Circular Motion and Centripetal Force
An object moving in a circle at a steady speed is accelerating. That sentence sounds wrong the first time you read it, and it is the reason this chapter goes badly for so many students. It is correct because velocity includes direction, and the direction is changing every instant. A change in velocity is an acceleration — and by Newton's second law, an acceleration requires a force. This article works out how big that acceleration is, what supplies the force in each real situation, and where students go wrong.
The formulas
v = ω r = 2πr / T (linear speed)
ac = v² / r = ω² r = 4π²r / T² (centripetal acceleration, pointing at the centre)
Fc = m ac = m v² / r = m ω² r
| Symbol | Meaning | SI unit |
|---|---|---|
| r | Radius of the circular path | m |
| T | Time for one complete revolution | s |
| f | Revolutions per second | Hz |
| ω | Angle swept per second | rad/s |
| v | Speed along the circle (tangential) | m/s |
| ac | Centripetal acceleration — directed inward | m/s² |
| Fc | Net inward force required | N |
The single most important point
Centripetal force is not a new kind of force. It is a job description, not an entry in a list of forces. Something already present must do the job, and in the exam you should always name it:
| Situation | What actually supplies the inward force |
|---|---|
| Stone whirled on a string | Tension in the string |
| Car turning on a flat road | Friction between the tyres and the road |
| Car on a banked track | The horizontal component of the normal reaction (plus friction) |
| Moon orbiting the Earth | Gravitational attraction |
| Electron in a Bohr orbit | Electrostatic attraction to the nucleus |
Because the inward force is always perpendicular to the motion, θ = 90° in W = Fs cosθ — so a centripetal force does no work at all. That is exactly why the speed can stay constant while the direction keeps changing.
Worked example 1 — a stone on a string
Question: A 0.5 kg stone is whirled on a 1.2 m string, making 2 complete revolutions per second. Find the speed, the centripetal acceleration and the tension in the string.
Angular velocity: f = 2 Hz, so T = 1/f = 0.5 s and ω = 2πf = 2 × 3.14159 × 2 = 12.566 rad/s
Speed: v = ωr = 12.566 × 1.2 = 15.08 m/s
Check: v = 2πr/T = (2 × 3.14159 × 1.2) ÷ 0.5 =
7.5398 ÷ 0.5 = 15.08 m/s. ✔
Centripetal acceleration: a = v²/r = 15.08² ÷ 1.2 =
227.39 ÷ 1.2 = 189.5 m/s²
Check: a = ω²r = 12.566² × 1.2 = 157.91 × 1.2 =
189.5 m/s². ✔
Tension: F = ma = 0.5 × 189.5 = 94.75 N
Notice how large that is — over nineteen times the stone's own weight of 0.5 × 9.8 = 4.9 N. Whirling something fast on a short string demands a surprisingly strong string, which is why the question is worth asking.
Worked example 2 — a car on a flat bend
Question: A 1000 kg car takes a bend of radius 50 m at 15 m/s. Find the centripetal force needed. If the coefficient of friction between tyres and road is 0.6, is the car safe, and what is its maximum safe speed? Take g = 9.8 m/s².
Force needed: Fc = mv²/r = (1000 × 15²) ÷ 50 = (1000 × 225) ÷ 50 = 225,000 ÷ 50 = 4500 N
Force available from friction: fmax = μmg = 0.6 × 1000 × 9.8 = 5880 N
5880 N is more than the 4500 N required, so the car holds the bend.
Maximum safe speed: set mv²/r = μmg. The mass cancels, which is
worth noticing — a loaded lorry and an empty car skid at the same speed on the same
bend.
v² = μgr = 0.6 × 9.8 × 50 = 294
v = √294 = 17.15 m/s, which is 17.15 × 3.6 =
61.7 km/h
Check: at 17.15 m/s the required force is (1000 × 294) ÷ 50 = 5880 N, exactly the friction available. ✔ Above that speed the road simply cannot push hard enough and the car slides outward along the tangent.
Worked example 3 — the top of a vertical loop
Question: A 0.2 kg ball is swung in a vertical circle of radius 0.8 m. (a) What is the minimum speed at the highest point for the string to stay taut? (b) If the ball passes the top at 4 m/s, what is the tension there?
(a) At the top, both the weight and the tension point downward, i.e.
towards the centre:
T + mg = mv²/r
The string goes slack when T = 0, so mg = mv²/r. The mass cancels:
vmin = √(gr) = √(9.8 × 0.8) = √7.84 =
2.8 m/s
Below 2.8 m/s at the top, gravity is more than enough to bend the path and the ball falls inward — the string slackens and the circular motion breaks.
