An object moves in a circle at a constant speed. Which statement is true?

It is accelerating towards the centre

It is accelerating along the tangent

The string in the last question breaks. The ball then moves:

Along the tangent to the circle

Circular Motion | HSC Physics Module 5

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, $F_{c} = \frac{m v ^{2}}{r}$ .

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). $f = \frac{1}{T}$ .
Speed: one circle is a distance of $2 π r$ , so

$v = \frac{2 π r}{T}$

Angular velocity $ω$ ("omega"): the angle swept out per second, in radians per second. One full circle is $2 π$ radians.

$ω = \frac{2 π}{T} = \frac{Δ θ}{t} v = ω r$

Centripetal acceleration and force

$a_{c} = \frac{v ^{2}}{r} F_{c} = \frac{m v ^{2}}{r}$

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.

What changes the force?

From $F = \frac{m v ^{2}}{r}$ :

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:

$tan θ = \frac{v ^{2}}{r g}$

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

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.

$τ = r F sin θ$

$τ$ = torque (N m)
$r$ = distance from the pivot to where the force acts (m)
$θ$ = 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.

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 11Why the centripetal force does no work

Work is $W = F s cos θ$ , where $θ$ 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: $\frac{m v ^{2}}{r} = μ m g$ , so $v_{m a x} = μg r$ . A smaller radius or a slippery road (smaller $μ$ ) lowers the safe speed.

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

Scaling friction. On a flat corner $\frac{m v ^{2}}{r} = F_{m a x}$ , so $v_{m a x} = \frac{r F _{m a x}}{m}$ . A truck with 4 times the mass and twice the friction has $v_{m a x}$ multiplied by $\frac{2}{4} \approx 0.71$ .

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

A coin on a turntable. Friction provides the force: $μ m g = m ω^{2} r$ , so the coin slides when $ω > \frac{μg}{r}$ . With $μ = 0.40$ and $r = 0.15$ m, $ω_{m a x} = 5.1$ rad/s, about 0.82 revolutions per second.

Year 12Vertical 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: $T + m g = \frac{m v ^{2}}{r}$ . The smallest speed that keeps the string tight (or keeps a cart on the track) is when $T = 0$ : $v_{m i n} = g r$ .
At the bottom: $T - m g = \frac{m v ^{2}}{r}$ , so the tension is greatest at the bottom.

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

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

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

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

Leaving the ground. Over the top of a hill of radius $r$ , the ground can only push up. If $\frac{m v ^{2}}{r} > m g$ , that is $v > g r$ , 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 $θ$ to the vertical: $T cos θ = m g$ and $T sin θ = \frac{m v ^{2}}{r}$ , so $tan θ = \frac{v ^{2}}{r g}$ , exactly like a banked track. For a pendulum of length $L$ , $r = L sin θ$ , which gives the period $T = 2 π \frac{L cos θ}{g}$ .

Rotor rides. The wall's normal force provides the centripetal force, $N = \frac{m v ^{2}}{r}$ , and friction must hold up the weight: $μ N \geq m g$ . So $v \geq \frac{g r}{μ}$ . 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.
$v = \frac{2 π r}{T}$ , $ω = \frac{2 π}{T}$ and $v = ω r$ .
$a_{c} = \frac{v ^{2}}{r}$ and $F_{c} = \frac{m v ^{2}}{r}$ . 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: $tan θ = \frac{v ^{2}}{r g}$ .
Torque $τ = r F sin θ$ is the turning effect of a force.

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Try 3 questions now

An object moves in a circle at a constant speed. Which statement is true?
1. A Its velocity is constant
2. B It is accelerating towards the centre
3. C It has no net force on it
4. D It is accelerating along the tangent
Show answer
B: It is accelerating towards the centre
The direction keeps changing, so the velocity changes. The acceleration points to the centre.
A ball on a string is swung in a horizontal circle. What provides the centripetal force?
1. A Gravity
2. B Air resistance
3. C Tension in the string
4. D Friction
Show answer
C: Tension in the string
The string pulls the ball towards the centre.
The string in the last question breaks. The ball then moves:
1. A Straight out from the centre
2. B Along the tangent to the circle
3. C In a smaller circle
4. D Straight down
Show 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.

Circular Motion

Moving in a circle at a steady speed

Period, speed and angular velocity

Centripetal acceleration and force

What changes the force?

Banked tracks

Torque: turning effect of a force

Quick summary

Quiz

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