Uniform circular motion

Why an object moving in a circle at constant speed is accelerating, angular velocity, period, centripetal acceleration and force, what actually provides the centripetal force, and the whirling-stopper investigation.

An elite hammer thrower swings a 7.26 kg ball on a wire around a circle of radius about 1.8 m (wire plus outstretched arms), reaching about 29 m/s at release. To keep that ball on its circle, the thrower has to pull inwards with a force of around 3400 N, the weight of about 350 kg. Yet the ball is moving at constant speed. So what is all that force doing?

The answer is that the force is changing the ball's direction. Velocity is a vector, so a change in direction is a change in velocity, and a change in velocity is an acceleration. This page builds the equations for that acceleration and the force that causes it.

Describing circular motion

Uniform circular motion means moving in a circle at constant speed. One full revolution takes the period $T$ and covers the circumference $2\pi r$, so

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

The frequency $f = 1/T$ is the number of revolutions per second. Engines and turbines are usually rated in revolutions per minute (rpm): divide by 60 to get hertz.

The angular velocity $\omega$ is the angle swept out per second, in radians per second:

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

Every point on a rotating object has the same $\omega$, but points further from the axis move faster. The tip of a wind turbine blade is far faster than the hub.

Worked example: wind turbine

A wind turbine turns at 15 rpm. Its blades are 60 m long. Find the angular velocity and the speed of a blade tip.

15 revolutions per minute, so $T = 60/15 = 4.0$ s.

period

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4.0} = 1.57\ \text{rad s}^{-1}$.

angular velocity

$v = r\omega = 60 \times 1.571 = 94.2\ \text{m s}^{-1}$, about 340 km/h.

tip speed

Centripetal acceleration

Between two nearby points the velocity changes direction. The change in velocity, Δv, points towards the centre.

Take two velocity vectors a short time apart. They have the same length but point in different directions. Subtracting them gives $\Delta \vec v$, which points towards the centre of the circle. So the acceleration points towards the centre at every instant: it is called centripetal ("centre-seeking") acceleration. Its size is

$a_c = \dfrac{v^2}{r}$

Going twice as fast on the same circle needs four times the acceleration. A tighter circle at the same speed needs more.

At every point the velocity is along the tangent, and both the acceleration and the net force point to the centre.

Centripetal force

By Newton's second law, an acceleration needs a net force in the same direction:

$F_c = \dfrac{mv^2}{r}$

"Centripetal force" is not a new kind of force. It is the name for whatever net force points to the centre. Always ask: what is actually providing it?

Situation	Force providing the centripetal force
Hammer throw, stopper on a string	tension in the wire or string
Car turning on a flat road	friction between tyres and road
Moon orbiting Earth	gravitational attraction
Electron in a magnetic field (Module 6)	magnetic force
Clothes in a spin dryer	normal force from the drum wall

On a flat road, sideways friction on the tyres is the only horizontal force, so it must provide the centripetal force.

If the centripetal force disappears, the object doesn't fly outwards. It carries on in a straight line along the tangent, exactly as Newton's first law says. The hammer flies off along the tangent at the moment of release; mud flies off a bike tyre the same way.

And the "outward force" you feel in a turning car? There isn't one. Your body tries to keep going straight; the car turns underneath you, and the door pushes you inwards. The so-called centrifugal force only appears if you describe motion from inside the rotating car, which is not an inertial frame. In HSC answers, don't use it.

Investigating centripetal force

Investigation: the whirling stopper

Setup: A rubber stopper is tied to a string that passes through a smooth glass tube. A hanging mass on the other end of the string provides the tension. A paper clip on the string just below the tube marks a fixed radius.

Method: Whirl the stopper in a horizontal circle, adjusting the speed so the paper clip stays just below the tube. Time 10 revolutions and divide by 10 to get $T$. Calculate $v = 2\pi r/T$. The centripetal force is the weight of the hanging mass, $F_c = Mg$. Repeat with different hanging masses, keeping $r$ and the stopper mass the same.

Analysis: plot $F_c$ against $v^2$. The model predicts a straight line through the origin with gradient $m/r$.

Limitations: the string isn't quite horizontal (it droops, so part of the tension holds the stopper up), friction at the tube, and the radius drifts if the speed isn't steady. Timing 10 revolutions rather than one reduces the percentage error from reaction time.

Safety: wear safety glasses, check the string is secure, and keep clear of other students.

Worked example: analysing a stopper trial

A 25 g stopper is whirled at a radius of 0.60 m. Ten revolutions take 7.2 s. Find the speed, the centripetal acceleration and the tension needed.

$T = 7.2/10 = 0.72$ s.

period

$v = \dfrac{2\pi(0.60)}{0.72} = 5.24\ \text{m s}^{-1}$.

speed

$a_c = \dfrac{5.24^2}{0.60} = 45.7\ \text{m s}^{-2}$, almost 5 times g.

acceleration

$F_c = ma_c = 0.025 \times 45.7 = 1.14$ N, so the hanging mass should be about $1.14/9.8 = 0.117$ kg.

force

InteractiveWhirling stopper

Try this

With the 25 g stopper, a 100 g hanging mass and a radius of 0.50 m, the readouts show $F_c$ = 0.980 N, $v$ = 4.43 m/s and a period of 0.710 s. Check these with $F_c = mv^2/r$.
Press Time 10 revolutions. Your stopwatch time includes a small reaction error. How far is the speed it gives from 4.43 m/s?
Record at least five results with different hanging masses (keep the radius fixed). The gradient of the $F$ against $v^2$ graph should come out close to $m/r = 0.025/0.50 = 0.05$ kg/m.
Press Let go. Which way does the stopper travel?

Test yourself

A car travels around a roundabout at a constant 30 km/h. Which statement is correct?
A stopper's speed on a fixed circle is doubled. The tension needed becomes
A 0.50 kg ball moves in a circle of radius 0.80 m with a period of 1.2 s. The centripetal force on it is closest to
A ball on a string is whirled in a horizontal circle and the string breaks when the ball is at the most northerly point of the circle, moving east. The ball then moves

Label the vectors for uniform circular motion.

Which force provides the centripetal force?

In uniform circular motion the speed is constant but the changes, because its direction changes. The acceleration points towards the and is called centripetal acceleration. Its size is speed squared divided by the . The time for one revolution is the . If the string breaks, the object moves off along the .

5 marks

In a whirling-stopper experiment, a 30 g stopper moves in a horizontal circle of radius 0.75 m. A 150 g mass hangs from the other end of the string. (a) Calculate the speed the stopper must have. (b) Calculate the time for 10 revolutions. (c) Explain one reason the measured period may differ from your answer.

Your answer