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The framework F = mv²/r = mω²r is short to write but rich in implication. Banked motorway curves and aircraft turns, the minimum speed at the top of a roller-coaster loop, the radius of an electron's track in a magnetic field, and the orbital speed of a satellite all fall out of the same equation set, with the only variation being which real force plays the centripetal role. This lesson works through four canonical scenarios — banked tracks, vertical loops, charged particles in magnetic fields, and satellite circular orbits — that between them anchor more than half of the AQA exam questions on circular motion. By the end, students should be able to read a problem, identify the geometry, draw the free-body diagram, and write the centripetal-force equation in fewer than three lines.
Spec mapping: This lesson sits under AQA 7408 section 3.6.1.1 (applications) and synoptically engages 3.7.2 (gravitational fields — satellite orbits) and 3.7.5 (magnetic fields — charged particles). It covers banked tracks (no-friction optimal angle), vertical-loop minimum speeds at the top and apparent weights at the bottom, charged-particle circular motion in uniform magnetic fields (r = mv/QB), and uniform circular-orbit motion in gravitational fields (v² = GM/r, T² = 4π²r³/GM). (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.7.2 (gravitational fields): Kepler's third law T² = 4π²r³/GM is derived directly by equating gravitational and centripetal forces — the algebra in this lesson is reused there.
- Section 3.7.5 (magnetic fields): the mass spectrometer, cyclotron, and electron-beam deflection experiments all use r = mv/(QB) from the centripetal-force balance. The momentum dependence p = QBr is the working formula in Aston-style mass spectrometers.
- Section 3.6.1.2 (SHM): vertical-loop and pendulum problems share the same gravity-versus-tension balance algebra; recognising the structural similarity speeds up SHM-system analysis later.
A flat road relies entirely on friction between the tyres and the surface to supply the centripetal force needed to round a bend. If the road is wet or icy, the available friction drops and the car skids outward. Engineers solve this problem at speed by banking the road: tilting the surface so that the horizontal component of the normal contact force supplies (some or all of) the centripetal force.
Consider a car of mass m on a road banked at angle θ to the horizontal. Treat the car as a point mass and ignore friction for now. Two forces act:
The car moves horizontally on a circle of radius r at speed v. There is no vertical acceleration; there is a horizontal centripetal acceleration v²/r towards the centre of the curve.
Resolve vertically: the vertical component of N balances the weight.
N cos θ = mg → N = mg/cos θ.
Resolve horizontally (towards the centre): the horizontal component of N supplies the centripetal force.
N sin θ = mv²/r.
Divide horizontal by vertical:
tan θ = v²/(rg) → v = √(rg tan θ).
This is the speed at which the bank is "perfectly tuned" — at this speed no friction is needed and the car would corner safely on perfect ice.
Equivalently, the optimal bank angle for a known design speed v on a curve of radius r is θ = arctan(v²/(rg)).
A motorway bend has radius 200 m and is designed for a speed of 30 m s⁻¹ (108 km h⁻¹). Calculate the optimal bank angle. Take g = 9.81 m s⁻².
tan θ = v²/(rg) = (30)²/(200 × 9.81) = 900/1962 = 0.459.
θ = arctan(0.459) = 24.6°.
For comparison, the velodrome at Lee Valley is banked at 42° on its tightest curve, optimal for cyclists at about v = √(rg tan 42°) ≈ 20 m s⁻¹ (about 72 km h⁻¹) at typical radius — a fair match to elite track-cycling speeds.
Above the optimal speed, the no-friction analysis fails: the required centripetal force exceeds N sin θ and the car would slide outward. Static friction acts down the bank to make up the shortfall. Below the optimal speed, N sin θ exceeds the required centripetal force and friction acts up the bank to prevent sliding inward.
The presence of friction broadens the safe-speed window. Real motorways are banked for design-speed comfort but rely on friction across a range of speeds.
flowchart TD
A["Car on banked curve<br/>angle θ"] --> B{"Speed v relative to<br/>v_opt = √(rg tan θ)?"}
B -->|"v = v_opt"| C["N sin θ = mv²/r<br/>No friction needed"]
B -->|"v > v_opt"| D["Friction acts down slope<br/>(prevents outward slip)"]
B -->|"v < v_opt"| E["Friction acts up slope<br/>(prevents inward slip)"]
style C fill:#27ae60,color:#fff
style D fill:#e67e22,color:#fff
style E fill:#3498db,color:#fff
When an object moves in a vertical circle — a roller-coaster cart, a bucket of water on a rope, a ball on a stiff rod — gravity points either towards or away from the centre, depending on the object's position on the loop. The geometry is now non-uniform: the speed itself varies around the loop because gravity does work. We focus on the two limiting positions: top and bottom.
At the top, both the weight (mg, downward) and the normal/tension force (also directed downward, towards the centre) act in the same direction — towards the centre of the loop. The centripetal-force equation is:
N_top + mg = mv_top²/r
If the object is on a string or sits on the underside of a track, the contact/tension force has a minimum value of zero (the string can only pull, not push; the track can only push outward on the cart). Setting N_top = 0 gives the minimum speed at the top for the object to maintain contact:
mg = mv_min²/r → v_min = √(rg).
Below v_min, the object loses contact with the track or the string goes slack and the path is no longer circular.
At the bottom, the weight points away from the centre (downward, while the centre is above), and the normal/tension force points towards the centre (upward). The centripetal-force equation is:
N_bottom − mg = mv_bottom²/r → N_bottom = mg + mv_bottom²/r.
