HookYour washing machine pulls 550g on every spin cycle

A domestic washing machine spinning at \(1{,}400\) revolutions per minute turns its drum through \(1400\times 2\pi\div 60\approx 147\) radians every second. For a typical drum of radius \(0.25\ \text{m}\), the acceleration of a sock pressed against the drum wall is \(a=r\omega^2=0.25\times 147^2\approx 5{,}400\ \text{m s}^{-2}\) — roughly \(550\) times the acceleration of gravity. Nothing is speeding up: the drum turns at a steady rate. All of that violence comes from changing direction, and it points, at every instant, towards the centre.

The spin cycle also settles the oldest argument in mechanics: the water is not 'flung outwards'. The drum wall pushes the sock inwards, forcing it round the circle; where there is a hole in the drum, the water behind it gets no inward push, so it does what Newton's first law says and carries straight on — through the hole. Nothing pulls it out; something merely stops pushing it in. This section builds that idea properly: angular speed in radians, the relations \(v=r\omega\) and \(a=r\omega^2=\frac{v^2}{r}\), the vector proof AQA can ask you to reproduce, and then the two great set-pieces of the Further Mechanics option — the conical pendulum and the vertical circle.

ModelAngular speed — the radian is not optional

A particle moving on a circle of radius \(r\) is located by the angle \(\theta\) its radius makes with a fixed direction. Its angular speed is \(\omega=\dfrac{d\theta}{dt}\), measured in radians per second. Radians earn their place through the arc-length formula \(s=r\theta\): differentiate with respect to time and the speed along the circle drops out immediately, \[v=\frac{ds}{dt}=r\frac{d\theta}{dt}=r\omega.\] That one-line result is only true because \(s=r\theta\) is only true in radians — quote \(v=r\omega\) with \(\omega\) in degrees per second or rev per minute and everything downstream is wrong.

Real problems rarely hand you radians. Convert first, every time: one revolution is \(2\pi\) radians, so \(1{,}400\ \text{rev min}^{-1}=\dfrac{1400\times 2\pi}{60}\approx 147\ \text{rad s}^{-1}\). Two companions complete the toolkit: the period \(T=\dfrac{2\pi}{\omega}\) (time for one lap) and the frequency \(f=\dfrac{1}{T}\), with \(\omega=2\pi f\). AQA is explicit that you must be able to work in both radians and revolutions per unit time — the marks are in the conversion being written down, not done silently on the calculator.

ModelConstant speed is not constant velocity

Velocity is a vector. On a circle the speed can be constant while the direction changes continuously — so the velocity is changing, and the particle is accelerating even though the speedometer never moves. The acceleration has magnitude \[a=r\omega^2=\frac{v^2}{r},\] and it points towards the centre of the circle at every instant (the two forms are interchangeable via \(v=r\omega\); use whichever matches the data given).

Newton's second law then demands a resultant force of magnitude \(\dfrac{mv^2}{r}\) directed towards the centre. This is the point students misfile: 'centripetal force' is a job description, not a new force. Something real must apply it — the tension in a string, friction from the road, the normal reaction from a drum wall, gravity on a satellite — and your force diagram should contain only those real forces. The exam method is always the same two lines: resolve towards the centre and set the resultant equal to \(\dfrac{mv^2}{r}\) (or \(mr\omega^2\)); resolve perpendicular to the plane of the circle and set that resultant to zero if the circle is horizontal.

Worked example

A car of mass \(900\ \text{kg}\) drives round a flat roundabout of radius \(30\ \text{m}\). The coefficient of friction between tyres and road is \(0.7\). Find the greatest speed at which it can take the roundabout without skidding.

Vertically there is no acceleration: \(N=mg\). The only horizontal force is friction, and it must supply the whole centripetal demand. At the point of skidding, friction is limiting: \(F=\mu N=\mu mg\). Setting supply equal to demand: \[\mu mg=\frac{mv^2}{r}\quad\Rightarrow\quad v=\sqrt{\mu gr}=\sqrt{0.7\times 9.8\times 30}=\sqrt{205.8}\approx 14.3\ \text{m s}^{-1},\] about \(52\ \text{km h}^{-1}\). Notice \(m\) cancelled: a loaded lorry and a motorbike skid at the same speed on the same bend — more weight means more friction available but exactly proportionally more centripetal force needed. Writing that sentence is an AO2 mark, not padding.

