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 \tau from forces
- Moment of inertia I (how “hard” it is to spin)
- Angular acceleration \alpha, angular speed \omega
- Angular momentum \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): \sum \tau_{\text{ext}} = I\alpha
- General torque–angular momentum relation: \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)
- Choose a coordinate system (sign convention for rotation too: CCW positive is common).
- Draw a clean FBD for every object that moves (blocks + rotating body).
- Pick an axis for torques (often the rotation axis). Decide what torques are positive.
- Write translation equations for each mass: \sum F = ma along the motion direction.
- Write the rotation equation for the rotating body:
- Fixed axis: \sum \tau = I\alpha
- Use \tau = rF\sin\phi (or vector \vec\tau = \vec r \times \vec F)
- Add kinematic/constraint relations:
- No-slip string on pulley: a = \alpha R
- Rolling without slipping: v_{\text{cm}} = \omega R and a_{\text{cm}} = \alpha R
- Solve the system (you usually have as many equations as unknowns).
- Check limiting cases (e.g., if I\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 R.
- Translational: mg - T = ma
- Rotational about pulley axle: TR = I\alpha
- No slip: a = \alpha R
Combine: TR = 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 \sum\tau = I\alpha. If the axis moves or you’re asked about angular momentum conservation, use \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)
| Quantity | Formula | When to use | Notes |
|---|---|---|---|
| Torque magnitude | \tau = rF\sin\phi | Force at distance from axis | \phi is angle between \vec r and \vec F |
| Torque (vector) | \vec\tau = \vec r \times \vec F | Direction/sign + 3D | Right-hand rule |
| Rotational Newton’s 2nd (fixed axis) | \sum \tau_{\text{ext}} = I\alpha | Rigid body about fixed axis | Requires constant axis + rigid body |
| Angular momentum (general) | \vec L = \vec r \times \vec p | Point particles | For rigid bodies sum/integrate |
| Angular momentum (rigid body, fixed axis) | L = I\omega | Rotation about symmetry/fixed axis | Direction along axis |
| Torque–angular momentum | \sum \vec\tau_{\text{ext}} = \frac{d\vec L}{dt} | Conservation, changing axes | If \sum\vec\tau_{\text{ext}}=\vec 0 then \vec L constant |
| Moment of inertia | I = \sum m_ir_i^2 or I=\int r^2\,dm | “How hard to spin” about an axis | Axis choice matters a lot |
Units & angles:
- I in \text{kg}\cdot\text{m}^2, \tau in \text{N}\cdot\text{m}, \omega in \text{rad/s}, \alpha in \text{rad/s}^2.
- Use radians in kinematics/energy. (Degrees break formulas.)
B. Rotational Kinematics (constant \alpha)
| Relationship | Formula | Notes |
|---|---|---|
| Angular velocity | \omega = \omega_0 + \alpha t | Constant \alpha |
| Angle | \theta = \theta_0 + \omega_0 t + \tfrac12\alpha t^2 | Constant \alpha |
| “No time” form | \omega^2 = \omega_0^2 + 2\alpha(\theta-\theta_0) | Great with energy-style setups |
| Average angular velocity | \omega_{\text{avg}} = \tfrac{\omega+\omega_0}{2} | Constant \alpha only |
Linear–angular links (for a point at radius r):
- Tangential speed: v = \omega r
- Tangential acceleration: a_t = \alpha r
- Centripetal acceleration: a_c = \omega^2 r
C. Work, Energy, and Power in Rotation
| Concept | Formula | When to use | Notes |
|---|---|---|---|
| Rotational kinetic energy | K_{\text{rot}} = \tfrac12 I\omega^2 | Spinning rigid body | Add translation separately |
| Total kinetic energy (rolling) | K = \tfrac12 mv_{\text{cm}}^2 + \tfrac12 I_{\text{cm}}\omega^2 | Rolling objects | Use v_{\text{cm}}=\omega R if no slip |
| Work by torque | W = \int \tau\, d\theta | Variable torque | If constant, W=\tau\Delta\theta |
| Power (rotation) | P = \tau\omega | Motors, instantaneous power | Sign matters |
| Work–energy theorem | W_{\text{ext}} = \Delta K | Often easiest path | Include 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 \omega.
D. Rolling Without Slipping (high-yield)
| Condition | Formula | Notes |
|---|---|---|
| No-slip kinematic constraint | v_{\text{cm}} = \omega R | Must be true at all times |
| No-slip acceleration constraint | a_{\text{cm}} = \alpha R | Along the rolling direction |
| “Effective inertia” trick (down incline) | a = \frac{g\sin\beta}{1 + \frac{I_{\text{cm}}}{mR^2}} | Rigid body rolling down an incline angle \beta |
Common moments of inertia ratios \frac{I_{\text{cm}}}{mR^2}:
- Solid disk/cylinder: \tfrac12
- Hoop/thin ring: 1
- Solid sphere: \tfrac25
- Thin spherical shell: \tfrac23
E. Standard Moments of Inertia (know these cold)
| Object | Axis | I |
|---|---|---|
| Point mass | distance r from axis | mr^2 |
| Thin hoop/ring | center, perpendicular to plane | mR^2 |
| Solid disk/cylinder | center, perpendicular to face | \tfrac12 mR^2 |
| Solid sphere | through center | \tfrac25 mR^2 |
| Thin spherical shell | through center | \tfrac23 mR^2 |
| Thin rod (length L) | through center, perpendicular | \tfrac{1}{12}mL^2 |
| Thin rod (length L) | about one end, perpendicular | \tfrac13 mL^2 |
Theorems:
- Parallel-axis theorem: I = I_{\text{cm}} + md^2 (shift axis by distance d)
- Perpendicular-axis theorem (planar lamina): I_z = I_x + I_y (only for flat objects in the xy-plane)
F. Angular Impulse & Momentum Conservation
| Idea | Formula | When to use | Notes |
|---|---|---|---|
| Angular impulse | \int \tau\,dt = \Delta L | Collisions/short pushes | Choose axis to kill unknown impulses |
| Conservation of angular momentum | L_i = L_f | If \sum\tau_{\text{ext}}=0 about chosen axis | Works even if forces are huge but internal |
G. Common Torque Setups (fast recognition)
- Force applied tangentially at radius R: \tau = FR
- Weight on a rod pivoted at one end (COM at L/2): \tau_g = mg\left(\tfrac{L}{2}\right)\sin\theta (where \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 r_\perp: \tau = Fr_\perp.
