# Circular Motion

> Why anything moving in a circle needs a force towards the centre, from cars on banked roads to spinning rides, plus torque.

From HSC Physics on WhizzNotes, free study notes for NSW students: https://whizznotes.com/physics/circular-motion/. Updated 2 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The bucket that did not spill
At the school fair, Mr Okafor fills a bucket with water and swings it in a big vertical circle over his head. The students at the front back away, ready to get soaked.

Not a drop falls out, even when the bucket is upside down at the top.

"The water wants to go in a straight line," he says. "The bucket keeps pushing it inwards, round the circle. As long as I swing fast enough, the bucket has to push on the water even at the top."

Then he slows down. *Splash.* The front row gets wet after all. This chapter explains exactly how fast is fast enough.

**In short:** In uniform circular motion an object moves at constant speed, but its direction keeps changing, so it accelerates towards the centre. The net force that keeps it on the circle is the centripetal force, $F_c = \dfrac{mv^2}{r}$.

## Moving in a circle at a steady speed

**Uniform circular motion** means moving around a circle at a **constant speed**.

The speed is steady, but the **direction keeps changing**. Velocity is a vector, so the velocity is changing. A changing velocity means the object is **accelerating**, even though its speed stays the same.

- The velocity at any moment points along the **tangent** to the circle.
- The acceleration points towards the **centre** of the circle. It is called **centripetal acceleration** ("centre-seeking").

## Period, speed and angular velocity

- **Period** $T$: the time for one full circle (seconds).
- **Frequency** $f$: circles per second (hertz). $f = \dfrac{1}{T}$.
- **Speed**: one circle is a distance of $2\pi r$, so

$$v = \frac{2\pi r}{T}$$

- **Angular velocity** $\omega$ ("omega"): the angle swept out per second, in **radians per second**. One full circle is $2\pi$ radians.

$$\omega = \frac{2\pi}{T} = \frac{\Delta\theta}{t} \qquad v = \omega r$$

### Worked example: A spinning ride
A ride carries people in a circle of radius 6.0 m once every 4.0 s.

- Speed: $v = 2\pi(6.0) \div 4.0 = 9.4$ m/s
- Angular velocity: $\omega = 2\pi \div 4.0 = 1.57$ rad/s

## Centripetal acceleration and force

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

The **centripetal force** is not a new kind of force. It is the name for the **net force** towards the centre. Something real has to provide it:

| Situation | What provides the centripetal force |
|---|---|
| Ball on a string | Tension in the string |
| Car turning on a flat road | Friction between tyres and road |
| Moon orbiting Earth | Gravity |
| Electron in a magnetic field | Magnetic force |
| Rider on a rotor ride | Normal force from the wall |

If the force stops, the object flies off **along the tangent**, in a straight line. It does not fly outwards from the centre.

### Worked example: A car on a flat corner
A 1200 kg car takes a flat corner of radius 40 m at 12 m/s. What friction force is needed?

$$F = \frac{mv^2}{r} = \frac{1200 \times 12^2}{40} = 4320 \text{ N}$$

If the road is icy and friction cannot reach 4320 N, the car slides off along the tangent.

### Watch out: "Centrifugal force throws you outwards"
In a turning car you feel pushed outwards, but no outward force acts on you. Your body is trying to keep going straight, and the car turns underneath you. The real force (from the seat or door) pushes you **inwards**.

## What changes the force?

From $F = \dfrac{mv^2}{r}$:

- Double the **mass**: double the force.
- Double the **speed**: **four times** the force.
- Double the **radius** at the same speed: half the force.

## Banked tracks

Roads and cycling tracks are **tilted** (banked) on corners. The road pushes on the car at right angles to its surface. On a banked road, this **normal force** leans towards the centre, so part of it provides the centripetal force. The car can then turn without relying on friction.

At the **design speed**, no friction is needed at all:

$$\tan\theta = \frac{v^2}{rg}$$

The design speed does **not** depend on the mass of the car.

### Worked example: Banking a velodrome
A velodrome bend has radius 25 m. Riders travel at 12 m/s. What banking angle needs no friction?

$\tan\theta = \dfrac{12^2}{25 \times 9.8} = 0.588$, so $\theta = 30°$.

## Torque: turning effect of a force

A **torque** makes something **rotate**. Pushing a door near its handle is easy; pushing near the hinges is hard. The turning effect depends on the force and on how far from the pivot it acts.

$$\tau = rF\sin\theta$$

- $\tau$ = torque (N m)
- $r$ = distance from the pivot to where the force acts (m)
- $\theta$ = angle between $r$ and $F$

The torque is biggest when you push **at right angles** ($\sin 90° = 1$) and **far from the pivot**. That is why spanners are long and door handles are far from the hinges.

**Mind map**

```text
Circular Motion
  Uniform circular motion
    Constant speed
    Changing direction
    Accelerating to centre
  Describing it
    T and f
    v = 2πr / T
    ω = 2π / T, v = ωr
  Centripetal force
    F = mv² / r
    Net force to centre
    Tension, friction, gravity
    No force: flies off along tangent
  Banked tracks
    Normal force leans inwards
    tan θ = v² / rg
    No friction at design speed
  Torque
    τ = rF sin θ
    Far from pivot
    At right angles
```

### Year 11: Why the centripetal force does no work
Work is $W = Fs\cos\theta$, where $\theta$ is the angle between force and displacement. In uniform circular motion, the force points to the centre and the motion is along the tangent, so they are always at 90°. $\cos 90° = 0$, so **no work is done**. That is why the kinetic energy, and the speed, stay constant.

