Must-know sheet

Physics C: Mechanics must-know sheet

The real AP Physics C: Mechanics exam gives you an equation sheet with constants, the exam's default assumptions and the main equations, and you can use a calculator on both sections. Start with the recent changes at the top. The rest covers what the equation sheet doesn't tell you: when each law applies, the results you derive with calculus, sign rules, graph shapes and free-response habits.

What's new: recent changes to the course and exam

Units 1, 2, 3, 4, 5, 6, 7

May 2027: 42 multiple-choice questions in 85 minutes: That's up from 40 questions in 80 minutes. The free-response section is now 95 minutes instead of 100, still for 4 questions, so the exam is still 3 hours. Each section is half your score, and you have a calculator and the equation sheet on both.
A hybrid digital exam: You answer the multiple-choice questions in the Bluebook app. You also read the free-response questions in Bluebook, but you write your answers by hand in a paper booklet. Try the Bluebook test preview before exam day so the format isn't a surprise.
Since 2024–25: four answer choices and four free-response types: Multiple-choice questions now have 4 choices (A–D) instead of 5. The free-response section has 4 questions, always in this order: Mathematical Routines (10 points), Translation Between Representations (12), Experimental Design and Analysis (10) and Qualitative/Quantitative Translation (8). The old exam was 90 minutes with 3 free-response questions, so older practice tests are good physics practice but not good timing practice.
Since 2024–25: center of mass and gravitation moved into Unit 2: Center of mass is now Topic 2.1 (it used to be in Unit 4), and the law of gravitation is Topic 2.6. Gravitation no longer has a unit of its own: circular orbits are in Topic 2.10 and orbit energy and angular momentum are in Topic 6.6.
Since 2024–25: rotation is split into two units: Unit 5 covers rotational kinematics, torque, rotational inertia, equilibrium and Newton's second law for rotation. Unit 6 covers rotational energy, angular momentum, rolling and satellite orbits. Oscillations moved from Unit 6 to Unit 7, so older videos and books number these units differently.

Units, constants and exam conventions

Units 1, 2, 3, 4, 5, 6, 7

Values the exam uses: When a number is needed, the course framework uses g = 10 m/s² near Earth's surface, but you won't lose credit for the more precise 9.8 or 9.81 m/s² (the equation sheet lists 9.8). The gravitational constant is G = 6.67 × 10⁻¹¹ N·m²/kg². The equation sheet gives you G and g; any planet's mass or radius you need will be in the question.
Derived SI units: Newton: N = kg·m/s². Joule: J = N·m = kg·m²/s². Watt: W = J/s. Momentum and impulse: kg·m/s = N·s. Torque: N·m (never written as joules, even though the units match). Angular momentum: kg·m²/s. Rotational inertia: kg·m². Spring constant: N/m. The constant k in a drag force $F = -kv$ : kg/s.
Angles in radians: Use radians whenever you use ω, α, $s = r\theta$ or $v = r\omega$ . One revolution is 2π rad = 360°. To turn rev/min into rad/s, multiply by $\frac{2\pi}{60}$ . Radians have no unit, so r·ω in (m)(rad/s) is just m/s.
Conversions to do first: Convert to kg, m and s before you plug in: 1 g = 10⁻³ kg, 1 cm = 10⁻² m, 1 cm² = 10⁻⁴ m², and km/h ÷ 3.6 = m/s. Prefixes: milli = 10⁻³, kilo = 10³, mega = 10⁶. A kilowatt-hour is 3.6 × 10⁶ J.
What the exam assumes unless it says otherwise: The equation sheet lists three: the reference frame is inertial (not accelerating), air resistance is negligible, and springs and strings are ideal. An ideal string is massless and doesn't stretch, and an ideal spring is massless. Pulleys aren't on that list: an ideal pulley is massless and frictionless, but if a problem gives a pulley a mass or rotational inertia, or gives a drag force, use it.
Derive with letters, then plug in: When a question says derive, start from a basic principle (Newton's second law, conservation of energy, the definition of torque), write it with the symbols the question gives you, and solve step by step. Only substitute numbers at the end, and include units with every numerical answer.
Check units and limiting cases: Test any expression before you trust it: the units must come out right, and extremes should make sense. Set the pulley's mass to zero and you should get the massless-pulley answer; make two Atwood masses equal and a should be zero; at t → ∞ a drag answer should give terminal velocity.

