Analyze uniform circular motion with a_c = v²/r and identify the force that provides the centripetal force

Learning goal: Explain why an object moving at constant speed in a circle is accelerating, with the acceleration directed toward the center; calculate centripetal acceleration from a_c = v²/r and the required net force from Newton's second law; identify, for cars on curves, satellites, planes in loops and objects on strings, which real force supplies the centripetal force; and interpret data relating the force on an object in circular motion to its speed.

Changing direction is accelerating. Velocity is a vector, so a change in direction is a change in velocity even when the speed is constant. An object moving in a circle at constant speed therefore has an acceleration, and by Newton's second law a net force must be acting. The force must be directed at right angles to the velocity, toward the center of the circle: any component along the velocity would change the speed, and a force toward the center is the only way to keep turning the velocity without changing its size. This inward net force is called the centripetal force. It is not a new kind of force; "centripetal" describes the direction of whatever real force or combination of forces is doing the job. If that force is removed and nothing else exerts a net force, the object does not fly outward: it continues in a straight line along the tangent, which is exactly what Newton's first law says, and is what a puck whirled on a frictionless table does when its string breaks. A stone whirled in the air also leaves along the tangent, but gravity still acts on it, so its path then curves downward like any projectile's.

The acceleration. For an object moving at speed v in a circle of radius r, the centripetal acceleration is a_c = v²/r, directed toward the center. Doubling the speed quadruples the acceleration; doubling the radius halves it. The net force is then F_net = m a_c = mv²/r. The speed is linked to the period T, the time for one revolution, by v = 2πr/T, which is how speed is usually measured in the laboratory.

A fully worked model: a car on an unbanked curve. A 1200 kg car rounds a level curve of radius 50 m at 15 m/s. Centripetal acceleration: 15² ÷ 50 = 4.5 m/s². Required net force: 1200 × 4.5 = 5400 N, directed toward the center of the curve. What provides it? On a level road the only horizontal force on the car is friction between the tires and the road, acting sideways toward the center; because the tires roll without sliding sideways, it is static friction. If the road's coefficient of static friction is 0.70 (example value), the maximum friction is 0.70 × 1200 × 9.8 = 8232 N, more than enough. The maximum safe speed is where mv²/r = μ_s mg, so v_max = √(μ_s g r) = √(0.70 × 9.8 × 50) = 18.5 m/s, independent of the car's mass. On ice with μ_s = 0.10 the limit drops to 7.0 m/s, and a car entering the curve faster than that cannot follow the curve: its tires slip, the smaller kinetic friction can no longer supply mv²/r, and it runs wide toward the outside of the curve, heading off roughly along the tangent rather than being thrown outward along the radius.

Identifying the centripetal force. In each situation, draw the free-body diagram and ask which forces, or components, point toward the center. A ball whirled on a string in a horizontal circle on a frictionless table: the string's tension. A plane flying a level, horizontal circle banks its wings: the lift on the wings tilts toward the center, its horizontal component supplies mv²/r and its vertical component balances the weight. A satellite in orbit: gravity alone; the nearly circular orbits of planets and many satellites are held by gravity acting as the centripetal force. A car on a level curve: static friction. A rider at the bottom of a vertical loop: the seat's normal force minus the weight. Clothes in a spin dryer: the drum wall pushes inward on the clothes, while the water leaves through the holes because nothing pushes inward on it, so it carries on almost straight ahead along the tangent. There is no outward force on the object; the sensation of being pushed outward is the body continuing straight while the car or seat turns under it.

Gravity as the centripetal force. A satellite at r = 6.77 × 10⁶ m from Earth's center (about 400 km up) is held in a circular orbit by gravity, so GMm/r² = mv²/r and v = √(GM/r) = √(3.98 × 10¹⁴ ÷ 6.77 × 10⁶) = 7.7 × 10³ m/s. The period is 2πr/v = 2π × 6.77 × 10⁶ ÷ 7.67 × 10³ = 5.5 × 10³ s, about 92 minutes. The satellite's acceleration v²/r = 8.7 m/s² equals the gravitational field strength there, computed in the gravitation lesson; the satellite is in free fall, but its sideways speed is so large that it falls around Earth rather than into it. The mass of the satellite canceled, so every object at that radius orbits at the same speed.

Vertical circles. A pilot of mass 70 kg pulls out of a dive at the bottom of a vertical circle of radius 500 m at 100 m/s. Centripetal acceleration: 100² ÷ 500 = 20 m/s², upward, toward the center. Forces: the seat's normal force N upward and weight 686 N downward; the net force is N − 686 = 70 × 20 = 1400, so N = 2086 N, about three times the pilot's weight. At the top of a loop at the same speed and radius the center is below the pilot, so N + 686 = 1400 and N = 714 N; the seat pushes less hard because gravity is helping. If the speed at the top were such that v²/r = g, the seat would push with zero force and the pilot would feel weightless; any slower and the pilot would rise off the seat unless the harness held them down, while the plane itself would have to use its wings to pull less hard toward the center.

