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How does uniform circular motion work?

Investigate uniform circular motion, including the centripetal acceleration $a = v^2 / r$ and the net force required to maintain circular motion ($F_c = m v^2 / r$)

Q: Year 11 SAC HSC: A $0.50$ kg ball is swung in a horizontal circle of radius $1.2$ m by a string. The ball completes one revolution every $2.0$ s. Find (a) the speed of the ball, (b) the tension in the string.

**(a) Speed.** Circumference is $2\pi r = 2\pi(1.2) = 7.54$ m. $v = 7.54/2.0 = 3.77$ m s$^{-1}$. **(b) Tension.** For a horizontal circle, the net horizontal force (string tension) provides the centripetal force. $F_c = m v^2 / r = (0.50)(3.77)^2/1.2 = (0.50)(14.21)/1.2 = 5.92$ N. Tension $T = F_c = 5.92$ N. Markers reward circumference-based speed calculation, the substitution into $F_c = mv^2/r$, and identification of tension as the source of the net force.

A focused answer to the VCE Physics Unit 2 dot point on uniform circular motion. Derives the centripetal acceleration $a_c = v^2/r$ and centripetal force $F_c = mv^2/r$, identifies the source of the net force in named situations (string tension, friction, gravity), and works the VCAA SAC-style turntable and banked-curve problems.

Generated by Claude OpusReviewed by Better Tuition Academy5 min answerUpdated 2026-05-19

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What this dot point is asking

VCAA wants you to recognise uniform circular motion as motion in a circle at constant speed, to derive the centripetal acceleration from the changing velocity vector, and to identify the source of the centripetal force in named physical situations.

Why circular motion requires acceleration

In uniform circular motion the speed is constant but the velocity vector changes direction continuously. By Newton's second law, this changing velocity requires a net force directed toward the centre of the circle.

The acceleration produced is the centripetal acceleration:

a_c = \frac{v^2}{r}

directed toward the centre. Magnitude only; the direction continuously changes.

Centripetal force

Newton's second law applied to uniform circular motion:

F_c = m a_c = \frac{m v^2}{r}

This is not a separate kind of force. It is whatever real force (or net force) happens to be directed toward the centre. Common sources:

Situation	Source of centripetal force
Ball on a string in horizontal circle	Tension
Car turning on a flat road	Friction (between tyres and road)
Car on a banked road	Component of normal force
Satellite in orbit	Gravity
Conical pendulum	Horizontal component of string tension
Charged particle in magnetic field	Magnetic force (Year 12)

Period, frequency, angular speed

Period $T$ : time for one revolution. Frequency $f = 1/T$ . Angular speed $\omega = 2\pi/T = 2\pi f$ .

Speed: $v = 2\pi r / T = \omega r$ .

Substituting: $a_c = \omega^2 r$ (equivalent form for centripetal acceleration).

Worked example (banked curve)

A car of mass $m$ rounds a banked curve of radius $r$ and bank angle $\theta$ , designed so that no friction is needed. Find the design speed.

Free-body diagram: weight $mg$ down, normal force $N$ perpendicular to the road surface.

Horizontal: $N \sin\theta = m v^2 / r$ .

Vertical: $N \cos\theta = mg$ .

Divide: $\tan\theta = v^2 / (rg)$ .

Design speed: $v = \sqrt{rg \tan\theta}$ .

A banked curve of radius $50$ m at $20°$ has design speed $\sqrt{50 \cdot 9.8 \cdot \tan 20°} = \sqrt{50 \cdot 9.8 \cdot 0.364} = \sqrt{178.4} = 13.4$ m s $^{-1}$ .

Common traps

Treating centripetal force as a "new" force. It is the net inward force from the actual forces acting (tension, friction, gravity, normal).

Pointing centripetal force outward. Always toward the centre. The "centrifugal" effect is a perceived inertia, not a force in an inertial frame.

Forgetting that speed is constant. $a_c$ has constant magnitude but changing direction. KE is constant in uniform circular motion; net work done by centripetal force is zero (force perpendicular to velocity).

Using $a_c = v^2/r$ when motion is not uniform. If speed is also changing, there is a tangential acceleration component in addition. Year 11 problems use uniform circular motion.

In one sentence

Uniform circular motion has constant speed but changing direction, requiring a centripetal acceleration $a_c = v^2/r$ directed toward the centre and a net force $F_c = mv^2/r$ supplied by whatever real force (tension, friction, gravity, normal component) acts inward.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksA $0.50$ kg ball is swung in a horizontal circle of radius $1.2$ m by a string. The ball completes one revolution every $2.0$ s. Find (a) the speed of the ball, (b) the tension in the string.

Show worked answer →

(a) Speed. Circumference is $2\pi r = 2\pi(1.2) = 7.54$ m.

$v = 7.54/2.0 = 3.77$ m s $^{-1}$ .

(b) Tension. For a horizontal circle, the net horizontal force (string tension) provides the centripetal force.

$F_c = m v^2 / r = (0.50)(3.77)^2/1.2 = (0.50)(14.21)/1.2 = 5.92$ N.

Tension $T = F_c = 5.92$ N.

Markers reward circumference-based speed calculation, the substitution into $F_c = mv^2/r$ , and identification of tension as the source of the net force.