3.5 Circular motion — Learn

Prerequisites: Newton's second law from Section 3.2; resolving a vector into components from Section 2.2; the ratio of an arc to a radius; rearranging a formula.

By the end of this section you can: measure an angle in radians; find the angular velocity of an object moving in a circle; use $v = \omega r$ ; and apply the equations for centripetal acceleration and centripetal force.

1. Angles in radians

An angle can be measured in radians instead of degrees. One radian is the angle at the centre of a circle that cuts off an arc equal in length to the radius.

\theta = \frac{s}{r}

Here $s$ is the arc length and $r$ is the radius. Because the ratio of two lengths has no unit, an angle in radians is a pure number.

A full circle has an arc length of $2\pi r$ , so

2\pi\ \text{rad} = 360^\circ

To change degrees to radians, multiply by $\dfrac{\pi}{180}$ . To change radians to degrees, multiply by $\dfrac{180}{\pi}$ .

Worked example 1. Write $150^\circ$ in radians.

150 \times \frac{\pi}{180} = \frac{5\pi}{6} = 2.62\ \text{rad}

2. Angular velocity

Angular velocity $\omega$ is the rate of change of angle.

\omega = \frac{\Delta\theta}{\Delta t}

It is measured in radians per second, $\text{rad}\,\text{s}^{-1}$ . For an object that goes round a circle once in a time $T$ , the angle changes by $2\pi$ in one period, so

\omega = \frac{2\pi}{T} = 2\pi f

where $f = 1/T$ is the frequency of rotation.

The speed of the object along the circle is the arc length per unit time. Over one period it travels $2\pi r$ , so

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

The speed is larger for a point farther from the centre, because it travels a larger circle in the same time.

Worked example 2. A wheel spins at 300 revolutions per minute. Find its angular velocity in radians per second.

300\,\text{rev}\,\text{min}^{-1} = \frac{300}{60} = 5.0\,\text{rev}\,\text{s}^{-1}

\omega = 2\pi f = 2\pi \times 5.0 = 31.4\,\text{rad}\,\text{s}^{-1}

3. Centripetal acceleration

An object moving in a circle at constant speed still has a changing velocity, because its direction changes. A change in velocity means an acceleration.

The acceleration points towards the centre of the circle. It is called the centripetal acceleration, and its size is

a = \frac{v^2}{r} = \omega^2 r

The direction is always towards the centre, even though the velocity is along the tangent. If the string holding a ball breaks, the ball flies off along the tangent, because that is the direction of the velocity at the instant the centripetal force disappears.

Worked example 3. A car goes round a roundabout of radius 40 m at a constant $12\,\text{m}\,\text{s}^{-1}$ . Find its centripetal acceleration.

a = \frac{v^2}{r} = \frac{12^2}{40} = \frac{144}{40} = 3.6\,\text{m}\,\text{s}^{-2}

The acceleration is $3.6\,\text{m}\,\text{s}^{-2}$ towards the centre of the roundabout.

4. Centripetal force

Newton's second law says that a resultant force is needed to produce the centripetal acceleration. That force points towards the centre and is called the centripetal force.

F = ma = \frac{mv^2}{r} = m\omega^2 r

The centripetal force is not a new kind of force. It is the name for whatever resultant force happens to point towards the centre. It may be provided by tension, friction, gravity, a normal reaction, or a combination.

Situation	What provides the centripetal force
Ball on a string	tension in the string
Car on a roundabout	friction between the tyres and the road
Satellite in orbit	gravitational force
Conical pendulum	horizontal component of the tension
Car on a banked track	horizontal component of the normal reaction

Worked example 4. A ball of mass 0.50 kg is whirled in a horizontal circle of radius 0.80 m at $3.0\,\text{m}\,\text{s}^{-1}$ . Find the tension in the string.

