Newton's Second Law for Rotation
- Apply the rotational form of Newton's second law, τ_net = Iα
- Solve for the angular acceleration of a rigid body under applied torques
- Analyze a mass hanging from a pulley that has rotational inertia
The rotational second law
Just as a net force produces linear acceleration, a net torque produces angular acceleration: τ_net = Iα. The moment of inertia I is the rotational mass, resisting changes in spin. Given the torques on a rigid body and its moment of inertia, this single equation delivers the angular acceleration — the rotational twin of a = ΣF/m.
When rotation and translation couple
A block hanging from a string wrapped around a real (massive) pulley links the two laws. The block obeys ΣF = ma, while the pulley obeys τ_net = Iα, and the rolling string ties them with a = Rα. Solving the pair shows the tension is no longer equal on both sides of an ideal pulley — the pulley "steals" some torque to spin itself up, so the block accelerates more slowly than in free fall.
A uniform solid disk (I = ½MR²) of mass 4 kg and radius 0.5 m can spin about its center. A rope wrapped around its rim is pulled with a constant tangential force of 6 N. Find the disk's angular acceleration.
- 1.The force is tangent to the rim, so the torque is τ = FR = 6 × 0.5 = 3 N·m.
- 2.Compute the moment of inertia: I = ½MR² = ½(4)(0.5²) = ½(4)(0.25) = 0.5 kg·m².
- 3.Apply τ_net = Iα: 3 = 0.5 α.
- 4.Solve: α = 3/0.5 = 6 rad/s².
A rigid body with moment of inertia 2 kg·m² experiences a net torque of 8 N·m. What is its angular acceleration?
A block hangs from a string wrapped around a pulley that has rotational inertia. Compared with an ideal massless pulley, the block's downward acceleration is:
With a massive pulley you can no longer assume equal tension on both sides or ignore the pulley in energy accounting. Write ΣF = ma for the hanging mass, τ_net = Iα for the pulley, and link them with a = Rα before solving.
Answer the 2 checkpoints as you read.
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