Circular Motion: theory | Physics Circle

CONCEPT

Angular Variables

In one line
On a circle the radius stays fixed, so we follow the angle turned. v = rω links the angle to the speed.

DISPLACEMENT

3.1.1 Displacement and distance

The position vector keeps its size r but turns as the body moves.
Displacement is the chord AB. Distance is the arc AB.

Body moves from A to B through angle θ

Δr = 2r sin θ2

s = rθ

Half a turn: displacement 2r, distance πr. Full turn: displacement 0, distance 2πr.
Distance ÷ displacement = θ2 cosec θ2.
A wheel rolling through θ also moves forward by rθ.

ANGULAR DISPLACEMENT

3.1.2 Angular displacement

The angle turned from a reference line. Unit: radian. 2π rad = 360° = 1 revolution.
Dimension: [M⁰L⁰T⁰].
Its direction is along the axis, given by the right-hand rule.

ANGULAR VELOCITY

3.1.3 Angular velocity, period and frequency

ω = dθdt

ω = 2πT = 2πn

T = 1n

Unit rad s⁻¹, dimension [T⁻¹]. ω points along the axis.
Same period ⇒ same ω, whatever the mass or radius.
Given θ(t): differentiate once for ω, twice for α.

CLOCK HANDS

3.1.4 Hands of a clock

Hand	Period	ω (rad s⁻¹)
second	60 s	π30
minute	3600 s	π1800
hour	12 h = 43200 s	π21600
earth (spin)	24 h	π43200

ω_rel = ω₁ − ω₂, T_rel = T₁T₂T₂ − T₁

Minute and hour hands meet every 1211 h.
Equal ω (geostationary satellite and earth): no relative motion.

LINEAR SPEED

3.1.5 Linear speed: v = rω

v = rω = 2πrn = 2πrT

As vectors v = ω × r, so v ⟂ r at every point.
All points of a rotating body have the same ω, so v ∝ r.
At latitude λ on the earth: r = R cos λ, so v = V cos λ.

ANGULAR ACCELERATION

3.1.6 Angular acceleration

α = dωdt = d²θdt²

Linear (tangential) acceleration a = αr.
Uniform circular motion: α = 0.

CHANGE IN VELOCITY

3.1.7 Change in velocity

v₁ and v₂ have the same size; Δv joins their tips

Δv = 2v sin θ2

angle between Δv and v₂ = 90° − θ2

Half a turn: Δv = 2v. Quarter turn: Δv = v√2. 60°: Δv = v.
Average acceleration = Δvt. Average velocity = Δrt.
As θ → 0, Δv turns towards the centre: this is the centripetal acceleration.

Practise PYQs on this topicAngular Variables · 35 questions with solutions

CONCEPT

Centripetal Acceleration & Force

In one line
Motion on a circle needs a force towards the centre. It only turns the velocity, so it does no work.

ACCELERATION

3.2.1 Centripetal acceleration

v is along the tangent; F and a point to the centre

a = v²r = ω²r = vω

a = 4π²n²r = 4π²rT²

It points along the radius, to the centre. Its direction keeps turning.
Same ω: a ∝ r. Same v: a ∝ 1r.
At P(R, θ): a = −v²R(cos θ î + sin θ ĵ).

WHAT IS CONSTANT

3.2.2 What stays constant in UCM

Quantity	In UCM
speed, KE, ω, size of a, angular momentum	constant
velocity, momentum, acceleration (as vectors)	change: their direction turns

v ⟂ r, a ⟂ v, and a is opposite to r.

CENTRIPETAL FORCE

3.2.3 Centripetal force

F = mv²r = mω²r = 4π²mrT²

It is not a new kind of force: tension, friction, gravity, a normal reaction or a magnetic force does this job.
F ⟂ displacement, so the work done = 0 and KE stays constant.
With angular momentum L = mvr: KE = L²2mr². Same L, r halved ⇒ v doubles.

