Lessons · rotation

Rotational work–energy: ½Iω² is the area under the torque curve

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A flywheel (moment of inertia $I = 0.5$ kg·m²) starts from rest and is driven by a torque that builds with angle, $\tau = c\theta$ ( $c = 8$ N·m/rad), through $\theta = 1.5$ rad. How fast is it spinning at the end? The work done is the area under the torque–angle curve — and it is exactly the rotational kinetic energy $\tfrac12 I\omega^2$ . Drag the cursor and watch the shaded work and the spin energy grow together: the rotational twin of $\int F\,dx = \tfrac12 mv^2$ .

Rotational work

W = \tfrac12 c\,\theta^2

(the shaded area)

9 J

Rotational kinetic energy

\tfrac12 I\omega^2

9 J

Final angular speed

\omega = \sqrt{c/I}\,\theta

6 rad/s

Maximum torque

\tau_{\max} = c\,\theta

12 N·m

Rotational work is the area under the torque–angle curve

W = \int \tau\,d\theta

Rotational work is the integral of torque over the angle turned — the rotational twin of $\int F\,dx$ . For a constant torque this integral is a rectangle (and gives $W = \tau\,\Delta\theta$ ); for any torque it is the area under the $\tau$ – $\theta$ curve.

check $\tfrac{d}{d\theta}W = \tau$ ; check $W = \int_0^\theta \tau\,d\theta'$ ; back-substitute into $\tfrac12 I\omega^2 = W$ ; collapse the constant-torque case to $\tau_0\,\Delta\theta$

✓ The work's slope is the torque: $W'(\theta) = \tau(\theta)$ — the area's rate of growth is the curve's height. [structural]
✓ The accumulated work is exactly the area: $W(\theta) = \int_0^\theta \tau\,d\theta'$ . [structural]
✓ The rotational kinetic energy equals the work: $\tfrac12 I\omega^2 = W$ — the memorized $\tfrac12 I\omega^2$ is the area. [structural]
✓ For a constant torque the integral collapses to $W = \tau_0\,\Delta\theta$ — the area is a rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “Double the angle you drive a flywheel through and you double its final spin rate.”

The work you do is the area under the torque–angle curve, and the energy goes as $\tfrac12 I\omega^2$ — so the speed follows the area, not the angle. With a constant torque, doubling the angle doubles the work but raises the speed only by $\sqrt{2}$ . With a torque that grows as $\tau = c\theta$ (as here), doubling the angle quadruples the area, so the work quadruples and the speed doubles. Either way $\omega \propto \sqrt{W}$ , never linearly in the angle.

Modeling assumptions — author-asserted, disclosed not discharged

Rotation about a fixed axis; $\tau = c\theta$ is the net torque on the flywheel, so all the work goes into rotational kinetic energy (no friction or load).
Rigid flywheel of constant moment of inertia $I$ ; it starts from rest ( $\omega_0 = 0$ ).

The τ–θ graph, fully annotated

A static rendering (Matplotlib): the shaded area under τ is the accumulated integral W, and the slope of W is τ. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):