Centripetal Acceleration and Centripetal Force
An object moving in a circle at constant speed accelerates toward the center of the circle with centripetal acceleration a = v2 ÷ r = ω2r, and the net force that causes this acceleration, the centripetal force, is F = mv2 ÷ r = mω2r, directed toward the center. In brief: centripetal force is not a new kind of force: it is the name for the net inward force, supplied by tension, friction, gravity, a normal force or a magnetic force. If the force disappears, the object moves off along a tangent in a straight line. The “centrifugal force” is a fictitious force felt in a rotating frame.
A car turning a corner, a satellite in orbit and clothes in a spinning washing machine all need an inward force. This guide derives the centripetal acceleration, explains what provides the force in different situations, compares examples in a table, works through problems on a curve and a vertical circle, and corrects common mistakes.
Why is there an acceleration?
Acceleration is the rate of change of velocity, a vector. In uniform circular motion, the speed v is constant, but the direction of the velocity changes continuously. Over a short time Δt, the object turns through a small angle Δθ = vΔt ÷ r, and the velocity vector changes by Δv ≈ vΔθ, pointing toward the center. Dividing by Δt gives the magnitude of the acceleration a = v2 ÷ r. With v = ωr, it is also a = ω2r = 4π2r ÷ T2. The acceleration is always perpendicular to the velocity, so it changes the direction but not the speed. A larger speed or a smaller radius (a tighter curve) means more acceleration.
What provides the centripetal force?
By Newton’s second law, a net force Fc = ma = mv2 ÷ r toward the center is required. It is not an additional force to be drawn in a free-body diagram: it is the sum of the real forces (or their components) pointing toward the center. Examples:
- Ball on a string: the tension.
- Car on a flat curve: static friction between the tires and the road. The maximum speed is v = √(μgr).
- Banked curve: the horizontal component of the normal force (and friction).
- Planet or satellite: gravity.
- Electron in a magnetic field: the magnetic force qvB.
- Passenger in a loop-the-loop: the sum of the normal force and the weight components.
- Clothes in a spin dryer: the normal force from the drum wall (the water escapes through the holes because there is no force to keep it moving in the circle).
What about “centrifugal force”?
When a car turns left, passengers seem to be thrown outward to the right. In the frame of the ground (an inertial frame), there is no outward force: the passengers tend to continue in a straight line (Newton’s first law) while the car turns under them. In the rotating frame of the car, it appears as if an outward force acts; this “centrifugal force” is a fictitious (inertial) force which arises from using an accelerating frame. In an inertial frame, it should not be included in the force diagram.
Sources of centripetal force compared
| Situation | Force providing the centripetal force |
|---|---|
| Stone whirled on a string | Tension in the string |
| Car on a flat bend | Friction between tires and road |
| Moon orbiting Earth | Gravitational attraction |
| Charged particle in a magnetic field | Magnetic force |
| Rider on a vertical loop | Normal force plus the component of weight toward the center |
Worked example: a curve and a vertical circle
- Centripetal acceleration and force. A 1,200 kg car takes a curve of radius 50 m at 20 m/s. a = v2 ÷ r = 400 ÷ 50 = 8.0 m/s2. F = ma = 1,200 × 8.0 = 9,600 N, supplied by friction.
- Maximum speed. With a coefficient of static friction of 0.70, the maximum friction force is μmg. Setting μmg = mv2 ÷ r gives v = √(μgr) = √(0.70 × 9.8 × 50) = √343 = 18.5 m/s (67 km/h). At 20 m/s the car skids. If the road is wet (μ = 0.30): v = √(0.30 × 9.8 × 50) = 12.1 m/s.
- Ball in a vertical circle. A 0.50 kg ball on a string of length 1.0 m is swung in a vertical circle with a speed of 6.0 m/s (assume constant for simplicity). Required centripetal force: mv2 ÷ r = 0.50 × 36 ÷ 1.0 = 18 N. At the bottom, T − mg = 18, so T = 18 + 4.9 = 22.9 N. At the top, T + mg = 18, so T = 18 − 4.9 = 13.1 N. (In reality the speed is less at the top; the speed must be at least √(gr) = 3.1 m/s there for the string to stay taut.)
- Orbit. The Moon (r = 3.84 × 108 m, T = 27.3 days = 2.36 × 106 s) has a centripetal acceleration a = 4π2r ÷ T2 = 39.5 × 3.84 × 108 ÷ 5.57 × 1012 = 2.7 × 10−3 m/s2, which is close to g ÷ 602: this was Newton’s check of gravitation (the Moon is 60 Earth radii away).
How was it discovered?
Christiaan Huygens derived the formula for centripetal acceleration in 1659. Newton, around 1666, used it with Kepler’s third law to show that the force on a planet must vary as the inverse square of the distance, and compared the acceleration of the Moon with g to test the law of gravitation. The term “centripetal” (“center-seeking”) was coined by Newton.
Which mistakes are common?
Mistake 1: Drawing centripetal force as an additional force
It is the resultant of the real forces. Do not add it on top of tension, friction or gravity.
Mistake 2: Believing that an outward force acts on the object in an inertial frame
The object tends to move along a tangent. The “outward” force is fictitious.
Mistake 3: Thinking that the object flies off radially when the string breaks
It moves along the tangent at the point of release.
Mistake 4: Forgetting to square the speed
a = v2 ÷ r. Doubling the speed quadruples the force required.