Torque and Rotational Dynamics

What this unit covers

The topics below follow the published Physics C: Mech course framework for Unit 5. This unit is worth 10–15% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• Rotational Kinematics13 min · 3 objectivesDefine angular velocity and acceleration as derivatives ω = dθ/dt and α = dω/dt · Apply the constant-angular-acceleration equations · Relate linear and angular quantities with v = rω and a_t = rα
• Torque & Moment of Inertia15 min · 3 objectivesDefine torque as τ = rF sin θ and identify the lever arm · Define the moment of inertia I = ∫r² dm and compute it for a simple body · Explain how the distribution of mass, not just its amount, sets the rotational inertia
• Newton's Second Law for Rotation14 min · 3 objectivesApply 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
• Rolling Without Slipping14 min · 3 objectivesState the rolling constraint v_cm = Rω and a_cm = Rα · Analyze an object rolling down an incline using both force and torque equations · Compare the accelerations of different shapes rolling down the same incline
• Moment of Inertia by Integration & the Parallel-Axis Theorem14 min · 3 objectivesCompute moments of inertia by integrating over a continuous mass distribution · Apply the parallel-axis theorem to shift the axis of rotation · Explain why moment of inertia depends on the axis, not only on the object
• Static Equilibrium: Ladders, Beams & Hinges14 min · 3 objectivesApply the two conditions for static equilibrium to extended bodies · Choose a pivot that eliminates unknown forces from the torque equation · Solve ladder and beam problems including the required coefficient of friction
• Torque as a Cross Product14 min · 3 objectivesCompute torque magnitude using rF sin θ and identify the moment arm · Determine torque direction with the right-hand rule · Explain why torque is a vector and what its direction represents

Formulas in Unit 5

Every term in Unit 5

All 38 terms we publish for Torque and Rotational Dynamics, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Parallel axis theorem

I = I_cm + Md². Moving the axis a distance d from the center of mass always increases the moment of inertia.

Rolling without slipping

v_cm = ωR and a_cm = αR, with static friction (which does no work) supplying the torque. Down an incline, a = g sin θ/(1 + I/MR²).

Torque as a cross product

τ = r × F, with magnitude rF sin θ. Direction is given by the right-hand rule along the rotation axis.

Moment of inertia by integration

I = ∫r² dm. The integral runs over the mass distribution, with r measured perpendicular to the axis.

Standard moments of inertia

Rod about center ML²/12, rod about end ML²/3, disk ½MR², hoop MR², solid sphere ⅖MR², spherical shell ⅔MR².

Rotational form of Newton's second law

Στ = Iα about a fixed axis, or about the center of mass for a body that is also translating.

Rolling down an incline

a = g sin θ/(1 + I/MR²). Objects with mass concentrated near the axis accelerate faster, independent of mass and radius.

Torque about different axes

Choosing the axis through an unknown force eliminates its torque, which reduces the number of unknowns in equilibrium problems.

Static equilibrium conditions

ΣF = 0 and Στ = 0 simultaneously. If ΣF = 0, then Στ has the same value about every axis.

Deriving a moment of inertia

Set up I = ∫r² dm, express dm through the mass distribution, and choose limits matching the geometry. For a rod, dm = (M/L)dx.

Why the axis matters

A rod is ML²/12 about its center but ML²/3 about its end — four times larger, because more mass sits far from the axis.

Torque from a distributed force

Gravity on an extended body acts effectively at the center of mass, which is why a beam's weight is drawn at its midpoint.

Combining translation and rotation

Write ΣF = Ma_cm and Στ = I_cm α, then apply the rolling constraint a_cm = Rα to link them.

Direction of friction in rolling

For an object rolling down an incline, static friction acts up the incline and supplies the torque that produces angular acceleration.

Slipping vs rolling

Rolling requires the needed static friction to stay below μsN. Above that the object slips, kinetic friction acts, and v ≠ Rω.

Angular kinematics with calculus

ω = dθ/dt and α = dω/dt. The constant-α equations are the special case, exactly as in linear motion.

Equilibrium of a leaning ladder

Three unknown forces require all three equilibrium equations; taking torques about the base eliminates two of them at once.

