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Circular Motion

Section 01
Interactive Simulation
Circular Motion Simulator — Centripetal Force Engine
v = 0.00 m/s
ω (rad/s)
0.00
rad/s
v (m/s)
0.00
m/s
a_c (m/s²)
0.00
m/s²
T (s)
0.00
s
F_c (N)
0.00
N
θ (°)
0.00
°
Controls
Parameters
Radius r1.00m
Mass m1.00kg
Angular vel. ω2.00rad/s
Gravity g9.81m/s²
Display
Trail
Vectors
Grid
Labels
Section 02
The Idea, Step by Step

Tie a ball to a string and whirl it over your head. Your hand has to keep pulling inward the whole time — the instant you let go, the ball doesn't fly straight out, it shoots off in a straight line. That constant inward tug is the whole secret of circular motion: to bend a path into a circle, something must pull toward the center, every single moment.

Putting numbers on the pull

Two things decide how hard that pull must be: how fast the object moves (its speed $v$) and how tight the circle is (its radius $r$). Go faster, or turn tighter, and you must pull harder. The inward pull is called the centripetal force, and the inward acceleration it produces is the centripetal acceleration:

Centripetal force (simplest form)
$$F_c = \frac{mv^2}{r}$$

Worked example: spin a $0.20\text{ kg}$ ball on a $0.50\text{ m}$ string at $v = 3\text{ m/s}$. Then $F_c = \dfrac{0.20\times 3^2}{0.50} = 3.6\text{ N}$ — about the weight of a small apple, pulling steadily inward through the string. Notice the speed is squared: double the speed and the pull quadruples to $14.4\text{ N}$.

Why there must be acceleration at all

Here is the subtle part. Even when the speed never changes, the direction of the velocity is turning constantly. Velocity is a vector, so a changing direction is a change in velocity — and any change in velocity is an acceleration. That acceleration always points straight at the center, and its size is

Centripetal acceleration
$$a_c = \frac{v^2}{r} = \omega^2 r,\qquad v = \omega r$$

where $\omega$ is the angular velocity (radians turned per second). Because this acceleration is always perpendicular to the motion, the inward force does zero work — it changes the object's direction but never its speed. The sim's sliders map straight onto these symbols: r sets the radius, m the mass, and ω the angular velocity, with $v = \omega r$ following automatically. The "force" $F_c$ is never a new force of its own — it is whatever real agent points inward: string tension, gravity, friction, or a track's normal force.

Try this in the sim above

Push ω higher and watch $F_c$ climb far faster than $\omega$ does — that's the squared term. Then double r and see how the readouts shift. Finally switch to Vertical Loop mode and lower ω until the green normal-force vector at the top just vanishes: you've found the critical speed $v_{\min}=\sqrt{gr}$, the slowest a coaster can crest a loop without falling.

Section 03
Equations & Derivation
Centripetal Acceleration
$$a_c = \frac{v^2}{r} = \omega^2 r$$
Centripetal Force
$$F_c = ma_c = \frac{mv^2}{r} = m\omega^2 r$$
Period, Frequency & Speed
$$T = \frac{2\pi}{\omega} = \frac{2\pi r}{v},\quad f = \frac{1}{T},\quad v = \omega r$$
Vertical Loop — Min Speed at Top
$$v_{\min,\text{top}} = \sqrt{gr},\qquad v_{\text{bottom}} = \sqrt{5gr}$$

Symbol Definitions

SymbolQuantitySI Unit
$r$Radius of circular pathm
$v$Tangential (linear) speedm s⁻¹
$\omega$Angular velocityrad s⁻¹
$a_c$Centripetal acceleration (toward centre)m s⁻²
$F_c$Centripetal force (toward centre)N
$T$Period — time for one full revolutions
$f$Frequency of revolutionHz
$m$Mass of the objectkg

