Aim for FivePhysics 1

Unit 2 · Topic 2.9

2.9 Circular Motion

An object moving in a circle is always accelerating, because its direction keeps changing. The acceleration toward the center is v²/r, and real forces (tension, gravity, friction or the normal force) must supply the net inward force. The same idea explains loops, curves and orbits, including Kepler's third law.

Key terms

  • centripetal acceleration
  • centripetal (net inward) force
  • tangential acceleration
  • period and frequency
  • circular orbit
  • Kepler's third law

Centripetal acceleration

In uniform circular motion, an object moves around a circle at constant speed. Its velocity is always tangent to the circle, and because the velocity's direction keeps changing, the object accelerates toward the center. This is centripetal acceleration: a_c = v²/r.

Doubling the speed makes a_c four times as large. For the same speed, a tighter circle (smaller r) needs more acceleration.

What provides the inward force

Newton's second law toward the center: ΣF_inward = mv²/r. "Centripetal force" isn't a new kind of force and never goes on a free-body diagram. It's a name for the net inward force made of real forces:

  • Ball whirled on a string: tension.
  • Car turning on a flat road: static friction from the road.
  • Car on a banked curve: part of the normal force (plus friction, if needed).
  • Conical pendulum (a mass swinging in a horizontal circle on a string): the horizontal component of tension.
  • Moon or satellite: gravity.

Vertical circles and loops

At the top of a loop, both gravity and the track's normal force point down, toward the center: F_N + mg = mv²/r. The slower the car, the smaller F_N. At the minimum speed, F_N = 0 and gravity alone provides the inward force: mg = mv²/r, so v_min = √(gr).

At the bottom, the normal force points up (inward) and gravity points down (outward): F_N − mg = mv²/r. That's why riders feel heaviest at the bottom.

Banked curves are tested quantitatively only when no friction is needed. Then the horizontal part of the normal force supplies mv²/r and the vertical part balances mg, which gives tan θ = v²/(gr). If a car goes faster or slower than that speed, friction has to help, and you only describe that case in words.

A conical pendulum works the same way with tension in place of the normal force. With the string at angle θ from vertical, the vertical part of the tension balances gravity (F_T cos θ = mg) and the horizontal part points to the center of the circle (F_T sin θ = mv²/r).

Speeding up while turning

If the speed is changing too, there's also a tangential acceleration along the path, equal to the rate at which the speed changes. The total acceleration is the vector sum of the centripetal and tangential parts, so it points partly inward and partly forward (or backward).

Period, frequency and orbits

The period T is the time for one full revolution, in seconds. The frequency f is the number of revolutions per second, in hertz (Hz). T = 1/f. At constant speed, one lap covers the circumference, so v = 2πr/T.

For a circular orbit, gravity is the only inward force: GMm/r² = mv²/r. The satellite's mass cancels, so v = √(GM/r). Substituting v = 2πr/T gives Kepler's third law: T² = (4π²/GM) r³. Farther orbits are slower and take longer, and the period depends on the central mass M, not the satellite's mass. Kepler's first and second laws aren't on the AP Physics 1 exam.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Maximum speed on a flat curve

    A car rounds a flat, unbanked curve of radius 50 m. The coefficient of static friction between tires and road is 0.80. Use g = 9.8 m/s². What is the maximum speed for which the car won't skid?

    Show the solution
    1. Step 1: Static friction is the only horizontal force, and it points toward the center. At maximum speed it's at its maximum, μ_s F_N = μ_s mg.
    2. Step 2: Second law toward the center: μ_s mg = mv²/r. Mass cancels: v = √(μ_s g r).
    3. Step 3: v = √(0.80 × 9.8 × 50) = √392 ≈ 19.8 m/s.

    Answer: About 20 m/s, regardless of the car's mass

  2. Example 2Calculator allowed

    Top of a roller-coaster loop

    A 500 kg coaster car goes over the top of a vertical loop of radius 10 m on the inside of the track. Use g = 9.8 m/s². (a) What is the minimum speed at the top? (b) If its speed at the top is 14 m/s, what normal force does the track exert?

    Show the solution
    1. Step 1: (a) At the minimum speed, F_N = 0 and gravity alone gives the inward force: mg = mv²/r, so v_min = √(gr) = √(9.8 × 10) ≈ 9.9 m/s.
    2. Step 2: (b) At the top both forces point down, toward the center: F_N + mg = mv²/r.
    3. Step 3: F_N = m(v²/r − g) = 500 × (14²/10 − 9.8) = 500 × (19.6 − 9.8) = 4900 N, pointing down.

    Answer: (a) about 9.9 m/s; (b) 4900 N downward

  3. Example 3Calculator allowed

    Satellite in a circular orbit

    A satellite orbits Earth in a circle of radius 7.0 × 10⁶ m (measured from Earth's center). Earth's mass is 5.97 × 10²⁴ kg. Find the satellite's orbital speed and period.

    Show the solution
    1. Step 1: Gravity provides the inward force: GMm/r² = mv²/r, so v = √(GM/r).
    2. Step 2: v = √[(6.67 × 10⁻¹¹)(5.97 × 10²⁴) ÷ (7.0 × 10⁶)] ≈ 7.5 × 10³ m/s.
    3. Step 3: T = 2πr/v = 2π(7.0 × 10⁶) ÷ (7.54 × 10³) ≈ 5.8 × 10³ s, about 97 minutes.
    4. Step 4: Kepler's third law gives the same period: T = √(4π²r³/GM) ≈ 5.8 × 10³ s. The satellite's mass was never needed.

    Answer: v ≈ 7.5 km/s; T ≈ 5.8 × 10³ s (about 97 min)

Common mistakes

  • Drawing a "centripetal force" on a free-body diagram. Draw only real forces; their net inward part equals mv²/r.
  • Thinking an object at constant speed in a circle has zero acceleration. Its direction changes, so it accelerates toward the center.
  • Saying an object released from a circle flies outward along the radius. It moves off along the tangent, in the direction it was already going.
  • Using the orbit's altitude instead of its radius from the planet's center in v²/r or Kepler's third law.

On the exam

  • Expect free-body diagrams at the top and bottom of a vertical circle, followed by a ΣF = mv²/r equation toward the center. Take the direction toward the center as positive.
  • Derivations are common: minimum speed at the top of a loop, maximum speed on a curve, or orbital speed. Show the starting equation and keep everything in symbols until the end.
  • Lab questions may ask you to linearize data, such as graphing T² against r³ for orbits; the slope is 4π²/GM.

Connected topics

Videos

Check yourself

4 questions on 2.9 Circular Motion. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A ball on a string is swung in a horizontal circle on a frictionless table. The string breaks when the ball is at the north side of the circle, moving east. Which describes the ball's path right after the string breaks?

Question 2 of 4Calculator allowed

A roller coaster car goes around the inside of a vertical loop of radius 0.90 m in a model. What is the minimum speed at the top of the loop for the car to stay on the track? Use g = 10 m/s².

Question 3 of 4Calculator allowed

A ball on a string is swung in a vertical circle. When the ball is at the very top of the circle, which forces act on it?

Question 4 of 4Calculator allowed

A 1000 kg car drives over the top of a rounded hill whose radius is 40 m, at 10 m/s. What normal force does the road exert on the car at the top? Use g = 10 m/s².

0 of 4 answered