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⚙️AP Physics C: Mechanics (2025) Unit 5 Review

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5.6 Newton's Second Law in Rotational Form

⚙️AP Physics C: Mechanics (2025)
Unit 5 Review

5.6 Newton's Second Law in Rotational Form

Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 examWritten by the Fiveable Content Team • Last updated September 2025
⚙️AP Physics C: Mechanics (2025)
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Torque and Angular Acceleration

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Net Torque and Rotational Inertia

Newton's Second Law for rotational motion establishes a fundamental relationship between torque, rotational inertia, and angular acceleration, similar to how force relates to mass and linear acceleration in linear motion.

τ=Iα\sum \tau = I \alpha

This equation states that the net torque (Στ) acting on an object equals the product of its rotational inertia (I) and angular acceleration (α). This relationship allows us to predict how objects will rotate when subjected to various torques.

  • Rotational inertia (I) quantifies an object's resistance to changes in rotational motion, analogous to how mass resists changes in linear motion
  • Unlike mass, rotational inertia depends not only on the amount of mass but also on how that mass is distributed relative to the axis of rotation
  • Torque (τ) is the rotational equivalent of force, causing angular acceleration just as force causes linear acceleration

Direction of Angular Acceleration

The relationship between torque and angular acceleration includes directional components that follow the right-hand rule for rotational quantities.

  • Net torque and angular acceleration vectors always point in the same direction
  • Counterclockwise torques (positive by convention in 2D problems) produce counterclockwise angular accelerations
  • Clockwise torques (negative by convention in 2D problems) result in clockwise angular accelerations
  • For complex situations with multiple torques, determine the net torque by vector addition before calculating angular acceleration

Mass Distribution Effects

The distribution of mass in an object significantly impacts its rotational inertia and consequently its rotational dynamics.

Rotational inertia depends on the distance of mass elements from the axis of rotation, with the mathematical relationship:

  • For discrete masses: I=miri2I = \sum m_i r_i^2
  • For continuous mass distributions: I=r2dmI = \int r^2 dm

This distance dependence creates several important effects:

  • Objects with mass concentrated farther from the axis have greater rotational inertia
  • A hollow cylinder resists angular acceleration more than a solid cylinder of equal mass because its mass is distributed farther from the center
  • Figure skaters exploit this principle by pulling their arms close to their body to reduce rotational inertia and spin faster
  • The same torque applied to objects with different mass distributions will produce different angular accelerations

🚫 Boundary Statement

The AP Physics C: Mechanics exam does not require analyzing rotation in multiple planes simultaneously.

Practice Problem 1: Torque and Angular Acceleration

A uniform solid disk with mass 2.0 kg and radius 0.15 m is free to rotate about its center. A constant force of 3.0 N is applied tangentially at the rim of the disk. Calculate the angular acceleration of the disk.

Solution

To solve this problem, we need to find the torque applied to the disk and its rotational inertia, then apply Newton's Second Law in rotational form.

Step 1: Calculate the torque applied to the disk. The torque is the product of the force and the perpendicular distance from the axis of rotation: τ=F×r=3.0 N×0.15 m=0.45 N\cdotpm\tau = F \times r = 3.0 \text{ N} \times 0.15 \text{ m} = 0.45 \text{ N·m}

Step 2: Calculate the rotational inertia of the solid disk. For a uniform solid disk rotating about its center, the rotational inertia is: I=12MR2=12×2.0 kg×(0.15 m)2=0.0225 kg\cdotpm2I = \frac{1}{2}MR^2 = \frac{1}{2} \times 2.0 \text{ kg} \times (0.15 \text{ m})^2 = 0.0225 \text{ kg·m}^2

Step 3: Apply Newton's Second Law in rotational form to find the angular acceleration. α=τI=0.45 N\cdotpm0.0225 kg\cdotpm2=20 rad/s2\alpha = \frac{\tau}{I} = \frac{0.45 \text{ N·m}}{0.0225 \text{ kg·m}^2} = 20 \text{ rad/s}^2

Therefore, the disk experiences an angular acceleration of 20 rad/s².

Practice Problem 2: Mass Distribution Effects

A solid cylinder and a hollow cylinder (thin-walled hoop) have the same mass M and radius R. Both are released from rest at the top of an inclined plane and roll without slipping. Which one reaches the bottom first, and why?

Solution

To determine which cylinder reaches the bottom first, we need to compare their angular accelerations, which depend on their rotational inertias.

Step 1: Identify the rotational inertias of both objects.

  • Solid cylinder: Isolid=12MR2I_{\text{solid}} = \frac{1}{2}MR^2
  • Hollow cylinder (hoop): Ihollow=MR2I_{\text{hollow}} = MR^2

Step 2: For an object rolling down an incline without slipping, the angular acceleration is: α=mgsinθI/R+M\alpha = \frac{mg\sin\theta}{I/R + M}

Where θ\theta is the angle of the incline.

