# Rotational Kinetic Energy - Translational Kinetic Energy

Kinetic energy is the energy associated with the motion of the objects.

Motion of an object can be categorized as

- pure translatory motion,
- pure rotatory motion,
- mixed translatory and rotatory motion (general plane motion).

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## Kinetic Energy of a Particle

An object is made up of many small point particles. Thus, to understand the total kinetic energy possessed by a body, first ponder upon the kinetic energy of a single particle.

A particle can either move in a straight line or on a curved path. In both cases, the kinetic energy represents the energy of motion, and thus depends upon the speed with which the particle is moving. The higher the speed, the greater is the kinetic energy. The amount of motion not only depends upon the speed, but also on the mass. Suppose a truck and a car is moving with the same speed, then the truck (due to its higher mass) is more difficult to stop and has a greater kinetic energy. The kinetic energy of a particle is therefore given as

$K{E_\text{particle}} = \frac{1}{2}m{v^2}.$

## Kinetic Energy in Pure Translatory Motion

$v$) and the total kinetic energy can be calculated by the sum of the kinetic energies of individual particles. Thus, $\begin{aligned} KE_\text{translation} &= \frac{1}{2}{m_1}{v^2} + \frac{1}{2}{m_2}{v^2} +\cdots+ \frac{1}{2}{m_N}{v^2}\\ &= \frac{1}{2}({m_1} + {m_2} + \cdots + {m_N}){v^2}\\ &= \frac{1}{2}{M_\text{body}}{v^2}. \end{aligned}$

In a pure translatory motion, all the particles in the body, at any instant of time, have equal velocity and acceleration. Kinetic energy is a scalar quantity with no direction associated with it. If we consider a rigid body in pure translational motion, then all the particles must be moving with equal speed (say## Kinetic Energy in Pure Rotation

$v,$ radius $r,$ and angular velocity $\omega$ is related as

$v = r\omega.$
Thus, if the rigid body is thought of being made up of many small particles, then for a small particle $dm$ at a distance $r$ from the axis moving with speed $v = r\omega ,$ the kinetic energy is given by,

$d(KE) = \frac{1}{2}dm{(r\omega )^2}.$

The total kinetic energy of the body can be calculated by summing the kinetic energies of all such elemental particles. For summation, integration can be used. Thus,

$KE_\text{Pure Rotation} = \int_{}^{} {\frac{1}{2}dm{{(r\omega )}^2}}.$
As the angular speed of all the particles is same, it can be taken out of the integration,

$KE_{Pure Rotation} = \frac{1}{2}{\omega ^2}\left( {\int_{}^{} {dm\, {r^2}} } \right),$
where $r$ is the distance of the particle from the axis of rotation, and thus the term in the bracket is the moment of inertia of the body about the axis of rotation. Hence,
$KE_{Pure Rotation} = \frac{1}{2}{I_\text{rot}}{\omega ^2},$
where ${I_\text{rot}}$ is the moment of inertia of the body about the axis of rotation.

## Kinetic Energy in General Plane Motion

When a rigid body is rotating and translating as well (that means the axis of rotation is not fixed, like the motion of a wheel), then the motion of the body can be seen as the combination of pure translation of the center of mass and pure rotation of the body about the center of mass.

Consider a body whose center of mass is moving with velocity $\vec {V_\text{cm}}$ and which has an angular velocity $\omega$. Now in the frame of the center of mass the center of mass appears to be at rest and the whole body appears to be rotating about the center of mass with angular velocity $\omega$. Now let us take a small particle of mass $dm$ distance $r$ away from the axis passing through the center of mass. The velocity of this point with respect to the center of mass is $\vec \omega \times \vec r$. In ground frame the velocity will be $\vec \omega \times \vec r + {\vec V_\text{cm}}.$

To calculate the total kinetic energy,

$\begin{aligned} KE &= \int_{}^{} {\frac{1}{2}dm\, \big(\vec \omega \times \vec r + {{\vec V}_\text{cm}}\big)\cdot \big(\vec \omega \times \vec r + {{\vec V}_\text{cm}}\big)} \\ &= \int_{}^{} {\frac{1}{2}dm\left[ {\big(\vec \omega \times \vec r\big)\cdot \big(\vec \omega \times \vec r\big) + 2{{\vec V}_\text{cm}}\cdot \big(\vec \omega \times \vec r\big) + {{\vec V}_\text{cm}}\cdot{{\vec V}_\text{cm}})} \right]}. \end{aligned}$

As the reference for distance measurement is taken as the axis passing through the center of mass, $\int_{}^{} {dm\, r} = 0$

Also, $\omega$ and $r$ are perpendicular to each other and $\int_{}^{} {dm\, {r^2}} = {I_\text{cm}},$ so

$\begin{aligned} KE &= \frac{1}{2}V_{_\text{cm}}^2\int_{}^{} {dm} + \frac{1}{2}{\omega ^2}\int_{}^{} {dm\, {r^2}} \\ &= \frac{1}{2}MV_{_\text{cm}}^2 + \frac{1}{2}{I_\text{cm}}{\omega ^2}. \end{aligned}$

The term $\frac{1}{2}MV_{_\text{cm}}^2$ is the kinetic energy when the whole body translates with the velocity equal to velocity of the center of mass, and the term $\frac{1}{2}{I_\text{cm}}{\omega ^2}$ is the kinetic energy when the body rotates with angular velocity $\omega$ about the center of mass.

**Cite as:**Rotational Kinetic Energy - Translational Kinetic Energy.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/rotational-kinetic-energy-translational-kinetic/