Rotational Kinetic Energy & Work-Energy Theorem

Summary

Rotational kinetic energy is the energy of rotating bodies

Key Equations:

Rotational kinetic energy
$K = \frac{1}{2} ω^{2} I_{0}$

Rotational work-kinetic energy theorem:
$W = Δ K$

Read Time

⏱ 3 mins

Definition

Rotational kinetic energy is the kinetic energy of rotating objects. It's the rotational work done on an object to keep it rotating or to cause rotation.

Derivation

Assumptions

To derive rotational kinetic energy, take the example to find the rotational kinetic energy of the rigid body below.

This can be found by taking the sum of very small mass elements and their respective displacement element from an axis.
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As well, assume the following:

The definition of angular velocity is $ω = \frac{v}{r} \Rightarrow v = r w$

Kinetic energy is defined as $K = \frac{1}{2} m v^{2}$

\begin{array}{c} \\ K_ = \sum _{i}\frac{1}{2}m_{i}v_{i}^2 \\ v_{i} = r_{i }\omega \\ \begin{align*} K &= \sum _{i}\frac{1}{2}m_{i}(r_{i}\omega)^2 \\ &=\frac{1}{2} \omega {#2} \underbrace{ \sum_{i} m_{i} r_{i} }_{ I_{0} } \\ &= \frac{1}{2}\omega^2 I_{0} \end{align*} \end{array}

Note

In this derivation we also define the moment of inertia for rotating bodies. $\sum_{i} m_{i} r_{i} = I_{0}$

Rotational Work-Energy Theorem

Assumptions

To derive an equation to relate rotational work and kinetic energy. Create an equation to describe work of a rotating body from point a to b. Assume the following:

Rotational work is defined as $W = \int_{a}^{b} τ \cdot d θ$
Assume torque and angular displacement are always in the same direction
Torque is defined as $τ = I α$
Angular acceleration is defined as $α = \frac{d ω}{d t}$
Angular velocity is defined as $ω = \frac{d θ}{d t} \Rightarrow d θ = ω d t$
The moment of inertia is the same through time $t_{1}$ and $t_{2}$
Rotational kinetic energy is defined as $\frac{1}{2} ω I^{2}$

\begin{align*} W &= \int_{t_{1}}^{t_{2}} \tau \cdot d\theta \\ &= \int_{t_{1}}^{t_{2}} \tau d\theta \\ &= \int_{t_{1}}^{t_{2}}I\alpha d\theta\\ &= I \int_{t_{1}}^{t_{2}} \frac{d\omega}{\cancel{ dt }} (\omega \cancel{ dt }) \\ &= I \int_{t_{1}}^{t_{2}} \omega d\omega \\ &= I \int_{t_{1}}^{t_{2}} \omega d \omega \\ &= \frac{1}{2}I[\omega {#2} (t_{2})-\omega {#2} (t_{1})] \\ &= \frac{1}{2} I (\omega_{f} - \omega_{i}) \\ &= \Delta K \end{align*}