Situation	Centripetal force
stone whirled on a string	tension
car turning on a level road	friction on the tyres
vehicle on a speed breaker	weight or its part
earth round the sun	gravity
electron round the nucleus	Coulomb attraction
charge in a magnetic field	magnetic force

RATIOS

3.2.4 Ratios and percentage change

F ∝ mv²r: change each factor and multiply.
m, v, r all × 1.2: F × 1.2 × 1.441.2 = 1.44, a rise of 44%.
Equal forces on two stones: m₁v₁²r₁ = m₂v₂²r₂.

SPRING OR STRING

3.2.5 Spring or string gives the force

A spring stretched by x pulls the mass to the centre

kx = mω²(l + x)

x = mω²lk − mω²

Mass through a hole, holding a hanging mass M: Mg = mω²L.
String of breaking tension T_max: T_max = mω²L gives the largest ω.

CENTRIFUGAL FORCE

3.2.6 Centrifugal force

Felt only in a rotating frame: a pseudo force mω²r, outward.
From the ground there is no such force. When the string breaks, the body flies off along the tangent.

RADIUS OF CURVATURE

3.2.7 Radius of curvature

R = v²a_⟂

Top of a projectile: v = u cos θ and a_⟂ = g, so R = u² cos²θg.
At any other instant, use the part of g perpendicular to v.
Central force F ∝ r⁻ⁿ on a circular orbit: set mω²r = F. F ∝ r⁻¹ gives T ∝ r; F ∝ r⁻³ gives T ∝ r².

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Charged particle in a magnetic field

r = mvqB

The magnetic force qvB is always ⟂ v, so it gives the centripetal force and the speed never changes.
Period T = 2πmqB: it does not depend on the speed.

Practise PYQs on this topicCentripetal Acceleration & Force · 52 questions with solutions

CONCEPT

Level Road & Rotating Platform

In one line
On a flat road only friction can give the centripetal force, so μ sets the top speed.

LEVEL ROAD

3.3.1 Skidding on a level road

Top view: friction on the tyres points to the centre

μmg ≥ mv²r

v_max = √μrg

The mass does not matter.
μ ∝ v²: half the speed on a wet road means one-fourth the μ.
Faster than √μrg: the car skids outward.

DRUM

3.3.2 Drum and wall of death

The wall pushes to the centre; friction holds the weight

N = mv²r = mω²r

μN ≥ mg

v_min = √rgμ, ω_min = √gμr

Here a minimum speed is needed, not a maximum.
Drum with a horizontal axle (concrete mixer): mixing needs ω²r ≤ g at the top.

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Coin on a rotating disc

ω_max = √μgr

The coin farthest from the centre slips first.
Just slipping: r ∝ 1ω². ω halved ⇒ the coin can sit 4 times farther out.

Racing car with downforce

μ(mg + F_L) = mv²r

A downward lift F_L raises the normal force, so the car can take the turn faster.

Practise PYQs on this topicLevel Road & Rotating Platform · 15 questions with solutions

CONCEPT

Banking of Roads & Bridges

In one line
Tilting the road (or the rider) lets part of the normal reaction point to the centre.

CYCLIST

3.4.1 Bending of a cyclist

The cyclist leans by θ; R sin θ gives the centripetal force

R sin θ = mv²r

R cos θ = mg

tan θ = v²rg

Sharper turn or higher speed ⇒ lean more.
A bob hanging in a turning car tilts by the same θ.

BANKED ROAD

3.4.2 Banking of a road

(A) forces on a car on a banked road; (B) width l and rise h

tan θ = v²rg = ω²rg = vωg

tan θ ≈ hl

h = v²lrg

R cos θ = mg and R sin θ = mv²r.
This speed needs no friction at all.
Same θ: v² ∝ r. Speed × 1.2 ⇒ radius × 1.44.