MechanismThe vector derivation — differentiate the circle

AQA expects you to treat position, velocity and acceleration in circular motion as vectors, and the whole theory falls out of one differentiation. Place the centre at the origin and let the particle move with constant angular speed \(\omega\), so \(\theta=\omega t\): \[\mathbf{r}(t)=r\left(\cos\omega t\,\mathbf{i}+\sin\omega t\,\mathbf{j}\right).\] Differentiate once: \[\mathbf{v}=\dot{\mathbf{r}}=r\omega\left(-\sin\omega t\,\mathbf{i}+\cos\omega t\,\mathbf{j}\right),\] which has magnitude \(|\mathbf{v}|=r\omega\) (confirming \(v=r\omega\)) and satisfies \(\mathbf{r}\cdot\mathbf{v}=r^2\omega(-\cos\omega t\sin\omega t+\sin\omega t\cos\omega t)=0\) — the velocity is perpendicular to the radius, i.e. tangent to the circle.

Differentiate again: \[\mathbf{a}=\dot{\mathbf{v}}=-r\omega^2\left(\cos\omega t\,\mathbf{i}+\sin\omega t\,\mathbf{j}\right)=-\omega^2\,\mathbf{r}.\] Everything the topic claims is in that last equality. The magnitude is \(r\omega^2\); the minus sign says the acceleration points along \(-\mathbf{r}\), from the particle back towards the centre; and substituting \(\omega=v/r\) converts the magnitude to \(v^2/r\). The relation \(\ddot{\mathbf{r}}=-\omega^2\mathbf{r}\) is also your first sighting of the simple harmonic motion equation — each coordinate of uniform circular motion performs SHM, a connection the pure core's differential equations section (I7) picks up.

CaseThe conical pendulum — resolve twice, divide once

Swing a bob on a string so it traces a horizontal circle and the string sweeps a cone: this is the conical pendulum, AQA's favourite horizontal-circle setting. Let the string have length \(L\) and make angle \(\theta\) with the vertical; the circle's radius is \(r=L\sin\theta\). Two directions, two equations. Vertically the bob does not accelerate: \(T\cos\theta=mg\). Horizontally the tension's component supplies the centripetal force: \(T\sin\theta=mr\omega^2\). Divide the second by the first: \[\tan\theta=\frac{r\omega^2}{g},\] and substituting \(r=L\sin\theta\) gives the cleaner \(\cos\theta=\dfrac{g}{L\omega^2}\). Spin faster and \(\cos\theta\) shrinks — the string rises. But it can never reach the horizontal: that would need \(\cos\theta=0\), i.e. infinite \(\omega\), because a horizontal string has no vertical component left to hold the weight up.

The two-string variant is the same machine with more bookkeeping: a particle attached by two strings to points on a rotating vertical pole has two tensions, each resolved vertically and radially, giving two simultaneous equations. The standard sting in the tail is a condition question — the lower string only stays taut while its tension is non-negative, so setting \(T_2\ge 0\) yields the minimum angular speed for the configuration to hold.

Worked example

A bob of mass \(0.2\ \text{kg}\) on a light inextensible string of length \(0.5\ \text{m}\) describes a horizontal circle at constant angular speed \(\omega=7\ \text{rad s}^{-1}\). Find the angle of the string to the vertical and the tension.

From \(\cos\theta=\dfrac{g}{L\omega^2}=\dfrac{9.8}{0.5\times 49}=0.4\), the angle is \(\theta=\cos^{-1}0.4\approx 66.4^\circ\). The vertical equation gives the tension directly: \(T=\dfrac{mg}{\cos\theta}=\dfrac{0.2\times 9.8}{0.4}=4.9\ \text{N}\). Check it against the radial equation, because checks are cheap: \(r=L\sin\theta=0.5\sin 66.4^\circ\approx 0.458\ \text{m}\), so \(mr\omega^2=0.2\times 0.458\times 49\approx 4.49\ \text{N}\), and indeed \(T\sin\theta=4.9\times\sin 66.4^\circ\approx 4.49\ \text{N}\). Both equations agree — the resolve-twice-divide-once machine is self-auditing, and showing the check is exactly the kind of communication the mark scheme's explanation marks reward.