Step 3: Substitute the rotational inertias and simplify.

  • For solid cylinder: αsolid=mgsinθ12MR2/R+M=mgsinθ12M+M=23gsinθ\alpha_{\text{solid}} = \frac{mg\sin\theta}{\frac{1}{2}MR^2/R + M} = \frac{mg\sin\theta}{\frac{1}{2}M + M} = \frac{2}{3}g\sin\theta
  • For hollow cylinder: αhollow=mgsinθMR2/R+M=mgsinθM+M=12gsinθ\alpha_{\text{hollow}} = \frac{mg\sin\theta}{MR^2/R + M} = \frac{mg\sin\theta}{M + M} = \frac{1}{2}g\sin\theta

Since 23gsinθ>12gsinθ\frac{2}{3}g\sin\theta > \frac{1}{2}g\sin\theta, the solid cylinder has greater angular acceleration and will reach the bottom first. This demonstrates how mass distribution affects rotational dynamics - the solid cylinder has more mass concentrated closer to the axis of rotation, giving it a smaller rotational inertia relative to its mass, allowing it to accelerate more quickly.

Frequently Asked Questions

What is Newton's second law for rotation and how is it different from regular F=ma?

Newton’s second law for rotation says the net torque on a rigid system produces angular acceleration: α = Στ / I (angular acceleration α is directly proportional to net torque Στ and inversely proportional to the rotational inertia I). That’s the rotational analog of F = ma, but key differences matter: - Quantities: F = ma uses force, mass, linear acceleration; τ = Iα uses torque, moment of inertia, angular acceleration. - Role of geometry: Torque τ = r × F depends on where the force acts (lever arm and angle). Mass is replaced by rotational inertia I, which depends on how mass is distributed. - Direction and vectors: Torques and angular acceleration follow the right-hand rule (they’re vectors about the rotation axis). - When to use both: For rolling or linked motion you often solve linear (F = ma) and rotational (τ = Iα) equations independently and combine them (CED 5.6.A.2, 5.6.A.3). For a quick study refresher see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK). For more practice, check the unit page (https://library.fiveable.me/ap-physics-c-mechanics/unit-5) and thousands of problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Why does angular velocity only change when there's a net torque?

Angular velocity ω changes only when there's a net torque because torque is the rotational analogue of force. For a rigid body, Newton’s second law in rotational form says α = τ_net / I (CED 5.6.A.2). If τ_net = 0, then α = 0, so ω is constant. Physically: torques produce angular acceleration by changing how fast mass elements rotate about the axis; opposing torques can cancel so there’s no net twisting effect and no change in ω (CED 5.6.A.1). Also note: if τ_net = 0, angular momentum L = Iω is conserved (useful on AP problems about isolated systems). When solving exam problems, check all torques (including friction, gravity component about the axis, normal forces at offsets) and compute τ_net before saying ω changes. For a quick review, see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

I'm confused about the equation α = τ/I - can someone explain what each variable means?

α = τ/I is the rotational version of Newton’s 2nd law. Here's what each symbol means and how to use it: - α (alpha): angular acceleration of the rigid system—how fast the angular velocity ω is changing (rad/s^2). - τ (tau): net torque acting on the object about the chosen axis (N·m). It’s the sum of all torques; signs/direction follow the right-hand rule. - I: moment of inertia (rotational inertia) of the rigid system about that axis (kg·m^2). It measures how mass is distributed relative to the axis. So α = τ/I says angular acceleration is directly proportional to net torque and inversely proportional to rotational inertia (CED 5.6.A.2). If τnet = 0 then α = 0 (no change in ω). Use τ = r × F or τ = rF sinθ to compute torques, and compute I from known formulas or the parallel-axis theorem (CED keywords). This relation is required on the AP exam for rotational dynamics problems (Topic 5.6). For a focused review and practice problems, see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and the AP Physics C practice set (https://library.fiveable.me/practice/ap-physics-c-mechanics).

How do you calculate rotational inertia and why does it matter for angular acceleration?

Rotational inertia (moment of inertia) I measures how mass is distributed about an axis and you calculate it so torque gives the right angular acceleration. For discrete masses: I = Σ miri^2 (each mass times its distance squared). For continuous bodies use an integral I = ∫ r^2 dm and standard results: hoop I = MR^2, solid disk I = (1/2)MR^2, solid sphere I = (2/5)MR^2. Use the parallel-axis theorem when the axis is offset: I = Icm + Md^2. Why it matters: Newton’s 2nd law in rotational form is α = τnet / Isys (CED 5.6.A.2). For the same net torque, a larger I gives a smaller angular acceleration. Example: τ = 10 N·m; if I = 2 kg·m^2 then α = 5 rad/s^2; if I = 0.5 kg·m^2 then α = 20 rad/s^2. Master these formulas and when to apply parallel-axis/integrals for AP problems (Topic 5.6 study guide: https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK). For more practice, check unit 5 and the AP practice bank (https://library.fiveable.me/ap-physics-c-mechanics/unit-5 and https://library.fiveable.me/practice/ap-physics-c-mechanics).