Vectors, signs and the calculus toolkit

Units 1, 2, 3, 4, 5, 6

Components: For a vector of size A at angle θ above the +x axis, $A_x = A\cos\theta$ and $A_y = A\sin\theta$ . Going back, $A = \sqrt{A_x^2 + A_y^2}$ and $\tan\theta = \frac{A_y}{A_x}$ , checking which quadrant the vector is in. If θ is measured from a different line, redraw the triangle instead of trusting cos for x.
Unit-vector notation: $\vec{r} = 3\hat{i} + 4\hat{j}$ means 3 m along x and 4 m along y. Add or subtract vectors one component at a time, and differentiate or integrate them one component at a time too.
Dot product (gives a scalar): $\vec{A}\cdot\vec{B} = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z$ . It's zero for perpendicular vectors and negative when the angle is over 90°. You use it for work, $W = \int \vec{F}\cdot d\vec{r}$ , and power, $P = \vec{F}\cdot\vec{v}$ .
Cross product (gives a vector): $\lvert \vec{A}\times\vec{B} \rvert = AB\sin\theta$ , perpendicular to both vectors by the right-hand rule. $\hat{i}\times\hat{j} = \hat{k}$ , $\hat{j}\times\hat{k} = \hat{i}$ , $\hat{k}\times\hat{i} = \hat{j}$ , and swapping the order flips the sign. You use it for torque, $\vec{\tau} = \vec{r}\times\vec{F}$ , and a particle's angular momentum, $\vec{L} = \vec{r}\times\vec{p}$ ; on the exam you'll usually just say clockwise or counterclockwise.
Pick a positive direction and keep it: Choose + for each axis (and a + sense of rotation) at the start, and give every vector quantity its sign. A negative velocity means moving in the − direction. A negative acceleration does not by itself mean slowing down: an object slows down when v and a have opposite signs, and speeds up when they match.
Derivatives go down the chain, integrals come back: Position → velocity → acceleration by taking $\frac{d}{dt}$ ; back the other way by integrating over time and adding the starting value. The same pattern links θ, ω and α; p and F; L and τ; and W and P.
Acceleration as a function of position: If a force or acceleration depends on x instead of t, use the chain rule $a = v\frac{dv}{dx}$ , so $\int v\,dv = \int a\,dx$ . Often it's faster to use the work–energy theorem, which is the same idea.
Separating variables: For $\frac{dv}{dt} = f(v)$ , move every v to one side and every t to the other, $\int_{v_0}^{v}\frac{dv'}{f(v')} = \int_0^t dt'$ , and put the starting values in the limits so you don't need a separate constant. This is how drag problems are solved.

Kinematics with calculus

Unit 1

Velocity and acceleration as derivatives: $v_x = \frac{dx}{dt}$ and $a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}$ . Average velocity is $\frac{\Delta x}{\Delta t}$ and average acceleration is $\frac{\Delta v}{\Delta t}$ over an interval; instantaneous values are the limits as Δt → 0.
Going back with integrals: $v(t) = v_0 + \int_0^t a(t')\,dt'$ and $x(t) = x_0 + \int_0^t v(t')\,dt'$ . The constant of integration is the starting value, so read v₀ and x₀ from the question.
Displacement versus distance: Displacement is $\int v\,dt$ , with signs. Distance traveled is $\int \lvert v \rvert\,dt$ : find where v = 0, split the interval there, and add the sizes of each piece. Speed is the size of velocity and is never negative.
Constant-acceleration equations: $v = v_0 + at$ , $x = x_0 + v_0t + \frac{1}{2}at^2$ , $v^2 = v_0^2 + 2a(x - x_0)$ and $\Delta x = \frac{1}{2}(v_0 + v)t$ . They hold only when a is constant. If a depends on time or position (a spring, drag, a given a(t)), integrate instead.
Free fall: With up as +, a = −g the whole time, going up, at the top and coming down. At the highest point the vertical velocity is zero but the acceleration is still g downward. On level ground the trip up takes as long as the trip down, and the object lands with its launch speed.
Projectile motion: Ignoring air resistance, horizontal velocity $v_0\cos\theta$ stays constant and vertical motion has a = −g; the only link between the two directions is time. Maximum height is $\frac{(v_0\sin\theta)^2}{2g}$ . On level ground only, the range is $\frac{v_0^2\sin 2\theta}{g}$ , largest at 45°, and angles that add to 90° give the same range.
Horizontal launch: Something launched horizontally from height h falls for $t = \sqrt{\frac{2h}{g}}$ , the same time as an object dropped from rest at that height. Its horizontal distance is $v_0t$ .
Motion in two dimensions from a function: Given $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$ , differentiate each component for $\vec{v}$ and again for $\vec{a}$ . Speed is $\sqrt{v_x^2 + v_y^2}$ . The velocity is always tangent to the path.
Relative velocity: $\vec{v}_{\text{A rel ground}} = \vec{v}_{\text{A rel B}} + \vec{v}_{\text{B rel ground}}$ , added as vectors. For a boat crossing a river, the crossing time depends only on the velocity component straight across. Observers in different inertial frames measure different velocities but the same acceleration.
Motion graphs: On x–t, the slope is v, and the curve bends up (concave up) when a > 0. On v–t, the slope is a and the area to the time axis is Δx (area below the axis counts as negative). On a–t, the area is Δv. Where x–t has a peak or valley, v = 0.

