AP Physics C: Mechanics — Cheatsheet
Formulas, exam-day tips, and key terms on one page.
Formulas & relationships
Velocity as a derivative
v(t) = dx/dt = lim (Δt→0) Δx / Δt
The slope of the position–time graph at an instant. Units: meters per second.
Acceleration as a derivative
a(t) = dv/dt = d²x/dt²
First derivative of velocity; second derivative of position. Units: meters per second².
Building motion back up by integration
v(t) = v₀ + ∫ a dt x(t) = x₀ + ∫ v dt
Each indefinite integral introduces a constant; v₀ and x₀ are those constants, set by the initial conditions.
Constant-a kinematics as integral results
v = v₀ + at x = x₀ + v₀t + ½at²
Valid ONLY when a is constant. If a depends on t, integrate a(t) directly instead.
Component form of the motion vectors
r = x î + y ĵ v = (dx/dt) î + (dy/dt) ĵ a = (dvₓ/dt) î + (dv_y/dt) ĵ
A vector of magnitude V at angle θ resolves as Vₓ = V cos θ and V_y = V sin θ; recombine with |V| = √(Vₓ² + V_y²).
Relative-velocity addition
v(A/C) = v(A/B) + v(B/C)
Read the subscripts like a chain: A-relative-to-C equals A-relative-to-B plus B-relative-to-C. The inner label B cancels. Reversing a pair flips the sign: v(B/A) = −v(A/B).
Slopes down, areas up
v = dx/dt a = dv/dt Δv = ∫a dt Δx = ∫v dt
A slope has units of the vertical axis divided by the horizontal; an area has units of their product. Checking units is the fastest way to confirm you read the right feature.
The identity that removes time
a = dv/dt = (dv/dx)(dx/dt) = v dv/dx
Use this whenever acceleration is given as a function of position and the question does not ask about time. It turns a differential equation in t into one relating v and x directly.
Centripetal acceleration and the kinematics of the circle
a_c = v²/r = ω²r v = ωr ω = 2π/T = 2πf
a_c always points toward the center, perpendicular to the velocity. In uniform circular motion the speed is constant, so there is no tangential component of acceleration.
Newton’s second law
ΣF = ma = m dv/dt
The net (vector) force equals mass times acceleration. Applied one axis at a time: ΣFₓ = maₓ and ΣF_y = ma_y.
Friction force
f_k = μ_k N f_s ≤ μ_s N
Kinetic friction has fixed magnitude μ_k N once sliding. Static friction adjusts up to a maximum μ_s N to prevent sliding; it equals whatever is needed below that cap.
Falling object with linear drag
m dv/dt = mg − bv
Down is positive. Weight mg pulls down; drag bv pushes up and grows as v grows. This first-order differential equation governs the whole descent.
Terminal velocity
v_t = mg / b (linear drag) v_t = √(mg / c) (quadratic drag)
Found by setting dv/dt = 0 so the resistive force equals the weight. No calculus needed for v_t itself — only the balance condition.
Atwood machine
a = (m₂ − m₁)g / (m₁ + m₂) T = 2 m₁ m₂ g / (m₁ + m₂)
Two masses hang over an ideal pulley. The heavier mass m₂ falls, the lighter m₁ rises, both with the same |a|. The tension is the same throughout the single string.
Universal gravitation
F = G m₁ m₂ / r²
Attractive, directed along the line between the masses. G = 6.67 × 10⁻¹¹ N·m²/kg². Doubling either mass doubles F; doubling r cuts F to one quarter.
Surface gravity
g = GM / R²
The free-fall acceleration at a spherical planet of mass M and radius R. At a distance r > R from the center, the local acceleration is GM/r².
The two friction laws
f_s ≤ μ_s N f_k = μ_k N with generally μ_s > μ_k
Both are empirical approximations, not fundamental laws. Neither depends on contact area, and the normal force N is rarely equal to mg except on a level surface with no vertical applied force.
Three standard results
banked curve: tan θ = v²/(rg) top of vertical loop: v_min = √(gr) conical pendulum: tan θ = v²/(rg)
The banked curve and the conical pendulum give the same relation because in both cases a tilted force (normal or tension) is resolved into an inward horizontal component and a vertical component balancing gravity.
