# Proof 35: Kinetic energy

Linear version of kinetic energy

A spaceship in outer space starts its rocket engines to propel itself forwards with a rocket force. That force does work on the spaceship, making it speed up.

In other words, this work $W$ increases the kinetic energy $K$. It causes a change in the kinetic energy $\Delta K$. This is the work-energy theorem, since we have a scenario with no other energies involved:

\begin{align} \Delta K&=W\\ ~& =F\Delta s \qquad\leftarrow \text{Work formula}\\ ~& =ma\Delta s\qquad\leftarrow \text{Newton’s 2nd law}\\ ~& =m\cancel a\left(\frac{v^2-v_0^2}{2\cancel a}\right)\quad\leftarrow \text{Motion equation, } v^2=v_0^2+2a\Delta s\Leftrightarrow \Delta s=\frac{v^2-v_0^2}{2a}\\ ~& =m\left(\frac 12v^2-\frac 12v_0^2\right)\\ ~& =\underbrace{\frac 12 mv^2}_{K_\text{after}}-\underbrace{\frac 12 m v_0^2}_{K_\text{before}} \end{align}

In the last step we see two equal expressions, only separated in time, subtracted from each other. Since this equals the difference in kinetic energy $\Delta K=K_\text{after}-K_\text{before}$, it makes sense to call each of these two expressions the kinetic energy at each moment in time.

Only translational/linear properties were used, so here we’ve found the translational version of the kinetic-energy formula:

$$K_\text{trans}=\frac 12 mv^2\qquad\blacksquare$$

Rotational version of kinetic energy

A piece of fireworks consisting of small explosive pieces located at the wing-tips of a fan-shape is nailed to a pole. The pieces fire and push clockwise. The work they do make the fan rotate and speed up its rotation.

So, this work $W$ causes a change in the rotational kinetic energy $\Delta K$, which is the rotational version of the work-energy theorem:

\begin{align} \Delta K&=W\\ ~& =\tau\Delta \theta \qquad\leftarrow \text{Rotational work formula}\\ ~& =I\alpha\Delta \theta\qquad\leftarrow \text{Rotational version of Newton’s 2nd law}\\ ~& =I\cancel \alpha \left(\frac{\omega^2-\omega_0^2}{2\cancel \alpha}\right)\quad\leftarrow \text{Rotational motion equation, } \omega^2=\omega_0^2+2\alpha\Delta \theta\Leftrightarrow \Delta \theta=\frac{\omega^2-\omega_0^2}{2\alpha}\\ ~& =I\left(\frac 12\omega^2-\frac 12\omega_0^2\right)\\ ~& =\underbrace{\frac 12 I\omega^2}_{K_\text{after}}-\underbrace{\frac 12 I \omega_0^2}_{K_\text{before}} \end{align}

As before, we at the end see two equal expressions representing the difference in kinetic energy, $\Delta K=K_\text{after}-K_\text{before}$. With only rotational/angular properties used, we’ve found the rotational version of the kinetic-energy formula:

$$K_\text{rot}=\frac 12 I\omega^2\qquad\blacksquare$$