Forces and Newton's laws

Unit 2

Newton's second law for a system: $\sum\vec{F}_{\text{ext}} = m_{\text{sys}}\vec{a}_{\text{cm}}$ : only external forces count, and they set the acceleration of the center of mass. Apply it separately along each axis. ma is the result of the forces, never a force on the diagram.
Newton's first law: Zero net force means constant velocity, which includes being at rest; it does not mean no forces act. Something moving at constant velocity in a straight line is in translational equilibrium. Forces can balance along one axis and not another.
Newton's third-law pairs: If A pushes on B, B pushes on A with an equal and opposite force of the same type. The two forces act on different objects, so they never cancel on one free-body diagram. Weight and the normal force on a book on a table are not a third-law pair: they're different types acting on the same object.
Free-body diagram rules: One dot for the object, one arrow per force starting at the dot, each labeled ( $F_g$ , $F_N$ , $F_T$ , $F_f$ , $F_s$ ). Draw whole forces, not components, and don't draw the net force or 'ma'. Tilt your axes so one runs along the acceleration (for example along an incline).
Weight and the normal force: Weight is $F_g = mg$ , toward Earth's center. The normal force is perpendicular to the surface and takes whatever size it needs; it equals mg only for a flat surface with no other vertical forces and no vertical acceleration. On an incline at angle θ with nothing else pushing, $F_N = mg\cos\theta$ .
Apparent weight: A scale reads the normal force. In an elevator accelerating at a (up taken +), $F_N = m(g + a)$ : heavier while speeding up going up or slowing down going down, and lighter in the opposite cases. In free fall it reads zero.
Inclines: Split gravity into $mg\sin\theta$ down the slope and $mg\cos\theta$ into the slope. Without friction, a = g sin θ down the slope whether the object is moving up or down it. A block just about to slip on a tilting surface has $\tan\theta = \mu_s$ .
Static friction: Static friction takes whatever size and direction keeps the surfaces from sliding, up to $F_{f,s} \le \mu_sF_N$ . Only use $\mu_sF_N$ when the problem says the object is just about to slip. It can point forward, as when it pushes a car or a walking person forward.
Kinetic friction: $F_{f,k} = \mu_kF_N$ , opposite the direction of sliding, and roughly independent of speed and contact area. Usually $\mu_k < \mu_s$ , which is why it's harder to start something sliding than to keep it going.
Ideal strings and pulleys: An ideal string has the same tension all along it and can only pull, along its own length. An ideal pulley just changes the tension's direction. Objects joined by a taut string share the same size of acceleration.
Connected objects: Write Newton's second law for each object along its own direction of motion, then solve together, or treat the whole set as one system to find a quickly. Atwood machine: $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$ and $F_T = \frac{2m_1m_2g}{m_1 + m_2}$ . Block m₁ on a table pulled by hanging m₂: $a = \frac{(m_2 - \mu_km_1)g}{m_1 + m_2}$ .
Springs and Hooke's law: $F_s = -k\Delta x$ : the force is proportional to the stretch or compression and always points back toward the relaxed length. A mass hanging at rest stretches a spring by $\frac{mg}{k}$ . Springs in parallel: $k_{\text{eq}} = k_1 + k_2$ . In series: $\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$ , less than either one.
Internal forces cancel: Forces between parts of a system come in third-law pairs, so they add to zero and can't change the motion of its center of mass. That's why you can't lift yourself by your own bootstraps, and why a system approach skips the tensions between connected blocks.

Center of mass

Units 2, 4

Separate particles: $x_{\text{cm}} = \frac{\sum m_ix_i}{\sum m_i}$ , and the same for y. The center of mass sits closer to the heavier masses. Measure every position from the same origin.
Solid objects by integration: $x_{\text{cm}} = \frac{1}{M}\int x\,dm$ , with $dm = \lambda\,dx$ for a rod and $M = \int \lambda\,dx$ . For a rod from 0 to L with $\lambda = cx$ , $M = \frac{cL^2}{2}$ and $x_{\text{cm}} = \frac{2L}{3}$ , toward the denser end.
Use symmetry first: A uniform object's center of mass is at its geometric center: the middle of a rod, the center of a disk, ring or sphere. It doesn't have to be inside the material; a ring's is at its empty center. Composite shapes: find each piece's center of mass and combine them as particles.
Velocity and acceleration of the center of mass: $\vec{v}_{\text{cm}} = \frac{\sum m_i\vec{v}_i}{\sum m_i} = \frac{\vec{p}_{\text{total}}}{M}$ and $\vec{a}_{\text{cm}} = \frac{\sum\vec{F}_{\text{ext}}}{M}$ . With no net external force, the center of mass moves at constant velocity, whatever the parts do.
Explosions and internal pushes: If a projectile explodes in midair, its center of mass keeps following the original parabola (until a piece hits something). If a person walks along a boat on frictionless water, the system's center of mass stays put, so the boat moves the other way.

