Uniform circular motion (horizontal and vertical): VCE Physics Unit 3

Syllabus dot point

“Investigate and analyse theoretically and practically the uniform circular motion of an object moving in a horizontal plane and on a vertical circle, including a quantitative analysis of the forces acting at the top and bottom of the vertical circle”

11 min readUpdated 21 May 2026

Quick answer

Uniform circular motion needs a net inward force $F_c = \frac{m v^2}{r}$ supplied by real forces (gravity, friction, tension, normal), with tension at the top of a vertical loop equal to $\frac{m v^2}{r} - mg$ and at the bottom $\frac{m v^2}{r} + mg$ .

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What this dot point is asking
The answer
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to analyse uniform circular motion in a horizontal plane (cars on bends, conical pendulums, balls on strings) and on a vertical circle (roller coasters, buckets of water, balls swung in a vertical loop), and to apply $F_c = \frac{mv^2}{r}$ quantitatively at the top and bottom of the vertical loop.

The answer

An object in uniform circular motion travels at constant speed $v$ around a circle of radius $r$ . Speed is constant, but velocity changes direction continuously, so the object is accelerating toward the centre.

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

Centripetal force is not a new force. It is the net inward force supplied by gravity, friction, tension, normal force, or any combination of real forces acting toward the centre.

Period, frequency and angular velocity

The period $T$ is the time for one revolution; the frequency $f = 1/T$ .

v = \frac{2 \pi r}{T} = 2 \pi r f, \quad \omega = \frac{2 \pi}{T}, \quad v = \omega r, \quad a_c = \omega^2 r

Horizontal circle: the conical pendulum

A mass on a string swung in a horizontal circle so that the string makes angle $\theta$ with the vertical. Two real forces act: tension $T$ along the string, and gravity $mg$ down. The vertical components must balance; the horizontal component of tension supplies the centripetal force.

T \cos\theta = mg, \quad T \sin\theta = \frac{m v^2}{r}

Dividing gives $\tan\theta = \frac{v^2}{rg}$ . As the speed increases, the string angle from vertical increases.

Horizontal circle: car on a flat bend

The only horizontal force is friction between tyres and road. For the bend to be completed without skidding, $\mu_s m g \geq \frac{m v^2}{r}$ , so $v_{\max} = \sqrt{\mu_s g r}$ .

Vertical circle: forces at the top and bottom

Consider a ball on a string of radius $r$ in a vertical circle at constant speed $v$ . Gravity always acts down. Tension acts along the string toward the centre.

At the top. Both tension $T_{top}$ and gravity point down (toward the centre).

T_{top} + mg = \frac{m v^2}{r}

The minimum speed for the string to stay taut at the top is when $T_{top} = 0$ : $v_{\min} = \sqrt{gr}$ .

At the bottom. Tension $T_{bot}$ points up (toward the centre); gravity points down (away).

T_{bot} - mg = \frac{m v^2}{r}

So tension at the bottom is greater than at the top by $2mg$ .

Roller coaster at the top of a loop. Replace tension with the normal force from the rail. The cart feels lightest (sometimes momentarily weightless) at the top, when $N + mg = \frac{m v^2}{r}$ .

Roller coaster at the bottom. Normal force points up; $N - mg = \frac{m v^2}{r}$ , so apparent weight (felt by the rider) is maximum: $N = mg + \frac{m v^2}{r}$ .

Worked example

A $1.5$ kg bucket of water is swung in a vertical circle of radius $0.80$ m. Find the minimum speed at the top of the circle so the water stays in the bucket, and the normal force from the bucket on the water at the bottom at that speed.

At the top, minimum speed is when normal force from the bucket equals zero, leaving gravity alone to supply the centripetal force on the water:

$mg = \frac{m v_{\min}^2}{r}$ , so $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 0.80} = 2.8$ m/s.

At the bottom, normal force from the bucket on the water points up:

$N - mg = \frac{m v^2}{r}$ , $N = m \left( g + \frac{v^2}{r} \right) = 1.5 \times \left( 9.8 + \frac{2.8^2}{0.80} \right) = 1.5 \times (9.8 + 9.8) = 29.4$ N.

Try it: Centripetal force calculator - enter mass, speed and radius and get $F_c$ , $a_c$ , $T$ and $\omega$ .

Common traps

Treating centripetal force as a separate force on a free-body diagram: Draw the real forces (gravity, normal, friction, tension), then state that their net inward component supplies $\frac{mv^2}{r}$ .
Forgetting that tension adds to gravity at the top: Both point toward the centre at the top, so $T_{top} + mg = \frac{mv^2}{r}$ . At the bottom they oppose, so $T_{bot} - mg = \frac{mv^2}{r}$ .
Confusing centripetal and centrifugal: Centrifugal force is fictitious; only use it in a rotating frame, which VCE does not require.
Using diameter instead of radius: $a_c = \frac{v^2}{r}$ , not $\frac{v^2}{d}$ .
Mixing up frequency and angular velocity: $\omega$ is in rad/s, $f$ is in Hz, related by $\omega = 2 \pi f$ .

