Speed Maths LAB
EXP-003
Module 5
OCR A H556 · Module 5 · 5.1 Thermal physics · 5.2 Circular motion

Newtonian World

Seven benches. Circular motion first, because that's where you are, then gases and internal energy. Every bench has a purple "The maths behind it" box that goes past the spec: the derivations, the calculus, the extra results. Colours:

velocity acceleration / resultant force individual real forces object, gas particles energy, temperature
What clicked

Centripetal force is a job, not a force.

Part

Circular motion

Bench 1
5.2.2(a)(c)

Constant speed, yet accelerating: where Δv points

Velocity is a vector, so turning it is changing it. Take the velocity at two moments and subtract: the change Δv points towards the centre. Shrink the gap between the two moments and \(\Delta v/\Delta t\) settles on exactly \(v^2/r\). That's the whole of centripetal acceleration.

The maths behind it

Geometry (what the picture shows). The two velocity vectors have the same length v and an angle Δθ between them, so they form an isosceles triangle:

\[|\Delta v| = 2v\sin\tfrac{\Delta\theta}{2}, \qquad \Delta t = \frac{\text{arc}}{v} = \frac{r\,\Delta\theta}{v}\]
\[\frac{|\Delta v|}{\Delta t} = \frac{v^2}{r}\left[\frac{\sin(\Delta\theta/2)}{\Delta\theta/2}\right] \;\longrightarrow\; \frac{v^2}{r} \quad \text{as } \Delta\theta \to 0\]

The bracket tends to 1 because \(\sin x/x \to 1\) (small-angle approximation, from your Pure).

Calculus (cleaner). Position on the circle: \(\mathbf{r}(t) = r(\cos\omega t,\ \sin\omega t)\). Differentiate twice:

\[\mathbf{v}(t) = r\omega\,(-\sin\omega t,\ \cos\omega t) \quad\Rightarrow\quad |\mathbf{v}| = r\omega,\ \text{at } 90^\circ \text{ to } \mathbf{r}\]
\[\mathbf{a}(t) = -r\omega^2(\cos\omega t,\ \sin\omega t) = -\omega^2\,\mathbf{r}(t)\]

The minus sign says "pointing back towards the centre", and \(|\mathbf a| = \omega^2 r = v^2/r\). You'll meet \(a = -\omega^2 x\) again in simple harmonic motion (5.3): SHM is the shadow of circular motion on a straight line.

In the exam
  • "Velocity / direction is always changing, so it is accelerating; acceleration is towards the centre."
  • \(a = v^2/r = \omega^2 r\); \(v = \omega r\); \(\omega = 2\pi/T = 2\pi f\).
Bench 2
Newton's first law

Cut the string: which way does it go?

Most people guess "outwards". Cut the string and watch. With no force, the ball carries on in a straight line along the tangent, the direction it was already moving. There is no outward force throwing it away; the only force was the inward pull, and you just removed it.

The maths behind it

"Centrifugal force" only appears if you sit in the rotating frame (on the roundabout), where you need a fictitious outward force to explain why you feel pushed outwards. From the ground, nothing pushes you out: your body tries to go straight (N1) and the roundabout keeps pulling it inwards. A-level answers are always in the ground frame, so never write "centrifugal force".

Bench 3
5.2.2(d)(i)

The conical pendulum: centripetal force is a job, not a force

Only two forces act on the bob: tension and weight. Neither is "the centripetal force". Spin faster and the string swings out until the horizontal part of the tension is exactly what's needed for the circle, while the vertical part still holds up the weight. The red arrow is the resultant: always horizontal, always towards the axis.

The maths behind it

Mass m, string length L, angle θ to the vertical, radius \(r = L\sin\theta\).

\[\text{horizontal: } T\sin\theta = m\omega^2 r = m\omega^2 L\sin\theta \quad\Rightarrow\quad T = m\omega^2 L\]
\[\text{vertical: } T\cos\theta = mg \quad\Rightarrow\quad \cos\theta = \frac{g}{\omega^2 L}\]

Two surprises. The tension is \(m\omega^2 L\) whatever the angle. And there's a threshold: if \(\omega^2 L < g\), then \(\cos\theta\) would be bigger than 1, so the bob just hangs straight down. It only swings out once \(\omega > \sqrt{g/L}\).

\[\text{period: } \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{h}{g}}\]

where \(h = L\cos\theta\) is the height of the pivot above the circle. Two conical pendulums at the same height have the same period, whatever their string lengths. Slide L and ω and watch h in the readout.

