Notes · PhysicsUK · A-Levels

Further mechanics and thermal physics

This chapter extends motion to the circular and the oscillatory, then turns to heat and gases. It develops circular motion and the centripetal force, defines simple harmonic motion and derives its solutions and energy, and treats resonance and damping. It then covers thermal energy transfer, the gas laws and, as its culmination, the kinetic-theory derivation that links the pressure of a gas to the motion of its molecules.

6 sections~17 min reading time3 competenciesLevel Standard 3 · Advanced 3

T·0666 / 13

Exam profile

AO1 · Define angular speed, centripetal acceleration and simple harmonic motion and state the gas laws and kinetic-theory assumptionsAO2 · Calculate centripetal force, SHM periods, speeds and energies, thermal energy transfers and gas-state quantitiesAO3 · Derive pV = (1/3)Nmc-rms-squared, analyse resonance and damping curves and evaluate the thermal required practicals

Operators:definederiveshow thatcalculateexplainsketch

basic level

This is A2 (full A-Level) content, though circular motion and SHM also appear on Paper 1. It requires centripetal force, the SHM equations, thermal energy transfer and the gas laws.

higher level

The full A-Level requires the kinetic-theory derivation of pV = (1/3)Nmc-rms-squared, the link to temperature via (1/2)mc-rms-squared = (3/2)kT, and quantitative treatment of resonance and damping.

Depth

Text

Contents · 6 sections

Further mechanics and thermal physics

Circular motion and centripetal force#

StandardAQA 7408 3.6.1.1DfE GCE Physics - circular motion

Centripetal force in circular motion

Fig. 1An object moving round a circle has a velocity along the tangent and a centripetal force directed towards the centre; the force is always perpendicular to the velocity.

An object moving in a circle at constant speed is continually changing direction, and therefore continually accelerating, even though its speed is constant - velocity is a vector. Its angular speed

\omega

is the angle swept per unit time, in radians per second, related to the linear speed by

v = \omega r

and to the period and frequency by

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

The acceleration points towards the centre of the circle and is called the centripetal acceleration,

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

. It is caused by a resultant force directed towards the centre - the centripetal force - of magnitude

F = \dfrac{mv^2}{r} = m\omega^2 r

. This is not a new kind of force; it is provided by whatever real force acts towards the centre: gravity for a satellite, tension for a whirled mass, friction for a cornering car.

Because the centripetal force is always perpendicular to the velocity, it does no work on the object and its speed does not change - only its direction. If the force is removed (the string breaks, the road ices over) the object continues in a straight line along the tangent, obeying Newton's first law. This is why cornering demands a sideways force and why that force is felt to be 'missing' when grip is lost.

In a vertical circle the required centripetal force is the same, but gravity now helps or hinders it. At the top of a loop, weight and tension both act downward towards the centre, so the minimum speed to maintain contact is where weight alone provides the centripetal force:

mg = \dfrac{mv^2}{r}

, giving

v_{\min} = \sqrt{gr}

. At the bottom the tension must exceed the weight to provide the centripetal force and support the object.

\omega = \dfrac{2\pi}{T} = 2\pi f, \qquad v = \omega r

Angular speed

Relates angular speed to period, frequency and linear speed.

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

Centripetal acceleration

Directed towards the centre of the circle.

F = \dfrac{mv^2}{r} = m\omega^2 r

Centripetal force

The resultant force towards the centre.

Worked example

Tension in a whirled string

A $0.15\,\text{kg}$ mass on a $0.60\,\text{m}$ string is whirled in a horizontal circle at $2.0\,\text{rev s}^{-1}$ . Find the tension, treating the string as horizontal.

Angular speed
$\omega = 2\pi f = 2\pi \times 2.0 = 12.6\,\text{rad s}^{-1}$ .
Centripetal force
The tension provides it: $F = m\omega^2 r = 0.15 \times (12.6)^2 \times 0.60$ .
$F = 0.15 \times 158 \times 0.60 = 14\,\text{N}$

Result: The string tension is about $14\,\text{N}$ .

Typical mistakes

Treating 'centrifugal force' as a real outward force - the only real force is the inward (centripetal) one.
Forgetting to include the weight when analysing motion in a vertical circle.

Active revision

A $0.20\,\text{kg}$ ball on a string of length $0.80\,\text{m}$ is whirled in a horizontal circle at $3.0\,\text{rev s}^{-1}$ . Find the tension in the string.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Simple harmonic motion: definition and solutions#

AdvancedAQA 7408 3.6.1.2AQA 7408 3.6.1.3DfE GCE Physics - simple harmonic motion

Displacement and velocity in SHM

Fig. 2Displacement follows a cosine; velocity is a negative sine a quarter-cycle ahead - zero at the extremes of displacement and greatest as the object passes through the centre.