RAILWAY TRACK

3.4.3 Railway track

The outer rail is raised by h. Gap between the rails = d (meter gauge: d = 1 m).
h = d tan θ = dv²rg.
Radius of the track: r = v²dgh.

BRIDGES

3.4.4 Concave and convex roads

Lowest point of a concave road and top of a convex bridge

concave: R = mg cos θ + mv²r

convex: R = mg cos θ − mv²r

θ is measured from the vertical; θ = 0 at the lowest or highest point.
Reaction: concave road > level road (mg) > convex bridge.
A car leaves the top of a convex bridge if v > √gr.

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Banked road with friction

v_max = √rg μ + tan θ1 − μ tan θ

v_min = √rg tan θ − μ1 + μ tan θ

At v_max friction acts down the slope; at v_min it acts up the slope.
Normal force at v_max: N(cos θ − μ sin θ) = mg.

Overturning of a vehicle

R₁ = 12M(g − v²hra), R₂ = 12M(g + v²hra)

v_max = √grah

2a = gap between the wheels, h = height of the centre of gravity.
The inner wheel leaves the road first.

Practise PYQs on this topicBanking of Roads & Bridges · 13 questions with solutions

CONCEPT

Non-Uniform Circular Motion

In one line
When the speed changes there is a tangential acceleration too: it changes the speed, a_c turns the direction.

TWO ACCELERATIONS

3.5.1 Tangential and centripetal acceleration

a_t along the tangent, a_c to the centre

a_t = αr = dvdt

a_c = ω²r = v²r

a = √a_c² + a_t²

a_ca_t = v²r²α.
Path length s(t) given: v = dsdt, a_t = dvdt.

a_c	a_t	Motion
0	0	uniform, straight line
0	≠ 0	accelerated, straight line
≠ 0	0	uniform circular
≠ 0	≠ 0	non-uniform circular

EQUAL ACCELERATIONS

3.5.2 When a_t = a_c

From rest with constant α: ω = αt.
a_t = a_c when ω² = α, so t = 1√α.
α = 4 rad s⁻² ⇒ t = 0.5 s.

EQUATIONS

3.5.3 Equations of circular motion

Speeding up	Slowing down
ω₂ = ω₁ + αt	ω₂ = ω₁ − αt
θ = ω₁t + 12αt²	θ = ω₁t − 12αt²
ω₂² = ω₁² + 2αθ	ω₂² = ω₁² − 2αθ
θ_n = ω₁ + α2(2n − 1)	θ_n = ω₁ − α2(2n − 1)

θ_n = angle turned in the n^th second.
From rest, angles in equal time intervals go 1 : 3 : 5.
A fan slows from ω₀ to ω₀3 in N turns: it makes N8 more turns before stopping.

AVERAGE VELOCITY

3.5.4 Average velocity from rest

v_av = rθt = r√αθ2

Small θ: displacement ≈ arc rθ, and θ = ½αt².

FORCE AND POWER

3.5.5 Forces, KE and power

Only the tangential force F_t = ma_t does work; the centripetal force does none.
Power P = F_tv. KE after a path length s: K = ma_ts.

Practise PYQs on this topicNon-Uniform Circular Motion · 21 questions with solutions

CONCEPT

Vertical Circle: Speed & Tension

In one line
In a vertical circle gravity changes the speed: the body is fastest at the bottom and slowest at the top, and the tension follows.

SPEED

3.6.1 Speed at any point

Body on a string of length l; θ is measured from the lowest point A

v = √u² − 2gl(1 − cos θ)

u = speed at the bottom; height risen h = l(1 − cos θ).
String horizontal: v² = u² − 2gl. Top: v² = u² − 4gl.

TENSION

3.6.2 Tension at any point

T − mg cos θ = mv²l

T = ml[u² − gl(2 − 3 cos θ)]

Point	Speed	Tension
A (0°)	u	mu²l + mg
B (90°)	√u² − 2gl	mu²l − 2mg
C (180°)	√u² − 4gl	mu²l − 5mg
D (270°)	√u² − 2gl	mu²l − 2mg

T_A − T_C = 6mg, T_A − T_B = 3mg, T_B − T_C = 3mg.
KE changes around a vertical circle; on a horizontal circle it stays the same.

