Circular motion is everywhere around us. Satellites maintain orbit, car wheels turn, planets sweep around the sun, and fairground rides spin their passengers — all governed by the same elegant mathematics. At its heart, circular motion requires a constant inward force to keep an object curving rather than flying off in a straight line. Understanding where that force comes from, and what happens when it changes, unlocks a powerful branch of Mechanics.
v = rω | a = v²/r = ω²r | F = mv²/r = mω²r | T = 2π/ω
In this module you will move from the basic vocabulary of angular velocity through to challenging vertical circle problems where energy conservation and Newton's second law work together. Every formula here links back to a physical reality you can picture.
Learning Objectives
Convert between linear speed v and angular velocity ω using v = rω
Calculate period T and frequency f from angular velocity
Calculate centripetal acceleration using a = v²/r = ω²r
Calculate centripetal force using F = mv²/r = mω²r
Analyse the conical pendulum by resolving forces vertically and horizontally
Analyse vertical circle problems using Newton's second law at specific points
Find minimum speed at the top of a vertical loop for maintained contact or taut string
Find tension and normal reaction at any point in a vertical circle using energy conservation
Explain why centripetal force is not a new force but the resultant of real forces
Topics in This Module
Angular Velocity
ω in rad/s, linking to linear speed v = rω and period T
Centripetal Acceleration
a = v²/r = ω²r always directed towards the centre
Centripetal Force
F = mv²/r — the required resultant of real forces
Conical Pendulum
Resolve vertically and horizontally to find T, ω, θ
Vertical Circles
Newton's 2nd law at top, bottom, and general position
Minimum Speed
Conditions for T ≥ 0 or N ≥ 0 at the top of a loop
Learn 1 — Angular Velocity and Circular Motion Basics
When an object moves in a circle, we describe its rotation using angular velocity ω (omega), measured in radians per second (rad/s). It tells us how quickly the angle swept out changes with time.
ω = θ/t (rad/s) | v = rω
Linear Speed and Angular Velocity
An object at radius r from the centre travels a distance rθ in time t. Its linear speed is therefore:
v = arc length / time = rθ/t = r × (θ/t) = rω
So v = rω links the linear speed v (m/s) to the angular velocity ω (rad/s) and radius r (m).
Period and Frequency
One complete revolution sweeps out an angle of 2π radians. The period T is the time for one full revolution.
T = 2π/ω = 2πr/v | f = 1/T = ω/(2π)
Derivation: In time T, the angle swept = 2π. Since ω = θ/t, we get ω = 2π/T, therefore T = 2π/ω.
Example — Satellite
A satellite orbits at radius r = 7.0 × 10⁶ m with period T = 5600 s. Find ω and v.
ω = 2π/T = 2π/5600 ≈ 1.12 × 10⁻³ rad/s
v = rω = 7.0 × 10⁶ × 1.12 × 10⁻³ ≈ 7840 m/s
Converting Units
Revolutions per minute (rpm) to rad/s: multiply by 2π/60 Revolutions per second (Hz) to rad/s: multiply by 2π
Example: 300 rpm = 300 × 2π/60 = 10π ≈ 31.4 rad/s
Tip: ω must be in rad/s for all formulas to work. Always convert from rpm or Hz before substituting.
Learn 2 — Centripetal Acceleration and Force
Even when an object moves at constant speed in a circle, its velocity is changing direction. By Newton's first law, this requires a net force. The acceleration — and the force producing it — always point towards the centre of the circle.
a = v²/r = ω²r (towards centre)
F = mv²/r = mω²r (towards centre)
Why There Is an Acceleration
Imagine two velocity vectors at two nearby points on the circle, both of magnitude v but different directions. The change in velocity Δv points inward. In the limit as the time interval → 0, this inward acceleration has magnitude v²/r. Using v = rω: a = (rω)²/r = ω²r.
Centripetal Force Is NOT a New Force
Key concept: "Centripetal force" is just the name we give to the net inward resultant of whatever real forces are acting. It is provided by:
• Tension in a string (ball on a string)
• Normal reaction (car on a banked bend, roller coaster loop)
• Gravity (satellite orbiting Earth)
• Friction (car on flat road round a bend)
Never draw a separate "centripetal force" arrow on a free-body diagram.