The rotational kinematic equations

ω = ω₀ + αt, θ = θ₀ + ω₀t + ½αt² and ω² = ω₀² + 2αΔθ, valid only for constant α. They are the linear equations with each symbol replaced by its angular counterpart.

Relating tangential and angular quantities

s = rθ, v = rω, a_t = rα — all requiring radians. Points farther from the axis have greater linear speed at the same angular speed, which is why the rim of a wheel moves fastest.

Torque definition and units

τ = rF sin θ, measured in N·m. Despite sharing units with the joule it is not energy, and the two must never be conflated on a rubric.

Moment arm

The perpendicular distance from the axis to the line of action of the force. Using r_⊥F is usually faster and safer than hunting for the correct angle in rF sin θ.

Rotational equilibrium

Στ = 0 is required in addition to ΣF = 0. Two equal and opposite forces at different points form a couple with zero net force and nonzero net torque.

The rotational second law

Στ = Iα, the angular counterpart of ΣF = ma, with moment of inertia playing the role of mass. Both are exact statements, not approximations.

Moment of inertia of a point mass

I = mr², the building block from which every other result is assembled by summation or integration.

Rod about its center

I = ML²/12, the smallest moment of inertia for a rod about any perpendicular axis, since it passes through the center of mass.

Rod about one end

I = ML²/3, four times the value about the center. The parallel-axis theorem gives it as ML²/12 + M(L/2)².

Solid disk or cylinder about its axis

I = ½MR². The mass is distributed from the center outward, so it is half the hoop value.

Hoop or thin cylindrical shell

I = MR², the maximum for a given mass and radius, because every element sits at the full radius.

Solid sphere

I = (2/5)MR². The smallest of the standard rolling shapes, which is why a solid sphere wins the rolling race.

Spherical shell

I = (2/3)MR², larger than the solid sphere because the mass is concentrated at the surface.

Perpendicular-axis theorem

For a flat lamina, I_z = I_x + I_y about perpendicular in-plane axes. It applies only to planar objects, which is the restriction usually forgotten.

Moments of inertia add

For a composite object about a common axis, simply sum the individual moments of inertia. A rod with a bob at the end is I_rod + I_bob.

Pulley with mass

A massive pulley requires net torque to angularly accelerate, so the tensions on its two sides differ: T₁ − T₂ = Iα/R. Assuming equal tensions is the standard error.

Falling spool or yo-yo

Tension provides both an upward force and a torque, giving a = g/(1 + I/MR²) — the same structure as rolling down an incline, with the incline replaced by a vertical string.

Choosing a pivot

A force acting at the pivot has zero moment arm and drops out of the torque equation. Since Στ = 0 about every point in equilibrium, choose the pivot where the unknowns act.

Sign convention for rotation

Counterclockwise positive is the usual choice, matching the right-hand rule with the axis out of the page. State it and keep it, exactly as with linear sign conventions.

Torque and angular acceleration are not proportional to force alone

The same force applied at a different point or angle produces a different angular acceleration. Where and how a force is applied matters as much as how large it is.

Rotational equilibrium of a leaning object

For a ladder against a frictionless wall, taking torques about the base eliminates both unknown floor forces at once and gives μ_min = 1/(2 tan θ), independent of mass.

What examiners penalize here

• Two objects with identical mass can have very different moments of inertia. Always ask where the mass sits relative to the axis: the r² weighting in I = ∫r² dm rewards mass at the rim and penalizes it near the center.
• For rolling problems, always bring in the constraint a_cm = Rα to connect the force equation to the torque equation. Remember that the shape factor I/MR² alone decides the race down an incline — mass and radius drop out.
• Free-response equilibrium problems award points for the **setup**: a labeled free-body diagram, an explicit statement of the pivot chosen, and the torque equation with moment arms identified. Students who go straight to numbers routinely lose those points even when the final answer is right.

Practice Physics C: Mech

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 7 units of AP Physics C: Mechanics

1. Unit 1 · Kinematics
2. Unit 2 · Force and Translational Dynamics
3. Unit 3 · Work, Energy, and Power
4. Unit 4 · Linear Momentum
5. Unit 5 · Torque and Rotational Dynamics
6. Unit 6 · Energy and Momentum of Rotating Systems
7. Unit 7 · Oscillations