CaseVertical circles — energy between points, Newton at a point

Tip the circle vertical and gravity starts doing work, so the speed is no longer constant and \(a=r\omega^2\) with fixed \(\omega\) is dead. The strategy that replaces it is a two-tool method. Tool one: conservation of energy links the speeds at any two heights: \(\frac{1}{2}mv_1^2+mgh_1=\frac{1}{2}mv_2^2+mgh_2\). Tool two: Newton's second law towards the centre, applied at the single point the question cares about — usually the top or bottom, where weight and tension are collinear and the algebra is clean.

The classic question is whether a particle on a string completes the circle. A string can only pull, so the motion survives the top only if the tension there is non-negative. At the top, both \(T\) and \(mg\) point down (towards the centre): \(T+mg=\dfrac{mv_{top}^2}{r}\), so \(T\ge 0\) forces \[v_{top}^2\ge gr.\] Energy then translates that into a condition at the bottom: \(\frac{1}{2}mv_{b}^2=\frac{1}{2}mv_{top}^2+mg(2r)\) gives \(v_b^2=v_{top}^2+4gr\ge gr+4gr=5gr\). The same logic with a normal reaction \(R\ge 0\) covers a body on the inside of a track. But swap the string for a light rod, or thread a bead on a smooth circular wire, and the support can push as well as pull — the particle then needs only \(v_{top}^2\ge 0\), i.e. \(v_b^2\ge 4gr\), and can crawl over the top arbitrarily slowly. Identifying which support you have is the first mark of the question.

Worked example

A ball of mass \(0.3\ \text{kg}\) is whirled in a vertical circle on a light inextensible string of length \(0.8\ \text{m}\). Find the minimum speed at the lowest point for it to complete the circle, and the string tension at that moment.

At the top, the limiting case is \(T=0\): \(mg=\dfrac{mv_{top}^2}{r}\), so \(v_{top}^2=gr=9.8\times 0.8=7.84\), i.e. \(v_{top}=2.8\ \text{m s}^{-1}\). Energy from bottom to top (a rise of \(2r=1.6\ \text{m}\)): \[\tfrac{1}{2}v_b^2=\tfrac{1}{2}v_{top}^2+g(2r)\quad\Rightarrow\quad v_b^2=7.84+4\times 9.8\times 0.8=39.2,\] so \(v_b=\sqrt{39.2}\approx 6.26\ \text{m s}^{-1}\). At the bottom, tension and weight oppose: \(T-mg=\dfrac{mv_b^2}{r}\), so \(T=m\left(g+\dfrac{v_b^2}{r}\right)=0.3\left(9.8+\dfrac{39.2}{0.8}\right)=0.3\times 58.8=17.6\ \text{N}\). That is exactly \(6mg\) — and it always is: \(v_b^2=5gr\) makes \(T=mg+\dfrac{m\cdot 5gr}{r}=6mg\), a result worth knowing so you can sanity-check any completing-the-circle answer in one glance.

VocabularyKey terms the mark scheme pays for

Radian

The angle for which arc length equals radius, so \(s=r\theta\). Every circular-motion formula (\(v=r\omega\), \(a=r\omega^2\)) assumes angles in radians; one revolution is \(2\pi\) radians.

Angular speed (ω)

The rate of change of the angle locating the particle, \(\omega=\frac{d\theta}{dt}\), in \(\text{rad s}^{-1}\). Convert from rev per minute by multiplying by \(2\pi\) and dividing by \(60\).

Period

The time for one complete revolution, \(T=\frac{2\pi}{\omega}\); the frequency is \(f=\frac{1}{T}\), with \(\omega=2\pi f\).

Centripetal acceleration

The acceleration of a particle moving in a circle, magnitude \(a=r\omega^2=\frac{v^2}{r}\), directed towards the centre — non-zero even at constant speed, because the velocity's direction changes.

Centripetal force

Not a separate force: the name for the resultant of the real forces (tension, friction, gravity, normal reaction) that must point to the centre with magnitude \(\frac{mv^2}{r}\) for circular motion to happen.