Applying Newton's Second Law
Step 1: Draw a free-body diagram. Identify all real forces. Step 2: Resolve towards the centre. The net inward component = mv²/r. Step 3: Solve for the unknown.
Tip: "Towards the centre" is the positive direction when applying F = mv²/r. Forces pointing away from the centre are negative in this equation.
Learn 3 — Horizontal Circular Motion
In horizontal circular motion, the object moves in a horizontal plane. The centripetal force is a horizontal force directed towards the centre. Gravity acts vertically and is balanced by a vertical component of another force (tension, normal reaction, etc.).
Stone on a String (Horizontal Circle)
A stone of mass m is attached to a string of length r and swings in a horizontal circle at angular velocity ω. The string makes no angle with the horizontal (idealised). Tension T provides centripetal force: T = mω²r
There is no vertical component, so gravity must be balanced by an upward support — in reality, the string must be slightly inclined. For exam problems where "horizontal circle" is stated, use T = mv²/r directly.
The Conical Pendulum
A ball of mass m hangs on a string of length L at angle θ to the vertical, moving in a horizontal circle of radius r = L sinθ.
Vertical: T cosθ = mg | Horizontal: T sinθ = mω²r = mω²L sinθ
From the vertical equation: T = mg/cosθ From the horizontal equation: T sinθ = mω²L sinθ → T = mω²L Dividing: mg/cosθ = mω²L → ω² = g/(L cosθ) → ω = √(g/(L cosθ)) Period: T_period = 2π/ω = 2π√(L cosθ / g)
Memorise: Conical pendulum — resolve vertically (T cosθ = mg) and horizontally towards centre (T sinθ = mω²r). These two equations let you find any unknown.
Car on a Flat Bend
A car of mass m takes a circular bend of radius r at speed v. Friction provides the centripetal force.
F_friction = mv²/r
Maximum speed: F_friction ≤ μmg → v_max = √(μgr)
Watch out: For the conical pendulum, it's T cosθ = mg (the vertical component). The angle θ is measured from the vertical. If the angle is measured from the horizontal, cosθ and sinθ swap — read the question carefully.
Learn 4 — Vertical Circles
When an object moves in a vertical circle, both its speed and the forces acting on it change continuously. We apply Newton's second law towards the centre at any position, and use energy conservation to relate speeds at different points.
At the Bottom of the Circle
N − mg = mv²/r → N = mg + mv²/r
The normal reaction N acts upward (towards centre), weight mg acts downward (away from centre). Net inward force = N − mg = mv²/r. Note: N > mg always at the bottom — you feel heavier at the bottom of a loop.
At the Top of the Circle
mg + N = mv²/r → N = mv²/r − mg
At the top, BOTH weight mg and normal reaction N (or tension T) point inward (downward, towards centre). Their sum provides centripetal force: mg + N = mv²/r.
Energy Conservation in Vertical Circles
If no friction, total mechanical energy is conserved. Taking the bottom as the reference level (h = 0):
½mv_bottom² = ½mv_top² + mg(2r)
At the top, height = 2r above the bottom.
Therefore: v_bottom² = v_top² + 4gr
Or in general between any two points: ½mv₁² + mgh₁ = ½mv₂² + mgh₂
General Position at Angle θ from Bottom
Height above bottom: h = r − r cosθ = r(1 − cosθ)
Energy: v² = v_bottom² − 2gr(1 − cosθ)
Centripetal equation: T − mg cosφ = mv²/r (where φ is angle from vertical through centre)
Strategy: (1) Use energy conservation to find v at the required position. (2) Apply Newton's 2nd law towards the centre at that position to find T or N.
Learn 5 — Minimum Speed Conditions
The most common vertical circle question asks: what is the minimum speed at the top of the loop? The answer comes from the physical constraint that the string cannot push and a surface cannot pull.
