Angular displacement in radians and degrees, and converting between them

Angular velocity ω, period and frequency; v = ωr and T = 2π/ω

Deriving the centripetal acceleration a = v²/r = rω² from a vector diagram

Why a resultant (centripetal) force is needed, and F = mv²/r = mrω²

Cars on flat bends and banked tracks (qualitatively)

Vertical loops, the conical pendulum and fairground rides

Testing F = mrω² with a graph, including combining percentage uncertainties

1. A body moving in a circle is always changing direction, so it is always accelerating towards the centre, and a real resultant force must supply that acceleration: F = mv²/r = mrω².

2. An angle in radians is arc length ÷ radius: θ = s/r. One full turn = 2π rad = 360°.

3. Angular velocity ω is the angle turned per second (rad s⁻¹): v = ωr, T = 2π/ω, ω = 2πf.

4. The centripetal acceleration is a = v²/r = rω². The force is provided by real forces: tension, friction, a normal reaction, weight, lift.

5. Remove the force and the object moves off along the tangent, not outwards.

6. In a vertical loop the minimum speed at the top is √(gr), when weight alone supplies the centripetal force.

From the definition, the angle in radians is the arc length divided by the radius:

with θ in rad, s the arc length (m) and r the radius (m). This is not in the formulae list; learn it, because it is the definition of the radian. A whole circle has arc length 2πr, so 360° = 2π rad. To convert: radians = degrees × π/180 and degrees = radians × 180/π; 1 rad ≈ 57.3°.

The radian is a ratio of two lengths, so it has no dimensions; "rad" is written to remind you which angle measure is in use. Useful values: 90° = π/2 rad, 180° = π rad, one revolution = 2π rad. Every equation on this page that contains θ or ω needs the angle in radians.

WORKED EXAMPLE · QUESTION

The minute hand of a station clock is 0.85 m long. Calculate (a) the angle, in radians, through which it turns in 25 minutes and (b) the distance moved by its tip in this time.

SOLUTION

(a) The hand turns 360° in 60 minutes: angle = (25 ÷ 60) × 360° = 150° = 150 × π/180 = 2.62 rad

(b) Arc length = radius × angle in radians: s = 0.85 × 2.618 = 2.23 m ≈ 2.2 m

Using 150 instead of 2.62 would give 128 m. An angle in degrees in s = rθ is the most common slip in this topic.

Here v is the linear speed (m s⁻¹), ω the angular velocity (rad s⁻¹), r the radius (m) and T the period (s). Both equations are in the formulae list. Why v = ωr: in time Δt the arc travelled is rΔθ, so the speed is rΔθ/Δt = rω. One revolution is 2π rad, so T = 2π/ω and therefore ω = 2πf.

WORKED EXAMPLE · QUESTION

The rotor of a wind turbine makes 18 revolutions per minute. Each blade is 42 m long. Calculate (a) the period, (b) the angular velocity and (c) the speed of a blade tip.

SOLUTION

(a) f = 18 ÷ 60 = 0.30 Hz, so T = 1 ÷ 0.30 = 3.33 s ≈ 3.3 s

(b) ω = 2π ÷ T = 2π × 0.30 = 1.885 rad s⁻¹ ≈ 1.9 rad s⁻¹

(c) v = ωr = 1.885 × 42 = 79.2 m s⁻¹ ≈ 79 m s⁻¹

A point halfway along the blade has the same ω but only half the speed, about 40 m s⁻¹.

Velocity is a vector. In a circle the direction of the velocity changes continuously, even if its magnitude (the speed) does not. A change of velocity in a time interval is an acceleration, so uniform circular motion is accelerated motion. The acceleration is directed towards the centre of the circle and is called the centripetal acceleration.

Deriving a = v²/r

The body moves from A to B at constant speed v while the radius turns through a small angle Δθ.

The velocities v_A and v_B are each perpendicular to their radius, so the angle between them is also Δθ.

Drawn tail to tail, they form an isosceles triangle.

The change in velocity Δv = v_B - v_A joins their tips and points towards the centre.

For small Δθ the short side is almost an arc of radius v, so |Δv| = vΔθ.

The time taken is Δt = arc AB ÷ speed = rΔθ/v.

So a = |Δv|/Δt = vΔθ×v/rΔθ = v²/r; with v = ωr this is also rω².

The centripetal acceleration (m s⁻²) is always directed towards the centre. The equation is in the formulae list, but you may be asked to derive it.

In uniform circular motion its magnitude is constant while its direction rotates with the body.

WORKED EXAMPLE · QUESTION

A sample tube in a centrifuge is 8.5 cm from the axis. The rotor turns at 7200 revolutions per minute. Calculate the centripetal acceleration of the sample and express it as a multiple of g.

SOLUTION

ω= 2π×7200 ÷60 = 2π×120 = 754 rad s⁻¹.

