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Science

Centripetal Force Calculator

When an object moves at constant speed along a circular path, a net inward force is required to keep it on that path — the centripetal force. Enter mass m, radius r and any one of linear speed v, angular speed ω or period T, and this tool instantly returns the centripetal force F = m·v²/r = m·ω²·r together with centripetal acceleration a = v²/r, tangential speed, angular speed, period and rotation frequency. The centripetal force is not a new fundamental force — it is supplied by a real interaction (tension, gravity, friction, normal force, …).

Centripetal force F

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Centripetal acceleration a

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Tangential speed v

Calculated

Angular speed ω

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Period T

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Frequency f

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Formula

F = m · v² / r = m · ω² · r | a = v² / r | T = 2π / ω

SI units: m in kg, r in m, v in m/s, ω in rad/s, T in seconds; outputs F in newtons (N), a in m/s², f in hertz (Hz). The centripetal force is the net inward force required to keep an object on a circular path — it is supplied by some real interaction (tension, gravity, friction, normal force, …), not by itself.

Formula

F_c = m · v² / r = m · ω² · r a_c = v² / r = ω² · r v = ω · r, ω = 2π / T, f = 1 / T

Frequently asked

What is the difference between centripetal and centrifugal force?

The centripetal force is the real, inward net force acting on a body in circular motion — it is always supplied by some physical interaction (tension, gravity, friction, normal force, electromagnetic force, …). The centrifugal force is not a real force in an inertial frame; it is a fictitious (pseudo) force that appears in a rotating reference frame so that Newton's second law keeps working there. Example: in a car going round a bend you feel pushed outward, but seen from the road (an inertial frame) you only have inertia carrying you tangentially while the door supplies an inward normal force that bends your trajectory.

Does the centripetal force do any work?

No. Work is W = F · d · cos θ, and the centripetal force always points to the centre — exactly perpendicular to the instantaneous velocity (θ = 90°), so cos θ = 0 and the work it does is identically zero at every instant. That is why the kinetic energy ½mv² (and the speed v) stay constant in uniform circular motion. If the object speeds up along the circle, that work is done by some tangential force (an engine, thruster, electromagnetic accelerator, …) — not by the centripetal force.

What real forces typically supply the centripetal force?

It depends on the situation: (1) planets and satellites in orbit — gravity, F = G·M·m/r²; (2) an electron round a nucleus — electromagnetic (Coulomb) force; (3) a conical pendulum or a ball on a string — string tension; (4) a car going round a bend — static friction between tyre and road (with banking, the horizontal component of the normal force also contributes); (5) a roller-coaster loop — rails supply the normal force, gravity adds in at the top; (6) a centrifuge or washing-machine spin cycle — the normal force from the drum wall; (7) a charged particle in a uniform magnetic field — the Lorentz force q·v × B. This tool only computes the magnitude of the required net inward force; in engineering you then check whether the chosen mechanism (friction limited by μ·N, tension limited by breaking strength, …) can actually supply it.

Worked example: how much friction does a 1,500 kg car need to round a 50 m flat curve at 60 km/h?

Convert first: v = 60 km/h ≈ 16.67 m/s. Required centripetal acceleration is a = v² / r = 16.67² / 50 ≈ 5.56 m/s². For a 1,500 kg car the required centripetal force is F = m·a ≈ 1,500 × 5.56 ≈ 8,333 N. On a flat curve this whole inward force must come from static friction between the tyres and the road. Maximum available friction is μ_s · m·g; dry asphalt has μ_s ≈ 0.7–0.9, giving ≈ 1,500 × 0.7 × 9.81 ≈ 10,300 N — only a slim safety margin. On a wet surface (μ_s ≈ 0.4) the limit drops to ≈ 5,886 N and the car slides out. That is precisely why real-world curves are banked: the horizontal component of the normal force takes part of the centripetal load off the tyres.

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