Rotational Kinetic Energy

Suppose that we have an object rotating about some axis. The rotational kinetic energy is simply the kinetic energy possessed by the object due to its rotation.

To calculate this, we take small elements ($i$) on the body with mass $m_i$ and distance from the axis $r_i$ and add them all to find the total kinetic energy of the rotating body.

Thus by using the following formula for Kinetic Energy ($KE$) of a body, we may calculate the kinetic energy of the corresponding small piece: [ \begin{align}

KE &=\frac{1}{2}mv^2 \

\Rightarrow KE_i &=\frac{1}{2}m_iv_i^2\

\Rightarrow KE_i &=\frac{1}{2}m_i(\omega r_i)^2\

\Rightarrow KE_i &=\frac{1}{2}\omega^2 m_ir_i^2 \end{align} ] Summing this over all $i$, $$KE=\sum_i\frac{1}{2}\omega^2 m_ir_i^2$$ $$\Rightarrow KE=\frac{1}{2}\omega^2\sum_i m_ir_i^2$$ But wait! The summation of $m_ir_i^2$ over all of the small pieces of the object is simply $I$, the object's moment of inertia! Thus, we have the following simple formula for the rotational kinetic energy: $$KE=\frac{1}{2}I\omega^2$$ Keep in mind that the moment of inertia is dependent on the axis of rotation.

Notice first of all that this is dimensionally correct.

Also notice that this looks very similar to our formula for linear kinetic energy: $$KE=\frac{1}{2}mv^2$$ This is yet another of those cases in which one may swap common linear measurements for common rotational measurements and derive similar equations. In this case, swap the mass out for the moment of inertia, and swap the linear velocity out for the angular velocity.