T = \frac{mv^2}{r} = \frac{0.50 \times 3.0^2}{0.80} = \frac{4.5}{0.80} = 5.6\,\text{N}

5. Motion in a vertical circle

In a vertical circle the speed changes, because gravity does work on the object. At the top of the circle, both the weight and the force from the track or string act downwards, towards the centre. Their sum provides the centripetal force.

For a bucket of water swung in a vertical circle, the water stays in the bucket at the top when its weight is no more than the centripetal force needed.

mg = \frac{mv^2}{r} \quad\Rightarrow\quad v = \sqrt{gr}

This is the minimum speed at the top. Below it, the water would fall away from the bucket.

Worked example 5. A bucket of water is swung in a vertical circle of radius 0.60 m. Find the minimum speed at the top for the water to stay in the bucket.

v = \sqrt{gr} = \sqrt{9.81 \times 0.60} = 2.4\,\text{m}\,\text{s}^{-1}

Worked example 6. The roller-coaster car of mass 500 kg passes the top of the 12 m loop at $20.3\,\text{m}\,\text{s}^{-1}$ . Find the centripetal acceleration and the force the track exerts on the car. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

\begin{aligned} a &= \frac{v^2}{r} = \frac{20.3^2}{12} = \frac{412}{12} = 34.3\,\text{m}\,\text{s}^{-2} \\ F &= ma = 500 \times 34.3 = 1.72 \times 10^4\,\text{N} \end{aligned}

At the top of the loop the weight also acts downwards, so the track force is smaller than the total:

\begin{aligned} F_{\text{net}} &= mg + N \\ N &= F_{\text{net}} - mg \\ &= 500(34.3 - 9.81) \\ &= 1.23 \times 10^4\,\text{N} \end{aligned}

The track pushes down on the car with about $1.2 \times 10^4\,\text{N}$ .

Watch out. There is no outward centrifugal force acting on the object. It is tempting to invent an outward force to balance the inward one, because the object does not move towards the centre in the rotating frame of the object. But the acceleration is towards the centre, and Newton's second law needs a resultant force in that direction. If you add an outward force to make the forces balance, the resultant becomes zero and there is no reason for the object to move in a circle at all. Name the real force that provides the centripetal force, and check that it points towards the centre.

6. Try it: speed and radius

The figure shows a small object moving in a horizontal circle. Two sliders set the speed and the radius.

Predict first. Do not move the sliders yet. Write down your answers.

If you double the speed and keep the radius the same, what happens to the centripetal acceleration?
If you double the radius and keep the speed the same, what happens to the centripetal acceleration?
What happens to the centripetal acceleration if you double both the speed and the radius?

Test. Set the radius to 2.0 m and increase the speed. Read the acceleration and the force. Then keep the speed fixed and change the radius. Check each prediction.

Explain. Why is the acceleration larger when the speed is larger, and smaller when the radius is larger? Write two or three sentences. Use the ideas that the direction of the velocity changes faster at a higher speed, and that a larger circle bends the path more gently.

Progress check

Return to the chapter problem. The car and riders have a mass of 500 kg and pass the top of the 12 m loop at $20.3\,\text{m}\,\text{s}^{-1}$ .

Find the centripetal acceleration at the top of the loop.
Find the minimum speed the car needs at the top to stay on the track.
Find the force the track exerts on the car at the top of the loop.

If question 1 goes wrong, use $a = \dfrac{v^2}{r}$ with $r = 12\,\text{m}$ . If question 2 goes wrong, the minimum speed satisfies $v^2 = gr$ . If question 3 goes wrong, the track force plus the weight provide the centripetal force, so $N + mg = \dfrac{mv^2}{r}$ .

You should now be able to

measure an angle in radians and convert between radians and degrees;
find the angular velocity of an object from its period or frequency;
use $v = \omega r$ to relate speed to angular velocity;
calculate the centripetal acceleration and state its direction;
calculate the centripetal force and name the force that provides it;
explain why there is no centrifugal force on the object;
find the minimum speed at the top of a vertical circle.