Gravitation and circular motion

Units 2, 6

Newton's law of gravitation: $F_g = \frac{Gm_1m_2}{r^2}$ , attractive, along the line between centers, where r is center-to-center distance. Inverse square: double r and the force drops to one-fourth. Outside any spherically symmetric object, treat it as a point mass at its center. The mass in this law (gravitational mass) has been shown by experiment to equal the mass in F = ma (inertial mass).
Gravitational field: $g = \frac{GM}{r^2}$ outside a planet of mass M, so at height h above a surface of radius R, $g = \frac{GM}{(R + h)^2}$ . It's in N/kg, which equals m/s². Near the surface g is nearly constant, which is why mgh works there.
Gravity inside a uniform sphere: Only the mass closer to the center than you pulls on you, so $g = \frac{GMr}{R^3}$ for r ≤ R: zero at the center and growing linearly to its surface value. Outside it falls as $\frac{1}{r^2}$ .
Centripetal acceleration: $a_c = \frac{v^2}{r} = \omega^2r$ , toward the center. It isn't a new force: real forces like gravity, tension, normal force or friction (or their components toward the center) supply it, so write $\sum F_{\text{toward center}} = \frac{mv^2}{r}$ . Never draw a 'centripetal force' arrow on a free-body diagram. At constant speed, one trip around takes $T = \frac{2\pi r}{v}$ .
Speeding up or slowing down on a circle: A changing speed adds a tangential acceleration $a_t = \frac{dv}{dt}$ along the path. The total acceleration is $\sqrt{a_c^2 + a_t^2}$ , pointing partly inward. In uniform circular motion the speed is constant but the velocity isn't.
Vertical circles: At the top of a loop, $F_N + mg = \frac{mv^2}{r}$ (or $F_T + mg$ for a string), so the slowest speed that keeps contact is $\sqrt{gr}$ . At the bottom, $F_N - mg = \frac{mv^2}{r}$ , so the normal force or tension is greatest there.
Flat and banked curves: On a flat curve, static friction supplies the centripetal force, so $v_{\max} = \sqrt{\mu_sgr}$ . On a frictionless banked curve, the normal force's horizontal part does the job, and the right speed satisfies $\tan\theta = \frac{v^2}{rg}$ . A conical pendulum follows the same relationship, with θ measured from the vertical, because the string's tension plays the role of the normal force.
Circular orbits: Gravity alone supplies the centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ , so $v = \sqrt{\frac{GM}{r}}$ and $T^2 = \frac{4\pi^2}{GM}r^3$ (Kepler's third law). Neither depends on the satellite's mass, and a higher orbit is slower with a longer period. Kepler's first and second laws aren't expected by name.

Drag and other velocity-dependent forces

Unit 2

The drag model: A resistive force points opposite the velocity and grows with speed. The course usually uses $\vec{F}_r = -k\vec{v}$ , with k in kg/s. Newton's second law then becomes a differential equation, $m\frac{dv}{dt} = \sum F$ , that you solve by separating variables.
Falling from rest with drag: With down as +, $m\frac{dv}{dt} = mg - kv$ , giving $v(t) = \frac{mg}{k}\left(1 - e^{-kt/m}\right)$ and $a(t) = ge^{-kt/m}$ . The acceleration starts at g (no drag yet) and fades to zero.
Terminal velocity: The speed where drag balances the other forces, so a = 0: $v_T = \frac{mg}{k}$ for a falling object with linear drag. The object approaches it but never quite reaches it. Heavier objects (same k) have a higher terminal speed. If a problem gives a drag of size $bv^2$ instead, the same zero-net-force condition gives $v_T = \sqrt{\frac{mg}{b}}$ .
The time constant: $\tau = \frac{m}{k}$ sets how fast exponentials like $e^{-t/\tau}$ die out. After one τ a falling object reaches about 63% of its terminal velocity, and after about 5τ it's practically there.
Coasting with drag only: Sliding or coasting horizontally with only drag: $v(t) = v_0e^{-kt/m}$ and $x(t) = \frac{mv_0}{k}\left(1 - e^{-kt/m}\right)$ . It never quite stops, but it covers a finite total distance, $\frac{mv_0}{k}$ .
Thrown upward with drag: Going up, gravity and drag both point down, so the acceleration is bigger than g; coming down, drag points up, so it's smaller than g. The trip up takes less time than the trip down, and the object lands slower than it was thrown.
Graph shapes with drag: v–t for a falling object rises steeply at first (slope g) and levels off toward $v_T$ , approaching a horizontal asymptote without crossing it. a–t starts at g and decays toward zero. Sketch the asymptote as a dashed line.

