Circular Motion¶

Syllabus mapping

PY-12-01 — Advanced Mechanics. Covers: analysing the forces involved in uniform circular motion; relationships between period, frequency, speed and radius; centripetal acceleration and force; qualitative and quantitative analysis of horizontal circular motion (vehicles on curves, objects on strings).

Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.

You need to know: why an object moving in a circle at constant speed is still accelerating, how to calculate that acceleration and the force causing it, what happens the instant that force is removed, and how to apply all of this to real horizontal circular motion problems (cars on bends, masses on strings).

Assumed knowledge check¶

Before this page makes sense, you should be comfortable with:

Forces and Motion — Newton's second law, free-body diagrams
Quantities of Motion — vector subtraction, needed below to find the direction of a change in velocity

Core¶

Why constant speed still means acceleration¶

Velocity is a vector — magnitude and direction. An object moving in a circle at genuinely constant speed has a velocity that's constantly changing direction, so by definition (\(\vec{a}=\Delta\vec{v}/\Delta t\)) it is accelerating, even though a speedometer reading it would show no change at all.

This is a common sticking point: "constant speed" and "no acceleration" feel like the same thing, but they aren't. Acceleration only requires velocity to change somehow — direction alone is enough.

Worked example — finding the direction of the acceleration

An object moves anticlockwise around a circle. At one instant its velocity points due north; a short time later, having moved a small arc further round, its velocity points north-west.

The change in velocity is \(\Delta\vec{v} = \vec{v}_2 - \vec{v}_1\) — reverse \(\vec{v}_1\) and add tip-to-tail (see Quantities of Motion). Doing this geometrically for two velocity vectors of equal length, close together in direction, always gives a \(\Delta\vec{v}\) that points roughly toward the centre of the circle — not forward, not backward, inward. As the time interval shrinks, this becomes exact: the instantaneous acceleration of any object in uniform circular motion points directly at the centre.

This is why it's called centripetal ("centre-seeking") acceleration — the name describes its direction, not some new kind of force.

Centripetal acceleration and force¶

\[ a_c = \frac{v^2}{r} \qquad F_c = \frac{mv^2}{r} \]

Symbols: \(a_c\) — centripetal acceleration; \(F_c\) — centripetal force (the net force, not a separate force); \(v\) — speed; \(r\) — radius; \(m\) — mass.

\"Centripetal force\" is not a new force — it's a role, not a thing

This is the single most common misconception in this topic. There is no mysterious "centripetal force" pulling objects inward in addition to the real forces acting on them. \(F_c=mv^2/r\) is simply the net force required for circular motion at that speed and radius — and something real (tension, gravity, friction, the normal force, or some combination) has to supply it. When a question asks "what provides the centripetal force here?", the answer is always one of the four forces above, never "centripetal force" itself.

Period, frequency and speed¶

For an object completing one full revolution in time \(T\) (the period) around a circle of radius \(r\):

\[ v = \frac{2\pi r}{T} \]

Frequency \(f\) (revolutions per second) is the reciprocal of period, \(f=1/T\), exactly as in Wave Properties — same relationship, different context.

Worked example — period, speed, centripetal force together

A \(0.5\ \text{kg}\) ball on a string moves in a horizontal circle of radius \(0.8\ \text{m}\), completing one revolution every \(1.2\ \text{s}\). Find its speed and the tension in the string.

Speed:

\[ v = \frac{2\pi r}{T} = \frac{2\pi \times 0.8}{1.2} \approx 4.19\ \text{m s}^{-1} \]

The string is the only horizontal force present, so it alone supplies the centripetal force:

\[ F_c = \frac{mv^2}{r} = \frac{0.5 \times 4.19^2}{0.8} \approx 10.97\ \text{N} \]

What happens when the centripetal force is removed¶

If the force supplying \(F_c\) suddenly vanishes (a string snaps, friction is exceeded), the object does not fly directly outward, radially away from the centre — that's a common but incorrect intuition. Newton's first law applies: with no net force, it continues in a straight line tangent to the circle at the point it was released, in whatever direction it was already moving at that instant.

🖼️ Diagram needed: a circle with an object at one point, its velocity vector drawn tangent to the circle at that point, and a dashed straight line continuing in that tangent direction — contrasted with a (wrong) dashed arrow pointing straight outward from the centre, clearly labelled as the common misconception. Static, Excalidraw-style. TODO — placeholder until sourced/drawn.