4) Examples & Applications
Example 1: Block + Massive Pulley (classic AP C)
A block of mass m hangs from a string wrapped around a pulley (radius R, inertia I). Find acceleration magnitude a.
Setup:
- Block: mg - T = ma
- Pulley: TR = I\alpha
- Constraint: a = \alpha R
Key solve:
TR = I\frac{a}{R} \Rightarrow T = \frac{Ia}{R^2}
Plug into translation:
mg - \frac{Ia}{R^2} = ma \Rightarrow a = \frac{mg}{m + \frac{I}{R^2}}
Insight: Pulley inertia acts like “extra mass” \frac{I}{R^2}.
Example 2: Rolling Object Down an Incline
A rigid body (mass m, radius R, inertia I_{\text{cm}}) rolls without slipping down incline angle \beta.
Fast result (no need to solve for friction explicitly):
a = \frac{g\sin\beta}{1 + \frac{I_{\text{cm}}}{mR^2}}
Ranking speed at bottom (same drop height): smaller \frac{I}{mR^2} wins.
- Solid sphere \tfrac25 fastest
- Solid disk \tfrac12 next
- Hoop 1 slowest
Example 3: Door Torque (lever arm + angle trap)
You push on a door at distance r from hinges with force F at angle \phi to the door (in the plane).
Torque magnitude about hinge:
\tau = rF\sin\phi
Key insight: Pushing perpendicular to the door \Rightarrow \sin\phi = 1 gives max torque.
Common exam twist: same F but different push point: doubling r doubles \tau.
Example 4: Angular Momentum Conservation (person on stool)
A person on a frictionless rotating stool pulls arms in, changing inertia from I_i to I_f.
If external torque is negligible:
I_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):
K_{\text{rot}} = \tfrac12 I\omega^2 increases when I decreases.
5) Common Mistakes & Traps
Mixing up r vs. r_\perp (lever arm).
- Wrong: using \tau = rF automatically.
- Right: \tau = rF\sin\phi = Fr_\perp. Draw the perpendicular distance to the line of action.
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).
Using \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 \sum\vec\tau_{\text{ext}} = \frac{d\vec L}{dt} or use energy.
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.
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.
Forgetting the constraint a = \alpha R (strings and rolling).
- Wrong: solving translation and rotation separately.
- Fix: write constraints early; they are often the missing equation.
Using degrees instead of radians in kinematics/energy.
- Wrong: plugging \theta in degrees into \omega^2 = \omega_0^2 + 2\alpha\Delta\theta.
- Fix: convert to radians or keep everything symbolic.
Treating tension as the same on both sides of a massive pulley.
- Wrong: setting T_1=T_2 when pulley has nonzero I.
- Fix: use torque: (T_1-T_2)R = I\alpha.
6) Memory Aids & Quick Tricks
| Trick / mnemonic | Helps you remember | When to use |
|---|---|---|
| “Perp is power” | Torque uses perpendicular lever arm: \tau = Fr_\perp | Any torque problem |
| RHR (right-hand rule) | Direction of \vec\tau, \vec L, \vec\omega | Vector/sign direction |
| “ROLL” constraints | v_{\text{cm}}=\omega R and a_{\text{cm}}=\alpha R | Rolling/no-slip problems |
| Pulley inertia = extra mass | a = \frac{mg}{m + I/R^2}-style structure | Any string-on-pulley acceleration |
| Choose pivot to kill forces | Forces through axis give zero torque | Rods, doors, ladders, hinged objects |
| Energy shortcut for rolling | Use mgh \to \tfrac12 mv^2 + \tfrac12 I\omega^2 with v=\omega R | Find speed without time/forces |
7) Quick Review Checklist
- You can write and use \sum \tau = I\alpha (fixed axis) and \sum\vec\tau = \frac{d\vec L}{dt} (general).
- You consistently compute torque using \tau = rF\sin\phi = Fr_\perp with correct signs.
- You know the standard I formulas (disk, hoop, rod, sphere) and can use I = I_{\text{cm}} + md^2.
- You remember rotational kinematics (constant \alpha): \omega = \omega_0+\alpha t, \omega^2 = \omega_0^2+2\alpha\Delta\theta.
- You can switch between linear and angular: v=\omega r, a_t=\alpha r.
- You can do rotational energy: K_{\text{rot}}=\tfrac12 I\omega^2 and work by torque W=\int\tau\,d\theta, power P=\tau\omega.
- For rolling without slipping you immediately write v_{\text{cm}}=\omega R and (if needed) a_{\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.