Newton's Second Law for Rotation

The law

$\vec{\tau}_{net} = I\vec{\alpha}$

Net torque on a rigid body about a fixed axis equals the moment of inertia about that axis times the angular acceleration.

This is the rotational analogue of Newton's Second Law $\vec{F} = m\vec{a}$.

Derivation

For a rigid body rotating about a fixed axis, consider a mass element $m_i$ at distance $r_i$ from the axis.

The tangential force on $m_i$: $F_{t,i} = m_i a_{t,i} = m_i r_i \alpha$

Torque on $m_i$: $\tau_i = r_i F_{t,i} = m_i r_i^2 \alpha$

Sum over all elements (internal torques cancel by Newton's Third Law):

$\tau_{net} = \sum \tau_i = \sum m_i r_i^2 \cdot \alpha = I\alpha$

$\boxed{\tau_{net} = I\alpha}$

The complete analogy

Applying $\tau = I\alpha$

Step 1: Identify the axis of rotation.

Step 2: Draw free body diagram, identify all forces.

Step 3: Compute net torque about the axis (include sign).

Step 4: Set $\tau_{net} = I\alpha$ and solve for $\alpha$.

Example: A disc of mass $M$, radius $R$ has a force $F$ applied tangentially at the rim. Find $\alpha$.

$\tau = FR, \quad I = \frac{1}{2}MR^2$

$\alpha = \frac{\tau}{I} = \frac{FR}{\frac{1}{2}MR^2} = \frac{2F}{MR}$

Combined translation and rotation

For a body that both translates and rotates (like a rolling body):

$F_{net} = Ma_{cm} \quad \text{(translation of CM)}$

$\tau_{net,cm} = I_{cm}\alpha \quad \text{(rotation about CM)}$

These two equations, together with the rolling constraint $a_{cm} = R\alpha$, fully determine the motion.