Effective spring constants
parallel: k_eff = k₁ + k₂ series: k_eff = k₁k₂/(k₁ + k₂)
The series result is always smaller than the smaller of the two constants. If you compute a series k_eff larger than either spring, you have used the wrong formula.
Apparent weight in a vertically accelerating frame
N = m(g + a) for upward acceleration a N = m(g − a) for downward acceleration a
Both follow from ΣF = ma with up positive: N − mg = ma. In free fall a = g and N = 0, which is what apparent weightlessness means.
Work as a line integral
W = ∫ F·dx (constant force: W = Fd cos θ)
The dot product keeps only the force component along the displacement. Units: joules (1 J = 1 N·m). Area under an F-versus-x graph.
Work-energy theorem
W_net = ΔKE = ½mv_f² − ½mv_i²
The net work on an object equals its change in kinetic energy. Positive net work speeds it up; negative net work slows it down.
Potential energy functions
U_grav = mgh U_spring = ½kx²
Each is minus the work done by the conservative force. Potential energy is always measured relative to a chosen reference where U = 0.
Force from potential energy
F = −dU/dx
The force is the negative slope of the potential energy curve. Equilibrium occurs where dU/dx = 0; it is stable at a minimum of U, unstable at a maximum.
Conservation of mechanical energy
KE_i + U_i = KE_f + U_f
Valid when only conservative forces (gravity, springs) do work. Pick a reference level for U and apply it consistently at both instants.
Energy with a nonconservative force
KE_i + U_i + W_nc = KE_f + U_f
W_nc is the (usually negative) work of nonconservative forces. For friction over a distance d, W_nc = −f·d, draining mechanical energy into heat.
Average and instantaneous power
P_avg = W / Δt P_inst = dW/dt = F·v
Instantaneous power is the force dotted with the velocity, so a force perpendicular to the motion delivers zero power. Units: watts (W).
Work in general
W = ∫ F · dx (one dimension: W = ∫ F(x) dx from x₁ to x₂)
The dot product means only the component of force along the displacement does work. A force perpendicular to the motion — the normal force, or tension in a conical pendulum — does exactly zero work no matter how large it is.
The force–energy relationship and stability test
F = −dU/dx equilibrium: dU/dx = 0 stable if d²U/dx² > 0, unstable if d²U/dx² < 0
Stable equilibrium sits at a minimum of U — a valley. Unstable equilibrium sits at a maximum — a hilltop. Neutral equilibrium is a flat region where U is constant.
Gravitational potential energy and escape speed
U(r) = −GMm/r v_esc = √(2GM/R) E_circular orbit = −GMm/(2r)
U is negative because zero is defined at infinite separation and the force is attractive — bringing masses together releases energy, driving U below zero.
The generalized work–energy statement
K₁ + U₁ + W_applied = K₂ + U₂ + |ΔE_thermal| with ΔE_thermal = f_k d for friction
The distance d in f_k d is the actual path length traveled, not the displacement. That distinction matters whenever an object slides back and forth.
Momentum and Newton's second law
p = mv ΣF = dp/dt
Momentum is a vector (kg·m/s). The net force is the time rate of change of momentum; when m is constant this reduces to ΣF = ma.
Impulse-momentum theorem
J = ∫ F dt = Δp = m v_f − m v_i
Impulse (N·s) equals the change in momentum. Graphically it is the area under a force-versus-time curve; an average force gives J = F_avg Δt.
Conservation of momentum
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Total momentum before equals total momentum after. It is a vector equation — apply it component by component, keeping track of signs.
Perfectly inelastic collision
m₁v₁ + m₂v₂ = (m₁ + m₂) v_f
The objects stick and share one final velocity v_f. Momentum is conserved; kinetic energy is not.
Center of mass
x_cm = (Σ mᵢ xᵢ) / (Σ mᵢ) x_cm = (1/M) ∫ x dm
Discrete on the left, continuous on the right. For a continuous body, express dm through the density (dm = λ dx for a rod) and integrate.
