cm}+\sum\limits_{i}{\frac{1}{2}{{m}_{i}}{{v}^{2}}_{0}}+\left[ \sum\limits_{i}{{{m}_{i}}\overrightarrow{{{v}_{i,cm}}}} \right]\overrightarrow{{{v}_{0}}}$$.

Now,

$$\sum\nolimits_{i}{\frac{1}{2}}{{m}_{i}}{{v}^{2}}_{i,cm}$$, is the kinetic energy of the body in the centre of mass frame. In this frame, the body is making pure rotation with an angular velocity ω. Thus, this term is equal to $$\frac{1}{2}{{I}_{cm}}{{\omega }^{2}}$$. Also, $$\sum{\left( {{m}_{i}}\overrightarrow{{{v}_{i,cm}}} \right)}$$ is the velocity of the centre of mass in the centre of mass frame, hence $$\sum{\left( {{m}_{i}}\overrightarrow{{{v}_{i,cm}}}=0 \right)}$$ obviously is zero.

Thus, $$K=\frac{1}{2}{{I}_{cm}}{{\omega }^{2}}+\frac{1}{2}m{{v}_{0}}^{2}$$.