Circular Motion: Study Guide (Physics C: Mechanics 2.10)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses. Physics 1 teaches circular motion as Topic 2.9; this course teaches it as Topic 2.10, with calculus. Here you derive the centripetal acceleration from a position vector, find tangential acceleration as a derivative when the speed changes with time, and include friction on banked curves. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise. This topic follows Topic 2.9, Resistive Forces.

Why moving in a circle means accelerating

Velocity is a vector. On a circle its direction changes all the time, so the object accelerates even at constant speed. Calculus shows exactly how.

Put the origin at the centre of a circle of radius r. An object goes round at constant speed, once every period T. Write ω = 2π/T (you will meet ω again as angular speed in Unit 5). Its position vector is:

r(t) = r cos(ωt) î + r sin(ωt) ĵ

Differentiate once for velocity:

v(t) = dr/dt = −rω sin(ωt) î + rω cos(ωt) ĵ

Its size is rω, so v = rω = 2πr/T. Its dot product with r(t) is zero, so the velocity is tangent to the circle.

Differentiate again for acceleration:

a(t) = dv/dt = −rω² cos(ωt) î − rω² sin(ωt) ĵ = −ω² r(t)

The minus sign means the acceleration points opposite to the position vector: toward the centre. Its size is rω² = v²/r. This is the centripetal acceleration:

a_c = v²/r, directed toward the centre

Period and frequency

For uniform circular motion (constant speed):

• Period T: the time for one full revolution, in s. (The same word is used for one full rotation, and for one full cycle of an oscillation in Unit 7.)
• Frequency f: revolutions per second, in Hz (s⁻¹). T = 1/f.
• One revolution covers 2πr, so T = 2πr/v.

Substituting v = 2πr/T gives another useful form: a_c = 4π²r/T².

Tangential acceleration and net acceleration

If the speed changes, the object also has a tangential acceleration, along the velocity:

a_t = dv/dt (the rate of change of speed)

a_t points along v when the object speeds up and against v when it slows down. It is always perpendicular to a_c, so the net acceleration is the vector sum:

|a| = √(a_c² + a_t²), at angle φ from the inward radius, where tan φ = |a_t|/a_c.

If you know the distance travelled along the circle, s(t), then v = ds/dt and a_t = d²s/dt².

Where the inward force comes from

Newton’s second law along the radius, with + toward the centre:

ΣF_toward centre = m v²/r

“Centripetal force” is not a new kind of force. It is the name for the inward total of real forces. Never draw it on a free-body diagram. That total can come from:

• one force: gravity on a satellite, or tension on a puck whirled on a horizontal frictionless table;
• several forces: gravity and the normal force on a car at the top of a hill;
• components of forces: the normal force and friction on a banked curve, or tension in a conical pendulum.

Vertical loops: the minimum speed at the top

At the top of a vertical loop of radius r, a cart on the inside of the track feels gravity mg and the normal force N, both pointing down, toward the centre:

N + mg = mv²/r

N cannot be negative: the track can push but not pull. The smallest possible speed is when N = 0. Then gravity alone provides the centripetal acceleration:

mg = mv²/r, so v_min = √(gr)

For a toy track with r = 0.25 m, v_min = √(9.8 × 0.25) ≈ 1.6 m/s. Any slower, and the cart leaves the track before the top.

Banked curves with friction

A road banked at angle θ tilts the normal force toward the centre. Take +x horizontal, toward the centre, and +y up. Friction acts along the slope, and its direction depends on the speed.

At the greatest speed, the car tends to slide up and out, so static friction points down the slope, at its maximum f = μ_s N:

• x: N sin θ + μ_s N cos θ = mv²/r
• y: N cos θ − μ_s N sin θ − mg = 0

Divide one equation by the other to remove N and m:

v_max² = rg (sin θ + μ_s cos θ)/(cos θ − μ_s sin θ)

At the least speed, the car tends to slide down and in, so friction points up the slope. Swap the sign of every μ_s term:

v_min² = rg (sin θ − μ_s cos θ)/(cos θ + μ_s sin θ)

If tan θ ≤ μ_s, the top line is zero or negative: the car can even stay parked on the slope, so there is no minimum speed. With μ_s = 0, both reduce to v² = rg tan θ, the one speed that needs no friction at all.

Conical pendulums

A ball on a string of length L moves in a horizontal circle, with the string at angle θ to the vertical. Only the tension F_T and gravity act. The horizontal component of tension provides the centripetal acceleration:

• toward the centre: F_T sin θ = mv²/r, with r = L sin θ
• vertical: F_T cos θ = mg

Dividing gives tan θ = v²/(rg). Combining with T = 2πr/v gives the period T = 2π√(L cos θ/g). The mass cancels. For L = 1.2 m and θ = 40°, the period is about 1.9 s and the tension is mg/cos θ ≈ 1.3mg.

