Rotational Dynamics Formulas to Know for AP Physics C: Mechanics (2025)

1) What You Need to Know

Rotational dynamics is the “torque version” of Newton’s laws: forces cause linear acceleration, and torques cause angular acceleration. On AP Physics C: Mechanics, nearly every rotation problem is some mix of:

  • Torque τ\tau from forces
  • Moment of inertia II (how “hard” it is to spin)
  • Angular acceleration α\alpha, angular speed ω\omega
  • Angular momentum L\vec L and its conservation
  • Energy (work/rotational kinetic energy)
  • Rolling without slipping constraints

The two core “master equations” you must know:

  • Rotational Newton’s 2nd law (fixed axis): τext=Iα\sum \tau_{\text{ext}} = I\alpha
  • General torque–angular momentum relation: τext=dLdt\sum \vec\tau_{\text{ext}} = \frac{d\vec L}{dt}

Use rotational dynamics when:

  • A rigid body (disk, rod, pulley, wheel) is accelerating angularly.
  • Forces act at a distance from an axis (torques matter).
  • You need to connect translation + rotation (pulleys, rolling objects).

Big exam idea: Choose an axis strategically. Picking the right torque axis can eliminate unknown forces (e.g., hinge forces) and save you tons of algebra.


2) Step-by-Step Breakdown

A. Standard “Torque + Translation” Recipe (most AP C rotation problems)
  1. Choose a coordinate system (sign convention for rotation too: CCW positive is common).
  2. Draw a clean FBD for every object that moves (blocks + rotating body).
  3. Pick an axis for torques (often the rotation axis). Decide what torques are positive.
  4. Write translation equations for each mass: F=ma\sum F = ma along the motion direction.
  5. Write the rotation equation for the rotating body:
    • Fixed axis: τ=Iα\sum \tau = I\alpha
    • Use τ=rFsinϕ\tau = rF\sin\phi (or vector τ=r×F\vec\tau = \vec r \times \vec F)
  6. Add kinematic/constraint relations:
    • No-slip string on pulley: a=αRa = \alpha R
    • Rolling without slipping: vcm=ωRv_{\text{cm}} = \omega R and acm=αRa_{\text{cm}} = \alpha R
  7. Solve the system (you usually have as many equations as unknowns).
  8. Check limiting cases (e.g., if I0I\to 0 pulley becomes “massless,” does your result match intuition?).
B. Mini Worked Workflow (pulley constraint)

Suppose a mass pulls a string wrapped on a pulley of radius RR.

  • Translational: mgT=mamg - T = ma
  • Rotational about pulley axle: TR=IαTR = I\alpha
  • No slip: a=αRa = \alpha R
    Combine: TR=IaRT=IaR2TR = I\frac{a}{R} \Rightarrow T = \frac{Ia}{R^2}, then plug into translation.

Decision point: If the axis is fixed and the body is rigid, use τ=Iα\sum\tau = I\alpha. If the axis moves or you’re asked about angular momentum conservation, use τext=dLdt\sum\vec\tau_{\text{ext}} = \frac{d\vec L}{dt} and/or energy.


3) Key Formulas, Rules & Facts

A. Core Definitions (torque, angular momentum, inertia)
QuantityFormulaWhen to useNotes
Torque magnitudeτ=rFsinϕ\tau = rF\sin\phiForce at distance from axisϕ\phi is angle between r\vec r and F\vec F
Torque (vector)τ=r×F\vec\tau = \vec r \times \vec FDirection/sign + 3DRight-hand rule
Rotational Newton’s 2nd (fixed axis)τext=Iα\sum \tau_{\text{ext}} = I\alphaRigid body about fixed axisRequires constant axis + rigid body
Angular momentum (general)L=r×p\vec L = \vec r \times \vec pPoint particlesFor rigid bodies sum/integrate
Angular momentum (rigid body, fixed axis)L=IωL = I\omegaRotation about symmetry/fixed axisDirection along axis
Torque–angular momentumτext=dLdt\sum \vec\tau_{\text{ext}} = \frac{d\vec L}{dt}Conservation, changing axesIf τext=0\sum\vec\tau_{\text{ext}}=\vec 0 then L\vec L constant
Moment of inertiaI=miri2I = \sum m_ir_i^2 or I=r2dmI=\int r^2\,dm“How hard to spin” about an axisAxis choice matters a lot

