Physics I: Mechanics › Forces › full formula sheet

Fc = mv²/r = mω²rSay it: “centripetal force equals m v squared over r”

Centripetal force

The inward pull that bends motion into a circle — a job description, not a new force.

Notation on this page: Fc is the center-seeking net force in N, m is mass in kg, v is speed in m/s, r is the circle's radius in m, ω is angular speed in rad/s.

Before this lesson: Newton's second law, Centripetal acceleration

Where it comes from

Newton's first law says objects move in straight lines unless pushed. So a ball whirling on a string is a paradox: it moves in a circle, which means its velocity keeps turning — and turning velocity is acceleration, which demands a force. Cut the string and the paradox resolves: the ball flies off tangent, in a straight line, exactly as the first law always wanted.

Before reading on: the string breaks at some instant. Does the ball fly outward away from the center, continue along the circle, or leave along the tangent line?

The tangent wins — there is no outward fling, only the absence of the inward pull. While the string holds, it supplies a continuous center-seeking force that bends the path. “Centripetal” just means center-seeking: it is a job title, and the string (or friction, or gravity) is the employee. The size of the required pull:

Fc = mv²/r = mω²rthe net inward force that keeps the turn going — a role, not a new forceSay it: “F-c equals m v squared over r, which is m omega squared r”

Huygens worked out the circular-motion geometry in 1659; Newton folded it into the Principia. Note the v²: doubling the speed demands four times the inward force — the single most exam-tested fact on this page.

Derivation

The force half is one line — F = ma with a = v²/r. The real work is the acceleration's geometry: why a turning velocity vector produces an inward acceleration of v²/r.

|Δv|
≈
v · Δθ
Step 1 — the velocity turns. In time Δt the ball sweeps angle Δθ. The velocity vector rotates by the same Δθ; for small angles the chord |Δv| ≈ arc = v·Δθ.
Δθ
=
ωΔt = (v/r)Δt
Step 2 — angle per time. Angular speed ω = v/r, so the swept angle is Δθ = ωΔt.
ac
=
limΔt→0 |Δv|/Δt = v²/r
Step 3 — the acceleration. |Δv|/Δt = v·(v/r) = v²/r. As Δt → 0 the change Δv points inward — so the acceleration is center-seeking.
Fc
=
mac = mv²/r
Step 4 — Newton's second law. Multiply the inward acceleration by the mass: the net force must be mv²/r toward the center. With v = ωr this is mω²r. ∎

About “centrifugal force”: in the inertial frame there is no outward force — only inertia resisting the turn, which feels like an outward fling inside the rotating frame. On an exam, “centrifugal” as a real force is a trap answer.

How to use it

The procedure, every time:

  1. Ask: what provides the inward force? String → tension. Cornering car → static friction. Orbit → gravity. Banked curve → the normal force's horizontal component. Name the employee.
  2. Set it equal to mv²/r. That real force is the centripetal force — e.g. T = mv²/r, or μsmg = mv²/r.
  3. Solve. For max cornering speed: v = √(μsgr). For tension: T = mv²/r directly.
  4. Never add Fc to the free-body diagram. It is not an extra force — it is the name of the net inward force you already drew.
Common mistake: drawing tension plus a separate “centripetal force” arrow — double counting. The FBD shows real forces; “centripetal” is the label on their vector sum.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: 1000 kg car, v = 20 m/s, r = 50 m

  1. Identify. m = 1000 kg, v = 20 m/s, r = 50 m.
  2. Compute. Fc = mv²/r = 1000 × 400/50 = 8000 N toward the center of the turn.
  3. Provider. On a flat road this 8000 N must come from static friction between tires and road.
Common mistake: forgetting to square v (1000 × 20/50 = 400 N — twenty times too small). The v² is the whole character of the formula.
Your turn — 1200 kg car, v = 15 m/s, r = 60 m. Centripetal force?

Answer: 4500 N. Fc = 1200 × 225/60 = 270000/60 = 4500 N.

Before reading on: you whirl the ball twice as fast on the same string. Does the tension double, quadruple, or stay the same?