Simple harmonic motion (SHM) is defined by a single condition: the acceleration is proportional to the displacement from equilibrium and always directed back towards it,

a = -\omega^2 x

. The constant

\omega

is the angular frequency and the minus sign captures the restoring nature of the motion. Any system with a linear restoring force - a mass on a spring obeying

F = -kx

, a pendulum for small angles - performs SHM, and the negative-proportional relationship is the acid test in an exam.

Solving the defining equation gives sinusoidal motion. Taking the object released from its maximum displacement (amplitude

A

) at

t = 0

, the displacement is

x = A\cos(\omega t)

. Differentiating gives the velocity

v = -A\omega\sin(\omega t)

and a second differentiation gives

a = -A\omega^2\cos(\omega t) = -\omega^2 x

, confirming the definition. Eliminating time between displacement and velocity gives the useful relation

v = \pm\omega\sqrt{A^2 - x^2}

, so the speed is greatest (

v_{\max} = \omega A

) at the centre and zero at the extremes.

The period of SHM depends only on the system, not on the amplitude - a property called isochronism, which is why a pendulum keeps good time as its swing decays. For a mass

m

on a spring of stiffness

k

\omega = \sqrt{k/m}

, so

T = 2\pi\sqrt{m/k}

; for a simple pendulum of length

L

T = 2\pi\sqrt{L/g}

. The pendulum result gives an accurate laboratory measurement of

g

The graphs tell the story. Displacement is a cosine; velocity is a negative sine, a quarter-cycle ahead, zero at the extremes and maximum at the centre; acceleration is a negative cosine, in antiphase with displacement and maximum at the extremes. Recognising these phase relationships - and reading amplitude and period from them - is a routine exam task.

a = -\omega^2 x

Defining condition of SHM

Acceleration proportional to, and opposite to, displacement.

x = A\cos(\omega t), \qquad v = \pm\,\omega\sqrt{A^2 - x^2}

Displacement and velocity

Solutions of the SHM equation; speed is greatest at the centre.

T = 2\pi\sqrt{\dfrac{m}{k}}, \qquad T = 2\pi\sqrt{\dfrac{L}{g}}

Periods of standard oscillators

Mass-spring and simple pendulum.

Worked example

A mass on a spring

A $0.30\,\text{kg}$ mass on a spring of stiffness $12\,\text{N m}^{-1}$ oscillates with amplitude $0.040\,\text{m}$ . Find the period and the maximum speed.

Angular frequency
$\omega = \sqrt{k/m} = \sqrt{12/0.30} = \sqrt{40} = 6.32\,\text{rad s}^{-1}$ .
Period
$T = 2\pi/\omega = 2\pi/6.32 = 0.99\,\text{s}$ .
Maximum speed
$v_{\max} = \omega A = 6.32 \times 0.040$ .
$v_{\max} = 6.32 \times 0.040 = 0.25\,\text{m}\,\text{s}^{-1}$

Result: The period is $0.99\,\text{s}$ and the maximum speed is $0.25\,\text{m}\,\text{s}^{-1}$ .

Typical mistakes

Thinking the period of SHM depends on the amplitude - it does not (within the SHM approximation).
Losing the minus sign or the phase relationship between displacement, velocity and acceleration.

Active revision

A mass of $0.25\,\text{kg}$ on a spring of stiffness $40\,\text{N m}^{-1}$ oscillates with amplitude $0.050\,\text{m}$ . Find the period and the maximum speed.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Energy in SHM, resonance and damping#

AdvancedAQA 7408 3.6.1.3AQA 7408 3.6.1.4DfE GCE Physics - SHM energy and resonance

Energy against displacement in SHM

Fig. 3Potential energy (a parabola, minimum at the centre) and kinetic energy (an inverted parabola, maximum at the centre) always add to the constant total energy.

As an oscillator moves, energy interchanges continuously between kinetic and potential while the total stays constant (in the absence of damping). At the extremes all the energy is potential (the object is momentarily at rest); at the centre all of it is kinetic (maximum speed). The potential energy is

\tfrac{1}{2}k x^2

, the kinetic energy is

\tfrac{1}{2}k(A^2 - x^2)

, and their sum, the total energy

\tfrac{1}{2}k A^2

, is independent of position. A graph of energy against displacement shows two parabolas that always add to a horizontal line.

