Motion in a circle

Measuring angles in radians

An angle is a fraction of a full turn, and there are two ways to measure that fraction. Degrees divide a full turn into 360, which is a historical choice with no physical reason behind it. The radian is defined by the circle itself: it is the angle subtended at the centre of a circle by an arc equal in length to the radius of the circle. The circumference of a circle is 2 pi r, so a full turn of 360 degrees is 2 pi radians, and one radian is 360 / 2 pi = 57.3 degrees.

For an arc of length s on a circle of radius r, the angular displacement theta, measured in radians, is theta = s / r. The units cancel, because s and r are both lengths in metres, so theta is a pure number; the label 'rad' is written after it for clarity only. This is exactly why radians are used in physics. The arc length is s = r theta, a plain multiplication with no conversion factor, and every equation of circular motion, from omega = 2 pi / T to a = r omega^2, is correct only when the angle is in radians.

To convert, multiply degrees by pi / 180 to get radians, and multiply radians by 180 / pi to get degrees. An arc of length 0.45 m on a circle of radius 0.30 m subtends theta = 0.45 / 0.30 = 1.5 rad at the centre. For comparison, 30 degrees is 0.524 rad, 90 degrees is 1.57 rad and 180 degrees is 3.14 rad.

• theta = s / r gives angular displacement in radians, where s is the arc length and r is the radius.
• A full turn is 2 pi rad, so 360 degrees = 2 pi rad and one radian = 57.3 degrees.
• Degrees x pi / 180 gives radians; radians x 180 / pi gives degrees.

Angular speed and the link between v and omega

Angular speed omega is the rate at which angular displacement changes, so omega = theta / t, the angle turned through per unit time. It is measured in radians per second (rad s-1). For a body turning at a steady rate, the time for one complete revolution is the period T, and one revolution is an angle of 2 pi rad, so omega = 2 pi / T. If a body makes f revolutions each second then T = 1 / f, so omega = 2 pi f.

A fairground carousel that completes one revolution in 12.0 s turns at omega = 2 pi / 12.0 = 0.524 rad s-1. The London Eye takes about 30 minutes for one revolution, so T = 30 x 60 = 1800 s and omega = 2 pi / 1800 = 3.49 x 10^-3 rad s-1. Both values are small numbers because a radian is a large angle: one full turn is only 6.28 rad.

The linear speed and the angular speed are linked by v = r omega, where r is the distance of the point from the axis of rotation. The equation follows straight from the definition of the radian. The point travels an arc s = r theta in a time t, so v = s / t = r (theta / t) = r omega. Every point on a rigid rotating body has the same angular speed, because the whole body turns through the same angle in the same time, but the linear speed depends on the radius. A rider 6.5 m from the axis of the carousel moves at v = r omega = 6.5 x 0.5236 = 3.40 m s-1, while a passenger at the rim of the London Eye, 60 m from the centre, moves at v = 60 x 3.49 x 10^-3 = 0.21 m s-1. The rider is much faster even though the ride is only a few metres across, because the angular speed of the carousel is 150 times greater.

Why a body moving in a circle at a constant speed is accelerating

A body in uniform circular motion moves at a constant speed, but its velocity is not constant, because velocity is a vector and its direction changes continuously. In one revolution the velocity vector sweeps through a full turn, always along the tangent to the circle. The velocity is changing, so there is an acceleration, and an acceleration means a resultant force.

Draw the velocity at two positions very close together and subtract the vectors. The change in velocity points towards the centre of the circle, and dividing it by the short time interval shows that the acceleration is directed towards the centre, at right angles to the velocity. This is the centripetal acceleration, from the Latin for centre-seeking. A force of constant magnitude that is always perpendicular to the direction of motion therefore produces circular motion at a constant angular speed: it turns the velocity without changing its magnitude.

The same reasoning explains why the speed stays constant. The centripetal force is always at right angles to the displacement, so it does no work on the body. The kinetic energy is unchanged, so the speed and the angular speed are unchanged, and the force does nothing except steer the body round the circle. If the force is removed, the body carries on in a straight line along the tangent, which is Newton's first law, and that is what happens when the string of a whirled stone is cut.

Centripetal acceleration: a = v^2 / r and a = r omega^2

The size of the centripetal acceleration comes from two similar triangles. The triangle formed by the two velocity vectors and the change in velocity has two sides of length v with an angle theta between them; the triangle formed by the two radii and the arc has two sides of length r with the same angle theta between them. Both triangles are isosceles with equal apex angles, so they are similar, and the ratios of corresponding sides are equal: delta v / v = delta s / r, which rearranges to delta v = v (delta s / r).