Thrust and the equation of motion
Thrust = v_ex |dm/dt| m dv/dt = −v_ex dm/dt
v_ex is the exhaust speed relative to the rocket; dm/dt < 0 as fuel is expelled, so the thrust is a forward force of magnitude v_ex times the burn rate.
Ideal rocket equation
Δv = v_ex ln(m_i / m_f)
The speed gained depends on the exhaust speed and the natural log of the initial-to-final mass ratio. Valid with no external forces (deep space).
The impulse–momentum theorem
J = ∫F dt = Δp = mΔv F_avg = J/Δt
Impulse is a vector with the direction of the net force. The average force is defined so that F_avg Δt gives the same area as the true varying force — it is a derived quantity, not a measured one.
Two-dimensional momentum conservation
Σm v_x (before) = Σm v_x (after) Σm v_y (before) = Σm v_y (after)
Choose axes to make the algebra easy — usually along the initial velocity of one object. A well-chosen axis often makes one of the two initial components zero.
Center of mass position and velocity
x_cm = Σm_i x_i / M v_cm = Σm_i v_i / M = p_total/M
The second form is the useful one: the center-of-mass velocity is just the total momentum divided by the total mass, so a system with zero total momentum has a stationary center of mass.
Angular velocity and acceleration
ω = dθ/dt α = dω/dt = d²θ/dt²
Angular position θ in radians, ω in rad/s, α in rad/s². The direct analogs of v = dx/dt and a = dv/dt.
Linking linear and angular motion
v = rω a_t = rα a_c = ω²r = v²/r
Multiply an angular quantity by the radius to get the corresponding linear one at that point. The centripetal term a_c points toward the center.
Torque
τ = rF sin θ = r_⊥ F
r is the distance from the axis to the point of application; sin θ picks out the perpendicular component. Maximum when the force is applied perpendicular to the radius.
Moment of inertia
I = ∫ r² dm (hoop: MR²; disk: ½MR²; rod about center: 1/12 ML²; sphere: 2/5 MR²)
r is the distance of each mass element from the rotation axis. The same object has different I about different axes.
Newton's second law for rotation
τ_net = Iα
The net torque about an axis equals the moment of inertia about that axis times the angular acceleration. Solve for α = τ_net/I.
Rolling-without-slipping constraint
v_cm = Rω a_cm = Rα
The center-of-mass motion and the rotation are locked together by the wheel radius R. If the wheel slips, this constraint no longer holds.
Acceleration rolling down an incline
a_cm = g sin θ / (1 + I/MR²)
Only the shape factor I/MR² matters. Smaller I/MR² (mass nearer the axis) rolls faster; the plain block with no rotation (ratio 0) accelerates fastest of all at g sin θ.
Moment of inertia, discrete and continuous
I = Σ m_i r_i² I = ∫ r² dm with dm = λ dx, σ dA or ρ dV
The r in the integral is the perpendicular distance from the axis to the mass element, not the distance from the origin. Choosing dm correctly is the whole technique.
The conditions of static equilibrium
ΣF_x = 0 ΣF_y = 0 Στ = 0 about any point
If the net force is zero, the net torque is the same about every point. That freedom is what lets you choose the pivot strategically.
Torque as a vector product
τ = r × F |τ| = rF sin θ = r_⊥ F = r F_⊥
The three magnitude forms are equivalent: use the full angle, or the perpendicular distance from the pivot to the line of action (the moment arm), or the component of force perpendicular to r. Pick whichever the diagram makes easiest.
Rotational and total kinetic energy
KE_rot = ½Iω² KE_total = ½mv_cm² + ½Iω²
The rotational twin of ½mv². For rolling, add the translational and rotational parts, linked by v_cm = Rω.
Rotational work and power
W = ∫ τ dθ P = τω
Torque integrated over angle gives work; torque times angular velocity gives power. Both mirror their translational forms exactly.
Angular momentum
L = Iω (particle: L = rmv sin θ)
Rigid body on the left, single particle on the right. Units: kg·m²/s. A vector directed along the rotation axis.
Torque as the rate of change of angular momentum
τ_net = dL/dt
The rotational twin of F = dp/dt. When τ_net = 0, L is conserved.
Conservation of angular momentum
L = Iω = constant I₁ω₁ = I₂ω₂
Holds when the net external torque is zero. Reducing I (pulling mass inward) speeds up the spin; increasing I slows it.
