Uniform Circular Motion Calculator
Uniform Circular Motion Calculator
Give the radius and any one of speed, period, frequency or angular velocity, and the page returns the other three plus the centripetal acceleration and force. Uniform means the speed is constant, not the velocity — which is why there is an acceleration at all.
Circular motion: one quantity in, the whole circle out
A 200 g ball on the end of a 0.50 m string is whirled in a horizontal circle, completing one turn every 0.80 s. Find its centripetal acceleration and the tension in the string.
One angular velocity, and five expressions for the same acceleration
- ω
- angular velocity, in radians per second. Degrees per second and revolutions per minute are not usable in these formulae until converted.
- v
- speed along the circle, in m/s — constant in magnitude, continuously changing in direction. Sometimes called the linear or tangential speed.
- a
- centripetal acceleration, in m/s², directed towards the centre at every instant.
- F
- the resultant force needed to produce that acceleration, in N. It is not a new kind of force: it is whatever real force happens to be doing the job — tension, friction, gravity, a normal contact force.
- T, f
- period in s and frequency in Hz, with f = 1/T. Divide a revolutions-per-minute figure by 60 to get hertz.
Worked example
A 200 g ball on the end of a 0.50 m string is whirled in a horizontal circle, completing one turn every 0.80 s. Find its centripetal acceleration and the tension in the string.
Convert the period to an angular velocity. ω = 2π/T = 2π/0.80 = 7.854 rad/s. One turn is 2π radians, so this is the step the 2π belongs to.
Speed, if you want it. v = ωr = 7.854 × 0.50 = 3.927 m/s. The ball covers 2πr = 3.142 m per turn, and 3.142/0.80 = 3.927 m/s, which is the same thing said two ways.
Centripetal acceleration. a = ω²r = 7.854² × 0.50 = 61.685 × 0.50 = 30.84 m/s². Or a = v²/r = 15.42/0.50 = 30.84 m/s², identically.
Force. F = ma = 0.200 × 30.84 = 6.17 N. Convert the mass to kilograms first: using 200 instead of 0.200 gives 6169 N, a slip that survives a whole page of working because the number still looks like a force.
Put it in perspective. 30.84 m/s² is 3.14 times standard gravity, so the string is pulling with over three times the ball’s weight. That is why a conker on a string feels so much heavier when it is spinning.
A caution about the horizontal circle. A string genuinely cannot be horizontal while supporting a weight: the real tension also has to balance mg, so the string sags into a cone and the tension exceeds 6.17 N. The 6.17 N is the horizontal component. Questions that say “horizontal circle” usually intend you to ignore this; a conical-pendulum question does not.
The four routes in, and the one-line formula each one gives you
| If the question gives you | Get ω from | Then the acceleration is | Typical question |
|---|---|---|---|
| Speed v | ω = v/r | a = v²/r | A car on a bend at 15 m/s. |
| Period T | ω = 2π/T | a = 4π²r/T² | A satellite with a 90-minute orbit. |
| Frequency f | ω = 2πf | a = 4π²f²r | A centrifuge at 50 Hz. |
| Angular velocity ω | nothing to do | a = ω²r | A wheel at 7.85 rad/s. |
How the acceleration scales, starting from the worked example
| Change | r (m) | v (m/s) | T (s) | a (m/s²) | Factor |
|---|---|---|---|---|---|
| The worked example | 0.50 | 3.927 | 0.800 | 30.84 | — |
| Double the speed, same radius | 0.50 | 7.854 | 0.400 | 123.37 | ×4 |
| Double the radius, same speed | 1.00 | 3.927 | 1.600 | 15.42 | ÷2 |
| Double the radius, same period | 1.00 | 7.854 | 0.800 | 61.69 | ×2 |
| Halve the period, same radius | 0.50 | 7.854 | 0.400 | 123.37 | ×4 |
| Double the radius, same angular velocity | 1.00 | 7.854 | 0.800 | 61.69 | ×2 |
| Halve the radius, same speed | 0.25 | 3.927 | 0.400 | 61.69 | ×2 |
Why constant speed still means accelerating, and what centrifugal force is
The word uniform refers to the speed, not the velocity. An object going round a circle at a steady 3.9 m/s has a velocity whose magnitude never changes and whose direction changes continuously, and velocity is a vector, so it is changing. A changing velocity is an acceleration by definition. Students who answer “no acceleration, the speed is constant” have mixed up speed with velocity, and that is the single idea this topic exists to test.
