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Circular Motion

Angular Velocity

An object moving in a circle at constant speed has a constantly changing velocity (because the direction changes). The angular velocity ω is the rate of change of angle.

ω = 2πT = 2πf    (unit: rad s−¹)

Linear speed: v = ωr

Centripetal Acceleration & Force

The acceleration is always directed towards the centre of the circle. There is no outward force on the object — the centripetal force is the resultant force causing circular motion.

a = v²r = ω²r

F = mv²r = mω²r

Worked Example

A 0.15 kg ball on a 0.80 m string moves in a horizontal circle at 2.0 revolutions per second. Find the centripetal force.

ω = 2πf = 2π × 2.0 = 4π rad s−¹

F = mω²r = 0.15 × (4π)² × 0.80

F = 0.15 × 157.9 × 0.80 = 18.9 N

Key Facts

  • Centripetal force is not a new force — it is provided by tension, gravity, friction, normal force, etc.
  • The velocity is tangent to the circle; the acceleration is radial (towards the centre)
  • If the centripetal force is removed, the object moves in a straight line (tangent)

Simple Harmonic Motion (SHM)

Defining SHM

Simple harmonic motion is an oscillation in which the acceleration is:

  • Proportional to the displacement from the equilibrium position
  • Always directed towards the equilibrium position (restoring force)
a = −ω²x

The negative sign shows the acceleration opposes the displacement.

SHM Equations

Displacement: x = A cos(ωt)

Velocity: v = −Aω sin(ωt)   or   v = ±ω√(A² − x²)

Maximum velocity: vmax = Aω   (at x = 0)

Maximum acceleration: amax = Aω²   (at x = ±A)

Time Period of Oscillators

Simple pendulum: T = 2π√ (L/g) 

Mass–spring system: T = 2π√ (m/k) 

Note: for the pendulum, T does not depend on mass or amplitude (for small angles). For the spring, T does not depend on amplitude.

Energy in SHM

During SHM, energy is continuously exchanged between kinetic and potential energy. The total energy remains constant (in the absence of damping).

  • At equilibrium (x = 0): maximum KE, zero PE
  • At extremes (x = ±A): zero KE, maximum PE
  • Total energy ∝ A²

SHM Displacement–Time Graph

t x +A −A T/2 T x = A cos(ωt) v = 0, a = max v = max, a = 0 v = 0, a = max

Worked Example

A mass on a spring oscillates with amplitude 0.04 m and period 0.50 s. Find the maximum speed and maximum acceleration.

ω = 2π/T = 2π/0.50 = 4π = 12.57 rad s−¹

vmax = Aω = 0.04 × 12.57 = 0.50 m s−¹

amax = Aω² = 0.04 × 12.57² = 0.04 × 158.0 = 6.3 m s−²

Exam Tip

Always check whether the question uses x = A cos(ωt) or x = A sin(ωt). Cosine starts at maximum displacement; sine starts at equilibrium.

Forced Vibrations & Resonance

Damping

Damping is the loss of energy (and reduction in amplitude) of an oscillating system due to resistive forces.

  • Light damping: amplitude decreases gradually over many oscillations (e.g. pendulum in air)
  • Heavy damping: amplitude decreases rapidly; the system takes a long time to return to equilibrium without oscillating (e.g. door closer)
  • Critical damping: the system returns to equilibrium in the shortest possible time without oscillating (e.g. car shock absorbers)

Forced Vibrations & Resonance

Forced vibrations occur when a periodic external driving force is applied to an oscillating system. The system oscillates at the driving frequency, not its natural frequency.

Resonance occurs when the driving frequency equals the natural frequency of the system. At resonance:

  • Maximum energy transfer from driver to oscillator
  • Maximum amplitude of oscillation
  • The phase difference between driver and oscillator is π/2 (90°)

Resonance Curve

Driving frequency Amplitude f0 Light damping Medium damping Heavy damping

As damping increases:

  • The resonance peak becomes lower and broader
  • The peak shifts slightly to a lower frequency
  • The amplitude at all frequencies is reduced

Key Facts

  • Examples of resonance: Tacoma Narrows Bridge, microwave ovens, tuning a radio, MRI scanners
  • Resonance can be useful (e.g. musical instruments) or destructive (e.g. bridge oscillations)
  • Damping is used to reduce unwanted resonance effects

Thermal Energy & Ideal Gases

Thermal Energy Transfer

Specific heat capacity: Q = mcΔθ    (unit: J)

Specific latent heat: Q = mL    (unit: J)

Specific heat capacity (c): the energy required to raise the temperature of 1 kg of a substance by 1 K.

Specific latent heat (L): the energy required to change the state of 1 kg of a substance without changing its temperature. Lf for fusion (melting), Lv for vaporisation (boiling).

Worked Example

How much energy is needed to heat 0.50 kg of water from 20 °C to 100 °C? (c = 4200 J kg−¹ K−¹)

Q = mcΔθ = 0.50 × 4200 × (100 − 20)

Q = 0.50 × 4200 × 80 = 168 000 J = 168 kJ

Internal Energy

The internal energy of a system is the sum of the randomly distributed kinetic energies and potential energies of all its molecules.

  • Heating a substance increases the kinetic energy of its molecules (temperature rises)
  • During a change of state, energy goes into breaking/forming bonds (potential energy changes) at constant temperature

First Law of Thermodynamics

ΔU = Q − W

The change in internal energy (ΔU) equals the heat supplied to the system (Q) minus the work done by the system (W).

Ideal Gas Laws

pV = nRT     (n = number of moles, R = 8.31 J mol−¹ K−¹)

pV = NkT     (N = number of molecules, k = 1.38 × 10−²³ J K−¹)

The three gas laws (all at constant other variable):

  • Boyle's law: p ∝ 1/V at constant T   (pV = constant)
  • Charles's law: V ∝ T at constant p   (V/T = constant)
  • Pressure law: p ∝ T at constant V   (p/T = constant)

Temperature must always be in kelvin (K = °C + 273).

Molecular Kinetic Theory

The kinetic theory links the macroscopic properties of a gas (pressure, temperature) to the microscopic behaviour of its molecules.

pV = 13Nm<c²>

12m<c²> = 32kT

where <c²> is the mean square speed, m is the mass of one molecule, N is the number of molecules, and k is the Boltzmann constant.

The root mean square speed is crms = √<c²>.

Worked Example

Find the r.m.s. speed of nitrogen molecules (m = 4.65 × 10−²² kg) at 300 K.

½m<c²> = 3/2 kT

<c²> = 3kT/m = (3 × 1.38 × 10−²³ × 300) / (4.65 × 10−²²)

<c²> = 1.242 × 10−²° / 4.65 × 10−²² = 2.67 × 10&sup5; m² s−²

crms = √(2.67 × 10&sup5;) = 517 m s−¹

Assumptions of the Ideal Gas Model

  • Molecules have negligible volume compared with the volume of the container
  • No intermolecular forces act (except during collisions)
  • Collisions between molecules and with the walls are perfectly elastic
  • The time of a collision is negligible compared with the time between collisions
  • Molecules move randomly with a range of speeds

Exam Tip

Remember: temperature in gas law equations must be in kelvin. The most common error in gas law calculations is forgetting to convert from °C.

Further Mechanics & Thermal Flashcards

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Further Mechanics & Thermal Quiz

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Further Mechanics & Thermal — Mock Exam Questions

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