Newton's second law says a net force gives a mass an acceleration. The rotational version swaps each piece for its twin: net torque, rotational inertia and angular acceleration.
Why it works
For one bead of mass $m$ at radius $r$, a push along its circle gives $\tau = mr^2\alpha$. A rigid object is many beads that all share the same $\alpha$, so the torques add: $\Sigma\tau = (\Sigma mr^2)\alpha = I\alpha$. The forces the beads exert on each other come in third-law pairs along the lines between them, and their torques cancel, so only outside torques remain.
Key idea
$$\Sigma\tau = I\alpha.$$ The net torque about an axis equals the rotational inertia about that axis times the angular acceleration. Torque plays the role of force and $I$ plays the role of mass. Use signs: counterclockwise positive.
Worked examples
Worked example: A pulled string
A string wrapped around a $5.0\ \text{kg}$ solid disk of radius $0.20\ \text{m}$ is pulled with $10\ \text{N}$. The disk turns on a fixed, frictionless axle. Find $\alpha$.
- Torque: the string pulls along the rim, so $\tau = RF = (0.20)(10) = 2.0\ \text{N·m}$.
- $I = \tfrac{1}{2}(5.0)(0.20)^2 = 0.10\ \text{kg·m}^2$.
- $\alpha = \dfrac{\tau}{I} = \dfrac{2.0}{0.10} = 20\ \text{rad/s}^2$.
Answer: $20\ \text{rad/s}^2$
Worked example: Braking a wheel
A wheel with $I = 1.2\ \text{kg·m}^2$ turns at $120$ rpm. A brake stops it in $6.0\ \text{s}$. What steady torque does the brake apply?
- $\omega_0 = 120 \times \dfrac{2\pi}{60} = 12.6\ \text{rad/s}$.
- $\alpha = \dfrac{0 - 12.6}{6.0} = -2.09\ \text{rad/s}^2$.
- $\tau = I\alpha = (1.2)(-2.09) = -2.5\ \text{N·m}$: $2.5\ \text{N·m}$ against the turning.
Answer: $2.5\ \text{N·m}$
Worked example: Two ropes, opposite ways
A wheel ($I = 0.10\ \text{kg·m}^2$) has a rope at radius $0.10\ \text{m}$ pulling $20\ \text{N}$ counterclockwise and one at $0.25\ \text{m}$ pulling $10\ \text{N}$ clockwise. Find $\alpha$.
- $$\Sigma\tau = (0.10)(20) - (0.25)(10) = 2.0 - 2.5 = -0.5\ \text{N·m}.$$
- $\alpha = \dfrac{-0.5}{0.10} = -5.0\ \text{rad/s}^2$: clockwise, though the counterclockwise rope pulls twice as hard.
Answer: $-5.0\ \text{rad/s}^2$
Common mistake
Using the force instead of the torque, or the mass instead of the rotational inertia. Writing $\alpha = F/m$ gives units of $\text{m/s}^2$, not $\text{rad/s}^2$: a sign the formula doesn't fit. In rotation, every force enters through its torque ($rF\sin\theta$), and every mass through $I$. Check the units: $\text{N·m} \div \text{kg·m}^2 = \text{1/s}^2$, which is rad/s².