Objects moving in circles are constantly accelerating — even at constant speed. Discover why, and master the forces that keep satellites in orbit, cars on banked tracks, and conical pendulums spinning.
📐Define angular velocity ω and relate it to period T and linear speed v
⬅️Explain why centripetal acceleration always points toward the centre
🔢Apply a = v²/r = ω²r and F = mv²/r in calculations
🛣️Analyse forces on banked tracks and conical pendulums
🎡Solve problems involving vertical circles and minimum speed at the top
🔬Investigate circular motion experimentally with a whirling bung
Angular Velocity and Period
When an object moves in a circle, it is useful to measure how quickly it sweeps out angle rather than distance. Angular velocity ω is defined as the angle swept per unit time:
ω = θ / t (rad s⁻¹)
For one complete revolution, θ = 2π radians and t = T (the period), so:
ω = 2π / T and ω = 2πf
The linear (tangential) speed v at radius r is the arc length per unit time. Since arc length = rθ:
v = ωr
Notice that for a rigid rotating body every point has the same ω, but points farther from the centre have greater v. This is why the outer edge of a spinning disc moves faster than the centre.
Angular velocity ω — rate of change of angular displacement; SI unit rad s⁻¹. Always a scalar in magnitude but a vector (along the rotation axis) in full treatment.
Centripetal Acceleration
Even at constant speed, an object in circular motion is accelerating because velocity (a vector) continuously changes direction. This acceleration always points toward the centre of the circle — hence the name centripetal (Latin: "centre-seeking").
a = v² / r or equivalently a = ω²r
Both forms are derived from geometry: consider the velocity vector rotating through a small angle δθ in time δt. The change in velocity |δv| = v δθ, so a = v(dθ/dt) = vω = v(v/r) = v²/r.
⚠️ There is NO "centrifugal force" acting on the object. In an inertial frame, the only real force is the centripetal force directed inward. "Centrifugal" is a fictitious force felt only in a rotating reference frame.
The centripetal acceleration is provided by a real physical force. Examples: gravity for satellites, tension for a whirling bung, normal force for a car on a circular track.
Centripetal Force
By Newton's second law, the net force producing centripetal acceleration is:
F = ma = mv² / r = mω²r
This force must always be directed toward the centre. It is not a new type of force — it is the label we give to whatever real force (or resultant of forces) provides the inward acceleration.
Situation
Provider of centripetal force
Planet orbiting the Sun
Gravitational attraction
Car on flat circular road
Friction between tyres and road
Electron in magnetic field
Magnetic (Lorentz) force
Conical pendulum bob
Horizontal component of string tension
Object at top of vertical loop
Weight + normal reaction (both inward)
Always draw a free-body diagram first. Resolve forces toward the centre and set the resultant equal to mv²/r.
Banked Tracks and Conical Pendulum
Banked track: A road banked at angle θ allows a vehicle to travel in a circle without any friction if the horizontal component of the normal reaction provides the centripetal force:
N sinθ = mv²/r and N cosθ = mg
Dividing: tanθ = v²/(rg). This is the ideal banking angle for speed v and radius r.
Conical pendulum: A mass m on a string of length L making angle θ with the vertical moves in a horizontal circle of radius r = L sinθ. The vertical component of tension balances weight and the horizontal component provides centripetal force:
T cosθ = mg and T sinθ = mω²r
So: T sinθ / T cosθ = tanθ = ω²r/g. Since r = L sinθ, this simplifies to:
cosθ = g / (ω²L) → T = 2π√(L cosθ / g)
Vertical Circles
For an object moving in a vertical circle (e.g. a ball on a string or a roller-coaster loop), the centripetal force equation must account for gravity, which changes the tension at each point.
At the top: Weight and tension both point inward (downward):
T + mg = mv²top/r
For minimum speed at the top, T = 0 (string just taut or normal force just zero):
vmin,top = √(gr)
At the bottom: Tension points inward (upward), weight points outward (downward):
T − mg = mv²bottom/r → T = mg + mv²/r
The tension at the bottom is always greater than at the top — this is why riders feel heaviest at the bottom of a loop and lightest (or weightless) at the top. Using energy conservation between top and bottom: v²bottom = v²top + 4gr.