String (Tension T ≥ 0)
At the top of a vertical circle (string):
mg + T = mv²/r
For the string to remain taut: T ≥ 0
Minimum condition: T = 0 → mg = mv²/r → v_min = √(gr)
If v at the top < √(gr), the string goes slack and circular motion breaks down. The object becomes a projectile.
Surface (Normal Reaction N ≥ 0)
At the top of a loop (inside surface, e.g. roller coaster):
mg + N = mv²/r
For maintaining contact: N ≥ 0
Minimum condition: N = 0 → v_min = √(gr)
Finding Minimum Speed at the Bottom
Given minimum v_top = √(gr), use energy conservation to find the minimum speed needed at the bottom:
½mv_bottom² = ½mv_top² + mg(2r)
v_bottom² = gr + 4gr = 5gr v_bottom_min = √(5gr)
Outside of a Circular Surface (e.g. ball on top of a sphere)
At the top: mg − N = mv²/r (weight inward, N outward)
Leaves surface when N = 0: mg = mv²/r → v = √(gr)
Using energy from rest at the top: mgh = ½mv² → the object slides off at height h = r/3 above the centre... this leads to a geometric result on the sphere.
Important: v_min = √(gr) applies at the TOP. At the bottom the minimum is √(5gr). These are for a complete vertical circle with a string. Never confuse the two.
Worked Examples
Example 1 — Basic circular motion quantities
A particle moves in a circle of radius 0.5 m with angular velocity 4 rad/s. Find: (a) linear speed v, (b) centripetal acceleration a, (c) period T.
A child of mass 40 kg sits on a roundabout of radius 2 m, making 1 revolution every 4 seconds. Find (a) centripetal force, (b) friction force required.
Since N > 0, the particle maintains contact with the surface. ✓
Example 8 — Just maintaining contact at top of loop
At what speed must a ball move at the top of a loop of radius 0.6 m to just maintain contact with the track?
Just maintaining contact: N = 0
mg = mv²/r → v² = gr = 9.8 × 0.6 = 5.88 v = √5.88 ≈ 2.42 m/s2
Common Mistakes
Mistake 1 — Drawing "centrifugal force" on a diagram
Wrong: Drawing an outward arrow labelled "centrifugal force" on a free-body diagram in an inertial frame.
Correct: In an inertial reference frame, centrifugal force does not exist. Only real forces appear on the diagram. The net real force inward IS the centripetal force.
Mistake 2 — Wrong signs at the TOP of a vertical circle
Wrong: Writing T − mg = mv²/r at the top (treating weight as if it acts away from the centre).
Correct: At the TOP, weight acts downward = inward, tension/normal reaction also acts inward (downward). Both are in the same direction as centripetal acceleration: mg + T = mv²/r.
Mistake 3 — Wrong equation at the BOTTOM of a vertical circle
Wrong: Writing T = mv²/r and ignoring gravity, giving a tension that is too small.
Correct: At the BOTTOM, N (or T) acts upward = inward; weight mg acts downward = outward. Net inward = N − mg = mv²/r, so N = mg + mv²/r. Normal reaction is always greater than weight at the bottom.
Mistake 4 — Treating centripetal force as a separate force
Wrong: Listing three forces — tension, weight, and "centripetal force" — in a free-body diagram.
Correct: Centripetal force is the name for the required net inward force. It is already accounted for by the tension, weight, normal reaction, etc. Never add it as an extra force.
Mistake 5 — Not converting ω units
Wrong: Using ω in rpm directly in the formula F = mω²r, giving a wildly incorrect answer.
Correct: Always convert to rad/s first. 1 rpm = 2π/60 rad/s. Then substitute into F = mω²r.
Mistake 6 — Mixing up sin and cos in the conical pendulum
Wrong: Writing T sinθ = mg (vertical resolution) when θ is the angle with the vertical.
Correct: When θ is the angle the string makes with the vertical: T cosθ = mg (vertical) and T sinθ = mω²r (horizontal). The vertical component uses cosθ.
Mistake 7 — Forgetting energy conservation in vertical circle problems
Wrong: Trying to apply F = mv²/r at two different points with the same v, ignoring the speed change as height changes.