Work, energy and power

Unit 3

Kinetic energy: $K = \frac{1}{2}mv^2$ , a scalar that's never negative and has no direction. Its value depends on the reference frame, since speed does.
Work by a force: $W = \int_a^b\vec{F}\cdot d\vec{r}$ : only the force component along the displacement does work. For a constant force, $W = Fd\cos\theta$ . Work equals the signed area under a graph of that force component against position.
Signs of work: Positive work puts energy into the object, negative work takes it out, and a force perpendicular to the motion does none. So a normal force from a fixed surface, the tension in a swinging pendulum string and the net inward force in uniform circular motion do zero work. Kinetic friction on an object sliding over a fixed surface does negative work.
Work–energy theorem: $W_{\text{net}} = \Delta K$ , where $W_{\text{net}}$ is the work done by all forces on the object. It's often the fastest way to get a speed when a force varies with position, such as $F = -cx^3$ : integrate the force over position, then set it equal to ΔK.
Work by a spring: Stretching a spring from its relaxed length to x takes $\frac{1}{2}kx^2$ of work from you. The spring itself does $W = -\left(\frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2\right)$ , negative while it's being stretched or compressed further and positive as it relaxes.
Conservative forces and potential energy: Gravity and spring forces are conservative: their work doesn't depend on the path, and it's zero around a closed loop. For them, $\Delta U = -W_{\text{cons}} = -\int\vec{F}\cdot d\vec{r}$ and $F_x = -\frac{dU}{dx}$ . Friction and drag are not conservative; their work depends on the path length.
Potential energy formulas: Spring: $U_s = \frac{1}{2}k(\Delta x)^2$ , measured from the relaxed length. Gravity near the surface: $\Delta U_g = mg\Delta y$ , with zero wherever you choose. Gravity between spherical bodies: $U_G = -\frac{Gm_1m_2}{r}$ , zero at infinite separation, so always negative. mgh is the near-surface approximation of the general one. With three or more objects, add up U for every pair.
Reading a U(x) graph: The force is the negative of the slope, so it points downhill on the graph. Minimums are stable equilibrium points and maximums are unstable ones. Draw a horizontal line at the total energy E: the object can only be where E ≥ U, it turns around where E = U, and K = E − U everywhere.
Conservation of energy: $K_i + U_i + W_{\text{ext}} = K_f + U_f + \Delta E_{\text{thermal}}$ . If no outside work is done and nothing dissipates energy inside, mechanical energy K + U stays constant. Sliding friction turns mechanical energy into thermal energy equal to $F_{f,k}d$ , where d is the sliding distance.
Choose the system before you write energy: If Earth (or the spring) is in your system, count its potential energy and not its work. If it's outside, count the work it does and leave out the potential energy. Doing both counts the same energy twice.
Power: Average power is $P = \frac{\Delta E}{\Delta t}$ , and instantaneous power is $P = \frac{dW}{dt} = \vec{F}\cdot\vec{v}$ . Measured in watts, J/s. A motor pulling something at constant speed v against a resisting force F delivers P = Fv. The energy delivered is the area under a P–t graph.

Momentum, impulse and collisions

Unit 4

Momentum: $\vec{p} = m\vec{v}$ , a vector in the direction of motion, in kg·m/s. A system's momentum is the vector sum of its parts' momenta and equals $M\vec{v}_{\text{cm}}$ .
Newton's second law with momentum: $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$ : the net external force is the slope of a p–t graph. This form also handles changing mass.
Impulse: $\vec{J} = \int\vec{F}\,dt = \Delta\vec{p}$ , the area under a force–time graph, in N·s. Average force is $\frac{\Delta p}{\Delta t}$ , so stretching the stopping time (airbags, bending your knees) lowers the force for the same Δp. A ball that bounces back gets a bigger impulse than one that stops.
Conservation of momentum: If the net external force on a system is zero, its total momentum stays constant. In a brief collision or explosion the internal forces are so large that outside forces like gravity or friction barely change p, so you can conserve momentum just before and just after. Conserve each component separately; p can be conserved along x even when it isn't along y.
Kinds of collisions: Momentum is conserved in all of them (with no net external force). Elastic: total kinetic energy is also conserved. Inelastic: some kinetic energy becomes thermal energy, sound or deformation. Perfectly inelastic: the objects stick together, losing the most kinetic energy momentum allows. Explosions increase kinetic energy, using stored internal energy.
Perfectly inelastic collision: $v_f = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$ , with signs on the velocities. In two dimensions do this separately for x and y.
One-dimensional elastic collision: The relative velocity reverses: $v_1 - v_2 = -(v_1' - v_2')$ . With m₂ at rest at first, $v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1$ and $v_2' = \frac{2m_1}{m_1 + m_2}v_1$ . Equal masses swap velocities; a light object bounces off a heavy one at nearly its original speed.
Ballistic pendulum: two steps: Step 1, the collision: conserve momentum only (kinetic energy is lost). Step 2, the swing: conserve mechanical energy only. Together, $v = \frac{m + M}{m}\sqrt{2gh}$ . Never use energy conservation across a sticking collision.
Two-dimensional collisions: Write momentum conservation for x and for y, using components of each velocity. In an elastic glancing collision between equal masses with one at rest, the two move off at 90° to each other.
Changing mass at constant velocity: From $\vec{F} = \frac{d\vec{p}}{dt}$ , keeping a conveyor belt moving at constant v while sand drops onto it at rate $\frac{dm}{dt}$ takes a force $F = v\frac{dm}{dt}$ . The force speeds up each new bit of mass, not the belt.