In the exam
  • "(Horizontally) \(mv^2/r = T\sin\theta\) and (vertically) \(mg = T\cos\theta\)."
  • Faster: "T increases / angle θ increases to provide a larger centripetal force". Examiners penalise treating centripetal force as an extra real force.
  • Free-body diagram: real forces only, never an arrow labelled "centripetal force".
Bench 4
5.2.2(d)(i) and beyond

The banked bend: who supplies the push?

A car takes a bend of radius 50 m on a banked track. The normal force is at 90° to the road, so on a slope it leans inwards: its horizontal part helps provide the centripetal force. At the design speed it's enough on its own and no friction is needed. Go faster and friction must act down the slope to help; go slower and it acts up the slope to stop the car sliding in.

The maths behind it

Take friction f along the slope, positive down the slope. Resolve horizontally (towards the centre) and vertically:

\[N\sin\theta + f\cos\theta = \frac{mv^2}{r}, \qquad N\cos\theta - f\sin\theta = mg\]
\[\Rightarrow\; f = m\left(\frac{v^2}{r}\cos\theta - g\sin\theta\right), \qquad N = m\left(\frac{v^2}{r}\sin\theta + g\cos\theta\right)\]

No friction needed when \(f = 0\):

\[v_{\text{design}}^2 = rg\tan\theta\]

On a flat road (\(\theta = 0\)) all of \(mv^2/r\) must come from friction, which is why icy bends are dangerous.

In the exam
  • "The horizontal component of the normal force provides the centripetal force; the vertical component balances the weight."
  • Banked aircraft: same idea with lift: \(L\cos\theta = mg\), \(L\sin\theta = mv^2/r\). No marks for tan θ alone or for leaving out the trig.
Bench 5
vertical circles (beyond the spec's minimum)

The loop: why the top is the dangerous point

A ball on a string is whirled in a vertical circle, launched from the bottom. It slows as it rises (KE turns into GPE). At the top, the weight already points to the centre, so the string has the least work to do: tension is smallest at the top and largest at the bottom. Launch too slowly and the tension at the top would have to be negative, so the string goes slack and the ball falls inside the circle.

The maths behind it

Measure θ from the bottom. Energy conservation, then Newton's second law along the radius:

\[v^2 = u^2 - 2gr(1 - \cos\theta)\]
\[T - mg\cos\theta = \frac{mv^2}{r} \quad\Rightarrow\quad T = \frac{mu^2}{r} - mg(2 - 3\cos\theta)\]

At the bottom (\(\theta = 0\)): \(T = mu^2/r + mg\). At the top (\(\theta = \pi\)): \(T = mu^2/r - 5mg\).

\[T_{\text{top}} \ge 0 \iff u^2 \ge 5gr, \quad \text{and then } v_{\text{top}} \ge \sqrt{gr}\]

So the minimum speed at the top is \(\sqrt{gr}\), when the weight alone provides the centripetal force (that one is examinable), and you need \(\sqrt5 \approx 2.24\) times \(\sqrt{gr}\) at the bottom to get there. Below \(u = \sqrt{2gr}\) the ball never passes the horizontal, so it just swings like a pendulum and the string never goes slack.

In the exam
  • Minimum speed at the top: "when the weight provides / equals the required centripetal force"; \(mg = mv^2/r\), so \(v = \sqrt{gr}\).
  • Greatest force at the bottom: "it supports the weight and provides the upward resultant (centripetal) force".
Part

Thermal physics and ideal gases

Bench 6
5.1.2(c)

Brownian motion: what you see isn't what's moving it

Look into a smoke cell through a microscope and you see bright specks jiggling about at random. Those specks are smoke particles, lit from the side. Nothing pushes them on purpose. They're being hit, all the time and from every direction, by air molecules far too small to see. At any instant more hits land on one side than the other, so the speck lurches. Press the button to reveal the air.

In the exam
  • Answers about the smoke alone score zero. Every observation must be linked to the air molecules.
  • Smoke particles move randomly because they are hit by randomly moving air molecules.
  • The smoke is visible but the air molecules are not, so the air molecules must be very small.
  • The small, jerky movements come from large numbers of air molecules hitting from all sides, unevenly.
  • Spell Brownian correctly: it's a quality-of-written-communication mark.
Bench 7
5.1.4

A gas is just particles hitting walls

Every dot is a gas particle. Each time one bounces off a wall it changes momentum, and the wall feels a tiny kick: add up all the kicks per second per metre of wall and you have pressure. Heat the gas and every kick is harder and more frequent. Squash the box and the kicks are spread over less wall. Press "All the same speed": collisions between particles scramble the speeds into the Maxwell-Boltzmann shape on the right within seconds.