Free vibrations occur at the system's own natural frequency

f_0

once it has been displaced and released. Forced vibrations occur when a periodic driving force is applied; the system then oscillates at the driving frequency. When the driving frequency approaches the natural frequency, the amplitude grows dramatically - this is resonance, and it transfers energy into the oscillator most efficiently.

Damping is any process that removes energy from an oscillator, usually as heat through resistive forces, so the amplitude decays with time. Light damping gives a slowly decaying oscillation; heavy damping slows the return to equilibrium; critical damping returns the system to rest in the shortest time without overshooting - the design goal for car suspension and instrument needles. Increasing the damping lowers and broadens the resonance peak and shifts it slightly to a lower frequency.

Resonance is a double-edged phenomenon. It is exploited in tuned circuits, musical instruments and magnetic resonance imaging, but it must be avoided in structures: bridges, buildings and machinery are engineered so that their natural frequencies differ from the frequencies of likely driving forces (wind, footfall, engine vibration), and damping is added to limit the amplitude if resonance does occur.

E_{\text{total}} = \tfrac{1}{2}kA^2

Total energy in SHM

Constant, independent of displacement.

E_p = \tfrac{1}{2}kx^2, \qquad E_k = \tfrac{1}{2}k(A^2 - x^2)

Potential and kinetic energy

Two parabolas that sum to the constant total.

Resonance curves for different damping

Fig. 4As the driving frequency approaches the natural frequency the amplitude peaks; heavier damping lowers and broadens the peak.

Worked example

Total energy of an oscillator

The $12\,\text{N m}^{-1}$ spring oscillator has amplitude $0.040\,\text{m}$ . Find its total energy and its maximum kinetic energy.

Total energy
$E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(12)(0.040)^2$ .
$E = 0.5 \times 12 \times 1.6\times10^{-3} = 9.6\times10^{-3}\,\text{J}$
Maximum kinetic energy
At the centre all the energy is kinetic, so $E_{k,\max} = E = 9.6\times10^{-3}\,\text{J}$ .

Result: The total energy is $9.6\,\text{mJ}$ , entirely kinetic as the mass passes through the centre.

Typical mistakes

Thinking the total energy of an undamped oscillator varies through the cycle - it is constant; only the split between kinetic and potential changes.
Believing more damping raises the resonance peak - it lowers and broadens it.

Active revision

For the $0.30\,\text{kg}$ , $12\,\text{N m}^{-1}$ , $0.040\,\text{m}$ -amplitude oscillator above, find the total energy and sketch how kinetic and potential energy vary with displacement.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Thermal energy, specific heat capacity and latent heat#

StandardAQA 7408 3.6.2.1DfE GCE Physics - thermal energy transfer

The internal energy of a body is the sum of the randomly distributed kinetic and potential energies of its molecules. Raising the temperature increases the average molecular kinetic energy; changing state changes the molecular potential energy (the bonds) at constant temperature. Absolute (kelvin) temperature is proportional to the average kinetic energy per molecule, and

0\,\text{K}

(absolute zero) is the temperature at which molecular kinetic energy is a minimum.

To raise the temperature of a substance without changing its state, the energy needed is

Q = mc\,\Delta\theta

, where

c

is the specific heat capacity - the energy to raise one kilogram by one kelvin. Water's large specific heat capacity of about

4200\,\text{J kg}^{-1}\,\text{K}^{-1}

is why it is used as a coolant and why coastal climates are mild. Specific heat capacity is measured either electrically (an immersion heater) or by the method of mixtures.

To change the state of a substance at constant temperature, the energy needed is

Q = mL

, where

L

is the specific latent heat - of fusion for melting/freezing, of vaporisation for boiling/condensing. During a change of state the temperature stays constant because the energy goes into breaking or forming intermolecular bonds (changing potential energy) rather than raising kinetic energy. This is why steam at

100\,^{\circ}\text{C}

scalds far worse than water at

100\,^{\circ}\text{C}

A heating curve makes this visible: as energy is supplied at a steady rate, the temperature rises through each phase (slope set by

mc

) but plateaus at each melting and boiling point (flat, while

mL

is absorbed). Reading such a curve, and allowing for heat losses to the surroundings, is central to the thermal required practical.

Q = mc\,\Delta\theta

Specific heat capacity

Energy to change the temperature of a mass $m$ by $\Delta\theta$ .

Q = mL

Specific latent heat

Energy to change the state of a mass $m$ at constant temperature.