Rotation: kinematics, torque and rotational inertia

Unit 5

Angular quantities: Angular position θ in radians, $\omega = \frac{d\theta}{dt}$ in rad/s and $\alpha = \frac{d\omega}{dt}$ in rad/s². Graph rules match linear motion: slope of θ–t is ω, slope of ω–t is α, and the area under ω–t is Δθ. On the exam, directions are clockwise or counterclockwise.
Constant angular acceleration: $\omega = \omega_0 + \alpha t$ , $\theta = \theta_0 + \omega_0t + \frac{1}{2}\alpha t^2$ and $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$ , valid only when α is constant. Otherwise integrate, as in linear kinematics.
Linking linear and rotational motion: For a point at distance r from the axis: $s = r\theta$ , $v = r\omega$ , $a_t = r\alpha$ and $a_c = \omega^2r$ . Every point on a rigid object shares the same θ, ω and α, but points farther out move faster.
Torque: $\tau = rF\sin\theta = rF_\perp = r_\perp F$ , where θ is the angle between $\vec{r}$ (from the axis to where the force acts) and $\vec{F}$ , and $r_\perp$ is the lever arm. A force through the axis, or along the line from it, makes no torque. Torque depends on which axis you pick, so state it.
Torque from gravity: Gravity on a whole object acts at its center of mass, so $\tau_g = Mg\,r_\perp$ , where $r_\perp$ is the horizontal distance from the axis to the center of mass. A uniform rod's weight acts at its middle.
Rotational inertia: $I = \sum m_ir_i^2$ for particles and $I = \int r^2\,dm$ for a solid object, where r is the distance from the axis. Mass farther from the axis counts much more, and the same object has different I about different axes.
Rotational inertias to recognize: Point mass: $mr^2$ . Thin hoop or ring about its center: $MR^2$ . Solid disk or cylinder: $\frac{1}{2}MR^2$ . Thin rod about its center: $\frac{1}{12}ML^2$ ; about one end: $\frac{1}{3}ML^2$ . Solid sphere: $\frac{2}{5}MR^2$ . Thin spherical shell: $\frac{2}{3}MR^2$ . These aren't on the equation sheet. For any other shape, use $I = \int r^2\,dm$ , the parallel-axis theorem or the value the question gives.
Deriving I by integration: Rod about one end: $dm = \frac{M}{L}dx$ , so $I = \int_0^L x^2\frac{M}{L}\,dx = \frac{1}{3}ML^2$ . Disk: use thin rings, $dm = \frac{M}{\pi R^2}2\pi r\,dr$ , so $I = \frac{1}{2}MR^2$ . For a nonuniform rod, write dm = λ(x) dx with the given λ; for $\lambda = cx$ about the light end, $I = \frac{1}{2}ML^2$ .
Parallel-axis theorem: $I = I_{\text{cm}} + Md^2$ , where d is the distance between the new axis and a parallel axis through the center of mass. Check: a rod's $\frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2$ . The axis through the center of mass always gives the smallest I.
Newton's second law for rotation: $\alpha = \frac{\tau_{\text{net}}}{I}$ about a fixed axis (or the center of mass). Use the same axis and the same + sense for every torque and for α.
Pulleys with mass: When the pulley has rotational inertia, the tensions on its two sides differ, and that difference times R is what spins it. Write F = ma for each hanging mass and τ = Iα for the pulley, linked by a = Rα. Atwood with pulley: $a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/R^2}$ . One mass unwinding from a wheel: $a = \frac{mg}{m + I/R^2}$ .
Rod released from horizontal: A uniform rod pivoted at one end and let go from horizontal starts with $\alpha = \frac{\tau}{I} = \frac{Mg(L/2)}{\frac{1}{3}ML^2} = \frac{3g}{2L}$ . The free end's initial acceleration is $\frac{3g}{2}$ , more than g. As it swings down, the torque, and so α, shrinks.

Static equilibrium

Unit 5

Two conditions: For an object at rest (or moving and spinning steadily): $\sum\vec{F} = 0$ and $\sum\tau = 0$ . Zero net torque means constant angular velocity, not necessarily no rotation.
Pick the pivot to kill an unknown: In equilibrium the net torque is zero about every axis, so choose one where an unknown force acts (a hinge, a support): that force drops out of the torque equation. Then use the force equations for what's left.
Beams and supports: Put each weight at its own center of mass and each support force where the support touches. For a beam on two supports, the support closer to a heavy load carries more of it.
Hinges, cables and ladders: A hinge can push in any direction, so give it both a horizontal and a vertical component. A cable pulls only along itself, so split its tension into components. A ladder against a frictionless wall needs friction at the floor to balance the wall's push, and it slips when that friction would have to exceed $\mu_sF_N$ .