Units & angles:

  • II in kgm2\text{kg}\cdot\text{m}^2, τ\tau in Nm\text{N}\cdot\text{m}, ω\omega in rad/s\text{rad/s}, α\alpha in rad/s2\text{rad/s}^2.
  • Use radians in kinematics/energy. (Degrees break formulas.)

B. Rotational Kinematics (constant α\alpha)
RelationshipFormulaNotes
Angular velocityω=ω0+αt\omega = \omega_0 + \alpha tConstant α\alpha
Angleθ=θ0+ω0t+12αt2\theta = \theta_0 + \omega_0 t + \tfrac12\alpha t^2Constant α\alpha
“No time” formω2=ω02+2α(θθ0)\omega^2 = \omega_0^2 + 2\alpha(\theta-\theta_0)Great with energy-style setups
Average angular velocityωavg=ω+ω02\omega_{\text{avg}} = \tfrac{\omega+\omega_0}{2}Constant α\alpha only

Linear–angular links (for a point at radius rr):

  • Tangential speed: v=ωrv = \omega r
  • Tangential acceleration: at=αra_t = \alpha r
  • Centripetal acceleration: ac=ω2ra_c = \omega^2 r

C. Work, Energy, and Power in Rotation
ConceptFormulaWhen to useNotes
Rotational kinetic energyKrot=12Iω2K_{\text{rot}} = \tfrac12 I\omega^2Spinning rigid bodyAdd translation separately
Total kinetic energy (rolling)K=12mvcm2+12Icmω2K = \tfrac12 mv_{\text{cm}}^2 + \tfrac12 I_{\text{cm}}\omega^2Rolling objectsUse vcm=ωRv_{\text{cm}}=\omega R if no slip
Work by torqueW=τdθW = \int \tau\, d\thetaVariable torqueIf constant, W=τΔθW=\tau\Delta\theta
Power (rotation)P=τωP = \tau\omegaMotors, instantaneous powerSign matters
Work–energy theoremWext=ΔKW_{\text{ext}} = \Delta KOften easiest pathInclude both translational + rotational

Reminder: Static friction can do zero work in pure rolling (contact point instantaneously at rest), yet it can still provide a torque that changes ω\omega.


D. Rolling Without Slipping (high-yield)
ConditionFormulaNotes
No-slip kinematic constraintvcm=ωRv_{\text{cm}} = \omega RMust be true at all times
No-slip acceleration constraintacm=αRa_{\text{cm}} = \alpha RAlong the rolling direction
“Effective inertia” trick (down incline)a=gsinβ1+IcmmR2a = \frac{g\sin\beta}{1 + \frac{I_{\text{cm}}}{mR^2}}Rigid body rolling down an incline angle β\beta

Common moments of inertia ratios IcmmR2\frac{I_{\text{cm}}}{mR^2}:

  • Solid disk/cylinder: 12\tfrac12
  • Hoop/thin ring: 11
  • Solid sphere: 25\tfrac25
  • Thin spherical shell: 23\tfrac23

E. Standard Moments of Inertia (know these cold)
ObjectAxisII
Point massdistance rr from axismr2mr^2
Thin hoop/ringcenter, perpendicular to planemR2mR^2
Solid disk/cylindercenter, perpendicular to face12mR2\tfrac12 mR^2
Solid spherethrough center25mR2\tfrac25 mR^2
Thin spherical shellthrough center23mR2\tfrac23 mR^2
Thin rod (length LL)through center, perpendicular112mL2\tfrac{1}{12}mL^2
Thin rod (length LL)about one end, perpendicular13mL2\tfrac13 mL^2