UNIFORM SPEED

3.6.3 Whirled at a steady speed

T_max = mv²r + mg (bottom)

T_min = mv²r − mg (top)

n rpm: ω = 2πn60.
A thread breaks first at the bottom.

PENDULUM

3.6.4 Swinging pendulum

T_max = mg(3 − 2 cos θ₀)

T_min = mg cos θ₀

θ₀ = amplitude. T_max is at the bottom, T_min at the ends where v = 0.
Released from the horizontal (θ₀ = 90°): T_max = 3mg.

Practise PYQs on this topicVertical Circle: Speed & Tension · 16 questions with solutions

CONCEPT

Critical Speed & Looping

In one line
To just complete the loop the string must stay tight at the top: u = √5gl at the bottom.

CRITICAL SPEED

3.7.1 Just completing the circle

top: mg = mv²l ⇒ v_top = √gl

u_min = √5gl

Speeds and tensions when the body just completes the circle

Point	v	T	KE
A (bottom)	√5gl	6mg	52 mgl
B, D (side)	√3gl	3mg	32 mgl
C (top)	√gl	0	12 mgl
any θ	√gl(3 + 2 cos θ)	3mg(1 + cos θ)

Total energy = 52 mgl at every point.
a_c = g at the top, 3g at the sides, 5g at the bottom.
Apparent weight at the bottom = 6mg.

CONDITIONS

3.7.2 What happens for each speed

Speed at the bottom	What happens
u > √5gl	completes the circle; string never slack
u = √5gl	just completes; T = 0 at the top
√2gl < u < √5gl	string goes slack between B and C, then a parabola
u = √2gl	v and T both zero at B; swings in a semicircle
u < √2gl	swings about A below B

v = 0 at h = u²2g; T = 0 at h = u² + gl3g

BUCKET AND LOOP

3.7.3 Bucket of water and loop track

Water stays in the bucket if ω²r ≥ g at the top: period ≤ 2π√rg.
Smooth track with a loop of radius R: released from h ≥ 52R to go round.
Reaction N at the top of a loop: N + mg = mv²R, with v from energy.

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String going slack

T = 0 at angle φ above the horizontal: mg sin φ = mv²l.
Then use energy from the lowest point: u² = v² + 2gl(1 + sin φ).
After that the body moves on a parabola.

Body leaving a smooth hemisphere

Block slides from the top A and leaves the surface at B

mg cos θ = mv²r, v² = 2gr(1 − cos θ)

h = 23r, cos θ = 23

Released at θ₀ from the vertical: leaves where sin φ = 23 cos θ₀ (φ from the horizontal).
Inside a smooth bowl, released from the rim: N = 3mg sin α and F_c = 2mg sin α.

Practise PYQs on this topicCritical Speed & Looping · 13 questions with solutions

CONCEPT

Conical Pendulum & Funnel

In one line
The bob moves in a horizontal circle: T cos θ = mg, and T sin θ gives the centripetal force.

FORCES

3.8.1 Conical pendulum

String of length l at angle θ; the bob moves in a circle of radius r

T sin θ = mv²r

T cos θ = mg

tan θ = v²rg

v = √gr tan θ

TENSION

3.8.2 Tension in the string

T = mω²l

T = mgcos θ = mgl√l² − r²

T = mω²l for any θ: it depends only on how fast the bob turns.
Centripetal force = mg tan θ = mgr√l² − r².
String nearly horizontal (stone whirled round): T ≈ mv²l; the breaking tension gives the top speed.

PERIOD

3.8.3 Angular speed and period

ω = √gh = √gl cos θ

T_p = 2π√l cos θg = 2π√hg

h = l cos θ is the height of the support above the circle.