In vertical circle problems, always identify the point of interest, define which direction is "inward", and apply F_net(inward) = mv²/r at that point.
A car of mass 1200 kg travels at 20 m s⁻¹ around a circular bend of radius 50 m on a flat road. Calculate (a) the centripetal acceleration, (b) the centripetal force required, (c) the minimum coefficient of friction needed.
1Identify: v = 20 m s⁻¹, r = 50 m, m = 1200 kg, g = 9.81 m s⁻²
2Centripetal acceleration: a = v²/r = 20²/50 = 400/50 = 8.0 m s⁻²
A conical pendulum has a string of length 0.80 m. The string makes an angle of 30° with the vertical. Find (a) the radius of the circle, (b) the angular velocity, (c) the period of rotation.
1Radius: r = L sinθ = 0.80 × sin30° = 0.80 × 0.5 = 0.40 m
A ball on a string moves in a vertical circle of radius 0.5 m. What is the minimum speed at the top of the circle? If this speed is maintained, what is the tension in the string at the bottom?
1Minimum speed at top: v_top = √(gr) = √(9.81 × 0.5) = √4.905 = 2.21 m s⁻¹
v_min,top = 2.21 m s⁻¹; T_bottom = 11.8 N (for m = 0.2 kg)
A road is banked at 15°. What is the maximum speed at which a car can travel around a bend of radius 200 m without relying on friction?
1Ideal banking condition: tanθ = v²/(rg)
2v² = rg tanθ = 200 × 9.81 × tan15°
3tan15° = 0.2679, so v² = 200 × 9.81 × 0.2679 = 525.9 m² s⁻²
4v = √525.9 = 22.9 m s⁻¹ ≈ 82 km h⁻¹
v = 22.9 m s⁻¹ (82 km h⁻¹) — no friction needed at this speed
Q1. A stone is whirled in a horizontal circle of radius 1.5 m at 4.0 rad s⁻¹. What is its linear speed?
Q2. Which of the following correctly states the direction of centripetal acceleration?
Q3. A 0.5 kg ball moves in a circle of radius 2.0 m at 3.0 m s⁻¹. What centripetal force is needed?
Q4. For a car on a banked track at angle θ with no friction, which equation correctly gives the ideal speed?
Q5. What is the minimum speed needed at the top of a vertical circle of radius 0.8 m? (g = 9.81 m s⁻²)
Challenge 1. A satellite orbits Earth at altitude 400 km above the surface. Earth's radius is 6.4 × 10⁶ m and g at the surface is 9.81 m s⁻². The gravitational field strength at the satellite's orbit is 8.69 m s⁻². Calculate the orbital speed and period of the satellite.
✓ r = 6.4×10⁶ + 4×10⁵ = 6.8×10⁶ m. Gravity provides centripetal force: mg' = mv²/r → v = √(g'r) = √(8.69 × 6.8×10⁶) = √(5.91×10⁷) = 7690 m s⁻¹ ≈ 7.69 km s⁻¹. T = 2πr/v = 2π × 6.8×10⁶ / 7690 = 5550 s ≈ 92.5 minutes.
Challenge 2. A vertical circular loop of radius 12 m is part of a roller-coaster track. A car of mass 800 kg enters the loop at the bottom with speed 18 m s⁻¹. (a) Find the speed at the top using energy conservation. (b) Show whether the car maintains contact with the track at the top. (c) Calculate the normal reaction at the bottom.