Correct: Speed changes with height. Use ½mv₁² + mgh₁ = ½mv₂² + mgh₂ to relate speeds at different positions before applying Newton's second law at the required point.
Key Formulas — Motion in a Circle
Formula
Description
Notes
v = rω
Linear speed from angular velocity
r in m, ω in rad/s, v in m/s
T = 2π/ω
Period of circular motion
T in seconds, ω in rad/s
f = 1/T = ω/(2π)
Frequency
f in Hz (cycles per second)
a = v²/r = ω²r
Centripetal acceleration
Always directed towards the centre
F = mv²/r = mω²r
Centripetal force (net inward resultant)
Not a new force — resultant of real forces
T cosθ = mg
Conical pendulum — vertical resolution
θ = angle of string with vertical
T sinθ = mω²r
Conical pendulum — horizontal resolution
r = L sinθ where L = string length
ω = √(g / L cosθ)
Angular velocity of conical pendulum
Derived from combining the two above
At bottom: N − mg = mv²/r
Vertical circle — bottom
N acts up (inward), mg acts down (outward)
At top: mg + N = mv²/r
Vertical circle — top
Both mg and N act inward (downward)
v_min = √(gr)
Minimum speed at TOP of vertical circle
When T = 0 or N = 0 at the top
v_bottom_min = √(5gr)
Min speed at BOTTOM for complete loop
Derived using energy conservation: v_b² = v_t² + 4gr = 5gr
½mv₁² + mgh₁ = ½mv₂² + mgh₂
Energy conservation (no friction)
Use to relate speeds at different heights in vertical circle
Proof Bank
Proof 1 — Centripetal Acceleration a = v²/r
Consider a particle moving with constant speed v in a circle of radius r. At time t it is at point A, at time t + δt it is at point B. The angle swept is δθ = ω δt.
The velocity at A is v_A (tangential) and at B is v_B (also tangential, but rotated by δθ). Both have magnitude v.
The change in velocity: |Δv| ≈ v δθ (for small δθ, the chord of the velocity triangle ≈ arc = v δθ).
The direction of Δv points from A towards the centre (inward).
Acceleration: a = |Δv|/δt = v δθ/δt = v ω
Since v = rω: a = vω = (rω)ω = ω²r. Also a = vω = v(v/r) = v²/r.
Therefore a = v²/r = ω²r, directed towards the centre. ∎
Proof 2 — Minimum Speed at Top of Vertical Circle (v_min = √(gr))
Consider a particle of mass m moving in a vertical circle of radius r on the inside of a smooth loop (or on a string). At the top of the circle, both the tension T (or normal reaction N) and the weight mg act downward — both towards the centre.
Applying Newton's second law towards the centre (downward at the top):
T + mg = mv²/r
Rearranging: T = mv²/r − mg = m(v²/r − g)
For a string: T ≥ 0 (a string cannot push). The minimum condition is T = 0:
0 = m(v_min²/r − g) → v_min² = gr → v_min = √(gr)
If v at the top were less than √(gr), T would need to be negative — impossible for a string. The string goes slack before reaching the top unless the speed there satisfies v ≥ √(gr). ∎
Corollary (minimum speed at bottom): Using energy conservation between bottom and top:
½mv_b² = ½mv_t² + mg(2r)
v_b² = v_t² + 4gr. At minimum, v_t = √(gr), so v_b² = gr + 4gr = 5gr → v_b_min = √(5gr). ∎
Circular Motion Visualiser
Adjust radius and angular velocity. The particle moves in real time. The blue arrow shows the radius, the orange arrow shows the instantaneous velocity direction (tangential).
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1.0
Exercise 1 — Angular Velocity and Speed
Exercise 2 — Centripetal Acceleration
Exercise 3 — Centripetal Force
Exercise 4 — Vertical Circles
Exercise 5 — Conical Pendulum and Mixed
Practice — 30 Questions (Mixed Topics)
Challenge — 15 Hard Questions
Exam Style Questions
Question 1 [5 marks]
A particle P of mass 0.4 kg is attached to one end of a light inextensible string of length 0.9 m. The other end is fixed at point O. P moves in a horizontal circle of radius 0.9 m with angular velocity ω. The string makes an angle of 60° with the vertical.