Rotational energy, angular momentum and rolling

Unit 6

Rotational kinetic energy: $K = \frac{1}{2}I\omega^2$ . An object that moves and spins has $K = \frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$ . A wheel spinning in place has kinetic energy even though its center of mass isn't moving.
Work and power from torque: $W = \int\tau\,d\theta$ , or $\tau\Delta\theta$ for a constant torque: the area under a τ–θ graph. The net work by all torques equals $\Delta K_{\text{rot}}$ , and power is $P = \tau\omega$ .
Angular momentum: For a rigid object about a fixed axis, $L = I\omega$ , in kg·m²/s. For a particle, $\vec{L} = \vec{r}\times\vec{p}$ , with size $mvr_\perp$ , where $r_\perp$ is the perpendicular distance from the axis to the line it moves along. A particle moving in a straight line at constant velocity has constant angular momentum about any point.
Torque changes angular momentum: $\tau_{\text{net}} = \frac{dL}{dt}$ , and the angular impulse $\int\tau\,dt$ , the area under a τ–t graph, equals ΔL.
Conservation of angular momentum: With no net external torque about an axis, L about that axis stays constant: $I_i\omega_i = I_f\omega_f$ . A skater pulling in her arms lowers I and spins faster. Her kinetic energy, $\frac{L^2}{2I}$ , goes up because her muscles do work, so energy isn't conserved even though L is.
Something sticks to a pivoted object: When clay hits a rod on a pivot, conserve angular momentum about the pivot: $mvr_\perp = I_{\text{total}}\omega$ , with I including the clay. Linear momentum isn't conserved, because the pivot pushes on the rod. Kinetic energy is lost in the sticking, but you can use energy conservation for the swing afterward.
Rolling without slipping: $v_{\text{cm}} = R\omega$ and $a_{\text{cm}} = R\alpha$ . The contact point is momentarily at rest, and the top of the wheel moves at $2v_{\text{cm}}$ . Static friction acts at the contact point but does no work, so mechanical energy is conserved.
Rolling down an incline: Writing $I = \beta MR^2$ : $a = \frac{g\sin\theta}{1 + \beta}$ and the speed after dropping height h is $v = \sqrt{\frac{2gh}{1 + \beta}}$ . Mass and radius cancel; only the shape matters. Race order: solid sphere (β = 2/5), then solid disk (1/2), then hoop (1). Anything sliding without friction beats them all.
Friction on a rolling object: Going down a slope, static friction points up the slope: it slows the center's motion and supplies the torque that spins the object up. Rolling without slipping needs $\mu_s \ge \frac{\beta\tan\theta}{1 + \beta}$ . A frictionless ramp can't change an object's spin at all.
Rolling with slipping: When the object slides, kinetic friction ( $\mu_kF_N$ ) acts and $v_{\text{cm}} \ne R\omega$ . It takes mechanical energy out. A solid sphere thrown sliding with no spin slows down while spinning up until $v = R\omega$ ; it then rolls at $\frac{5}{7}v_0$ , which you can find by conserving angular momentum about a point on the floor.
Orbit energy: $U = -\frac{GMm}{r}$ , zero at infinity. In a circular orbit $K = \frac{GMm}{2r}$ and the total is $E = -\frac{GMm}{2r}$ : negative, which means bound. Moving to a higher orbit raises E (it takes work) even though the satellite ends up slower.
Escape speed: The launch speed that makes total mechanical energy zero: $v_{\text{esc}} = \sqrt{\frac{2GM}{r}}$ , which is $\sqrt{2}$ times the circular-orbit speed at that radius. It doesn't depend on the launched object's mass or the launch direction (ignoring air).
Elliptical orbits: Both total energy and angular momentum about the central body are conserved. At the closest and farthest points the velocity is perpendicular to r, so $mv_1r_1 = mv_2r_2$ : the satellite is fastest when closest. Gravity exerts no torque about the central body because it points straight toward it.

