Lesson 4: Uniform Circular MotionAP

An object moving in a circle at constant speed is accelerating — not because its speed changes, but because its direction does, and velocity is a vector. Uniform circular motion (CED Unit 3, developed further in Lesson 7) is where that idea becomes quantitative, and it underlies everything from cornering cars to orbiting moons.

Period, frequency and speed

One revolution covers the circumference 2πr. If the period (time per revolution) is T, the speed is v = 2πr/T. Frequency f = 1/T counts revolutions per second, so v = 2πrf. These are definitions, not new physics — but fluency with them is assumed in every circular-motion problem.

Centripetal acceleration

Consider the velocity vectors at two nearby points on the circle. Both have the same magnitude v, but they point in different directions, so Δv is not zero. Geometry (the velocity-change triangle is similar to the position triangle) shows that Δv points toward the centre and has magnitude v²Δt/r in time Δt, giving the centripetal acceleration: a_c = v²/r = 4π²r/T², directed toward the centre of the circle at every instant. Because a = Δv/Δt, an acceleration toward the centre is exactly what a continuously turning velocity vector requires.

Two consequences the exam probes hard: double the speed and a_c quadruples; double the radius (same speed) and a_c halves. And "centripetal" names a direction, not a kind of force — Lesson 7 shows how real forces (friction, tension, gravity) supply the required inward net force F_net = mv²/r.

Tangential versus centripetal

In uniform circular motion the speed is constant, so there is no tangential acceleration — the acceleration is purely centripetal. If the speed also changes (a swinging pendulum speeding up at the bottom, a car accelerating around a bend), a tangential component appears and the total acceleration is the vector sum. AP Physics 1 concentrates on the uniform case, plus recognising when a problem is not uniform.

Worked example 1 — car on a curve

A car rounds a flat curve of radius 50 m at a steady 20 m/s. Find (a) its centripetal acceleration, (b) the period of one full circle at this radius and speed. Given: v = 20 m/s, r = 50 m.

(a) a_c = v²/r = 20²/50 = 400/50 = 8.0 m/s², toward the centre of the curve
(b) T = 2πr/v = 2π(50)/20 ≈ 15.7 s

Check: 8.0 m/s² is about 0.8 g — a firm but plausible cornering acceleration for a car on tyres. Direction matters as much as magnitude: the acceleration is horizontal and inward, not forward.

Worked example 2 — whirling stopper

A rubber stopper moves in a horizontal circle of radius 0.80 m at 2.5 revolutions per second. Find (a) its speed, (b) its centripetal acceleration, in units of g. Given: r = 0.80 m, f = 2.5 Hz.

(a) v = 2πrf = 2π(0.80)(2.5) ≈ 12.6 m/s
(b) a_c = v²/r = (12.6)²/0.80 ≈ 198 m/s² ≈ 20 g  (using g ≈ 9.8 m/s²)

Check: the enormous acceleration is why the string must supply a large inward force — and why this classic lab is done with a light stopper and eye protection. The acceleration grows if either the frequency or radius rises, matching a_c = 4π²rf².

Real-world application: centrifuges and cornering limits

A laboratory centrifuge spinning samples at 3,000 rpm with r = 0.10 m produces a_c = 4π²rf² = 4π²(0.10)(50²) ≈ 9,900 m/s² — about 1,000 g — which is what separates blood components by density. At the other extreme, highway engineers assume tyre friction can deliver at most about 0.5–0.8 g of centripetal acceleration; combine that with v = √(a_c r) and you get the posted speeds for curves: for r = 50 m and a_c = 8 m/s², v = 20 m/s ≈ 72 km/h, exactly the situation of Worked example 1.

Experimental design note

The whirling-stopper investigation asks: is a_c proportional to v² at fixed r? Hang a known weight to set the string tension (the inward force), whirl the stopper at a measured radius, and time 20 revolutions for a reliable period. Plot the inward force against v²: a straight line through the origin with slope m/r confirms F ∝ v². The classic systematic error is the string's angle — the radius is the horizontal distance to the stopper, not the string length.

Common mistakes (from College Board scoring commentary)

• Drawing velocity tangent and acceleration tangent too — acceleration is toward the centre for uniform circular motion.
• Believing a "centrifugal force" throws objects outward in an inertial frame; the outward sensation is your own inertia.
• Using the diameter where the radius belongs in a_c = v²/r.
• Treating a_c = v²/r as a new force to add to free-body diagrams (see Lesson 7).

AP Exam Experience

MCQ 1. An object moves in a circle of radius r at speed v. If the speed is doubled and the radius is halved, the centripetal acceleration becomes:

• (A) 2 times larger
• (B) 4 times larger
• (C) 6 times larger
• (D) 8 times larger

Answer: (D). a_c = v²/r: doubling v multiplies a_c by 4, halving r multiplies by another 2 — total ×8. Proportional reasoning like this is the fastest route through many AP multiple-choice items.