(a) Find the tension in the string. [2]
(b) Find the angular velocity ω. [3]
(a) T cos60° = mg → T × 0.5 = 0.4 × 9.8 → T = 3.92/0.5 = 7.84 N [M1 A1] (b) r = L sin60° = 0.9 × (√3/2) = 0.9√3/2 m
T sin60° = mω²r → 7.84 × (√3/2) = 0.4 × ω² × 0.9 × (√3/2)
7.84 = 0.4 × ω² × 0.9 → ω² = 7.84/(0.36) = 21.78 → ω ≈ 4.67 rad/s [M1 M1 A1]
Question 2 [6 marks]
A ball of mass 0.5 kg moves in a vertical circle of radius 0.8 m on a string. At the bottom of the circle its speed is 6 m/s.
(a) Find the tension in the string at the bottom. [2]
(b) Using energy conservation, find the speed at the top. [2]
(c) Find the tension at the top and state whether the string is taut. [2]
(a) T − mg = mv²/r → T = 0.5 × 9.8 + 0.5 × 36/0.8 = 4.9 + 22.5 = 27.4 N [M1 A1] (b) ½mv_b² = ½mv_t² + mg(2r) → v_t² = 36 − 2 × 9.8 × 1.6 = 36 − 31.36 = 4.64 → v_t ≈ 2.15 m/s [M1 A1] (c) T_top = mv_t²/r − mg = 0.5 × 4.64/0.8 − 0.5 × 9.8 = 2.9 − 4.9 = −2.0 N. Since T < 0, the string would go slack before reaching the top. The ball does not complete a full vertical circle. [M1 A1]
Question 3 [4 marks]
A particle moves in a circle of radius 1.5 m. The centripetal acceleration is 24 m/s². Find (a) the speed v, and (b) the angular velocity ω.
(a) a = v²/r → 24 = v²/1.5 → v² = 36 → v = 6 m/s [M1 A1] (b) v = rω → 6 = 1.5ω → ω = 4 rad/s [M1 A1]
Question 4 [5 marks]
A car of mass 900 kg goes over the top of a hill, which is a circular arc of radius 40 m, at speed v m/s. Find (a) the normal reaction at speed 15 m/s, and (b) the maximum speed for which the car maintains contact with the road.
(a) At top of hill: mg − N = mv²/r (weight inward, N outward)
N = mg − mv²/r = 900 × 9.8 − 900 × 225/40 = 8820 − 5062.5 = 3757.5 N [M1 A1] (b) Contact lost when N = 0: mg = mv_max²/r → v_max² = gr = 9.8 × 40 = 392 → v_max = √392 ≈ 19.8 m/s [M1 M1 A1]
Question 5 [6 marks]
A particle of mass 0.3 kg is on the inside of a smooth circular track of radius 0.5 m in a vertical plane. The particle has speed 4 m/s at the top of the track.
(a) Find the normal reaction at the top. [2]
(b) Find the speed at the bottom using energy conservation. [2]
(c) Find the normal reaction at the bottom. [2]
(a) At top: mg + N = mv²/r → N = m(v²/r − g) = 0.3(16/0.5 − 9.8) = 0.3(32 − 9.8) = 0.3 × 22.2 = 6.66 N [M1 A1] (b) v_b² = v_t² + 4gr = 16 + 4 × 9.8 × 0.5 = 16 + 19.6 = 35.6 → v_b ≈ 5.97 m/s [M1 A1] (c) At bottom: N − mg = mv_b²/r → N = 0.3 × 9.8 + 0.3 × 35.6/0.5 = 2.94 + 21.36 = 24.3 N [M1 A1]
Question 6 [4 marks]
A conical pendulum consists of a particle of mass 0.2 kg on a string of length 0.6 m. The particle moves in a horizontal circle and the string makes an angle θ with the vertical such that cosθ = 0.75. Find (a) the tension T, and (b) the radius of the circle.
