Uniform circular motion is deceptive — an object moving at constant speed can still be accelerating. The force that causes this acceleration points inward, never forward, and understanding it unlocks planetary orbits, banked curves, and the spin of the cosmos.
Prerequisites Newton's Laws, Friction
Core equations a_c = v²/r, F_c = mv²/r
Sections 8
§ 01
Kinematics of Circular Motion
Before forces, we need the geometry. Circular motion introduces angular quantities that parallel every linear kinematic concept.
An object in uniform circular motion moves at constant speed along a circular path of radius $r$. The word "uniform" means the speed is constant — but since direction continuously changes, the velocity vector is not constant. This changing velocity implies acceleration.
Angular position
Angle θ
$\theta$ in radians
One full revolution = $2\pi$ rad = 360°. Radians are dimensionless ratios of arc length to radius: $\theta = s/r$.
Angular velocity
ω (omega)
$\omega = \dfrac{d\theta}{dt}$
Rate of angular change. Unit: rad/s. Relates to period: $\omega = 2\pi/T$ and to linear speed: $v = \omega r$.
Period & frequency
T and f
$T = 1/f = 2\pi/\omega$
Period $T$ is the time for one full revolution. Frequency $f$ is revolutions per second (Hz). These three are all interchangeable.
The key kinematic relations
Angular–Linear equivalence
Linear speed from ω
$$v = \omega r$$
A point further from the centre moves faster, even at the same $\omega$. This is why the tip of a spinning blade moves faster than the hub.
Period from speed
$$T = \frac{2\pi r}{v}$$
Circumference divided by speed. The time to complete one full circle.
Full angular equivalences
$$\omega = \frac{2\pi}{T} = 2\pi f$$
Three ways to express the same rotational rate.
Symbol reference
Symbol
Quantity
SI Unit
Linear analogue
$\theta$
Angular position
rad
$x$ (position)
$\omega$
Angular velocity
rad s⁻¹
$v$ (velocity)
$\alpha$
Angular acceleration
rad s⁻²
$a$ (acceleration)
$r$
Radius
m
—
$T$
Period
s
—
$f$
Frequency
Hz = s⁻¹
—
$v$
Tangential speed
m s⁻¹
$v$ (same)
§ 02
Centripetal Acceleration & Force
The acceleration of circular motion always points toward the centre. By Newton's second law, a net inward force must be responsible.
Even at constant speed, a circularly moving object has an acceleration — because its velocity direction is changing. This centripetal acceleration ("centre-seeking") always points perpendicular to the velocity, directly toward the centre of the circle.
Centripetal acceleration & force
Centripetal acceleration
$$a_c = \frac{v^2}{r} = \omega^2 r$$
Both forms are equivalent via $v = \omega r$. Points radially inward at every instant.
Centripetal force (Newton II)
$$F_c = \frac{mv^2}{r} = m\omega^2 r$$
Not a new type of force. This is whatever existing force (gravity, tension, friction, normal) is providing the inward net force.
✗ The biggest misconception in circular motion"Centrifugal force" — the sensation of being flung outward in a turning car — is not a real force in an inertial reference frame. It is a fictitious force that appears only in the non-inertial (rotating) frame. In the inertial frame, the car seat pushes you inward (centripetal). We return to this in §7.
📐 Direction is everything
Centripetal force is always directed radially inward — toward the centre. It is always perpendicular to the velocity. Therefore it does no work on the object (work = F·d cosθ, and θ = 90°). Speed stays constant; only direction changes.
§ 03
Derivation of $a_c = v^2/r$
The centripetal acceleration formula can be derived geometrically from the change in the velocity vector — no calculus required, just similar triangles.
Geometric derivation
Consider an object moving from point P to point Q on a circle of radius $r$ in a small time $\Delta t$. The arc length is $\Delta s = v\,\Delta t$. The velocity vectors at P and Q have the same magnitude $v$ but differ in direction by the same angle $\Delta\theta$ as the arc subtends at the centre.
$|\Delta\mathbf{v}| = v\,\Delta\theta$ for small $\Delta\theta$
The velocity vectors form an isoceles triangle of two sides $v$ and apex angle $\Delta\theta$. The chord length ≈ arc length = $v\,\Delta\theta$ for small angles.