MCQ 2. A ball on a string moves in a horizontal circle at constant speed. At the instant the ball is at the rightmost point of the circle moving toward the top of the page, the directions of its velocity and acceleration are:

• (A) velocity up the page, acceleration up the page
• (B) velocity up the page, acceleration toward the left (toward the centre)
• (C) velocity toward the left, acceleration up the page
• (D) velocity toward the right, acceleration away from the centre

Answer: (B). Velocity is always tangent to the path; at the rightmost point moving upward, the tangent points up the page. Acceleration is centripetal — toward the centre, which from the rightmost point is toward the left.

MCQ 3. Two satellites orbit at the same speed, one at radius 2R and one at radius R from a planet's centre. The ratio of their centripetal accelerations (a at 2R : a at R) is:

• (A) 1 : 4
• (B) 1 : 2
• (C) 1 : 1
• (D) 2 : 1

Answer: (B). At equal speeds, a_c = v²/r is inversely proportional to r, so doubling r halves a_c. (Real satellites at different radii do not share the same speed — Lesson 8 derives why — but the question fixes speed deliberately.)

FRQ. A 0.50 kg puck slides on a horizontal air table in a circle of radius 1.2 m, held by a string through the centre attached to a force sensor. The puck completes 10 revolutions in 25 s. (a) Calculate the puck's speed. (b) Calculate the tension in the string. (c) The string is lengthened so the radius doubles while the puck's speed is kept the same. State and explain how the tension changes. (d) Describe one change to the experiment that would test whether tension is proportional to the square of the speed, and state the graph you would plot. Scoring guideline. (a) 1 point: T = 2.5 s, v = 2π(1.2)/2.5 ≈ 3.0 m/s. (b) 1 point: tension supplies the centripetal net force: T = mv²/r = 0.50 × 9.0/1.2 = 3.8 N. (c) 1 point: tension halves, because T = mv²/r with m and v fixed is inversely proportional to r. (d) 2 points: vary the speed at fixed radius and mass by changing the rotation rate, record tension from the sensor for several speeds (1 point); plot tension against v² and expect a straight line through the origin with slope m/r (1 point).

Worked example 3 — the acceleration you are feeling now

Earth rotates once in about 86,400 s. A person standing on the equator moves in a circle of radius 6.37 × 10⁶ m. Find their centripetal acceleration.

a_c = 4π²r/T² = 4π² × 6.37 × 10⁶ / (86,400)² ≈ 2.51 × 10⁸ / 7.46 × 10⁹ ≈ 0.034 m/s²

Check: that is about 0.003 g — real, measurable with precision instruments, and the reason Earth bulges slightly at the equator, but far too small to feel. The same formula at a satellite's radius and period (Lesson 8) returns ~8.7 m/s²: the formula is the same, the numbers transform its meaning.

Non-uniform circular motion: the honest footnote

If an object speeds up or slows down while circling — a pendulum bob mid-swing, a car accelerating out of a bend — it has both components at once: centripetal (v²/r, toward the centre, from the direction change) and tangential (dv/dt, along the path, from the speed change). The total acceleration is their vector sum, tilted off the radial line. AP Physics 1 keeps most problems uniform, but watch for the pendulum: at the bottom of the swing the tangential acceleration is momentarily zero (speed is at its peak) while the centripetal is maximal — a clean test of whether you can separate the two ideas.

Worked example 4 — merry-go-round physics

A playground merry-go-round has radius 2.0 m and turns once every 4.0 s. Find (a) the speed of a child at the rim, (b) her centripetal acceleration, (c) the minimum coefficient of static friction between her shoes and the platform if she stands without holding on.

(a) v = 2πr/T = 2π × 2.0/4.0 ≈ 3.1 m/s
(b) a_c = v²/r = 9.9/2.0 ≈ 4.9 m/s² (about 0.5 g)
(c) Friction must supply mv²/r: μ_s mg ≥ mv²/r → μ_s ≥ 4.9/10 ≈ 0.49

Check: halving the period (spinning twice as fast) would quadruple a_c to ~20 m/s² and demand μ_s ≈ 2 — impossible for shoes — which is why merry-go-rounds fling riders who don't hold on, and why part (c)'s answer, while startling, is honest physics. Note also the bridge to Lesson 8: the space station's centripetal acceleration, v²/r with orbital numbers, is (7,670)²/(6.77 × 10⁶) ≈ 8.7 m/s² — and there the supplier of that acceleration is gravity itself, which is why the station's crew feel no push from any floor.