What's the difference between torque and force in rotational motion?

Force is a push or pull that changes an object's linear motion; net force gives linear acceleration a = ΣF/m. Torque is the rotational analogue: τ = r × F (magnitude τ = rF⊥ = rF sinθ), so the same force can produce different torques depending on the lever arm r and angle. Net torque causes angular acceleration: α = Στ / I (CED 5.6.A.2). Key differences: torque depends on where and in what direction the force is applied; force changes center-of-mass motion while torque changes rotation about an axis. In problems, you often need both analyses independently (CED 5.6.A.3): use ΣF = ma for translation and Στ = Iα for rotation. Remember the right-hand rule for torque direction and that rotational inertia I resists changes in ω. For AP review, see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Why do we need to do both linear and rotational analysis for some problems?

You need both linear and rotational analysis when an object’s motion includes translation of its center of mass and rotation about that center (or another axis). Newton’s second law for translation, ΣF = ma_cm, tells you how the center of mass moves; the rotational form, Στ = Iα, tells you how the object spins. Many AP problems (like rolling without slipping, a wheel plus hanging mass, or a body with an off-center force) couple the two: friction or torque links the translation and rotation (for rolling, a = αR). If you only do one type of analysis you’ll miss constraints or internal torques that change speeds, accelerations, or energy distribution. The CED explicitly says “To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently” (5.6.A.3) and uses Στ = Iα (5.6.A.2). For more worked examples and exam-style practice on when to combine both, see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and the Unit 5 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-5). For extra practice, try the problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Can you give me a real world example of when angular acceleration changes due to torque?

Think of pedaling a bicycle: when you push harder the chain delivers a larger net torque to the rear wheel, so the wheel’s angular acceleration increases. AP idea: αsys = τnet / Isys (CED 5.6.A.2). Example with numbers: if the wheel’s rotational inertia is 0.5 kg·m^2 and you increase net torque from 2.0 N·m to 6.0 N·m, α goes from 4.0 rad/s^2 to 12.0 rad/s^2—triple the torque → triple the α (assuming I constant). Keep in mind on real bikes Isys can change a little (tire/gear effects) and you may need separate rotational and translational analyses (CED 5.6.A.3) to connect wheel α to the bike’s forward acceleration. For more examples and AP-aligned practice, check the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

I don't understand how direction works with angular acceleration - is it the same as the torque direction?

Short answer: yes—the angular acceleration vector points the same way as the net torque vector (α = Στ / I). That’s exactly Essential Knowledge 5.6.A.2 in the CED. Why: torque and angular acceleration are vectors along the rotation axis. Use the right-hand rule: curl your fingers in the rotation direction produced by the torque; your thumb points the torque (and α) direction. If Στ = 0, α = 0 and the angular speed won’t change. In practice you usually choose a sign convention (e.g., CCW = +), so a torque that produces CCW rotation gives positive τ and positive α; an opposite torque gives negative τ and α. Example: a tangential force that tries to spin a disk CCW produces a torque vector “out of the page” and α out of the page → ω increases CCW. For more AP-aligned review, see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and extra practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

How do I know when to use τ = Iα versus F = ma in a physics problem?

Use τ = Iα when you’re analyzing rotation of a rigid body about an axis: there’s a net torque (sum of torques) producing an angular acceleration α and you need the object’s rotational inertia I about that axis. Use F = ma when you’re analyzing translation (center-of-mass motion) of a particle or whole object under net force. Quick rules: - If the question asks about angular acceleration, angular velocity, or torque about an axis → start with τnet = Iα (CED 5.6.A.2). - If it asks about linear acceleration, linear forces, or momentum of the center of mass → use ΣF = ma. - Often you need both: e.g., rolling without slipping or a pulley + masses problem—use ΣF = ma for linear pieces and τ = Iα for rotating pieces, linked by a = αR. AP tip: the CED expects you to perform rotational and linear analyses independently when needed (5.6.A.3). For worked examples and targeted review, see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and Unit 5 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-5). For practice, try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

What happens to a spinning object when the net torque becomes zero?

