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How does the net torque on a rigid body determine its angular acceleration, and how does rotational inertia mediate that relationship?

Topic 5.6 Newton's Second Law in Rotational Form: relate the net torque on a rigid body to its angular acceleration and rotational inertia through tau_net = I*alpha.

A focused answer to AP Physics 1 Topic 5.6, covering the rotational form of Newton's second law tau_net = I*alpha, its parallel with F_net = ma, how net torque produces angular acceleration mediated by rotational inertia, and solving rotational dynamics problems, with full worked examples.

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  1. What this topic is asking
  2. The rotational second law
  3. The two proportionalities
  4. Solving rotational dynamics problems
  5. Try this

What this topic is asking

The College Board (Topic 5.6) wants you to relate the net torque on a rigid body to its angular acceleration and rotational inertia through the rotational form of Newton's second law, τnet=Iα\tau_{net} = I\alpha. This is the rotational twin of Fnet=maF_{net} = ma: net torque causes angular acceleration, and rotational inertia mediates the relationship just as mass does for linear motion.

The rotational second law

This law completes the rotational picture begun in Topic 5.1. Where torque (Topic 5.3) is the rotational analogue of force and rotational inertia (Topic 5.4) is the analogue of mass, this law ties them to angular acceleration exactly as Fnet=maF_{net} = ma ties force and mass to linear acceleration.

The two proportionalities

These proportionalities let you reason qualitatively. A larger torque spins an object up more quickly; a heavier or more spread-out object (larger II) resists that spin-up. A flywheel with large rotational inertia barely speeds up under a modest torque, while a light spindle responds instantly.

Solving rotational dynamics problems

The solution routine parallels translational dynamics step for step. Identify the forces and where they act, compute the torque each produces about the axis (τ=rFsinθ\tau = rF\sin\theta), add them with signs to get the net torque, find the rotational inertia about that axis, and apply α=τnet/I\alpha = \tau_{net}/I. From the angular acceleration you can then use the rotational kinematic equations (Topic 5.1) to find angular velocities and displacements over time, just as you used linear kinematics after finding a linear acceleration. The analogy with Fnet=maF_{net} = ma is the organizing idea for the whole of rotational dynamics and is worth stating explicitly: translation and rotation obey the same logical structure, with torque, rotational inertia and angular acceleration as the rotational counterparts of force, mass and linear acceleration. This unifying parallel means every problem-solving habit from Unit 2, draw the diagram, find the net cause, divide by the resistance to get the effect, transfers directly. Many AP problems combine the two worlds: a falling mass on a string wrapped around a pulley exerts a torque that angularly accelerates the pulley while the mass linearly accelerates, and you solve by writing Fnet=maF_{net} = ma for the mass and τnet=Iα\tau_{net} = I\alpha for the pulley, linked by a=rαa = r\alpha from Topic 5.2. Recognizing τnet=Iα\tau_{net} = I\alpha as "the rotational F=maF = ma" is the single most useful insight for the dynamics half of Unit 5.

Try this

Q1. A net torque of 6.06.0 N\cdotm acts on a body of rotational inertia 1.51.5 kg\cdotm squared. Calculate its angular acceleration. [2 points]

  • Cue. α=τnet/I=6.0/1.5=4.0\alpha = \tau_{net}/I = 6.0/1.5 = 4.0 rad/s squared.

Q2. A body of rotational inertia 0.500.50 kg\cdotm squared angularly accelerates at 8.08.0 rad/s squared. Calculate the net torque on it. [1 point]

  • Cue. τnet=Iα=(0.50)(8.0)=4.0\tau_{net} = I\alpha = (0.50)(8.0) = 4.0 N\cdotm.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)5 marksSection II (short FRQ, quantitative). A solid disc of rotational inertia 0.400.40 kg\cdotm squared is free to rotate about its center. A constant tangential force of 1212 N is applied at its rim, 0.300.30 m from the axis. (a) Calculate the net torque on the disc. (b) Calculate its angular acceleration. (c) Explain how this calculation parallels finding the linear acceleration of a block from a net force.
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A 5-point FRQ on the rotational form of Newton's second law.

(a) Net torque (2 points): the force is tangential (perpendicular to the radius), so τ=rFsin90=(0.30)(12)(1)=3.6\tau = rF\sin 90^\circ = (0.30)(12)(1) = 3.6 N\cdotm.
(b) Angular acceleration (2 points): τnet=Iα\tau_{net} = I\alpha, so α=τnetI=3.60.40=9.0\alpha = \dfrac{\tau_{net}}{I} = \dfrac{3.6}{0.40} = 9.0 rad/s squared.
(c) Explain (1 point): this mirrors Fnet=maF_{net} = ma. Net torque plays the role of net force, rotational inertia plays the role of mass, and angular acceleration plays the role of linear acceleration. Dividing the net torque by the rotational inertia gives the angular acceleration, just as dividing net force by mass gives linear acceleration.

Markers reward the torque from the rim force, α=τnet/I\alpha = \tau_{net}/I, and a clear term-by-term analogy with Fnet=maF_{net} = ma.

AP 2023 (style)1 marksSection I (multiple choice). The same net torque is applied to two discs; disc X has twice the rotational inertia of disc Y. How do their angular accelerations compare? (A) X has twice the angular acceleration (B) X has half the angular acceleration (C) they are equal (D) X has four times the angular acceleration. Justify your reasoning.
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A 1-point MCQ on the rotational second law. The answer is (B).

From τnet=Iα\tau_{net} = I\alpha, the angular acceleration is α=τnet/I\alpha = \tau_{net}/I. With the same net torque, angular acceleration is inversely proportional to rotational inertia. Disc X has twice the rotational inertia, so it has half the angular acceleration. The trap is forgetting the inverse relationship; more rotational inertia means a smaller angular acceleration for the same torque.

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