The maths behind it

The full derivation of \(pV = \tfrac13 Nm\overline{c^2}\) (not required by the spec, but it's where everything comes from). One particle, mass m, velocity components \((c_x, c_y, c_z)\), in a cube of side L:

\[\text{each hit on the right wall: } \Delta p = 2mc_x; \qquad \text{hits per second} = \frac{c_x}{2L}\]
\[\text{average force} = 2mc_x \times \frac{c_x}{2L} = \frac{mc_x^2}{L}\]
\[\text{all } N \text{ particles: } F = \frac{Nm\langle c_x^2\rangle}{L}, \qquad p = \frac{F}{L^2} = \frac{Nm\langle c_x^2\rangle}{V}\]

No direction is special, so \(\langle c_x^2\rangle = \langle c_y^2\rangle = \langle c_z^2\rangle\), and since \(c^2 = c_x^2 + c_y^2 + c_z^2\) each is a third of \(\langle c^2\rangle\):

\[pV = \tfrac{1}{3}Nm\langle c^2\rangle \qquad \text{(the }\tfrac{1}{3}\text{ is “three directions share it”)}\]

Then the required one: \(pV = NkT\) gives \(\tfrac12 m\langle c^2\rangle = \tfrac32 kT\). Read it as \(\tfrac12 kT\) of energy per direction of motion (equipartition). This simulation is 2D, so it has two directions: \(pA = \tfrac12 Nm\langle c^2\rangle\) and \(\tfrac12 m\langle c^2\rangle = kT\). The readout checks that live.

Numbers for real air: N₂ has \(m = 4.65\times10^{-26}\ \text{kg}\). At 293 K, \(c_{\text{rms}} = \sqrt{3kT/m} \approx 510\ \text{m s}^{-1}\), faster than the speed of sound (343 m s⁻¹). That's no coincidence: sound is carried by these molecules.

In the exam
  • Pressure: momentum changes at the wall; force = rate of change of momentum (N2); equal and opposite force on the wall (N3); pressure = force / area.
  • Heating at constant volume: faster molecules, more frequent collisions, greater change of momentum per collision.
  • Maxwell-Boltzmann: hotter means the peak moves right and gets lower, the curve broadens, the area (number of particles) stays the same.
Bench 8
5.1.2-5.1.3

Where the energy goes: KE, PE and the flat bits

Heat 1 kg of ice from −20 °C at a steady rate. On the slopes, the energy speeds the particles up: KE rises, so the temperature rises. On the flat parts, the temperature can't change, because the energy goes into pulling particles apart: PE rises. Notice how long the boiling plateau is compared with melting: separating particles completely takes far more energy than loosening them.

Watch the particles on the right while you drag. On a slope, the vibration amplitude (or speed) grows, because that's what a temperature rise is. On a flat part it freezes: the dashed ring around the highlighted particle stops growing, because the temperature, and so the mean KE, isn't changing. What changes instead is the separation and arrangement: the lattice comes apart when melting, and the particles fly far apart when boiling. That's the PE going up.

The maths behind it

Each slope is \(E = mc\,\Delta\theta\), so on a temperature-energy graph the gradient is \(1/(mc)\): a smaller specific heat capacity gives a steeper slope. Ice (2100 J kg⁻¹ K⁻¹) climbs about twice as steeply as water (4200 J kg⁻¹ K⁻¹). On a temperature-time graph at constant power P the gradient is \(P/(mc)\), the same idea.

\[\text{ice } {-20}\to 0^\circ\text{C: } 1\times 2100\times 20 = 42\text{ kJ}; \qquad \text{melting: } 1\times 3.34\times10^5 = 334\text{ kJ}\]
\[\text{water } 0\to 100^\circ\text{C: } 1\times 4200\times 100 = 420\text{ kJ}; \qquad \text{boiling: } 1\times 2.26\times10^6 = 2260\text{ kJ}\]

Boiling takes almost 7 times as long as melting. Melting only loosens the structure; boiling pulls particles apart until there are essentially no forces between them, and in an ideal gas the PE is zero by assumption.

In the exam
  • Melting at constant temperature: "no change in KE because temperature is constant; PE increases; internal energy increases". "KE is not increasing" doesn't score.
  • Internal energy: "the sum of the randomly distributed KE and PE of the molecules".
  • \(c = E/(m\,\Delta\theta)\): energy per unit mass per kelvin. Gradient on a θ-t graph is inversely proportional to c ("not c = 1/gradient").
Bench 9
5.1 wording

Would this score?

Thermal physics marks are mostly lost on wording, not ideas. Every statement below is the kind of thing examiners report seeing. Decide whether it would get the mark, then check. Each one cites the mark scheme it comes from.

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