✓ (a) v²_top = v²_bottom − 4gr = 18² − 4×9.81×12 = 324 − 470.9 → v²_top = −147 m² s⁻². This is negative, which means the car does NOT have enough speed to maintain contact — it would leave the track before reaching the top. (b) Minimum v²_top needed = gr = 9.81×12 = 117.7 m² s⁻², requiring v_bottom ≥ √(5gr) = √(588) = 24.2 m s⁻¹. At 18 m s⁻¹ the car loses contact. (c) At the bottom with 18 m s⁻¹: N − mg = mv²/r → N = m(g + v²/r) = 800(9.81 + 324/12) = 800(9.81+27) = 800×36.81 = 29 448 N ≈ 29.4 kN.
Challenge 3. A conical pendulum string makes 30° with the vertical. The bob has mass 0.15 kg and the string length is 0.60 m. (a) Find the tension in the string. (b) Find the period of revolution. (c) If the angle increases to 60°, what happens to the period — does it increase or decrease? Justify your answer using the formula.
✓ (a) T cosθ = mg → T = mg/cosθ = (0.15×9.81)/cos30° = 1.4715/0.866 = 1.70 N. (b) Period formula: T_period = 2π√(L cosθ/g) = 2π√(0.60×0.866/9.81) = 2π√(0.0530) = 2π×0.230 = 1.45 s. (c) At θ=60°: T_period = 2π√(0.60×cos60°/9.81) = 2π√(0.60×0.5/9.81) = 2π√(0.0306) = 2π×0.175 = 1.10 s. The period DECREASES as θ increases because cosθ decreases, making the effective pendulum length L cosθ smaller — the bob rotates faster.
Challenge 4 (Synoptic). A motorcyclist rides over a hill that is approximately circular with radius of curvature 30 m. At what speed will the motorcyclist just leave the ground at the top of the hill? How does this relate to the concept of "weightlessness" experienced by astronauts in orbit?
✓ At the top of the hill, weight acts downward (inward) and normal reaction acts upward (outward). For circular motion: mg − N = mv²/r. The rider just leaves the ground when N = 0: mg = mv²/r → v = √(gr) = √(9.81×30) = √294.3 = 17.2 m s⁻¹ ≈ 61.8 km h⁻¹. Link to orbit: astronauts in orbit are in free fall — gravity provides all centripetal force, N=0, so they feel "weightless". Similarly the motorcyclist experiences zero normal force at this critical speed. Both involve the condition where gravity alone provides the centripetal acceleration, so the contact force vanishes.
Required Practical
Investigating Circular Motion — Whirling Bung
Aim
To investigate how the centripetal force on a bung moving in a horizontal circle depends on the radius, speed and mass, and to verify F = mv²/r.
Apparatus
Rubber bung (mass m), nylon string (≈1 m), glass or plastic tube (~15 cm), paper clip or tape marker, hanging masses (M), stopwatch, ruler, balance.
Setup
Thread the string through the tube. Attach the rubber bung to the top end and a set of hanging masses (M) to the bottom end. Place a paper clip on the string just below the tube as a radius marker — the clip should remain stationary during rotation, indicating constant radius r.
Method
Measure and record the radius r from the top of the tube to the bung.
Whirl the bung in a horizontal circle, keeping the paper clip stationary just below the tube (this ensures r is constant and the hanging weight provides the centripetal force: F = Mg).
Time 20 complete revolutions with a stopwatch. Calculate period T = total time / 20.
Calculate linear speed: v = 2πr / T.
Repeat for different radii (vary by moving the clip) and different hanging masses.
Analysis
The hanging mass provides the centripetal force: Mg = mv²/r. Plot mv²/r against Mg — a straight line through the origin with gradient 1 confirms the relationship. Alternatively, plot v² against r (at constant M) to verify v² ∝ r, or v² against 1/M (at constant r) to verify the inverse proportionality.
Safety
⚠️ Ensure a clear area around the experimenter. Keep other students well back. The bung can cause injury if released or if the string breaks. Use safety goggles. Start with slow rotation and increase gradually.
Sources of Error
Error
Minimisation
Radius r not truly horizontal — bung dips below horizontal plane
Whirl faster; accept small angle with horizontal as systematic error
Paper clip moves (r not constant)
Practice technique; watch clip carefully during timing