The complete lecture, side by side with a live panel. As you read each idea on the left, a real-life object on the right brings it to life — a stone whirled on a string, the string cut, a tilted motorway curve, a spinning washing drum, a chand-raat Ferris wheel, a bucket of water looping overhead, and a merry-go-round.
1 — Angular displacement & the radian
When a stone moves on a string of radius r, the most natural way to describe its motion is by the angle the string sweeps, not the distance it covers. The angular displacement θ is the angle swept by the radius line joining the stone to your hand at the centre.
- Angular displacement (θ) — the swept angle. SI unit: the radian (rad), a dimensionless ratio of two lengths.
- Radian — the angle at the centre when the arc length equals the radius (s = r). The SI unit of plane angle.
Angle, arc & conversionsθ (rad) = s / r → s = r θ
full circle: 360° = 2π rad · 180° = π rad · 1 rad ≈ 57.3°
worked example — fan blade
A 0.60 m fan blade turns through 1.5 revolutions. Find θ and the arc covered.
θ = 1.5 × 2π = 9.42 rad → s = rθ = 0.60 × 9.42 = 5.65 m
Exam point: the radian is arc ÷ radius — it is dimensionless, so it never appears in a dimensional check.
2 — Angular velocity ω, period, frequency & v = rω
- Angular velocity (ω) — radians swept per second: ω = Δθ/Δt (rad/s).
- Time period (T) — time for one revolution; frequency (f) = 1/T (rev/s, Hz).
The ω family & the bridge equationω = Δθ/Δt = 2π/T = 2πf · rpm × 2π/60 = rad/s
linear speed along the arc: v = r ω (radians only!)
On a merry-go-round every horse turns together, so all share the same ω. But the outer horse traces a bigger circle in the same time, so its linear speed v = rω is larger. The velocity vector is always tangent to the circle.
worked example — inner vs outer rider
A merry-go-round spins at ω = 10 rad/s. Compare a rider at r = 0.70 m with one at r = 0.35 m.
outer: v = rω = 0.70 × 10 = 7.0 m/s · inner: 0.35 × 10 = 3.5 m/s — same ω, half the r → half the speed.
Exam point: every point of a rigid rotating body — hub to rim — shares one ω, but each has a different speed v = rω.
3 — Centripetal force F = mv²/r — who provides it
Circular motion is a forced state. To keep the stone on its circle, your hand must pull it towards the centre every instant through the string's tension. By Newton's second law this centre-directed pull is the centripetal force:
Centripetal forceFₛ = m a = m v²/r = m r ω² — always towards the centre · unit newton (N)
It is not a new kind of force — it is the name of a job that some real force must do:
| Motion | Who provides the centripetal force? |
|---|
| stone whirled on a string | tension in the string |
| car on a flat curve | friction (tyres–road) |
| Moon / satellite | gravity |
| car on a banked road | component of the normal reaction |
worked example — tension in the string
A 0.5 kg stone whirls in a circle of r = 0.8 m at 2 rev/s. Find the tension.
ω = 2πf = 4π = 12.57 rad/s → F = mrω² = 0.5 × 0.8 × 12.57² = ≈ 63 N
4 — No inward force, no circle: the tangent
Remove the inward pull — cut the string — and Newton's first law takes over: the stone continues in a straight line along the tangent, carrying whatever velocity it had at that instant. It does not fly outward.
Centripetal acceleration (turning the velocity)a = v²/r = rω² = vω = 4π²r/T² — towards the centre
a ∝ v² (double v → 4× a) · a ∝ 1/r (double r → half a)
Exam point: "centrifugal force" is fictitious — felt only inside the rotating frame. In the ground frame nothing pushes outward: mud from a tonga wheel, sparks from a grinding stone and the cut stone all leave tangentially, never radially.
worked example — acceleration on a curve
A car takes a curve of r = 100 m at v = 20 m/s. Find the centripetal acceleration.
a = v²/r = 400/100 = 4.0 m/s² ≈ 0.4 g — the push you feel in your seat.
5 — Banking of roads: tan θ = v²/rg
On a flat curve only friction turns the car — unreliable in rain. Engineers bank (tilt) the road by an angle θ so the normal reaction N itself leans towards the centre and supplies the turning force.
Resolve N on a road banked at θvertical: N cos θ = m g (supports the weight)
horizontal: N sin θ = m v²/r (turns the car)
divide: tan θ = v² / (r g) → design speed v = √(r g tan θ)
θ is independent of mass — the same motorway bank serves a Mehran and a loaded truck. The same physics tilts a turning aeroplane and a leaning cyclist.
worked example — banking a motorway curve
A curve of r = 320 m is designed for 25 m/s. Find the banking angle (g = 9.8 m/s²).
tan θ = 625/(320 × 9.8) = 0.199 → θ ≈ 11.3°
6 — The centrifuge & the washing-machine spin
A centrifuge spins a mixture at high ω. Every particle needs F = mrω² towards the axis. Denser particles need more force than the liquid can supply, so they drift to the wall — separation by density (cream separator, blood centrifuge).
In a washing-machine spin dryer the perforated drum wall pushes the clothes inward onto the circle, but at each hole nothing pushes the water — so it continues along the tangent, out through the perforations. The clothes don't "throw the water out"; the drum simply stops steering it.
worked example — spin-dryer acceleration
A spin drum of r = 0.25 m turns at 1200 rpm. Find the centripetal acceleration at the wall, in g's.
ω = 1200 × 2π/60 = 125.7 rad/s → a = rω² = 0.25 × 125.7² ≈ 3 950 m/s² ≈ 400 g
7 — Constant speed, changing direction
On a Ferris wheel the cabins move at a steady speed, yet the direction of the velocity changes every instant. A change in velocity is an acceleration — so even at constant speed the motion is accelerated, and that acceleration always points to the centre.
Derivation sketch (similar triangles)Δv/v = Δs/r → Δv = (v/r)·vΔt → a = Δv/Δt = v²/r
a = v²/r = rω² — directed towards the centre
You feel it as a gentle press: heaviest at the bottom of the wheel, lightest at the top — your seat's normal force changes to keep supplying the centripetal pull.
worked example — Ferris-wheel acceleration
A Ferris wheel of r = 9 m turns once every 12 s. Find a rider's centripetal acceleration.
ω = 2π/T = 0.524 rad/s → a = rω² = 9 × 0.524² ≈ 2.5 m/s²
8 — Motion in a vertical circle · exam recap
Whirl a bucket of water in a vertical circle: gravity helps the centripetal pull at the top and opposes it at the bottom, so the tension changes around the loop.
Tension around a vertical circletop: T = m v²/r − m g (gravity helps → slackest)
bottom: T = m v²/r + m g (gravity opposes → tightest)
minimum top speed (T = 0): v = √(g r) · full loop: v ≥ √(5gr) at the bottom
worked example — bucket of water
A bucket is whirled in a vertical circle of r = 1.0 m. Minimum top speed so no water spills?
v = √(gr) = √9.8 = 3.13 m/s — an easy arm swing.
- θ = s/r (radian, dimensionless); ω = 2π/T = 2πf; v = rω.
- a = v²/r = rω² towards the centre; a ∝ v², a ∝ 1/r.
- F = mv²/r is a job — tension, friction, gravity or N's component does it; cut it → tangent.
- Banking: tan θ = v²/rg, mass-independent.
- Vertical circle: T = mv²/r ∓ mg (top/bottom); minimum top speed √(gr).
- Centrifuge/spin dryer: the unsteered leaves on the tangent; a = rω² can reach hundreds of g.