Circular Motion
Constant speed, but constantly changing direction — so there IS acceleration.
An object moving in a circle at constant speed is continuously changing direction — so its velocity vector changes, which means it IS accelerating. The centripetal acceleration always points toward the centre of the circle: a_c = v²/r = ω²r. 'Centripetal' means 'centre-seeking'. Because a_c is perpendicular to v at all times, it changes direction not speed.
Centripetal force is NOT a separate type of force — it is whatever real force happens to point toward the centre: Gravity → satellite in orbit. Tension → ball on a string. Friction → car turning on a flat road. Normal force → car in a loop-the-loop at the top. Always identify the real force in your FBD first, then set it equal to mv²/r.
Angular velocity ω measures how fast the angle changes: ω = 2π/T = 2πf (rad/s). It relates to linear speed by v = ωr. A larger radius means a larger linear speed for the same ω. In exam problems, be ready to convert: if given RPM, convert to rad/s first (1 rpm = 2π/60 rad/s).
Period T = time for one complete revolution (seconds). Frequency f = number of revolutions per second (Hz). They are reciprocals: T = 1/f. For a circle: T = 2πr/v = 2π/ω. In exam problems, frequency is sometimes given in rpm — convert to Hz by dividing by 60.
Speed is constant, but the velocity vector continuously changes direction. The object traces a circular path. v is always tangential (perpendicular) to the radius. a_c is always radial (pointing inward). No work is done by the centripetal force (force ⊥ displacement at every instant).
At BOTTOM of loop: centripetal direction is upward. N − mg = mv²/r → N = mg + mv²/r. Normal force is greater than weight — you feel 'heavier'. At TOP of loop: centripetal direction is downward. T + mg = mv²/r (string) or mg − N = mv²/r (track). Minimum speed at top: set T = 0 (string) or N = 0 (track) → v_min = √(gr). Car over a hill: mg − N = mv²/r → N = m(g − v²/r). Car loses contact when N = 0: v_max = √(gr). Key: always identify which forces point TOWARD the centre at that position.
A banked road allows a vehicle to turn without friction. At the ideal banking angle θ for speed v: horizontal component of N provides centripetal force; vertical component balances weight. N cosθ = mg (vertical). N sinθ = mv²/r (centripetal). Dividing: tanθ = v²/(rg). At this angle, no friction needed. Above ideal speed: friction acts inward. Below ideal speed: friction acts outward. Exam tip: derive by resolving N into components — do NOT try to memorise; derive it every time.
F–v² Graph
slope: Slope = m/rFor fixed mass and radius, F_c = (m/r)·v². So F vs. v² is a straight line through the origin. The slope = m/r, which can be used to determine mass or radius experimentally.
F–(1/r) Graph
slope: Slope = mv²For fixed mass and speed, F_c = mv²·(1/r). So F vs. (1/r) is a straight line. Slope = mv².
F_c = ma_c. Always identify the real force from dynamics (gravity, tension, friction, normal force) that provides the centripetal force.
💡Exam tip: Draw the FBD. Identify which force(s) point toward the centre. Set them equal to mv²/r.
The symbol ω (angular frequency) is shared between circular motion and SHM. SHM can be viewed as the projection of circular motion onto one axis.
💡Exam tip: ω = 2π/T appears in both topics with the same meaning — memorise it once.
Click any formula to see symbol definitions.
All formulas for this topic are in BINAS BINAS 35A3. In the exam you don't need to memorise the equations — but you must know which table to open and what every symbol means.