With r in metres: a = rω² = 0.085 ×754.0² = 4.83×10⁴ m s⁻² ≈ 4.8 ×10⁴ m s⁻²

a/g = 4.83 × 10⁴ ÷ 9.81 ≈ 4900 g

Check with the other form: v = ωr = 64.1 m s⁻¹, and v²/r gives the same 4.83 × 10⁴ m s⁻².

By Newton's first law, a body with no resultant force moves in a straight line at constant speed. To make it follow a circle, a resultant force must act, and by Newton's second law it is in the direction of the acceleration: towards the centre. It is called the centripetal force. The force is perpendicular to the velocity, so it does no work: it changes the direction of motion but not the speed or the kinetic energy.

F is the resultant force towards the centre (N) and m the mass (kg); the equation is in the formulae list. Use the v form when you know the speed and the ω form when you know the rotation rate or period. "Centripetal force" is a job description, not a new kind of force: some real force, or the resultant of several, does the job.

A flat bend

On a level road the only horizontal force on the car is friction between the tyres and the road, acting sideways towards the centre of the bend. Friction has a maximum value, while the force needed, mv²/r, rises with the square of the speed, so there is a maximum safe speed for each bend. If the speed is too high, friction cannot supply mv²/r; the car follows a path of larger radius and slides towards the outside of the bend.

WORKED EXAMPLE · QUESTION

A car of mass 1250 kg travels round a level bend of radius 45 m. (a) Calculate the sideways friction needed at 14 m s⁻¹. (b) The greatest sideways friction the tyres can provide on this road is 9.6 kN. Calculate the maximum speed at which the car can take the bend.

SOLUTION

(a) Friction is the only horizontal force, so it must equal mv²/r: F = (1250 ×14²)/45 = 5444 N ≈ 5.4 kN

(b) Set mv²/r equal to the maximum friction: vₘₐₓ = √(Fr/m) = √((9600 ×45)/1250) = 18.6 m s⁻¹ ≈ 19 m s⁻¹

Doubling the speed needs four times the friction: F ∝ v².

A banked track: the qualitative idea

Velodromes, race tracks and some motorway slip roads are banked: the outer edge is higher than the inner edge. The normal reaction acts perpendicular to the surface, so it is tilted towards the centre of the bend, and its horizontal component provides some or all of the centripetal force. At one particular speed the horizontal component of the normal reaction alone is exactly mv²/r, and no sideways friction is needed at all. Faster than this, friction acts down the slope as well; slower, it acts up the slope. Either way, banking reduces the reliance on friction, so the bend can be taken more safely at higher speeds. A banking aircraft or a cyclist leaning into a bend uses the same idea: a force that would otherwise be vertical is tilted so that it gains a horizontal component.

In a vertical circle the weight acts down at every point, so its role changes around the loop. Take towards the centre as positive at each point and write: resultant force towards centre = mv²/r.

At the top (rider on the inside of a loop, or a ball on a string), the weight and the reaction or tension both act down, towards the centre: N + mg = mv²/r.

At the bottom, the reaction or tension acts up, towards the centre, and the weight acts away from it: N - mg = mv²/r.

The speed is not constant: the body speeds up on the way down and slows on the way up, so use conservation of energy to link the speeds at different heights.

At the minimum speed at the top, the reaction (or tension) is zero, so weight alone provides the centripetal force: mg = mv²/r, giving

This is not in the formulae list, so derive it. Slower than this, the weight is more than enough to provide the centripetal force; the body falls inside the circle and the string goes slack or the rider leaves the track.

WORKED EXAMPLE · QUESTION

A rider of mass 65 kg goes round a vertical loop of radius 7.5 m on a fairground ride, sitting on the inside of the track. At the top of the loop the rider's speed is 11 m s⁻¹. Friction and air resistance are negligible. (a) Calculate the force exerted on the rider by the seat at the top. (b) Calculate the minimum speed at the top for the rider to stay in contact with the seat without a harness. (c) Calculate the force from the seat at the bottom of the loop.

SOLUTION

(a) At the top both forces act towards the centre: N + mg = mv²/r, so N = (65 ×11²)/7.5 - 65 ×9.81 = 1049 - 638 = 411 N ≈ 410 N

(b) Contact is just lost when N = 0: mg = mv²/r, so v = √(9.81 ×7.5) = 8.58 m s⁻¹ ≈ 8.6 m s⁻¹

(c) Speed at the bottom from energy, as the rider falls 2r = 15 m: v² = 11² + 2 ×9.81 ×15 = 415.3 m² s⁻² (v = 20.4 m s⁻¹).

At the bottom the reaction acts towards the centre and the weight away from it: N = mg + mv²/r = 638 + (65 ×415.3)/7.5 = 4237 N ≈ 4.2 kN

A conical pendulum is a mass on a string (or a chair on a chain) moving in a horizontal circle, with the string sweeping out a cone. Only two forces act: the tension T along the string and the weight mg.