Investigation: force vs. speed for a whirling stopper. Materials: a rubber stopper of mass 0.020 kg, strong string, a hollow tube (a smooth plastic tube or the barrel of a pen with the ends rounded), a set of washers or slotted masses, a paper clip, a stopwatch, safety goggles, and an open space. Steps: thread the string through the tube, tie the stopper on one end and hang a known mass on the other; mark the string so the radius from tube to stopper is 0.60 m and clip a paper clip just below the tube as a reference. Whirl the stopper in a circle so the clip stays just below the tube, meaning the hanging weight is supplying the tension; time 10 revolutions; compute T, then v = 2πr/T, with r the horizontal radius of the circle. Repeat with four hanging masses of about 85, 120, 170 and 220 g. The string cannot be exactly horizontal: its tension must also hold up the stopper's 0.196 N weight, so it tilts, and only the horizontal part of the tension is the centripetal force. With hanging weights at least four times the stopper's weight the tilt is at most about 14.5° and the horizontal part is within about 3 percent of the hanging weight, so the hanging weight can be used as the centripetal force. (A hanging weight smaller than the stopper's own weight cannot work at all.) Record: hanging weight (approximately the centripetal force), period, speed. Analyze: plot force against speed, and force against speed squared. Predicted with r = 0.60 m and m = 0.020 kg: speeds of 5.0, 6.0, 7.0 and 8.0 m/s need centripetal forces of 0.83, 1.2, 1.6 and 2.1 N. A result that supports the claim: a curve for F vs. v that bends upward and a straight line through the origin for F vs. v², with slope m/r = 0.033 kg/m. Safety: goggles for everyone in the room, a clear radius of 2 m, string checked for wear, no metal stoppers. What the model cannot show: the small tilt of the string makes the hanging weight slightly larger than the true centripetal force, and the radius must be measured horizontally, not along the string; and friction between string and tube means the tension in the upper string is not exactly the hanging weight.

Worked example 1 — tension in a string. A 0.50 kg ball on a 0.80 m string moves in a horizontal circle on a frictionless table at 4.0 m/s. Centripetal acceleration: 4.0² ÷ 0.80 = 20 m/s². Tension: 0.50 × 20 = 10 N, directed along the string toward the center. At 8.0 m/s the tension would be 40 N; if the string could take only 30 N it would break at √(30 × 0.80 ÷ 0.50) = 6.9 m/s, and the ball would move off along the tangent.

Worked example 2 — maximum speed on an unbanked curve. A curve of radius 30 m on a road with μ_s = 0.50 (example value). The friction that can be supplied is at most μ_s mg, and the friction required is mv²/r. Setting them equal, v_max = √(μ_s g r) = √(0.50 × 9.8 × 30) = 12 m/s. The mass canceled, so a loaded truck and a small car have the same limit on this curve; what differs between them is how much friction their tires provide, expressed in μ_s.

Worked example 3 — speed from the period. A stopper moving in a circle of horizontal radius 0.60 m completes 10 revolutions in 4.7 s. Period: 0.47 s. Speed: 2π × 0.60 ÷ 0.47 = 8.0 m/s. Centripetal force on a 0.020 kg stopper: 0.020 × 8.0² ÷ 0.60 = 2.1 N, which should match, within a few percent, the weight of a hanging mass of about 0.22 kg.

Worked example 4 — a plane at the top of a loop. A plane flies a vertical loop of radius 400 m. At the top, the center is below, and both the seat's push and the weight point downward toward it: N + mg = mv²/r. The seat force vanishes when v² = gr, so v = √(9.8 × 400) = 63 m/s. At that speed the pilot feels weightless for an instant; at 80 m/s the seat pushes down on a 70 kg pilot with 70 × (80² ÷ 400 − 9.8) = 70 × (16 − 9.8) = 434 N.

Representation task: four free-body diagrams with the centripetal force identified. Draw the free-body diagram for each of: a car rounding a level curve, a satellite in circular orbit, a stopper whirled on a string in a horizontal circle (include its weight and note that the string tilts slightly), and a rider at the bottom of a roller-coaster dip. For each, circle the force or combination of forces that points toward the center and label it "provides mv²/r". Under the four diagrams, write two sentences explaining why there is no outward force on the object and what happens to the object if the inward force suddenly stops.

Independent performance: force vs. speed analysis. Carry out the whirling-stopper investigation with at least four hanging weights. Success criteria: the data table lists hanging weight, time for 10 revolutions, period, speed and speed squared; both graphs are drawn with labeled axes and units; the F vs. v² graph is fitted with a straight line through the origin whose slope is compared with m/r as a percentage; the report identifies the force that provided the centripetal force in the experiment; and at least two limitations of the setup (string angle, tube friction, timing) are named with the direction in which each one shifts the slope.