# Derivation of the total Kinetic energy Equation for Combined Translation and Rotation | Rotation of a rigid body about a moving axis

Last updated on April 12th, 2021 at 04:07 am

Before going for the* Derivation of the equation of total Kinetic energy of a body under combined translation and rotation*, let’s do some analysis of the dynamics of *Rigid-Body Rotation About a Moving Axis* to some cases in which the axis of rotation moves.

When that happens, the motion of the body is a combination of translation and rotation.

The key to understanding such situations is this: *Every possible motion of a rigid body can be represented as a combination of translational motion of the center of mass and rotation about an axis through the center of mass.*

This is true even when the center of mass accelerates so that it is not at rest in any inertial frame.

## Example of combined translation and rotation | with diagram

Figure 1 illustrates this for the motion of a tossed baton: The center of mass of the baton follows a parabolic curve, as though the baton were a particle located at the center of mass. Other examples of combined translational and rotational motions include a ball rolling down a hill and a yo-yo unwinding at the end of a string.

## Kinetic energy Equation for Combined Translation and Rotation: Energy Relationships

We will discuss here the *kinetic energy *of a rigid body that has both translational and rotational motions.

In this case, the body’s kinetic energy is the sum of a part (1/2)Mv_{cm}^{2} associated with the motion of the center of mass and a part (1/2)**I _{cm}**ω

^{2}associated with rotation about an axis through the center of mass.

Kinetic energy Equation for Combined Translation and Rotation =

**KE = (1/2)Mv**

_{cm}^{2}+ (1/2)I_{cm}ω^{2}(for a rigid body with both translation and rotation)## Derivation of the total kinetic energy equation for combined translation and rotation of a rigid body about a moving axis

**Derive the formula for total Kinetic energy for Combined Translation and Rotation**: To prove this relationship, we again imagine the rigid body to be made up of particles. Consider a typical particle with mass m_{i} as shown in Fig. 3. The velocity **v**_{i} of this particle relative to an inertial frame is the vector sum of the velocity **v**_{cm} of the center of mass and the velocity **v**_{i}^{’} of the particle *relative to* the center of mass:

around the center of mass. Using the formula v = ωr, we can write the last term as (1/2)I_{cm}ω^{2} ,

where I_{cm} is the moment of inertia with respect to the axis through the center of mass and ω is the angular speed.So the equation becomes:

**KE = (1/2)Mv**

_{cm}^{2}+ (1/2)I_{cm}ω^{2}(for a rigid body with both translation and rotation)## Related Study for reference:

Here we have derived the formula of **Kinetic energy of a rigid body with both translational and rotational motion**. (KE for a rigid body having a combination of translation and rotation).

But there are motions that are purely rotational motions, like the rotation of a ceiling fan, rotation of the blades of a windmill, etc.

Now if you want to read about it and the KE equation for that, you must try this link to **derive the equation of Kinetic energy for pure rotational motion**.