Since $\omega = \Delta\theta/\Delta t$ (definition of angular velocity)
$a_c = v\omega = v \cdot \dfrac{v}{r}$
Substituting $\omega = v/r$ (from $v = \omega r$)
$$\boxed{a_c = \frac{v^2}{r}}$$
Direction: $\Delta\mathbf{v}$ points toward the centre of the circle (perpendicular to velocity, inward) — this can be seen from the geometry of the vector triangle.
Vector calculus derivation
Using position vector $\mathbf{r}(t) = r(\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j})$:
Differentiate velocity. Notice the result is $-\omega^2\mathbf{r}$: it points opposite to the position vector, i.e. toward the centre.
$$\mathbf{a} = -\omega^2\mathbf{r} \implies |\mathbf{a}| = \omega^2 r = \frac{v^2}{r}$$
The magnitude equals $\omega^2 r$ and it always points toward the centre. This is the centripetal acceleration, derived rigorously.
Live vector diagram — position, velocity & acceleration
§ 04
Free Body Diagrams for Circular Motion
The key question in any circular motion problem: which real force provides the centripetal force? The answer is always one of the forces already present.
Solving circular motion problems follows a clear procedure: draw the FBD, identify the net inward force, set it equal to $mv^2/r$. The centripetal force requirement is an equation to satisfy, not a force to draw.
Horizontal circle — ball on a string
Vertical circle — top of a loop
Vertical circle — bottom of a loop
Banked curve — no friction
Worked Example 1 — Ball on a horizontal string
Tension provides centripetal force
A 0.3 kg ball is swung in a horizontal circle on a 0.8 m string at 4 rev/s. Find the tension in the string.
Circular dynamics appears in virtually every domain of engineering and physics. The question is always: what real force is providing the centripetal requirement?
🏎
Banked curves
$\tan\theta = v^2/(rg)$
The normal force from a banked road has a horizontal component that provides centripetal force, reducing reliance on friction. Ideal banking angle depends only on speed and radius, not vehicle mass.
🎡
Vertical loops
$N = m(v^2/r \mp g)$
At the top: $N = mv^2/r - mg$ (normal and gravity both inward). At the bottom: $N = mv^2/r + mg$ (normal inward, gravity outward). You feel heaviest at the bottom.
🌍
Orbital motion
$v = \sqrt{GM/r}$
Gravity provides centripetal force for satellite orbits. The orbital speed depends only on the planet's mass and orbital radius — independent of satellite mass. LEO ≈ 7.9 km/s.
💫
Centrifuges
$a = \omega^2 r$ (up to $10^6\,g$)
Medical centrifuges spin samples at high $\omega$ to produce effective gravitational fields thousands of times Earth's. Denser particles migrate outward faster. Ultracentrifuges reach $10^6 g$.
✈️
Aircraft banking
$L\cos\theta = mg$, $L\sin\theta = mv^2/r$
A banked aircraft uses the horizontal component of lift as centripetal force. The vertical component of lift must still equal weight, so total lift increases during a turn — hence the sensation of extra weight.
⚗️
Conical pendulum
$T = 2\pi\sqrt{L\cos\theta/g}$
A mass on a string sweeping a horizontal circle. Tension resolves into vertical (balancing weight) and horizontal (centripetal). The period depends on the angle — not the mass or string length directly.
Banked curve — the full derivation
A banked curve at angle $\theta$ lets a car turn without any friction. The normal force alone provides the centripetal component:
Vertical: $\quad N\cos\theta = mg$
No vertical acceleration — vertical components balance
Newton's law of gravitation combined with circular motion gives us the orbital mechanics of every moon, planet, and satellite in the solar system.
Newton's Law of Gravitation
$$F_g = \frac{Gm_1m_2}{r^2}$$
$G = 6.674 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ (gravitational constant). Force is attractive, acts along the line joining the centres, and is mutual.
Orbital speed
For a satellite of mass $m$ orbiting a planet of mass $M$ at radius $r$, gravity provides the centripetal force:
$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$
Gravity = centripetal force requirement
$\dfrac{GM}{r} = v^2$
Cancel $m$ (the satellite mass drops out entirely — orbital speed is independent of satellite mass)
$$\boxed{v_\text{orb} = \sqrt{\frac{GM}{r}}}$$
Higher orbit → smaller orbital speed. Counterintuitive: farther out means moving more slowly.