Horizontal circular motion: vehicles on bends and conical-pendulum-style problems¶

Vehicle on a flat, horizontal curve: the centripetal force is supplied by friction between the tyres and the road, acting horizontally toward the centre of the curve. If the required \(F_c=mv^2/r\) exceeds the maximum available friction, the vehicle cannot maintain the curve and skids outward (tangent to its path at that instant, per the paragraph above — not straight sideways).

Mass on a string swung at an angle (conical pendulum): the string tension has both a vertical component (balancing weight) and a horizontal component (supplying \(F_c\)). Resolve the tension into components exactly as in resolving forces:

\[ T\cos\theta = mg \qquad T\sin\theta = \frac{mv^2}{r} \]

where \(\theta\) is the angle the string makes with the vertical.

Worked example — vehicle on a curve

A \(1200\ \text{kg}\) car rounds a flat curve of radius \(50\ \text{m}\) at \(15\ \text{m s}^{-1}\). Find the minimum coefficient of friction needed.

Required centripetal force:

\[ F_c = \frac{mv^2}{r} = \frac{1200 \times 15^2}{50} = 5400\ \text{N} \]

This must come entirely from friction, and maximum available friction is \(f=\mu F_N=\mu mg\) (flat road, so \(F_N=mg\)):

\[ \mu mg \geq F_c \implies \mu \geq \frac{F_c}{mg} = \frac{5400}{1200\times9.8} \approx 0.46 \]

Practical¶

Suggested prac: mass/radius/velocity/centripetal force relationship — a rubber bung on a string through a tube, with a hanging mass providing a known, constant centripetal force, varying the radius or speed of the whirling bung and checking the relationship against \(F_c=mv^2/r\). TODO — link this school's actual prac instructions once written.

Advanced

Banked curves. Real curves (racetracks, highway on-ramps) are often banked at an angle specifically so that, at the design speed, the normal force alone (not friction) supplies the full centripetal force — reducing reliance on tyre grip, especially useful in wet conditions. This uses the same resolved-components approach as the conical pendulum above, with \(F_N\) playing the role \(T\) played there.

Vertical circular motion (not on the syllabus core but a natural extension): when a circle is vertical rather than horizontal, gravity contributes to or opposes the required centripetal force depending on position — at the top of a loop, both gravity and the track's normal force point toward the centre (downward), so \(F_N + mg = mv^2/r\); at the bottom, they oppose, so \(F_N - mg = mv^2/r\). This is why minimum-speed-at-the-top problems (roller coasters, a bucket of water swung overhead) are a classic combined application.

Video/visual resources¶

🖥️ PhET Simulation — Ladybug Motion 2D (has a circular-motion preset) — TODO: confirm current PhET link
🎥 Khan Academy — TODO: source a centripetal force/acceleration explainer. Must directly address the "centripetal force is not a separate force" misconception and the tangent-line-on-release misconception, both flagged above as the two biggest sticking points. Essential.
🎥 Physics High — TODO: check for a NSW-syllabus-aligned circular motion video, ideally covering a worked vehicle-on-a-curve or conical-pendulum problem specifically. Nice-to-have alongside the Khan Academy one above.

Check yourself¶

A \(60\ \text{kg}\) skater moves in a circle of radius \(4\ \text{m}\) at \(3\ \text{m s}^{-1}\). Find the centripetal acceleration and the centripetal force.

Answer

\(a_c = \dfrac{v^2}{r} = \dfrac{3^2}{4} = 2.25\ \text{m s}^{-2}\)

\(F_c = ma_c = 60 \times 2.25 = 135\ \text{N}\)
A satellite completes one orbit every \(90\) minutes at a radius of \(6.8\times10^6\ \text{m}\). Find its orbital speed.

Answer

\(T = 90 \times 60 = 5400\ \text{s}\)

\(v = \dfrac{2\pi r}{T} = \dfrac{2\pi \times 6.8\times10^6}{5400} \approx 7913\ \text{m s}^{-1}\)
A ball on a string is being swung in a horizontal circle when the string suddenly snaps. Describe the ball's subsequent motion, and explain why a common answer ("it flies straight out, away from the centre") is wrong.

Answer

With the string gone, there's no horizontal force left, so by Newton's first law the ball continues in a straight line in whatever direction it was moving at the instant the string snapped — tangent to the circle at that point, not radially outward. "Flies straight out from the centre" would only be correct if the ball's velocity at that instant happened to point directly away from the centre, which it never does in circular motion (velocity is always tangent to the circle, i.e. perpendicular to the radius).