Divide both sides by the time interval delta t: delta v / delta t = (v / r)(delta s / delta t). The left-hand side is the magnitude of the acceleration, and delta s / delta t is the speed v, so a = v^2 / r, directed towards the centre of the circle. Substituting v = r omega into a = v^2 / r gives a = (r omega)^2 / r, so a = r omega^2.

Use whichever form matches the data you are given. If the question gives a speed and a radius, use a = v^2 / r. If it gives a period or an angular speed, find omega first and then use a = r omega^2. The two forms are the same quantity written two ways, so never add them together.

A cyclist riding round a bend of radius 24 m at 8.0 m s-1 has a = v^2 / r = (8.0)^2 / 24 = 2.7 m s-2 towards the centre of the bend. A rider 6.5 m from the axis of the carousel, with omega = 0.524 rad s-1, has a = r omega^2 = 6.5 x (0.524)^2 = 1.8 m s-2. The cyclist is accelerating about one and a half times as strongly as the rider, even though the ride turns much faster.

Centripetal force: what actually provides it

Multiplying the acceleration by the mass gives the force needed to keep the body moving round the circle. Newton's second law gives F = ma, so F = mv^2 / r and F = mr omega^2. The force points towards the centre of the circle and is perpendicular to the velocity at every instant, which is why it changes the direction of the motion and not the speed.

The centripetal force is not a new kind of force. It is a job description: it is the name given to the resultant force when the motion is circular. Whatever force happens to point towards the centre in that situation is the centripetal force, and in an examination you have to name it. For a satellite it is the gravitational attraction of the Earth. For a car or a cyclist on a bend it is friction between the tyres and the road. For a stone whirled on a string it is the tension in the string. For a rider in a fairground rotor it is the normal contact force from the wall of the drum. For an electron moving at right angles to a magnetic field it is the magnetic force on the moving charge.

The cyclist on the bend has a total mass of 85 kg including the bicycle. The force needed is F = mv^2 / r = 85 x (8.0)^2 / 24 = 227 N, directed horizontally towards the centre of the bend. That 227 N can only come from friction between the tyres and the road, so on a wet or gravelly surface, where the maximum friction is smaller, the cyclist has to slow down or lean further to keep the same path.

Satellites in circular orbits: gravity as the centripetal force

For a satellite in a circular orbit, the only force acting on it is the gravitational attraction of the Earth, directed towards the centre of the Earth, and that force is the centripetal force. Its angular speed follows from the period and nothing else. A geostationary satellite takes one day for each orbit, so T = 24 x 3600 = 86 400 s and omega = 2 pi / T = 7.27 x 10^-5 rad s-1.

The radius of the geostationary orbit is 4.2 x 10^7 m, so the speed is v = r omega = 4.2 x 10^7 x 7.27 x 10^-5 = 3.1 x 10^3 m s-1, which is about 3 km every second. Because the satellite turns in step with the Earth, it always stays above the same point on the equator, which is why it is used for television broadcasts and weather pictures.

A satellite in a lower orbit has a shorter period, so from omega = 2 pi / T it must have a larger angular speed. The International Space Station orbits at a radius of about 6.8 x 10^6 m with a period of about 93 minutes, giving omega = 2 pi / (93 x 60) = 1.13 x 10^-3 rad s-1 and v = r omega = 6.8 x 10^6 x 1.13 x 10^-3 = 7.7 x 10^3 m s-1, which is why its crew see a sunrise every 90 minutes or so.

Units, direction and significant figures: where the marks are lost

Three things go wrong most often in circular motion calculations. The first is the unit of the angle: omega = 2 pi / T works only because 2 pi is in radians, so if a question gives an angle or a rate in degrees you must convert before you divide. The second is the direction: the acceleration and the force point towards the centre of the circle, so name that direction in your answer and remember that the velocity is perpendicular to both.

The third is the choice of form. Write down what you are given and pick the equation that uses it: given a period, find omega = 2 pi / T and then use a = r omega^2; given a speed, use a = v^2 / r directly. Show the equation, then the substitution, then the answer, and put the unit on the answer.

Finally, check the size of the answer against something familiar. A car going round a roundabout at 10 m s-1 with r = 20 m has a = v^2 / r = 100 / 20 = 5.0 m s-2, about half the acceleration of free fall, which is reasonable for a bend. A value of 5.0 m s-1 for an acceleration, or 0.5 m s-2, is a signal that a rearrangement or a unit has gone wrong.