(a) T cosθ = mg → T = mg/cosθ = 0.2 × 9.8/0.75 = 2.613 N [M1 A1] (b) sinθ = √(1 − 0.75²) = √(1 − 0.5625) = √0.4375 ≈ 0.661
r = L sinθ = 0.6 × 0.661 ≈ 0.397 m [M1 A1]
Question 7 [5 marks]
A particle moves in a complete vertical circle of radius r on a light inextensible string. Show that the difference in tension between the bottom and the top of the circle is 6mg.
A particle of mass m is on the outside of a smooth sphere of radius R, initially at the top. It slides from rest. Find the angle θ from the vertical at which it leaves the surface. (Hint: N = 0 when it leaves.)
At angle θ from vertical, height fallen from top: h = R − R cosθ = R(1 − cosθ)
Energy: ½mv² = mgh = mgR(1 − cosθ) → v² = 2gR(1 − cosθ) [M1]
At angle θ, centripetal direction is along the radius (inward). Radial equation:
mg cosθ − N = mv²/R [M1]
At the point of leaving, N = 0: mg cosθ = mv²/R = m × 2gR(1 − cosθ)/R = 2mg(1 − cosθ)
cosθ = 2 − 2cosθ → 3cosθ = 2 → cosθ = 2/3, θ = arccos(2/3) ≈ 48.2° [M1 A1]
Past Paper Questions — Cambridge 9709 Style
Past Paper Q1 — 9709 M2 Style [6 marks]
A particle of mass 0.8 kg is attached to one end of a light inextensible string of length 1.2 m. The other end is attached to a fixed point O. The particle moves in a horizontal circle below O. The string makes an angle of 45° with the vertical.
(i) Find the tension in the string. [2]
(ii) Find the angular velocity of the particle. [2]
A particle of mass 0.6 kg moves in a vertical circle of radius 0.5 m. At the lowest point of the circle the tension in the string is 18 N.
(i) Find the speed of the particle at the lowest point. [2]
(ii) Find the speed at the highest point using energy conservation. [3]
(iii) Find the tension at the highest point. [2]
(i) T − mg = mv²/r → 18 − 0.6×9.8 = 0.6×v²/0.5 → 18 − 5.88 = 1.2v² → 12.12 = 1.2v² → v² = 10.1 → v = 3.18 m/s [M1 A1] (ii) v_t² = v_b² − 4gr = 10.1 − 4×9.8×0.5 = 10.1 − 19.6 = −9.5
Since v_t² < 0, the particle does not reach the top. The string becomes slack before the particle completes the circle. [M1 M1] — award full marks for correct calculation showing this conclusion. [A1] Note: If the question is altered to initial tension 30 N: v_b² = (30−5.88)/1.2 = 20.1; v_t² = 20.1−19.6 = 0.5; v_t ≈ 0.707 m/s (iii) T_t = mv_t²/r − mg = 0.6×0.5/0.5 − 0.6×9.8 = 0.6 − 5.88 (using v_t² = 0.5) ≈ −5.28 N → string slack [M1 A1]
Past Paper Q3 — 9709 M2 Style [5 marks]
A particle of mass 0.4 kg moves on the inside of a smooth circular surface of radius 2 m in a vertical plane. At the top of the circle the normal reaction is 3 N. Find (i) the speed at the top, and (ii) the normal reaction at the bottom.
A small ball of mass m is attached to a string of length r and moves in a complete vertical circle. Show that the minimum speed at the bottom of the circle for the ball to maintain a taut string throughout is √(5gr), and find the tension at the bottom under this minimum condition.
Minimum speed at top: v_t_min = √(gr) (T = 0 at top) [M1 A1]
Energy conservation from bottom to top (height gain = 2r):
½mv_b² = ½mv_t² + mg(2r) [M1]
v_b² = v_t² + 4gr = gr + 4gr = 5gr → v_b_min = √(5gr) ∎ [A1]
Tension at bottom: T − mg = mv_b²/r = m(5gr)/r = 5mg
T = mg + 5mg = 6mg [M1 A1]