Oscillations and simple harmonic motion

Unit 7

When motion is simple harmonic: SHM needs a restoring force (or torque) proportional to displacement from equilibrium and opposite to it: $F = -kx$ or $\tau = -\kappa\theta$ . The equilibrium position is where the net force is zero.
The SHM test: Use Newton's second law to get $\frac{d^2x}{dt^2} = -(\text{constant})\,x$ . If you can, it's SHM and $\omega = \sqrt{\text{constant}}$ , so $T = \frac{2\pi}{\omega}$ . This is how you derive the period of any new oscillator.
Position, velocity, acceleration: $x(t) = A\cos(\omega t + \phi)$ , $v(t) = -A\omega\sin(\omega t + \phi)$ and $a(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2x$ . A is the amplitude, written $x_{\max}$ on the equation sheet. Released from rest at x = +A, φ = 0. Starting at equilibrium and moving in the + direction, use $x = A\sin\omega t$ .
Period, frequency and angular frequency: T is the time for one cycle (s), $f = \frac{1}{T}$ (Hz) and $\omega = 2\pi f = \frac{2\pi}{T}$ (rad/s). In SHM the period doesn't depend on the amplitude.
Mass on a spring: $T = 2\pi\sqrt{\frac{m}{k}}$ , the same horizontal or vertical. A vertical spring oscillates about its new equilibrium, stretched $\frac{mg}{k}$ , and gravity doesn't change the period. Quadrupling the mass doubles T.
Simple pendulum: $T = 2\pi\sqrt{\frac{\ell}{g}}$ for small angles (a common rule of thumb is under about 15°), where $\sin\theta \approx \theta$ . It doesn't depend on the mass or (for small angles) the amplitude. At large angles the motion isn't SHM and the period is longer.
Physical pendulum: $T = 2\pi\sqrt{\frac{I}{mgd}}$ for small angles, where I is about the pivot and d is the distance from the pivot to the center of mass. A uniform rod swinging from one end: $T = 2\pi\sqrt{\frac{2L}{3g}}$ . A simple pendulum is the case $I = m\ell^2$ , $d = \ell$ .
Torsion pendulum: A twisted wire or rod pushes back with torque $\tau = -\kappa\theta$ , proportional to the twist angle, so $T = 2\pi\sqrt{\frac{I}{\kappa}}$ . Unlike a gravity pendulum, it's SHM even at large angles as long as the wire obeys this rule.
Maximums and where they happen: $v_{\max} = A\omega$ at equilibrium, where the acceleration is zero. $a_{\max} = A\omega^2$ at the turning points (x = ±A), where the velocity is zero. Acceleration always points toward equilibrium.
Energy in SHM: For a spring–object system, $E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{\max}^2$ , constant. Kinetic energy is greatest at equilibrium, potential energy at the turning points, and they're equal at $x = \frac{A}{\sqrt{2}}$ . Speed at any position: $v = \omega\sqrt{A^2 - x^2}$ .
Graphs of SHM: x, v and a are all sine or cosine curves with the same period. v is a quarter cycle out of step with x (v = 0 where x is at ±A), and a is the mirror image of x, $a = -\omega^2x$ . Doubling A doubles $v_{\max}$ and $a_{\max}$ , quadruples E, and leaves T unchanged.
Resonance: The natural frequency is the frequency a system oscillates at on its own after being displaced. Pushing it periodically at that frequency adds energy each cycle, so the amplitude grows. Pushing at other frequencies gives smaller amplitudes.

Graphs, labs and free-response habits

Units 1, 2, 3, 4, 5, 6, 7

Slopes to know: Slope of x–t: v. Slope of v–t: a. Slope of p–t: net force. Slope of L–t: net torque. Slope of E–t: power. Negative slope of U–x: force. Slope of spring force against stretch: k.
Areas to know: Under v–t: Δx. Under a–t: Δv. Under F–x: work. Under F–t: impulse (Δp). Under τ–θ: rotational work. Under τ–t: angular impulse (ΔL). Under P–t: energy transferred. Areas below the axis count as negative.
Linearize before you graph: Rearrange the relationship into y = (slope)x + intercept and graph those quantities. Spring: T² against m, slope $\frac{4\pi^2}{k}$ . Pendulum: T² against ℓ, slope $\frac{4\pi^2}{g}$ . Rolling down a ramp: v² against h, slope $\frac{2g}{1 + \beta}$ . Then solve the slope expression for the quantity you want.
Using a best-fit line: Draw one straight line with the points scattered on both sides, and don't force it through the origin unless the physics says so. Take the slope from two points on the line that are far apart, not from data points, and give it units. A nonzero intercept often points to something left out, like the mass of a spring or a hanger.
Designing an experiment: Name what you change (independent variable), what you measure (dependent variable) and what you keep the same. Say which tool measures each quantity, and take several trials over a wide range. To reduce timing error, time many cycles and divide (10 oscillations, then T = total ÷ 10).
Explaining with physics: Justify a claim by naming the principle (Newton's second law, conservation of energy, momentum or angular momentum) and connecting it to the specific quantities in the problem. 'It's faster because energy is conserved' isn't enough; say what energy goes where.
Sketching graphs and diagrams: Label axes, mark key values (intercepts, maximums, asymptotes) and get the shape right: straight for constant slopes, curving the right way for changing ones, leveling toward an asymptote for exponentials. Draw a free-body diagram first when a part asks for one, then use those same forces in your equations.
Checking that answers agree: Translation Between Representations and Qualitative/Quantitative Translation questions ask whether your equation, graph and words tell the same story. Check that your equation predicts the same trend you argued in words (bigger mass, faster or slower?), that its limiting cases make sense, and that your graph's slope and intercept match it.