Worked example

Heating and melting ice

Find the energy needed to turn $0.20\,\text{kg}$ of ice at $0\,^{\circ}\text{C}$ into water at $20\,^{\circ}\text{C}$ . Take $L_f = 3.3\times10^{5}\,\text{J kg}^{-1}$ and $c_{\text{water}} = 4200\,\text{J kg}^{-1}\,\text{K}^{-1}$ .

Melt the ice
$Q_1 = mL_f = 0.20 \times 3.3\times10^{5} = 6.6\times10^{4}\,\text{J}$ .
Warm the water
$Q_2 = mc\,\Delta\theta = 0.20 \times 4200 \times 20 = 1.68\times10^{4}\,\text{J}$ .
Total
Add the two stages.
$Q = 6.6\times10^{4} + 1.68\times10^{4} = 8.3\times10^{4}\,\text{J}$

Result: About $8.3\times10^{4}\,\text{J}$ (83 kJ) is required, most of it to melt the ice.

Typical mistakes

Expecting the temperature to rise during a change of state - it stays constant while latent heat is absorbed.
Using degrees Celsius for a temperature interval where kelvin is required (a temperature difference is the same in K and $^{\circ}$ C, but absolute temperatures are not).

Active revision

A $2.0\,\text{kW}$ kettle heats $0.50\,\text{kg}$ of water from $20\,^{\circ}\text{C}$ to $100\,^{\circ}\text{C}$ . Calculate the minimum time, taking $c = 4200\,\text{J kg}^{-1}\,\text{K}^{-1}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Ideal gases and the gas laws#

StandardAQA 7408 3.6.2.2AQA 7408 3.6.2.3DfE GCE Physics - gas laws

Pressure-volume isotherms

Fig. 5Each curve is an isotherm (

pV = \text{constant}

, Boyle's law); the higher-temperature isotherm lies further from the origin.

Three experimental gas laws describe how a fixed mass of gas behaves. Boyle's law: at constant temperature, pressure is inversely proportional to volume,

pV = \text{constant}

. Charles's law: at constant pressure, volume is proportional to absolute temperature,

V/T = \text{constant}

. The pressure law: at constant volume, pressure is proportional to absolute temperature,

p/T = \text{constant}

. All three require temperature in kelvin.

Combining the three gives the ideal gas equation

pV = nRT

, where

n

is the amount in moles and

R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}

is the molar gas constant. Written per molecule it becomes

pV = NkT

, where

N

is the number of molecules and

k = R/N_A = 1.38\times10^{-23}\,\text{J K}^{-1}

is the Boltzmann constant. An ideal gas is one that obeys these laws exactly - a good model for real gases at low pressure and well above their boiling point.

On a

p

V

diagram, each constant-temperature process (an isotherm) is a hyperbola

p \propto 1/V

; higher temperatures give isotherms further from the origin. Because

pV = nRT

, doubling the absolute temperature at constant volume doubles the pressure, and cooling a gas at constant pressure shrinks its volume in proportion to

T

- which, extrapolated, points to absolute zero.

Work is done when a gas expands or is compressed: at constant pressure the work done by the gas is

W = p\,\Delta V

, the area under the

p

V

curve. This links directly to the first law of thermodynamics and, in the engineering-physics option, to engine cycles. For an ideal gas the internal energy depends only on temperature, so an isothermal change involves no change in internal energy.

pV = nRT

Ideal gas equation (moles)

$R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ ; temperature in kelvin.

pV = NkT

Ideal gas equation (molecules)

$k = 1.38\times10^{-23}\,\text{J K}^{-1}$ is the Boltzmann constant.

Worked example

Moles and molecules in a flask

A flask of volume $1.5\times10^{-3}\,\text{m}^3$ holds gas at $2.0\times10^{5}\,\text{Pa}$ and $300\,\text{K}$ . Find the amount in moles and the number of molecules. Take $R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ and $N_A = 6.02\times10^{23}\,\text{mol}^{-1}$ .

Amount in moles
From $pV = nRT$ , $n = pV/RT = (2.0\times10^{5})(1.5\times10^{-3})/(8.31 \times 300)$ .
$n = \dfrac{300}{2493} = 0.120\,\text{mol}$
Number of molecules
$N = n N_A = 0.120 \times 6.02\times10^{23} = 7.2\times10^{22}$ .

Result: The flask holds $0.12\,\text{mol}$ , about $7.2\times10^{22}$ molecules.

Typical mistakes

Using Celsius instead of kelvin in the gas laws - always add 273 to convert.
Confusing the amount in moles $n$ with the number of molecules $N$ (they differ by $N_A$ ).

