Why Define Rotational Kinetic Energy?

When we apply a torque to a rigid object, changing its angular velocity, this torque does work on the object. Similar to translational work, we can ask the question: where does the work/energy go?

Derivation and Definition

Consider a wheel with moment of inertia $I$ experiencing a net torque $\tau_\text{net}$ as it rotates through some angle $\Delta \theta$.

Recall that work done by a constant torque is:

$$W_\text{net} = \tau_\text{net} \cdot \Delta\theta$$

Using Newton's second law for rotation, $\tau_\text{net} = I\alpha$, we substitute:

$$W_\text{net} = \tau_\text{net} \cdot \Delta \theta = I\alpha \cdot \Delta\theta$$

From rotational kinematics, $\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$, so:

$$\alpha \cdot \Delta\theta = \frac{\omega_f^2 - \omega_i^2}{2}$$

Substituting this into our expression for work yields:

$$W_\text{net} = I\alpha \Delta\theta=I\big(\frac{\omega_f^2 - \omega_i^2}{2}\big)=\frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$$

Since work equals the change in kinetic energy (Work-Kinetic Energy Theorem),

$$W_\text{net} = K_f-K_i=\Delta K$$

we identify the rotational kinetic energy:

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