Vertically there is no acceleration: T cos θ = mg.

Horizontally the component of tension towards the centre provides the centripetal force: T sin θ = mv²/r.

Dividing the two equations removes T and m: tan θ= v²/rg. Faster rotation gives a larger angle, and the mass of the rider does not matter.

The radius is the radius of the horizontal circle, not the length of the string.

WORKED EXAMPLE · QUESTION

On a fairground swing-chair ride, each chair hangs from a chain 5.0 m long attached to the end of an arm 3.0 m from the central axis. When the ride turns steadily, each chain makes 35° with the vertical. A chair and rider have a combined mass of 80 kg. Calculate (a) the radius of the chair's circle, (b) the speed of the chair, (c) the time for one revolution and (d) the tension in the chain.

SOLUTION

(a) Add the horizontal reach of the chain to the arm: r = 3.0 + 5.0 sin 35° = 3.0 + 2.87 = 5.87 m

(b) tan θ = v²/(rg): v² = 5.868 × 9.81 × tan 35° = 40.3, so v = 6.35 m s⁻¹ ≈ 6.3 m s⁻¹

(c) One revolution is a distance 2πr: period = 2π × 5.868 ÷ 6.349 = 5.81 s ≈ 5.8 s

(d) Vertical equilibrium: T = mg/cos 35° = 80 × 9.81 ÷ 0.8192 = 958 N ≈ 960 N

Check: T sin 35° = 550 N, and mv²/r = 80 × 40.3 ÷ 5.868 = 550 N, so the horizontal component is exactly the centripetal force.

Keep m and r fixed, vary ω, and measure the force needed. Since F = (mr)ω², a graph of F against ω² should be a straight line through the origin with gradient mr.

One method: a mass fixed to a force sensor on a motor-driven arm; a light gate counts revolutions so that the period, and hence ω = 2π/T, is found from the time for many turns.

Simpler version: a rubber bung whirled on a string that passes through a glass tube, with hanging masses providing the tension. The radius is kept constant with a small clip fixed to the string, and the period is timed over 10–20 revolutions.

Safety: safety glasses, a clear space, and a string checked for wear before whirling anything.

WORKED EXAMPLE · QUESTION

Use the graph above, for which r = 0.200 ± 0.002 m, to determine the mass on the arm. The uncertainty in the gradient is estimated as ±3%. Give the mass with its absolute uncertainty.

SOLUTION

Gradient from a large triangle on the line: (5.39 - 0.61)/(180 - 20) = 0.0299 kg m.

Gradient = mr, so m = 0.0299 ÷ 0.200 = 0.149 kg.

For a quotient, add percentage uncertainties: r has 0.002 ÷ 0.200 × 100 = 1.0%, so m has 3% + 1.0% = 4%. m = 0.149 ± 0.006 kg

The known mass, 0.150 kg, lies inside this range, so the results support F = mrω².

2 Forces and Newton's laws

Circular motion is F = ma with the acceleration pointing to the centre; the first law explains why a force is needed at all, and resolving forces handles the conical pendulum and banked track.

5 Magnetic fields and 4.7 Particle accelerators

A charged particle moving at right angles to a uniform magnetic field moves in a circle because the magnetic force BQv provides mv²/r; this gives r = p/BQ, the basis of cyclotrons and particle detectors.

7 Gravitational fields and orbits, and 5.5 Simple harmonic motion

For a satellite or planet, gravity is the centripetal force: GMm/r² = mv²/r. The same ω appears in SHM: the shadow of a point moving round a circle at constant ω moves with simple harmonic motion, and T = 2π/ω carries over unchanged.

1. Convert an angle of 2.5 rad to degrees.

2. A computer hard disk spins at 5400 revolutions per minute. Calculate its angular velocity.

3. A car turns sharply left. Explain why a passenger feels pressed against the right-hand door.

4. A 0.060 kg ball on a string 0.50 m long moves in a horizontal circle on a smooth table at 4.0 m s⁻¹. Calculate the tension.

5. A toy car runs round the inside of a vertical loop of track of radius 0.25 m. Calculate the minimum speed at the top of the loop for the car to stay on the track.

1. 2.5 × 180/π = 143°.

2. ω = 2π × 5400 ÷ 60 = 565 rad s⁻¹.

3. The passenger's body tends to continue in a straight line. The car turns left underneath them until the right-hand door reaches them; the door then pushes them to the left, towards the centre of the turn, providing their centripetal force. There is no outward force on the passenger.

4. T = mv²/r = (0.060 ×4.0²)/0.50 = 1.9 N.

5. v = √(gr) = √(9.81 ×0.25) = 1.57 m s⁻¹ ≈ 1.6 m s⁻¹. At this speed the track exerts no force on the car and its weight alone provides the centripetal force.