Circular orbits and Kepler’s third law

For a satellite of mass m in a circular orbit of radius r around a central body of mass M, gravity is the only force. It provides all the centripetal acceleration:

GMm/r² = mv²/r, so v² = GM/r.

Substitute v = 2πr/T and rearrange:

T² = (4π²/GM) r³

This is Kepler’s third law for circular orbits. The satellite’s mass cancels, so T depends only on r and the central mass M. Measure T and r for any moon, and you can find the mass of its planet. The course does not expect Kepler’s first or second laws.

Worked example 1: a toy train speeding up

Question. A toy train starts from rest on a circular track of radius 0.60 m. The distance it travels along the track is s(t) = (0.30 m/s²)t². Find, at t = 2.0 s, (a) its speed, (b) its tangential acceleration, (c) its centripetal acceleration and (d) the size and direction of its net acceleration. (e) How long does it take to finish the first lap?

1. (a) v = ds/dt = 0.60t. At 2.0 s, v = 1.2 m/s.
2. (b) a_t = dv/dt = 0.60 m/s², along the velocity (speeding up). It is constant here.
3. (c) a_c = v²/r = (1.2)² ÷ 0.60 = 2.4 m/s², toward the centre.
4. (d) |a| = √(2.4² + 0.60²) = 2.5 m/s². Direction: tan φ = 0.60/2.4, so φ = 14° from the inward radius, tilted forward (the direction of travel), as in Figure 1.
5. (e) One lap is 2πr = 3.77 m. Solve 0.30t² = 3.77: t = 3.5 s. The speed then is 2.1 m/s and a_c has grown to 7.5 m/s².

Interpretation. a_t stays constant, but a_c grows as v², so the net acceleration swings closer and closer to the inward radius as the train speeds up.

Worked example 2: the greatest speed on a banked curve

Question. A motorway exit curve has radius 90 m and is banked at 12°. The coefficient of static friction between tyres and road is 0.40. Take +x horizontal toward the centre and +y up. Find (a) the speed that needs no friction and (b) the greatest speed at which a car can take the curve without sliding. (c) Is there a minimum speed?

1. (a) v² = rg tan θ = 90 × 9.8 × tan 12° = 187.5 m²/s², so v = 14 m/s (about 49 km/h).
2. (b) Friction points down the slope (Figure 2): v_max² = rg (sin 12° + 0.40 cos 12°)/(cos 12° − 0.40 sin 12°) = 90 × 9.8 × 0.5992/0.8950 = 590.5 m²/s². So v_max = 24 m/s (about 87 km/h).
3. Check with forces for a 1000 kg car at that speed: N = mg/(cos θ − μ_s sin θ) = 1.09 × 10⁴ N and f = 4.4 × 10³ N. Horizontal total N sin θ + f cos θ = 6.56 × 10³ N, which equals mv²/r = 6.56 × 10³ N.
4. (c) tan 12° = 0.21 is less than μ_s = 0.40, so the bracket in v_min² is negative. There is no minimum speed: friction can hold a parked car on this slope.

Comparison. On a flat road with the same friction, v_max = √(μ_s g r) = 19 m/s. Banking raises the limit by about 30%.

Worked example 3: the period of a low Earth orbit

Question. A satellite moves in a circular orbit 600 km above Earth’s surface. Find its period and speed.

1. r = 6.37 × 10⁶ m + 6.0 × 10⁵ m = 6.97 × 10⁶ m. Always measure r from the centre of Earth.
2. T = 2π√(r³/GM) = 2π√((6.97 × 10⁶)³ ÷ (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)) = 5.79 × 10³ s ≈ 97 minutes.
3. v = 2πr/T = 7.6 × 10³ m/s (7.6 km/s).

Factor of change. An orbit with 4 times the radius has a period 4^(3/2) = 8 times as long.

Common misconceptions

• “Constant speed means no acceleration.” The direction changes, so a_c = v²/r even at constant speed.
• “Centripetal force is an extra force on the diagram.” It is the inward net force. Draw only real forces: gravity, normal, tension, friction.
• “Objects are flung outward by a centrifugal force.” In an inertial frame there is no outward force. A released object moves off along the tangent, by Newton’s first law.
• “Net acceleration always points to the centre.” Only when the speed is constant. With a_t ≠ 0 it tilts forward or backward.
• “Friction on a banked curve always points down the slope.” It points down at high speed and up at low speed, and is zero at v² = rg tan θ.
• “At the top of a loop, the normal force points up.” On the inside of a loop the track is above the cart, so N points down, toward the centre.
• Using altitude as the orbit radius. Add Earth’s radius first.

Where this leads

This topic ends Unit 2. Next, Topic 3.1, Translational Kinetic Energy, opens Unit 3; energy methods later give the speed at each point of a loop. Unit 5 describes the same motion with angular quantities, and orbits return in Unit 6. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.