Conical pendulum

A bob on a string sweeping a horizontal circle: \(T\cos\theta=mg\) vertically and \(T\sin\theta=mr\omega^2\) radially, giving \(\cos\theta=\frac{g}{L\omega^2}\).

Critical speed at the top

For a particle on a string (or inside a track), the top of a vertical circle requires \(v_{top}^2\ge gr\) so that \(T\ge 0\) (or \(R\ge 0\)); by energy this means \(v^2\ge 5gr\) at the bottom.

Tangential velocity

The velocity of a particle in circular motion is always tangent to the circle: differentiating \(\mathbf{r}=r(\cos\omega t\,\mathbf{i}+\sin\omega t\,\mathbf{j})\) gives \(\mathbf{r}\cdot\mathbf{v}=0\), with \(|\mathbf{v}|=r\omega\).

TrapsMisconceptions that cost marks

“Objects in circular motion are flung outwards by centrifugal force.”

Actually: In an inertial frame nothing pulls outwards. The water in the spin cycle leaves through the holes because there the drum stops pushing it inwards, and it travels in a straight line — a tangent, not a radius. What you feel in a cornering car is the door pushing you towards the centre.

“Centripetal force is an extra force to draw on the diagram.”

Actually: Draw only real forces — weight, tension, normal reaction, friction — then set their resultant towards the centre equal to \(\frac{mv^2}{r}\). Adding a separate 'centripetal force' arrow double-counts and typically loses the method mark for the radial equation.

“If the speed is constant, the acceleration must be zero.”

Actually: Acceleration is the rate of change of velocity, and velocity is a vector. On a circle the direction changes continuously, so \(a=\frac{v^2}{r}\) is non-zero at constant speed — \(550g\) in a \(1{,}400\) rpm washing machine drum.

“Spin a conical pendulum fast enough and the string becomes horizontal.”

Actually: \(\cos\theta=\frac{g}{L\omega^2}\) is positive for every finite \(\omega\), so \(\theta\) approaches but never reaches \(90^\circ\). A horizontal string would have no vertical tension component to balance the weight — the bob would simply fall.

“To get round the top of a vertical circle a particle just needs to still be moving.”

Actually: On a string it needs \(v_{top}^2\ge gr\) so the tension stays non-negative — slower than that and the string goes slack, and the particle becomes a projectile before reaching the top. Only a bead on a wire or a bob on a rigid rod, where the support can push, may pass the top arbitrarily slowly.

ExamWhat examiners want

Convert the angular speed to \(\text{rad s}^{-1}\) before touching anything else, and show the conversion — 'rev per minute' data is AQA's standard opening trap, and the conversion line is often a mark in itself. Then earn the radial method mark the same way every time: a force diagram with only real forces, followed by 'resolving towards the centre: (resultant) \(=\frac{mv^2}{r}\)' or \(=mr\omega^2\), choosing the form that matches the data. For horizontal circles the perpendicular equation (usually \(T\cos\theta=mg\)) is the second method mark; state in words which real force is providing the centripetal resultant, because that sentence is where the AO2 communication credit sits.

Vertical-circle questions are a fixed two-step dance and the mark scheme is built around it: an energy equation between two named points, and Newton's second law towards the centre at the specific point of interest. Write the completing-the-circle condition before substituting numbers — \(T\ge 0\) for a string, \(R\ge 0\) on a track, but only \(v^2>0\) at the top for a rod or a bead on a wire — since choosing the wrong condition forfeits the whole final part. Quote the modelling assumptions when asked (particle, light inextensible string, smooth wire): Further Maths reserves roughly a fifth of its marks for AO3 modelling and interpretation, and these are its currency.

Use \(g=9.8\ \text{m s}^{-2}\) unless the paper says otherwise, keep exact multiples like \(v_b^2=5gr\) and \(T=6mg\) as long as possible, and round only at the end to three significant figures. The derivation of \(\mathbf{a}=-\omega^2\mathbf{r}\) by differentiating \(\mathbf{r}=r(\cos\omega t\,\mathbf{i}+\sin\omega t\,\mathbf{j})\) twice is a legitimate 'show that' target — practise writing it in four lines, magnitudes and direction statement included.