The direction of that acceleration is towards the centre at every instant — hence centripetal, centre-seeking — and it is therefore always at right angles to the velocity. That perpendicularity is why the speed stays constant: a force at right angles to the motion does no work, changes no kinetic energy, and can only turn the velocity. The moment an acceleration acquires a component along the motion, the speed starts changing and the motion is no longer uniform.
Centripetal force is not a force. It is a requirement. If an object of mass m is to travel a circle of radius r at speed v, the resultant of all the real forces on it must come to mv²/r pointing at the centre. The real force doing that job is different in every problem: string tension for a whirled stone, friction for a car on a flat bend, the normal contact force for a rider at the top of a loop, gravity for a satellite. Writing “centripetal force” in a free-body diagram as though it were a fifth force alongside weight and tension is a mark-losing error, because it double-counts.
And centrifugal force does not exist in the ground frame. The outward push you feel on a roundabout is your own inertia: your body continues in a straight line while the seat turns underneath you, and the seat pushes you inwards. In the ground frame there is only the inward force. In the rotating frame of the roundabout there is an outward term, and it is legitimate there — rotating frames are not inertial, and the centrifugal term is the price of pretending they are. School papers almost always want the ground-frame answer, so the safe habit is to name only real forces and let the resultant be the centripetal requirement.
Radians are compulsory here, not conventional. v = ωr is true only when ω is in radians per second, because the radian is defined as arc length over radius and that definition is what makes the arc length rθ and hence the speed rω. An angular velocity quoted in degrees per second or in revolutions per minute is not usable until converted: multiply revolutions per minute by 2π/60 to get radians per second, so 3000 rpm is 314.16 rad/s. The radian is the coherent SI unit of plane angle with dimension one, which is also why rad/s and s−1 have the same dimensions and the radian can appear and disappear from a unit check without anything being wrong.
The direction of the centripetal acceleration is the part that needs care in an exam, because it is not constant. Its magnitude is constant; its direction rotates with the object, always pointing at the centre. So the acceleration vector changes continuously even though the speed and the radius do not, and a question asking for the change in velocity over a quarter or a half turn is asking you to subtract two vectors of equal length pointing different ways. Over half a turn the velocity reverses, so the change in velocity has magnitude 2v even though the speed never altered — a result that looks wrong until you draw the two arrows head to tail.
Worth a sanity check on any answer: the acceleration must come out as a speed squared divided by a length, so a figure in m/s² should be roughly v²/r in size. A 0.5 m circle at 4 m/s gives about 30 m/s²; a 500 m motorway curve at 30 m/s gives about 1.8 m/s²; a 6 400 km planet turning once a day gives 0.034 m/s². Small circles taken quickly give large accelerations and large circles taken slowly give small ones, and if your answer does not fit that pattern you have almost certainly mixed a radius with a diameter or left a 2π out of the angular velocity.
Where the formula comes from, briefly, because it makes the r in the denominator believable. Over a small time the velocity vector turns through the same angle the object does, so the change in velocity has magnitude v Δθ and points towards the centre. Dividing by the time gives v times the rate of turning, which is vω; substituting ω = v/r turns that into v²/r. So the acceleration is large when the object is fast and when the turn is tight, which is exactly what a passenger feels, and the whole derivation is one similar-triangles argument about two velocity arrows rather than any calculus.
One honest limitation. This page describes uniform circular motion only: a constant speed, a constant radius, and therefore a constant-magnitude acceleration that merely rotates. A ball on a vertical circle does not qualify, because gravity has a component along the motion and the speed varies round the loop; nor does a car accelerating out of a bend, which has a tangential acceleration as well as a radial one and a total acceleration that is the vector sum of the two. Those are the next topic, and the equations here are the radial half of them.