Kepler's Third Law: the square of the orbital period is proportional to the cube of the orbital radius. This held empirically before Newton — now we understand why.
🌌 Geostationary orbit
A geostationary satellite has $T = 24\ \text{h}$ and orbits directly above the equator. Solving Kepler's Third Law for $r$ with $T = 86400\ \text{s}$ and $M_\text{Earth} = 5.97 \times 10^{24}\ \text{kg}$ gives $r \approx 42{,}164\ \text{km}$ — about 35,786 km above the surface. All TV and weather satellites at this altitude appear stationary from the ground.
Orbital data — solar system
Body
Orbital radius (AU)
Period (years)
T²/r³ check
Orbital speed (km/s)
Mercury
0.387
0.241
≈ 1.00
47.9
Venus
0.723
0.615
≈ 1.00
35.0
Earth
1.000
1.000
1.00
29.8
Mars
1.524
1.881
≈ 1.00
24.1
Jupiter
5.203
11.86
≈ 1.00
13.1
Saturn
9.537
29.46
≈ 1.00
9.7
Worked Example 5 — ISS orbital speed
Orbital mechanics
The ISS orbits at approximately 420 km above Earth's surface. Find its orbital speed and period. ($M_E = 5.97\times10^{24}$ kg, $R_E = 6.371\times10^6$ m)
altitude h = 420 kmM_E = 5.97×10²⁴ kgR_E = 6371 km
Orbital radius: $r = R_E + h = 6371 + 420 = 6791\ \text{km} = 6.791 \times 10^6\ \text{m}$
v ≈ 7.66 km/s | T ≈ 92.8 min (≈ 15.5 orbits per day)
§ 07
Non-Inertial Frames & Pseudo-Forces
In a rotating reference frame, objects appear to experience outward "centrifugal" forces and sideways "Coriolis" forces. These are not real forces — but they are essential for analysing rotating systems.
Recall from Module 1 that Newton's laws only hold in inertial frames. A rotating frame is non-inertial — things appear to accelerate without a real force. To apply Newton II in such a frame, we add fictitious forces equal and opposite to the frame's own acceleration effects.
Centrifugal force (fictitious)
In a frame rotating at angular velocity $\omega$, every object of mass $m$ at distance $r$ from the axis appears to experience an outward centrifugal force:
Centrifugal pseudo-force (rotating frame only)
$$\mathbf{F}_\text{cf} = m\omega^2 r \quad (\text{directed radially outward})$$
This is exactly equal and opposite to the centripetal acceleration term. In the rotating frame, this pseudo-force balances the real inward forces, explaining why objects appear stationary in the rotating frame even without net force.
Coriolis force (fictitious)
An object moving with velocity $\mathbf{v}'$ in a rotating frame also experiences a Coriolis force perpendicular to its velocity:
$\boldsymbol{\omega}$ is the rotation vector (pointing along the rotation axis). This force causes deflection of moving air masses (winds, ocean currents) and the rotation of Foucault pendulums. On Earth's surface, it deflects moving objects rightward in the Northern Hemisphere.
🌀 Coriolis in everyday life
The Coriolis effect causes hurricanes to rotate counterclockwise in the Northern Hemisphere (cyclonically) and clockwise in the Southern. The drainage direction of a sink is not reliably determined by Coriolis — the forces are too small relative to the initial rotation imparted while filling. Large-scale atmospheric and oceanic circulation, however, is dominated by it.
⚖️ Inertial vs rotating frame — same physics
Both approaches give the same observable predictions. In the inertial frame: the object accelerates inward, centripetal force is real. In the rotating frame: the object appears stationary, centrifugal and Coriolis forces are fictitious but mathematically valid. Neither is "more correct" — they are different computational tools.
Worked Example 6 — Apparent weight on the equator
Rotating frame / effective gravity
By how much does Earth's rotation reduce apparent weight at the equator? ($R_E = 6.371\times10^6$ m, $T = 86400$ s)
Reduction: $70 \times 0.034 = 2.4\ \text{N}$ or about 0.34% of weight
Apparent weight reduced by ≈ 2.4 N (0.34%) at the equator
§ 08
Interactive Simulation
All dynamics computed from $F = ma$ each frame. Force vectors drawn at true instantaneous magnitudes. Explore circular motion, vertical loops, orbits, and conical pendulums.