Step-by-Step Derivation

1
Velocity changes direction continuously. In uniform circular motion, speed $|v|$ is constant but the direction of $\vec{v}$ rotates. Since velocity is a vector, $d\vec{v}/dt \neq 0$ — there must be acceleration even at constant speed.
2
Magnitude of centripetal acceleration. In time $\delta t$, the particle moves arc $\delta s = v\delta t$ and the velocity rotates by $\delta\theta = \delta s/r$. The change in velocity vector has magnitude $|\delta\vec{v}| = v\,\delta\theta$, so $a_c = |\delta\vec{v}|/\delta t = v^2/r$. Using $v = \omega r$: $a_c = \omega^2 r$.
3
Direction. $\vec{a}_c$ always points toward the centre of the circle — perpendicular to $\vec{v}$. It never has a component along the direction of motion, so it does zero work and does not change speed.
4
Newton's 2nd Law. The net inward force $F_c = ma_c = mv^2/r$. This is not a new type of force — it is provided by tension (string), gravity (orbit), normal force (loop), or friction (car turning).
5
Vertical loop minimum speed. At the top of the loop, gravity and normal force both act inward. For minimum speed, $N=0$: $mg = mv^2/r$, giving $v_{\min} = \sqrt{gr}$. By energy conservation from the bottom: $\frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + mg(2r)$, so $v_{\text{bottom,min}} = \sqrt{5gr}$.
6
Conical pendulum. A mass on a string of length $L$ making angle $\theta$ with vertical: $T\sin\theta = m\omega^2 r$ (horizontal), $T\cos\theta = mg$ (vertical). Dividing: $\tan\theta = \omega^2 r/g = \omega^2 L\sin\theta/g$, so $\cos\theta = g/(\omega^2 L)$.
Ref: Halliday, Resnick & Walker, 10th Ed., §6-2 to §6-4 & §13-3; Serway & Jewett, 8th Ed., §6.1–6.3; Kleppner & Kolenkow, 2nd Ed., §7.1.
Section 04
Frequently Asked Questions
Uniform: constant $\omega$ — shows centripetal force, velocity, and acceleration vectors. Vertical Loop: speed varies due to gravity — watch the ball speed up at the bottom and slow at the top; normal force changes sign at the critical speed. Conical Pendulum: mass on a string swings in a horizontal circle — angle $\theta$ increases with $\omega$. Orbit: gravitational attraction provides the centripetal force — observe Kepler's period relation.
Key takeaway: All four modes share the same centripetal force equation $F_c = mv^2/r$ — only the source of that force differs.
No. "Centripetal force" is the name given to the net inward force that any real agent provides. In a car turn: static friction. For a satellite: gravity. Ball on a string: tension. Never draw a separate centripetal force on a free-body diagram — identify the real force that acts inward and label it as providing centripetal acceleration.
Key takeaway: Centripetal force is a role, not a new fundamental force. The real force (tension, gravity, friction) fills that role.
No. Centripetal force is always perpendicular to the velocity ($\vec{F}_c \perp \vec{v}$), so $W = \vec{F}\cdot d\vec{r} = 0$. Speed (and kinetic energy) stay constant in uniform circular motion. This is why a satellite in a circular orbit needs no engine — gravity does zero work on it yet keeps it moving.
Key takeaway: Centripetal force does zero work and does not change the object's speed — only its direction.
Cars turning on banked curves (normal force + friction), roller coaster loops (normal force and gravity), planetary and satellite orbits (gravity), centrifuges for separating blood components or uranium isotopes, spin cycles of washing machines, gyroscopes in aircraft and phones, and particle accelerators where magnetic force guides charged particles in circles.
Key takeaway: From GPS satellites to centrifuges, circular motion is fundamental to modern technology.
You are in a non-inertial (accelerating) reference frame. From the ground frame, your body tends to continue in a straight line (Newton's 1st Law) while the car turns inward. The apparent outward "centrifugal force" is a fictitious force that exists only in the rotating frame — no physical agent produces it. Engineers use it as a convenient fiction when analyzing problems in the rotating frame.
Key takeaway: Centrifugal force is fictitious — it has no physical agent. In an inertial frame, only the real inward centripetal force exists.
At the top of a vertical loop of radius $r$, both gravity ($mg$) and normal force ($N$) point downward (inward toward the centre). Newton's 2nd Law: $mg + N = mv^2/r$. The minimum speed occurs when $N = 0$ (the track just barely pushes — any slower and the track would need to pull, which it cannot). This gives $v_{\min} = \sqrt{gr}$. At the bottom, energy conservation: $\frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + 2mgr$ gives $v_b = \sqrt{5gr}$.
Key takeaway: Minimum speed at loop top: $\sqrt{gr}$. Minimum launch speed at bottom: $\sqrt{5gr}$.
Resources: Khan Academy — Circular Motion (khanacademy.org); HyperPhysics — Centripetal Force (hyperphysics.phy-astr.gsu.edu); MIT OCW 8.01 Lectures 7–8.
Section 05
Common Misconceptions
❌ Centripetal force must be drawn as a separate arrow on every free-body diagram.
✅ A correct FBD shows only real forces. One real force (or their resultant) provides the centripetal acceleration. Applying $\sum F_{\text{inward}} = mv^2/r$ gives the constraint equation — never add a phantom "centripetal force" arrow.
📖 Halliday, Resnick & Walker, 10th Ed., §6-3; Serway & Jewett, 8th Ed., §6.1.
❌ An object in circular motion is in equilibrium because its speed is constant.
✅ Equilibrium means zero net force and zero acceleration. Objects in circular motion have continuous centripetal acceleration $a_c = v^2/r \neq 0$. Speed is constant, but velocity (a vector) changes every instant — this is not equilibrium.
📖 HRW 10th Ed., §6-1 & Newton's 2nd Law (§5-2).
❌ When a string breaks, the ball flies outward radially.
✅ By Newton's 1st Law, the ball continues in a straight line tangent to the circle at the release point. It does not fly outward — there is no outward force. Students confuse the sensation of centrifugal force (fictitious) with an actual outward trajectory.
📖 HRW 10th Ed., §6-3; Kleppner & Kolenkow, 2nd Ed., §7.1.
❌ A faster satellite orbits at a higher altitude.
✅ The opposite: orbital speed $v = \sqrt{GM/r}$ decreases as $r$ increases. Lower orbits are faster with shorter periods. The ISS at ~400 km altitude moves at ~7.7 km/s with a 90-minute period. Geostationary satellites at 36,000 km move at only ~3.07 km/s with a 24-hour period.
📖 HRW 10th Ed., §13-3: Satellites — Orbits and Energy.
Misconception research: Halloun & Hestenes (1985), Am. J. Phys. 53, 1043 (Force Concept Inventory); Arons — A Guide to Introductory Physics Teaching, Ch. 5.