FUNNEL

3.8.4 Particle in a funnel

Particle on the smooth inner surface of a cone

N sin θ = mg, N cos θ = mv²r

h = v²g

θ = half-angle of the cone; h = height of the circle above the vertex.
h depends only on v: a faster particle runs higher.

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Corner

Bead on a rotating wire

tan θ = ω²xg

θ = slope of the wire where the bead sits, x = its distance from the axis.
Parabola y = kx²: slope 2kx, so ω² = 2kg.
Ring turning about a vertical diameter: cos θ = gω²r (θ from the downward vertical).

Practise PYQs on this topicConical Pendulum & Funnel · 16 questions with solutions

EXAM-DAY CARD

Just see this before the exam

v = rω

angular variables

a = v²r

centripetal

v = √μrg

level road

tan θ = v²rg

banking

u = √5gl

vertical circle

T cos θ = mg

conical pendulum

FORMULA SHEET

Whole chapter on one page

Angular variables

s = rθ, Δr = 2r sin θ2

ω = 2πT = 2πn

v = rω

Δv = 2v sin θ2

Centripetal

a = v²r = ω²r

F = mv²r = 4π²mrT²

W = 0

spring: kx = mω²(l + x)

Level road & drum

v_max = √μrg

drum: v_min = √rgμ

μ ∝ v²

coin: ω = √μgr

Banking & bridges

tan θ = v²rg ≈ hl

cyclist: tan θ = v²rg

concave: R = mg + mv²r

convex: R = mg − mv²r

Non-uniform

a = √a_c² + a_t²

a_t = αr

ω² = ω₀² + 2αθ

a_t = a_c: t = 1√α

Vertical & conical

v² = u² − 2gl(1 − cos θ)

T_A − T_C = 6mg

u_min = √5gl

T_p = 2π√hg, funnel h = v²g

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Half a turn: displacement = πr

In UCM the velocity is constant

Centripetal force does work

Critical speed at the bottom = √gl

CORRECT ✓

Displacement 2r; πr is the distance

Speed is constant; velocity turns

F ⟂ displacement, so W = 0

√gl at the top, √5gl at the bottom

Circular Motion

Angular Variables

3.1.1 Displacement and distance

3.1.2 Angular displacement

3.1.3 Angular velocity, period and frequency

3.1.4 Hands of a clock

3.1.5 Linear speed: v = rω

3.1.6 Angular acceleration

3.1.7 Change in velocity

Centripetal Acceleration & Force

3.2.1 Centripetal acceleration

3.2.2 What stays constant in UCM

3.2.3 Centripetal force

3.2.4 Ratios and percentage change

3.2.5 Spring or string gives the force

3.2.6 Centrifugal force

3.2.7 Radius of curvature

Corner

Level Road & Rotating Platform

3.3.1 Skidding on a level road

3.3.2 Drum and wall of death

Corner

Banking of Roads & Bridges

3.4.1 Bending of a cyclist

3.4.2 Banking of a road

3.4.3 Railway track

3.4.4 Concave and convex roads

Corner

Non-Uniform Circular Motion

3.5.1 Tangential and centripetal acceleration

3.5.2 When at = ac

3.5.3 Equations of circular motion

3.5.4 Average velocity from rest

3.5.5 Forces, KE and power

Vertical Circle: Speed & Tension

3.6.1 Speed at any point

3.6.2 Tension at any point

3.6.3 Whirled at a steady speed

3.6.4 Swinging pendulum

Critical Speed & Looping

3.7.1 Just completing the circle

3.7.2 What happens for each speed

3.7.3 Bucket of water and loop track

Corner

Conical Pendulum & Funnel

3.8.1 Conical pendulum

3.8.2 Tension in the string

3.8.3 Angular speed and period

3.8.4 Particle in a funnel

Corner

Just see this before the exam

Whole chapter on one page

Avoid these = easy marks

3.5.2 When a_t = a_c