Circular orbit
GMm/r² = mv²/r → v = √(GM/r) T² = (4π²/GM) r³
Orbital speed falls as 1/√r. The period relation T² ∝ r³ is Kepler's third law, obtained by combining v = √(GM/r) with T = 2πr/v.
Angular momentum at the apses
L = mvr = constant → v_p r_p = v_a r_a
At perihelion and aphelion the velocity is perpendicular to the radius, so L = mvr directly. The closer approach forces the faster speed.
Angular momentum and its rate of change
L = r × p |L| = rp sin θ = mvb Στ = dL/dt
For a rigid body rotating about a fixed axis this reduces to the familiar L = Iω. The point-particle form is the more general one and the two must agree wherever both apply.
Rolling without slipping
v = ωR Mgh = ½Mv² + ½Iω² a = g sin θ / (1 + I/MR²)
The rolling constraint v = ωR is what couples the two motions and lets one equation determine both. It holds only while the object rolls without slipping.
The three laws in Newtonian form
ellipse with sun at a focus dA/dt = L/2m = constant T² = (4π²/GM)a³
In the third law a is the semi-major axis of the ellipse, which reduces to the radius for a circular orbit. The constant depends only on the mass of the central body, so it is the same for every planet in a system.
The SHM equation and its solution
d²x/dt² = −ω²x x(t) = A cos(ωt + φ)
A is the amplitude, ω the angular frequency, φ the phase constant set by initial conditions. The period is T = 2π/ω.
Period, frequency, and angular frequency
T = 2π/ω f = 1/T ω = 2πf
ω is in rad/s, f in hertz, T in seconds. Together they describe how fast the oscillation repeats, independent of amplitude.
Mass-spring oscillator
ω = √(k/m) T = 2π√(m/k) f = 1/T
Angular frequency rises with stiffness and falls with mass. The period depends only on m and k — not on how far the spring is pulled.
Simple pendulum
ω = √(g/L) T = 2π√(L/g)
Valid in the small-angle limit sin θ ≈ θ. Remarkably, the period depends on length and g only — not on the bob's mass or (for small swings) the amplitude.
Physical pendulum
T = 2π√(I / (mgd))
I is the moment of inertia about the pivot; d is the pivot-to-center-of-mass distance. Reduces to the simple-pendulum result when the mass is concentrated at distance L.
Energy in SHM
E = ½kA² = ½kx² + ½mv² v_max = Aω = A√(k/m)
Total energy is fixed by the amplitude. The speed is greatest at equilibrium (v_max = Aω) and zero at the turning points.
Physical, simple and torsional pendulums
T = 2π√(I/Mgd) simple: T = 2π√(L/g) torsional: T = 2π√(I/κ)
d is the distance from the pivot to the center of mass. The simple pendulum is the special case I = ML² and d = L, which reduces the first formula to the second.
Damped oscillation
x(t) = A₀e^(−bt/2m) cos(ω′t + φ) ω′ = √(ω₀² − (b/2m)²) E ∝ A² so E decays as e^(−bt/m)
Because energy goes as amplitude squared, it decays at twice the rate of the amplitude. The damped frequency ω′ is always below the natural frequency ω₀ and falls to zero at critical damping.
On the exam
- On the AP exam, given x(t) you should be able to produce v(t) and a(t) instantly by differentiating, and read the reverse too: where the x–t slope is zero the object is momentarily at rest, and where the v–t slope is zero the acceleration is zero.
- Keep the subscript bookkeeping strict: write every velocity as v(object/frame). The chain v(A/C) = v(A/B) + v(B/C) only works when the inner labels match, and swapping any pair introduces a sign flip. Getting the subscripts right turns most relative-motion questions into simple addition.
- Read the question for **displacement** versus **distance** before computing anything. If any part of the velocity graph lies below the axis the two answers differ, and the difference is usually the entire point of the item.
- On the AP exam, draw the free-body diagram first and tilt your axes to match the motion. Write ΣF = ma along each axis separately. Getting the components of weight right — mg sin θ along a slope, mg cos θ into it — is where most points are won or lost.