(b) At 4 m/s:
Required inward force = mv²/r = (0.2 × 16) ÷ 0.8 = 3.2 ÷ 0.8 =
4.0 N
Weight already supplies mg = 0.2 × 9.8 = 1.96 N
Tension = 4.0 − 1.96 = 2.04 N
Check the limiting case: put v = 2.8 m/s into the same working. Required force = (0.2 × 7.84) ÷ 0.8 = 1.96 N, which equals the weight exactly, so T = 0. ✔ That is the definition of the minimum speed.
Worked example 4 — the Moon
Question: The Moon orbits the Earth at an average radius of 3.84 × 108 m with a period of about 27.3 days. Find its orbital speed and centripetal acceleration.
Period in seconds: T = 27.3 × 24 × 3600 = 2,358,720 s
Speed: v = 2πr / T = (2 × 3.14159 × 3.84 ×
108) ÷ 2,358,720
= (2.4127 × 109) ÷ (2.3587 × 106) =
1023 m/s
Acceleration: a = v²/r = (1023)² ÷ (3.84 × 108) = (1.0465 × 106) ÷ (3.84 × 108) = 2.72 × 10−3 m/s²
Check by the period formula: a = 4π²r / T² =
(39.478 × 3.84 × 108) ÷ (2.3587 ×
106)²
= (1.5160 × 1010) ÷ (5.5636 × 1012) =
2.72 × 10−3 m/s². ✔
Now compare that with g = 9.8 m/s² at the Earth's surface. The Moon is about 60 Earth-radii away, and 9.8 ÷ 60² = 9.8 ÷ 3600 = 2.7 × 10−3 m/s² — the same number. That inverse-square agreement is the calculation Newton himself used to argue that the force holding the Moon is the very same gravity that makes an apple fall.
Common mistakes that cost marks
- Adding "centripetal force" to a free-body diagram as an extra arrow. It is not extra. Draw the real forces — tension, weight, normal reaction, friction — and set their inward resultant equal to mv²/r.
- Writing centrifugal force in an ordinary solution. The outward "force" you feel in a turning car is not a force in the ground frame; it is your body continuing in a straight line while the car turns. Centrifugal force exists only as a correction inside a rotating frame, and school questions are almost never set in one.
- Saying acceleration is zero because the speed is constant. Speed is constant; velocity is not. This is the definitional trap of the chapter.
- Using degrees for ω. Angular velocity is in radians per second. One revolution is 2π rad, not 360 in these formulas.
- Forgetting to square ω or v. Doubling the speed quadruples the force needed — which is why a bend that is comfortable at 40 km/h is dangerous at 80 km/h.
- Mixing up r and diameter. If a question gives a 1.6 m diameter loop, r is 0.8 m. Using 1.6 halves the answer.
- Treating the top and bottom of a vertical loop the same way. At the top, weight and tension both point inward (T + mg = mv²/r); at the bottom, they oppose (T − mg = mv²/r). The tension at the bottom is much larger.
Where circular motion appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Uniform circular motion, centripetal acceleration, banking of roads, the conical pendulum |
| NEET | Direct substitution into ac = v²/r and F = mv²/r, plus vertical-circle minimum-speed questions |
| JEE Main & Advanced | Banked tracks with friction, non-uniform circular motion, combined circular and energy problems |
| Class 11–12 Chemistry | The Bohr model — the electron's orbit is set by equating electrostatic attraction to mv²/r |
| GATE (engineering papers) | Rotating machinery, balancing and dynamics of rigid bodies |
That chemistry row is worth reading twice. When you derive the Bohr radius, the very first line is "electrostatic force = centripetal force". A student who has not understood this chapter cannot follow that derivation, which is why circular motion quietly decides how well Structure of Atom goes.
The method to use every time
- Draw the object and mark the centre of its circular path.
- Draw only the real forces acting on it — no centripetal arrow.
- Take the direction towards the centre as positive and add up the components along that line.
- Set that sum equal to mv²/r and solve.
Every question in this chapter, from a conical pendulum to a banked curve to a satellite, answers to those four steps.
Check your working in seconds. The suite has no dedicated circular-motion panel, so this one opens the Scientific Calculator that a plain visit already shows. Use it for the squares, the square roots and the 2πr/T divisions above — and always run the second route (ω²r against v²/r) as your check.
Open the ABC Chemistry Calculator Suite →Circular motion is the bridge between Class 11 physics and the Bohr model in Class 11 chemistry, and students who skip it feel the gap later. ABC Chemistry runs Class 11–12 coaching at the Gurugram centre and online classes across India — details at abcchemistry.in. For one-to-one help at home in Delhi, Noida or Gurgaon, see delhihometutor.com.