Examples in context

Example 1. Luna Park Melbourne Scenic Railway vertical loop. Luna Park's heritage rollercoaster includes a vertical-circle loop of $8.0$ m radius. At the top of the loop, the minimum speed for the car to maintain contact is found from $mg = mv^2/r$ , giving $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 8.0} = 8.85$ m s $^{-1}$ . Operating above this minimum, normal force from the track adds to weight providing extra centripetal force. At the bottom of the loop at $14$ m s $^{-1}$ , normal force is $N - mg = mv^2/r$ , so $N = m(g + v^2/r) = m(9.8 + 24.5) = 34.3 m$ , equivalent to $3.5g$ on a $70$ kg passenger ( $2400$ N).

Example 2. Bathurst Mountain Straight bowl turn. Mt Panorama's "The Cutting" includes a near-horizontal $90$ m radius curve. A $1300$ kg Supercar travelling at $40$ m s $^{-1}$ requires centripetal force $F_c = mv^2/r = 1300 \times 1600/90 = 2.31 \times 10^4$ N pointing horizontally toward the curve centre. On level tarmac this must come entirely from tyre friction. With $\mu_s = 0.9$ for racing slicks, max friction $\mu_s mg = 0.9 \times 1300 \times 9.8 = 1.15 \times 10^4$ N is insufficient: the car would slide out. This is why Mt Panorama's racing line uses the entire width of the track, effectively increasing the curve radius.

Try this

Q1. Derive the minimum speed at the top of a vertical circle for an object to maintain contact with the track. [2 marks]

Cue. At the minimum, $N = 0$ , so $mg = mv^2/r$ , giving $v_{\min} = \sqrt{gr}$ .

Q2. A $0.5$ kg ball is swung in a vertical circle of radius $1.2$ m at constant speed $4.0$ m s $^{-1}$ . Calculate (a) the tension at the top, and (b) the tension at the bottom. [4 marks]

Cue. (a) $T_{\rm top} = m(v^2/r - g) = 0.5(13.3 - 9.8) = 1.75$ N. (b) $T_{\rm bot} = m(v^2/r + g) = 0.5(13.3 + 9.8) = 11.6$ N.

Q3. Refer to Luna Park's $8$ m loop. (a) Calculate the minimum speed at the top. (b) Determine the normal force at the bottom for a $70$ kg passenger at $14$ m s $^{-1}$ . (c) Explain why passengers do not fall out at the top of the loop. [2+3+2 marks]

Cue. (a) $8.85$ m s $^{-1}$ . (b) $2400$ N. (c) Centripetal acceleration is provided by gravity plus normal force; the passenger is in free fall toward the centre of the circle.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

A 0.50 kg ball on a 1.2 m string is swung in a vertical circle at constant speed 5.0 m/s. Calculate the tension in the string at the top and at the bottom of the circle.

Show worked answer →

Centripetal acceleration $a_c = \frac{v^2}{r} = \frac{5.0^2}{1.2} = 20.83$ m/s squared.

Required centripetal force $F_c = m a_c = 0.50 \times 20.83 = 10.42$ N. This must be the net force directed toward the centre.

At the top. Both tension $T$ and gravity $mg$ point downward (toward the centre).

$T + mg = F_c$ , so $T = F_c - mg = 10.42 - (0.50)(9.8) = 10.42 - 4.9 = 5.5$ N.

At the bottom. Tension $T$ points upward (toward the centre), gravity points downward (away from the centre).

$T - mg = F_c$ , so $T = F_c + mg = 10.42 + 4.9 = 15.3$ N.

Markers reward identifying the centre direction, signing each force correctly, and noting that the tension is highest at the bottom.

Practice3 marks

A 1200 kg car travels around a flat horizontal bend of radius 60 m at 15 m/s. Calculate the minimum coefficient of friction required between the tyres and the road for the car to complete the bend.

Show worked answer →

The horizontal friction supplies the centripetal force.

$F_c = \frac{m v^2}{r} = \frac{1200 \times 15^2}{60} = \frac{1200 \times 225}{60} = 4500$ N.

Normal force on flat ground $N = mg = 1200 \times 9.8 = 11760$ N.

Maximum static friction is $\mu_s N$ . For the bend to be completed, $\mu_s N \geq F_c$ :

$\mu_s \geq \frac{F_c}{N} = \frac{4500}{11760} = 0.38$ .

Minimum $\mu_s = 0.38$ .

Markers reward identifying friction as the centripetal force, correct $N = mg$ on flat ground, and a dimensionless final answer.