On a **flat** corner, the maximum safe speed comes from the maximum friction: $\dfrac{mv^2}{r} = \mu mg$, so $v_{\max} = \sqrt{\mu g r}$. A smaller radius or a slippery road (smaller $\mu$) lowers the safe speed.

**Wheels.** A wheel rolling without slipping has $v = \omega r$, so a car at 25 m/s (90 km/h) with wheels of radius 0.30 m turns them at $\omega = \frac{25}{0.30} \approx 83$ rad/s.

**Scaling friction.** On a flat corner $\dfrac{mv^2}{r} = F_{\max}$, so $v_{\max} = \sqrt{\dfrac{rF_{\max}}{m}}$. A truck with 4 times the mass and twice the friction has $v_{\max}$ multiplied by $\sqrt{\frac{2}{4}} \approx 0.71$.

**Force from kinetic energy.** Since $\frac{1}{2}mv^2 = K$, the centripetal force is $F = \dfrac{2K}{r}$. Halving $r$ and doubling $K$ makes it four times bigger.

**A coin on a turntable.** Friction provides the force: $\mu mg = m\omega^2 r$, so the coin slides when $\omega > \sqrt{\dfrac{\mu g}{r}}$. With $\mu = 0.40$ and $r = 0.15$ m, $\omega_{\max} = 5.1$ rad/s, about 0.82 revolutions per second.

### Year 12: Vertical circles and banked forces
**Vertical circles.** The speed changes as the object goes up and down, but at the top and bottom the net force still points to the centre.

- At the **top** of a loop: $T + mg = \dfrac{mv^2}{r}$. The smallest speed that keeps the string tight (or keeps a cart on the track) is when $T = 0$: $v_{\min} = \sqrt{gr}$.
- At the **bottom**: $T - mg = \dfrac{mv^2}{r}$, so the tension is greatest at the bottom.

A car going over a hump-backed bridge of radius $r$ has $mg - N = \dfrac{mv^2}{r}$. At $v = \sqrt{gr}$ the normal force is zero and the car just leaves the road.

**Banked tracks, derived.** At the design speed the forces are weight $mg$ and normal force $N$. Vertical: $N\cos\theta = mg$. Horizontal: $N\sin\theta = \dfrac{mv^2}{r}$. Divide to get $\tan\theta = \dfrac{v^2}{rg}$. Faster than this, friction acts down the slope; slower, friction acts up the slope.

**Feeling heavier or lighter.** At the top of a loop, $N + mg = \dfrac{mv^2}{r}$. For a normal force equal to the driver's weight, $2mg = \dfrac{mv^2}{r}$, so $v = \sqrt{2gr}$. At the bottom, $N - mg = \dfrac{mv^2}{r}$, so riders feel heavier.

**Starting height for a loop.** A ball rolling without friction needs at least $\sqrt{gr}$ at the top of a loop of radius $r$. Energy from height $h$: $mgh = mg(2r) + \frac{1}{2}m(gr)$, so $h = 2.5r$.

**Leaving the ground.** Over the top of a hill of radius $r$, the ground can only push up. If $\dfrac{mv^2}{r} > mg$, that is $v > \sqrt{gr}$, the vehicle leaves the surface. On the Moon $g$ is small, so this happens at low speeds.

**Conical pendulums and banked planes.** With the string (or lift force) at angle $\theta$ to the vertical: $T\cos\theta = mg$ and $T\sin\theta = \dfrac{mv^2}{r}$, so $\tan\theta = \dfrac{v^2}{rg}$, exactly like a banked track. For a pendulum of length $L$, $r = L\sin\theta$, which gives the period $T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$.

**Rotor rides.** The wall's normal force provides the centripetal force, $N = \dfrac{mv^2}{r}$, and friction must hold up the weight: $\mu N \geq mg$. So $v \geq \sqrt{\dfrac{gr}{\mu}}$. Mass cancels, so every rider sticks at the same speed.

## Quick summary

- In uniform circular motion the speed is constant but the velocity changes, so the object accelerates towards the centre.
- $v = \dfrac{2\pi r}{T}$, $\omega = \dfrac{2\pi}{T}$ and $v = \omega r$.
- $a_c = \dfrac{v^2}{r}$ and $F_c = \dfrac{mv^2}{r}$. The centripetal force is the net force, provided by tension, friction, gravity and so on.
- Without the force, the object moves off along the tangent.
- Banked tracks use the normal force to help turn: $\tan\theta = \dfrac{v^2}{rg}$.
- Torque $\tau = rF\sin\theta$ is the turning effect of a force.

## Quiz

Try these questions from the chapter quiz.

1. An object moves in a circle at a constant speed. Which statement is true?
   - A. Its velocity is constant
   - B. It is accelerating towards the centre
   - C. It has no net force on it
   - D. It is accelerating along the tangent
   - Answer: B. It is accelerating towards the centre. The direction keeps changing, so the velocity changes. The acceleration points to the centre.

2. A ball on a string is swung in a horizontal circle. What provides the centripetal force?
   - A. Gravity
   - B. Air resistance
   - C. Tension in the string
   - D. Friction
   - Answer: C. Tension in the string. The string pulls the ball towards the centre.

3. The string in the last question breaks. The ball then moves:
   - A. Straight out from the centre
   - B. Along the tangent to the circle
   - C. In a smaller circle
   - D. Straight down
   - Answer: B. Along the tangent to the circle. With no force towards the centre, the ball keeps the velocity it had, which points along the tangent.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/physics/circular-motion/#quiz