The normal force at the bottom is much greater than mg. For a roller-coaster cart, the rider experiences this as "apparent weight" — a feeling of being pressed into the seat. Conservation of energy links v_top and v_bottom: ½mv_top² + mg(2r) = ½mv_bottom², so v_bottom² = v_top² + 4rg.
A small ball of mass 0.15 kg is attached to a string of length 0.40 m and swung in a vertical circle. Calculate (a) the minimum speed at the top of the circle for the string to remain taut, and (b) the tension in the string at the bottom assuming the speed there is the value that just maintains v_min at the top.
(a) At the top with T = 0: mg = mv_min²/r → v_min = √(rg) = √(0.40 × 9.81) = √3.924 = 1.98 m s⁻¹.
(b) Energy conservation between top and bottom (height difference 2r = 0.80 m):
½v_bottom² = ½v_top² + g(2r) → v_bottom² = v_top² + 4rg = 3.924 + 4 × 0.40 × 9.81 = 3.924 + 15.696 = 19.62
v_bottom = √19.62 = 4.43 m s⁻¹.
T_bottom = mg + mv_bottom²/r = 0.15 × 9.81 + (0.15 × 19.62)/0.40 = 1.47 + 7.36 = 8.83 N.
The tension at the bottom is roughly six times the weight of the ball, a striking ratio that explains why fairground rides feel so much heavier at the lowest point. The rider's apparent weight follows the same arithmetic.
A charged particle of charge Q and mass m moving at speed v through a uniform magnetic field B, with v perpendicular to B, experiences a magnetic force F = BQv. This force is always perpendicular to v⃗ — it does no work and so does not change the speed. It does, however, continuously deflect the particle, and the only path on which a perpendicular constant-magnitude force generates uniform circular motion is, of course, a circle.
The magnetic force supplies the centripetal force:
BQv = mv²/r → r = mv/(QB) = p/(QB).
The radius depends on the momentum p = mv of the particle, not on the charge-to-mass ratio alone. A mass spectrometer exploits this: ions accelerated to the same kinetic energy (½mv² = QV, so v = √(2QV/m)) and entering a perpendicular B-field at the same speed each curve to a radius r = √(2mV/Q)/B that depends on m/Q. Lighter or more highly charged ions curve more tightly; heavier or singly charged ions curve more widely. By measuring r for each ion, the device infers m/Q.
The period of the circular motion is T = 2πr/v = 2π × [mv/(QB)]/v = 2πm/(QB), so the angular frequency is
ω = 2π/T = QB/m (the cyclotron angular frequency)
and the corresponding linear frequency is f = QB/(2πm). Remarkably, both are independent of v: at non-relativistic speeds, all particles of the same Q/m circulate at the same rate, regardless of energy. This is the founding principle of the cyclotron particle accelerator.
A proton (m = 1.67 × 10⁻²⁷ kg, charge e = 1.60 × 10⁻¹⁹ C) is accelerated from rest through a potential difference of 500 V and then enters a region of uniform magnetic field 0.20 T perpendicular to its velocity. Calculate (a) its speed entering the field and (b) the radius of its circular path.
(a) Energy gained = work done by field: ½mv² = eV → v = √(2eV/m) = √(2 × 1.60 × 10⁻¹⁹ × 500 / 1.67 × 10⁻²⁷) = √(9.58 × 10¹⁰) = 3.10 × 10⁵ m s⁻¹.
(b) r = mv/(eB) = (1.67 × 10⁻²⁷ × 3.10 × 10⁵)/(1.60 × 10⁻¹⁹ × 0.20) = (5.18 × 10⁻²²)/(3.20 × 10⁻²⁰) = 0.0162 m ≈ 1.62 cm.
A 500-V acceleration gives a proton a cyclotron radius of about 1.6 cm in a modest 0.2 T field. Doubling V scales r by √2; doubling B halves r; doubling m scales r by √2 (because v also drops by √2). These scalings are the workhorse of practical mass-spectrometry design.
A satellite of mass m in a circular orbit of radius r around a central body of mass M is held on its path by gravitational attraction:
GMm/r² = mv²/r → v² = GM/r.
The orbital speed decreases with increasing orbit radius. Higher orbits are slower, not faster. Combining v = 2πr/T:
(2πr/T)² = GM/r → T² = 4π²r³/GM.
This is Kepler's third law: the square of the period is proportional to the cube of the orbital radius, with the constant of proportionality 4π²/(GM) shared by every satellite of the same central body.
A geostationary satellite orbits the Earth (M = 5.97 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m² kg⁻²) once per sidereal day (T ≈ 86,164 s). Calculate the orbital radius and orbital speed.
T² = 4π²r³/(GM) → r³ = GMT²/(4π²) = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (86,164)²)/(4π²)
GM = 3.98 × 10¹⁴; T² = 7.42 × 10⁹; product = 2.953 × 10²⁴; ÷ 4π² (= 39.48) = 7.482 × 10²².
r = (7.482 × 10²²)^(1/3) = 4.22 × 10⁷ m (about 42,200 km from the centre of Earth, or 35,800 km above the surface — the familiar geostationary altitude).
v² = GM/r = 3.98 × 10¹⁴ / 4.22 × 10⁷ = 9.43 × 10⁶ → v = 3,070 m s⁻¹ ≈ 3.07 km s⁻¹.
This is markedly slower than a low-Earth-orbit satellite, which orbits at v ≈ 7.8 km s⁻¹ at r ≈ 6.78 × 10⁶ m. The slowdown with altitude is the consequence of v² ∝ 1/r.
Specimen question modelled on the AQA paper format. Nine marks.
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