Four standard set-ups, and which real force is doing the centripetal job in each. A car on a flat bend: friction between tyre and road, so the maximum safe speed is √(μgr) and is independent of mass. A car on a banked bend: the horizontal component of the normal contact force, which is why a banked track works even with no friction at all. A rider at the top of a vertical loop: weight plus the normal force, with the minimum speed set by the point where the normal force reaches zero and gravity alone supplies mv²/r. A satellite: gravity alone, which is why its orbital speed depends on the radius and not on its own mass. In every case the method is identical — list the real forces, resolve towards the centre, set the total equal to mv²/r.
Frequently asked questions
If the speed is constant, how can there be an acceleration?
Because acceleration is the rate of change of velocity, and velocity includes direction. Going round a circle, the direction changes every instant, so the velocity changes every instant even though its magnitude does not. The acceleration points towards the centre and is exactly perpendicular to the velocity, which is why it changes the direction without changing the speed. This is the one idea the topic is built on.
Is centripetal force a real force I should draw on a free-body diagram?
No. It is the resultant that the real forces must add up to if the object is to keep going round. Draw the real forces — tension, weight, friction, normal contact — and then set their resultant towards the centre equal to mv²/r. Adding a separate arrow labelled centripetal force counts the same force twice and is a standard way to lose marks.
What about centrifugal force? I can feel it.
What you feel is the seat, the door or the harness pushing you inwards, plus your own inertia trying to carry you straight on. In the ground frame, which is inertial, there is no outward force. In the rotating frame of the vehicle there is a centrifugal term, and it is perfectly valid there — rotating frames are not inertial and the term is what you add to make Newton’s second law work in them. School questions almost always want the ground-frame answer.
Why do I have to use radians for the angular velocity?
Because v = ωr is derived from the definition of the radian as arc length divided by radius. The arc swept in angle θ is rθ only in radians, so the speed is rω only in radians per second. Convert revolutions per minute by multiplying by 2π/60: 3000 rpm is 314.2 rad/s, not 3000 and not 50. Using degrees per second understates ω by a factor of 57.3.
Does a bigger circle mean a bigger or a smaller acceleration?
It depends what you hold constant, and that is why the question is set so often. At fixed speed, a = v²/r falls as the radius grows — a gentle bend needs less grip than a tight one at the same speed. At fixed period, a = 4π²r/T² rises as the radius grows, because the object has further to go in the same time and must move faster. The second table shows both, starting from the same circle.
Why does my horizontal-circle answer disagree with the conical-pendulum formula?
Because a string holding up a weight cannot be truly horizontal. The tension has to supply the inward mv²/r and balance the weight mg, so the string sags into a cone and the real tension is √((mv²/r)² + (mg)²), larger than the horizontal component this page reports. Questions that say “horizontal circle, neglect the sag” want the simpler figure; a conical-pendulum question wants the full one.
What is the change in velocity over half a turn, if the speed never changes?
2v, directed opposite to the original velocity. After half a turn the object is moving the other way at the same speed, so the velocity has gone from +v to −v and the change has magnitude 2v — 7.85 m/s for the worked example. The speed changed by nothing and the velocity changed by twice it, which is the clearest possible demonstration that the two are different quantities. Over a quarter turn the change is v√2, at 45° to both.
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References
- BIPM, The International System of Units (SI), 9th edition (2019). Definitions of the metre and the second, and the convention that the radian is the coherent SI unit of plane angle with dimension one — which is why every trigonometric formula in mechanics is written in radians.
- NIST, CODATA Internationally Recommended Values of the Fundamental Physical Constants — complete listing. The standard acceleration of gravity is tabulated as 9.806 65 m s−2, marked exact rather than carrying an uncertainty.
physics.nist.gov/cuu/Constants/ - OCR, Data, Formulae and Relationships Booklet for AS and A Level GCE Physics A (H156 and H556), 8 pages. Lists the acceleration of free fall as g = 9.81 m s−2 alongside the thirteen other physical constants the papers assume.
- NCERT, Physics: Textbook for Class XI, Part I, Chapter 2 “Motion in a Straight Line”, §2.4 “Kinematic Equations for Uniformly Accelerated Motion”. Derives the three equations writing v0 for the initial velocity and x for the displacement, and takes g as 9.8 m s−2 (10 m s−2 in Example 2.3).