- Keep the two formulas straight: F = Gm₁m₂/r² is the force between two masses, while g = GM/R² is the field (acceleration) one mass produces at distance R. Both are inverse-square — whenever a distance changes, square the ratio before scaling.
- N = mg only on a horizontal surface with no vertical component of applied force. On an incline N = mg cos θ; with a push at an angle, N changes; in an accelerating elevator, N changes. Deriving N from the perpendicular equilibrium condition every time costs ten seconds and prevents a whole family of errors.
- On any apparent-weight question, write ΣF = ma with a stated sign convention before substituting numbers. Deciding "heavier or lighter" by intuition works about as often as it fails; the equation N = m(g + a) with a signed correctly never does.
- Move fluently in both directions: integrate a conservative force to get U (U = −∫F dx), and differentiate U to get the force (F = −dU/dx). On a U-versus-x graph, remember the force points downhill and equilibria sit where the slope is zero.
- Use P = W/Δt when you know a total amount of work over an interval, and P = Fv (or F·v) when you want the power at a specific instant or speed. A force perpendicular to the velocity — like the centripetal force on a satellite — does zero work and delivers zero power.
- A force perpendicular to displacement does zero work. That single fact disposes of the normal force on a level surface, the tension in a conical pendulum, the magnetic force on a charge, and the centripetal force in any circular orbit — all common exam targets.
- Free-response energy questions are graded on the **accounting**, not the arithmetic. Write an explicit initial-equals-final statement naming every term before substituting numbers. A correct number arrived at without a stated energy equation routinely loses the setup points; a stated equation with a small numerical slip usually keeps most of them.
- For a continuous body, the recipe is always the same: write dm using the density, integrate x dm for the numerator and dm for the total mass M, then divide. Never average the endpoints — that only works for a uniform object.
- The rocket equation comes from momentum conservation with changing mass, not from F = ma with constant mass. Watch the logarithm: the payoff for carrying more fuel diminishes, since Δv grows only as ln(m_i/m_f).
- Set up a small table with columns for x and y and rows for before and after. Filling it in before doing any algebra prevents the dominant error on these problems, which is not conceptual but bookkeeping — a dropped component or a sign flipped on one object.
- 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.
- Keep L = Iω and τ = dL/dt distinct: one is the angular momentum, the other is the rule for how torque changes it. The single most useful consequence is that zero net external torque forces angular momentum to stay constant.
- For circular orbits set gravity equal to the centripetal force (GMm/r² = mv²/r). For elliptical orbits, use conservation of angular momentum (v_p r_p = v_a r_a) at the apses and conservation of energy between them — energy and angular momentum together pin down the whole orbit.
- When a rolling problem appears, immediately identify k = I/MR² and reach for a = g sin θ/(1 + k). Deriving it from scratch each time is a valid but slow route, and the derivation is worth doing once carefully so the formula can be used with confidence thereafter.
- State the small-angle approximation explicitly on free-response: sin θ ≈ θ is what turns d²θ/dt² = −(g/L)sin θ into SHM. Remember the simple-pendulum period is independent of mass, and use the physical-pendulum formula T = 2π√(I/mgd) whenever the swinging object is an extended body.
- Resonance is defined by the **driving** frequency matching the natural frequency, not by any property of the driving force's strength. A weak force applied at resonance can produce a larger response than a strong force applied far from it — which is the whole reason resonance is worth a name.
How to get a 5
- Always define your coordinate system and direction of positive rotation. Consistency between linear (a) and rotational (α) signs is crucial in rolling problems.
- Don't memorize specific derived formulas; practice deriving them from Newton's Laws and Energy Conservation. You will need to show the steps on the FRQ.
- In variable mass or density problems, always relate dm to dx (or dr) using density (λ, σ, ρ). e.g., dm = λ dx.
- Check your boundary conditions and limits of integration carefully when doing work or center of mass integrals.
- Check first whether a quantity is constant. If a, F, or λ depends on t, x, or r, then the constant-acceleration formulas and W = Fd are off the table — write the derivative or the integral instead, and say explicitly what variable you are integrating over.