Theorems:

  • Parallel-axis theorem: I=Icm+md2I = I_{\text{cm}} + md^2 (shift axis by distance dd)
  • Perpendicular-axis theorem (planar lamina): Iz=Ix+IyI_z = I_x + I_y (only for flat objects in the xyxy-plane)

F. Angular Impulse & Momentum Conservation
IdeaFormulaWhen to useNotes
Angular impulseτdt=ΔL\int \tau\,dt = \Delta LCollisions/short pushesChoose axis to kill unknown impulses
Conservation of angular momentumLi=LfL_i = L_fIf τext=0\sum\tau_{\text{ext}}=0 about chosen axisWorks even if forces are huge but internal

G. Common Torque Setups (fast recognition)
  • Force applied tangentially at radius RR: τ=FR\tau = FR
  • Weight on a rod pivoted at one end (COM at L/2L/2): τg=mg(L2)sinθ\tau_g = mg\left(\tfrac{L}{2}\right)\sin\theta (where θ\theta is angle between rod and vertical if you define it that way—be consistent)
  • Multiple forces: sum torques with sign.

Critical: Torque depends on the perpendicular lever arm rr_\perp: τ=Fr\tau = Fr_\perp.


4) Examples & Applications

Example 1: Block + Massive Pulley (classic AP C)

A block of mass mm hangs from a string wrapped around a pulley (radius RR, inertia II). Find acceleration magnitude aa.

Setup:

  • Block: mgT=mamg - T = ma
  • Pulley: TR=IαTR = I\alpha
  • Constraint: a=αRa = \alpha R

Key solve:
TR=IaRT=IaR2TR = I\frac{a}{R} \Rightarrow T = \frac{Ia}{R^2}
Plug into translation:
mgIaR2=maa=mgm+IR2mg - \frac{Ia}{R^2} = ma \Rightarrow a = \frac{mg}{m + \frac{I}{R^2}}
Insight: Pulley inertia acts like “extra mass” IR2\frac{I}{R^2}.


Example 2: Rolling Object Down an Incline

A rigid body (mass mm, radius RR, inertia IcmI_{\text{cm}}) rolls without slipping down incline angle β\beta.

Fast result (no need to solve for friction explicitly):
a=gsinβ1+IcmmR2a = \frac{g\sin\beta}{1 + \frac{I_{\text{cm}}}{mR^2}}
Ranking speed at bottom (same drop height): smaller ImR2\frac{I}{mR^2} wins.

  • Solid sphere 25\tfrac25 fastest
  • Solid disk 12\tfrac12 next
  • Hoop 11 slowest

Example 3: Door Torque (lever arm + angle trap)

You push on a door at distance rr from hinges with force FF at angle ϕ\phi to the door (in the plane).

Torque magnitude about hinge:
τ=rFsinϕ\tau = rF\sin\phi
Key insight: Pushing perpendicular to the door sinϕ=1\Rightarrow \sin\phi = 1 gives max torque.

Common exam twist: same FF but different push point: doubling rr doubles τ\tau.


Example 4: Angular Momentum Conservation (person on stool)

A person on a frictionless rotating stool pulls arms in, changing inertia from IiI_i to IfI_f.

If external torque is negligible:
Iiωi=Ifωfωf=IiIfωiI_i\omega_i = I_f\omega_f \Rightarrow \omega_f = \frac{I_i}{I_f}\,\omega_i
Energy is not conserved here (muscles do internal work):
Krot=12Iω2K_{\text{rot}} = \tfrac12 I\omega^2 increases when II decreases.


5) Common Mistakes & Traps

  1. Mixing up rr vs. rr_\perp (lever arm).

    • Wrong: using τ=rF\tau = rF automatically.
    • Right: τ=rFsinϕ=Fr\tau = rF\sin\phi = Fr_\perp. Draw the perpendicular distance to the line of action.
  2. Forgetting that torque depends on axis choice.