Example 2 — tension: 2 kg ball, 1.5 m string, v = 3 m/s

  1. Provider. The string's tension is the only horizontal force — so T = Fc.
  2. Compute. T = mv²/r = 2 × 9/1.5 = 12 N.
  3. The v² rule: twice the speed (6 m/s) would need 48 N — quadruple. Strings snap on the square.
Common mistake: “double speed, double tension.” No — v² means quadruple. This is the most expensive misconception on the page.
Your turn — 0.5 kg ball, r = 2 m, v = 4 m/s. Tension?

Answer: 4 N. T = 0.5 × 16/2 = 4 N.

Example 3 — max cornering speed: μs = 0.8, r = 40 m

  1. Provider. Static friction, max value μsmg, supplies the inward force.
  2. Equate. μsmg = mv²/r — mass cancels!
  3. Solve. v = √(μsgr) = √(0.8 × 9.8 × 40) = √313.6 ≈ 17.7 m/s (≈ 64 km/h).
  4. Faster than this: required friction exceeds the max — the tires slide and the car runs wide.
Common mistake: using μk. Cornering tires roll without slipping — the contact patch is instantaneously at rest, so it's static friction (the stronger kind).
Your turn — μs = 0.5, r = 25 m. Max speed?

Answer: about 11.1 m/s. v = √(0.5 × 9.8 × 25) = √122.5 ≈ 11.07 ≈ 11.1 m/s.

Example 4 — the ω form: 0.2 kg, r = 0.8 m, 2 rev/s

  1. Convert. ω = 2 rev/s × 2π rad/rev = 4π ≈ 12.57 rad/s.
  2. Compute. Fc = mω²r = 0.2 × 157.9 × 0.8 ≈ 25.3 N.
  3. Check via v: v = ωr = 10.05 m/s; mv²/r = 0.2 × 101.1/0.8 ≈ 25.3 N ✓ — both forms agree.
Common mistake: plugging revolutions per second straight into mω²r. ω must be in radians per second — multiply rev/s by 2π first.
Your turn — 0.3 kg, r = 0.5 m, 3 rev/s. Centripetal force?

Answer: about 53.3 N. ω = 6π ≈ 18.85 rad/s; Fc = 0.3 × 355.3 × 0.5 ≈ 53.29 ≈ 53.3 N.

Memorization tips

  • Say it aloud: “F-c equals m v squared over r.” Stress the squared — it is the exam's favorite trap.
  • Job title, not a force: “centripetal” describes what a force does. Your FBD lists tension, friction, gravity — never a separate Fc arrow.
  • Inward, always: the force points at the center. The feeling of being flung outward is inertia, not a force.
  • v = ωr bridges the forms: mv²/r = mω²r. Given rev/s, convert to rad/s first (multiply by 2π).
  • Cornering shortcut: vmax = √(μsgr) — mass cancels, so a truck and a bike share the same limit on the same curve.
  • Units check: kg·(m/s)²/m = kg·m/s² = N ✓. If your “force” isn't in newtons, the v² probably went missing.

Final challenge

Five mixed questions — providers, the v² rule, and the traps, all in one. Score 5/5 and centripetal force is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is centripetal force?

The center-seeking net force required for circular motion: Fc = mv²/r = mω²r. “Centripetal” names a role, not a new force — tension, friction, gravity, or the normal force plays the part.

Which way does centripetal force point?

Toward the center of the circle, always. It is what continuously bends the velocity vector inward; without it the object would fly off tangent in a straight line.

What is the difference between centripetal and centrifugal force?

Centripetal is the real inward force (in an inertial frame) causing circular motion. “Centrifugal” is the apparent outward push felt inside a rotating frame — an inertia effect, not a real force.

Why does doubling the speed quadruple the force?

Because Fc = mv²/r has v squared: the velocity vector must turn twice as fast AND each turn bends a faster-moving object, so the required inward force grows quadratically.

What provides the centripetal force for a car turning a corner?

Static friction between the tires and the road (for an unbanked turn). The maximum speed is v = √(μsgr) — exceed it and the tires slide outward.

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