- Show every integral with its limits and its dm or dt element before evaluating. Rubrics award points for a correct integrand and correct limits even when the arithmetic then goes wrong, while a bare numerical answer earns nothing.
- Turn “derive” prompts into named steps: state the principle (Newton’s second law, conservation of angular momentum), write it symbolically for this system, apply the constraint (a = αR, v = ωR, inextensible string), then solve for the requested symbol, leaving the answer in the given variables.
- Check limits and units on every symbolic result. For a = mg/(m + M/2), confirm a → g as M → 0 and a → 0 as M → ∞. A limit that misbehaves reveals an algebra error faster than re-reading each line.
- Graph-sketching questions want features, not artistry: correct intercept, correct initial slope, correct curvature, and a labeled asymptote. For exponential approaches, mark the value at one time constant (63% of the total change) and show the curve flattening rather than touching the asymptote.
Key terms
Kinematic Equations (Calculus) — v(t) = ∫a(t)dt, x(t) = ∫v(t)dt, a = dv/dt, v = dx/dt. Also a = v(dv/dx).
Work (Variable Force) — W = ∫ F(x) dx. The area under the Force vs. Position graph.
Potential Energy from Force — F = -dU/dx. Force is the negative gradient (slope) of potential energy.
Center of Mass (Continuous) — x_cm = (1/M) ∫ x dm. Use a linear mass density λ = dm/dx to integrate.
Moment of Inertia (Continuous) — I = ∫ r² dm. Common results: Solid disk = ½MR², Hoop = MR², Solid sphere = (2/5)MR², Rod (center) = (1/12)ML².
Parallel axis theorem — I = I_cm + Md². Moving the axis a distance d from the center of mass always increases the moment of inertia.
Torque (Cross Product) — τ = r × F. Magnitude is rF sinθ. Direction given by right-hand rule.
Angular Momentum (Particle) — L = r × p. Magnitude is mvr sinθ (or mv_perp r).
Conservation of angular momentum — Στ_ext = dL/dt, so L is conserved when no external torque acts. A collapsing star spins faster for exactly this reason.
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).
Simple Harmonic Motion Equation — d²x/dt² = -ω²x. The general solution is x(t) = A cos(ωt + φ).
Newton's Law of Universal Gravitation — F_g = G(m₁m₂)/r². Potential energy U_g = -G(m₁m₂)/r.
Calculus definitions of the kinematic quantities — v = dx/dt and a = dv/dt = d²x/dt². In reverse, v = v₀ + ∫a dt and x = x₀ + ∫v dt, where the constants of integration are the initial conditions.
When may the constant-acceleration formulas be used? — Only when a is constant. If a depends on t, x, or v, you must integrate or separate variables instead.
Acceleration as a function of position — Use a = v dv/dx (from the chain rule) to get v(x) without ever finding t: ∫a dx = ½v² + C.
Displacement vs. distance by integration — Displacement = ∫v dt; distance = ∫|v| dt. Split the integral wherever v changes sign.
Linear drag equation of motion — m dv/dt = mg − bv gives v(t) = v_t(1 − e^(−t/τ)) with v_t = mg/b and τ = m/b. After one τ the object has 63% of terminal speed.
Quadratic drag — m dv/dt = mg − cv² gives terminal speed v_t = √(mg/c). The power of v in the drag law changes both the terminal speed and the shape of the approach.
Work done by a variable force — W = ∫F·dx, the area under an F-versus-x graph. W = Fd applies only when F is constant and parallel to the displacement.
Force from a potential energy function — F_x = −dU/dx, so the force points downhill on the U curve. In three dimensions F = −∇U.
Classifying equilibrium — Equilibrium wherever dU/dx = 0. Stable if d²U/dx² > 0 (a minimum), unstable if d²U/dx² < 0 (a maximum), neutral if it is zero.
Small oscillations about a potential minimum — k_eff = d²U/dx² evaluated at the minimum, so ω = √(k_eff/m). Any smooth minimum behaves locally as a harmonic oscillator.
Instantaneous power — P = dW/dt = F·v. Average power is total work divided by total time, and the two differ whenever force or speed varies.
Center of mass by integration — x_cm = (1/M)∫x dm. Express dm using linear density λ dx for a rod, then integrate over its length.