    • Wrong: computing torque about the wrong point, then wondering why hinge forces appear.
    • Fix: choose an axis that eliminates unknown forces (e.g., about a pivot so pivot forces give zero torque).
  3. Using τ=Iα\sum \tau = I\alpha when the axis isn’t fixed / body isn’t a simple rigid rotation.

    • Wrong: applying it blindly in situations with moving axes.
    • Fix: if unsure, fall back to τext=dLdt\sum\vec\tau_{\text{ext}} = \frac{d\vec L}{dt} or use energy.
  4. Sign errors (CW vs CCW) when summing torques.

    • Wrong: mixing sign conventions between translation and rotation.
    • Fix: declare “CCW positive” (or CW), then stick to it across the problem.
  5. Assuming friction always opposes motion (rolling friction confusion).

    • Wrong: claiming static friction must point uphill on an incline.
    • Truth: static friction opposes relative slipping at the contact point. Its direction depends on the tendency to slip.
  6. Forgetting the constraint a=αRa = \alpha R (strings and rolling).

    • Wrong: solving translation and rotation separately.
    • Fix: write constraints early; they are often the missing equation.
  7. Using degrees instead of radians in kinematics/energy.

    • Wrong: plugging θ\theta in degrees into ω2=ω02+2αΔθ\omega^2 = \omega_0^2 + 2\alpha\Delta\theta.
    • Fix: convert to radians or keep everything symbolic.
  8. Treating tension as the same on both sides of a massive pulley.

    • Wrong: setting T1=T2T_1=T_2 when pulley has nonzero II.
    • Fix: use torque: (T1T2)R=Iα(T_1-T_2)R = I\alpha.

6) Memory Aids & Quick Tricks

Trick / mnemonicHelps you rememberWhen to use
“Perp is power”Torque uses perpendicular lever arm: τ=Fr\tau = Fr_\perpAny torque problem
RHR (right-hand rule)Direction of τ\vec\tau, L\vec L, ω\vec\omegaVector/sign direction
“ROLL” constraintsvcm=ωRv_{\text{cm}}=\omega R and acm=αRa_{\text{cm}}=\alpha RRolling/no-slip problems
Pulley inertia = extra massa=mgm+I/R2a = \frac{mg}{m + I/R^2}-style structureAny string-on-pulley acceleration
Choose pivot to kill forcesForces through axis give zero torqueRods, doors, ladders, hinged objects
Energy shortcut for rollingUse mgh12mv2+12Iω2mgh \to \tfrac12 mv^2 + \tfrac12 I\omega^2 with v=ωRv=\omega RFind speed without time/forces

7) Quick Review Checklist

  • You can write and use τ=Iα\sum \tau = I\alpha (fixed axis) and τ=dLdt\sum\vec\tau = \frac{d\vec L}{dt} (general).
  • You consistently compute torque using τ=rFsinϕ=Fr\tau = rF\sin\phi = Fr_\perp with correct signs.
  • You know the standard II formulas (disk, hoop, rod, sphere) and can use I=Icm+md2I = I_{\text{cm}} + md^2.
  • You remember rotational kinematics (constant α\alpha): ω=ω0+αt\omega = \omega_0+\alpha t, ω2=ω02+2αΔθ\omega^2 = \omega_0^2+2\alpha\Delta\theta.
  • You can switch between linear and angular: v=ωrv=\omega r, at=αra_t=\alpha r.
  • You can do rotational energy: Krot=12Iω2K_{\text{rot}}=\tfrac12 I\omega^2 and work by torque W=τdθW=\int\tau\,d\theta, power P=τωP=\tau\omega.
  • For rolling without slipping you immediately write vcm=ωRv_{\text{cm}}=\omega R and (if needed) acm=αRa_{\text{cm}}=\alpha R.
  • You avoid traps: tension differs across massive pulleys; static friction direction depends on slip tendency.

You’ve got this—if you set up torques cleanly and lock in the constraints, the algebra usually falls into place.