If the net torque on a spinning rigid object is zero, its angular acceleration α = Στ / I is zero, so its angular velocity ω does not change—the object rotates at constant ω (this is rotational equilibrium / Newton’s second law in rotational form; see CED 5.6.A.1–5.6.A.2). That means no change in rotational speed or direction of the spin axis unless an external torque appears. Two useful consequences for AP problems: - If there are no external torques, total angular momentum L is conserved (useful for collision or collapse problems on the exam). - If no torques do work about the rotation axis, rotational kinetic energy stays constant as well. Remember: “net torque = 0” doesn’t require each torque to be zero—opposing torques can cancel. For a quick review, check the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Why is rotational inertia in the denominator of the angular acceleration equation?

Because rotational inertia I is the rotational equivalent of mass, it appears in the denominator for the same reason mass does in α = F/m for translation. Torque τ is the “twisting” cause (like force), and I measures how hard the object is to change its rotation—how that mass is distributed around the axis. So for the same net torque, a larger I (mass farther from the axis or more mass) gives a smaller angular acceleration α = τ/I. Physically: if most mass sits far from the pivot, it has more rotational inertia and “resists” angular change, so you need more torque to get the same α. This is exactly the CED essential idea (5.6.A.2) connecting net torque, angular acceleration, and moment of inertia. For practice applying this idea to problems and FRQs, check the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and more problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

How does changing the mass distribution of an object affect its angular acceleration?

Angular acceleration depends on net torque and the system’s rotational inertia: α = Στ / I (CED 5.6.A.2). Changing how mass is distributed changes I (rotational inertia) even if total mass stays the same. Moving mass farther from the rotation axis increases I (for a point mass I = m r^2), so for the same net torque α gets smaller. Moving mass inward decreases I and increases α. Use the parallel-axis theorem when the axis shifts: I = Icm + M d^2. So if you apply a fixed torque (same Στ), putting more mass toward the rim slows angular acceleration; concentrating mass near the center speeds it up. That’s exactly the reasoning AP wants—connect torque, moment of inertia, and α (Topic 5.6). For more examples and quick practice, check the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK), the Unit 5 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-5), and lots of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

What's the rotational equivalent of Newton's first law and how does it relate to torque?

Newton’s first law in rotation: if the net external torque on a rigid body is zero, its angular velocity stays constant (including staying at zero). In AP terms this is rotational equilibrium: Στ = 0 ⇒ no change in ω. That’s the direct rotational analogue of “no net force ⇒ no change in linear velocity.” How this connects to torque and Topic 5.6: when Στ ≠ 0 you get angular acceleration α given by the rotational form of Newton’s second law, α = Στ / I (CED 5.6.A.2). So torque is the rotational “cause” that changes angular velocity; if net torque is zero there’s no angular acceleration. Remember direction: the torque vector and α follow the right-hand rule, and for full analysis you may need separate translational and rotational equations (CED 5.6.A.3). For revising, see the topic study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and more practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

I missed the lab on rotational motion - what experiments show Newton's second law for rotation?

Good question—labs that directly show τnet = Iα are ones where you can control torque and measure angular acceleration. Try these classroom-friendly experiments: - Hanging-mass on a pulley/rotary disk (rotational dynamics apparatus): vary hanging mass (thus applied torque τ = rT) and measure α with a rotary sensor. Plot α vs τ; you should get a straight line with slope 1/I. - Torque-vs-angle on a turntable: attach known weights at different lever arms to change τ while keeping I constant; measure α from angular velocity vs time and verify α ∝ τ. - Rolling objects down an incline: measure acceleration for different shapes (hoop, solid sphere) to show larger I gives smaller α and longer times—connects τ = Iα to moment of inertia. - Pulley + two masses over a massive wheel: use tensions T1, T2 and net torque = (T2−T1)r to predict α and compare to measured values. Those map to AP lab skills (Design & Analysis FRQ)—you can use the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and Unit 5 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-5) for procedures and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

How do you solve problems that involve both translational and rotational motion at the same time?

When a problem has both translation and rotation, treat them as two linked systems and write the appropriate laws for each: 1. Draw a clear diagram and free-body diagram. Identify forces, the axis of rotation, and any contact/constraint (e.g., no slipping). 2. Write Newton’s 2nd law (translational): ΣF = m a (for the center of mass). 3. Write rotational Newton’s 2nd law about the chosen axis: Στ = I α (or α = Στ / I). Use the parallel-axis theorem if needed. 4. Add kinematic constraints that connect a and α. For rolling without slipping: a = α r (or a = r α with sign conventions). 5. Solve the simultaneous equations for the unknowns (a, α, tensions, friction, etc.). Check directions with the right-hand rule and sign consistency. Common pitfalls: pick the easiest axis for torques, don’t forget static friction can provide torque without doing work, and remember EK 5.6.A: you may need separate linear and rotational analyses. For worked examples and topic review, see the Topic 5.6 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-5/6-newtons-second-law-in-rotational-form/study-guide/VuXCF8dn0RYwBEaK) and lots of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).