Active revision

A sealed flask of volume $2.0\times10^{-3}\,\text{m}^3$ contains gas at $1.0\times10^{5}\,\text{Pa}$ and $27\,^{\circ}\text{C}$ . Calculate the number of moles and the number of molecules.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Molecular kinetic theory#

AdvancedAQA 7408 3.6.2.4DfE GCE Physics - kinetic theory

The kinetic theory explains the pressure of a gas as the result of countless molecules colliding with the container walls. It rests on assumptions: the molecules are identical, move randomly and rapidly, are far apart compared with their size (negligible volume), exert no forces except during collisions, and undergo perfectly elastic collisions in a time negligible compared with the time between them. These assumptions define the ideal gas.

The central result derives the pressure from the molecular motion. Consider

N

molecules of mass

m

in a cube of side

L

. A molecule moving with velocity component

u

towards one wall rebounds elastically, reversing its momentum by

2mu

; it returns to the same wall every

2L/u

, so the average force it exerts is

\dfrac{2mu}{2L/u} = \dfrac{mu^2}{L}

. Summing over all molecules, dividing by the wall area

L^2

to get pressure, and using the fact that the mean square speed is shared equally among the three directions (

\overline{u^2} = \tfrac{1}{3}\overline{c^2}

), gives

pV = \tfrac{1}{3}Nm\overline{c^2}

, where

\overline{c^2}

is the mean square speed and the root of it,

c_{\text{rms}}

, is the root-mean-square speed.

Comparing this derived result with the experimental ideal gas equation

pV = NkT

links the microscopic and the macroscopic:

\tfrac{1}{3}Nm\overline{c^2} = NkT

, which rearranges to

\tfrac{1}{2}m\overline{c^2} = \tfrac{3}{2}kT

. This is a profound statement - the average translational kinetic energy of a gas molecule depends only on the absolute temperature. Temperature is, quite literally, a measure of molecular kinetic energy.

Two consequences follow. First, the root-mean-square speed is

c_{\text{rms}} = \sqrt{\dfrac{3kT}{m}}

, so lighter molecules move faster at the same temperature (hydrogen far faster than carbon dioxide), and raising the temperature raises the speed only as

\sqrt{T}

. Second, the internal energy of

N

molecules of an ideal monatomic gas is entirely kinetic,

U = \tfrac{3}{2}NkT

- it depends on temperature alone, confirming that an isothermal change of an ideal gas leaves its internal energy unchanged.

pV = \tfrac{1}{3}Nm\overline{c^2}

Kinetic-theory pressure

Pressure from the mean square speed of the molecules.

\tfrac{1}{2}m\overline{c^2} = \tfrac{3}{2}kT

Kinetic energy and temperature

Average molecular translational kinetic energy depends only on temperature.

c_{\text{rms}} = \sqrt{\dfrac{3kT}{m}}

Root-mean-square speed

Lighter molecules are faster; speed rises as the square root of temperature.

Worked example

Root-mean-square speed of oxygen

Find the root-mean-square speed of oxygen molecules (mass $5.3\times10^{-26}\,\text{kg}$ ) at $290\,\text{K}$ . Take $k = 1.38\times10^{-23}\,\text{J K}^{-1}$ .

Use the rms-speed formula
$c_{\text{rms}} = \sqrt{3kT/m}$ .
Substitute
$c_{\text{rms}} = \sqrt{\dfrac{3 \times 1.38\times10^{-23} \times 290}{5.3\times10^{-26}}}$ .
$c_{\text{rms}} = \sqrt{2.27\times10^{5}} = 476\,\text{m}\,\text{s}^{-1}$

Result: The oxygen molecules have an rms speed of about $480\,\text{m}\,\text{s}^{-1}$ - faster than the speed of sound in air.

Typical mistakes

Confusing the root-mean-square speed with the mean speed - $c_{\text{rms}} = \sqrt{\overline{c^2}}$ , not the simple average.
Forgetting the factor of $\tfrac{1}{3}$ that comes from sharing the mean square speed among the three directions.

Active revision

Calculate the root-mean-square speed of nitrogen molecules (mass $4.7\times10^{-26}\,\text{kg}$ ) at $300\,\text{K}$ . Take $k = 1.38\times10^{-23}\,\text{J K}^{-1}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Further mechanics and thermal physics

Key points

Tension in a whirled string

Key points

A mass on a spring

Key points

Total energy of an oscillator

Key points

Heating and melting ice

Key points

Moles and molecules in a flask

Key points

Root-mean-square speed of oxygen

Sources