physics

Introduction

a few basic equations

velocity
- velocity = distance travelled / unit time
- 100kph = 100kph x 1000m / (3600secs) = 27.8m/s
acceleration
- acceleration = velocity change / unit time
- a car accelerating uniformly from 0 to 100kph in 10secs will have:
  - average acceleration = 100kph x 1000m / (3600secs x 10secs) = 2.78m/second²
momentum
- linear momentum = mass x velocity
- angular momentum = the moment of inertia (I) x angular velocity (ω)
  - moment of inertia (I) = mass x radius² where radius is distance to the centre of motion
    - hence a ballerina will spin faster when limbs are close to the body as moment of inertia is reduced and thus angular velocity must increase as momentum remains constant
- NB. momentum is conserved during collisions, explosions, and other events involving objects in motion, however, in the real world, some kinetic energy is lost to heat during a collision, especially if the collision is inelastic, and thus overall momentum will fall.
- net system momentum is constant if the net external force (or for angular momentum, external torque) is zero
force
- force = mass x acceleration = change in momentum / unit time
- 1 newton (N) = 1 kilogram X meter/second²
- lifting a 1kg object to overcome gravity requires a force of 1kg x acceleration due to gravity = 9.8N
- force to move a 2000kg car from 0 to 100kph in 10secs = 2000 x 2.78 = 5556N
kinetic energy
- energy = 1/2 x mass x velocity²
- 1 Joule = 1 kg m²/sec²
gravitational potential energy
- energy = mass x gravitational constant x height
mechanical energy (kinetic + gravitational) is CONSTANT unless there are non-conservative works done such as creation of heat or sound via friction, etc
work
- work = force x distance = change in kinetic energy
- 1 Joule = 1Nm
- work done to lift a 10kg object 1m off the ground = 10kg x 9.8 x 1m = 100J
- work done to accelerate a 2000kg car from 0 to 100kph (excl. frictional and wind resistance work) = 0.5 x 2000 x (27.8m/s)² = 27,800J
- work done in an initially stationery object sliding down to the bottom of a ramp at an incline of theta and a friction coefficient of mu from height h and a length of the ramp L:
  - L = height / sin(theta)
  - work done = gravity work - frictional work = mgh - (mu x mgCos(theta) x L ) = change in KE of the object hence can calculate the end speed of the object at the bottom of the incline as 1/2 x mv² = total work done value
power
- power = work / time taken = energy used / time
- 1W = 1 Joule/sec
- power to accelerate a 2000kg car from 0 to 100kph in 10sec = 27,800J / 10sec = 2780W
- note that combustion engines are only about 40% efficient so the engine would actually need to use 2780/0.4 =7kW of power
- petrol has a energy density of 44MJ/kg thus over 10sec at 7000W requires 70000J = 70kJ and thus ~44000/70 = 1.6mL of fuel IF there were no frictional forces to overcome such as wind resistance, etc.
rotating bodies
- torque = moment of Inertia x angular acceleration
- moment of Inertia for a solid disk = mass x radius²/2
- eg. a falling single disk yo-yo suspended by the string contacting it's outer ring:
  - the torque of a falling yo-yo (assuming pure rolling without sliding) = tension in the string x radius of yo-yo (as the yo-yo rotates around the centre which is r distance from the string contact)
  - angular acceleration = linear acceleration / radius this gives tension = mass x linear acceleration / 2
  - now considering vertical forces of gravity and the opposing string tension:
    - (mass x gravity) - string tension = mass x linear acceleration, solving for string tension gives string tension = (mass x gravity) - (mass x linear acceleration)
    - substituting in tension = mass x linear acceleration / 2 gives mass x linear acceleration / 2 = (mass x gravity) - (mass x linear acceleration) gives linear acceleration = 2/3 x gravity
- eg. cylinder of mass m and radius r rolling down a smooth incline of angle theta and unknown surface friction coefficient (mu) and assuming no slipping or sliding
  - to find the linear acceleration down the slope (a)
  - torque = moment of Inertia x angular acceleration and this is produced by the friction of the surface of the incline and the cylinder
  - vertical downwards gravitation force = mass x gravity
  - gravitation force in direction of the slope = mass x gravity x sin(theta)
  - gravitation force perpendicular to the slope = mass x gravity x cos(theta)
    - frictional force in direction opposite to the slope = surface friction coefficient (mu) x gravitation force perpendicular to the slope = surface friction coefficient (mu) x mass x gravity x cos(theta)
    - torque on the cylinder = frictional force x radius = surface friction coefficient (mu) x mass x gravity x cos(theta) x radius
    - moment of Inertia for a solid disk = mass x radius²/2
    - angular acceleration = linear acceleration / radius
    - substituting these in gives frictional force = 1/2 x mass x linear acceleration = surface friction coefficient (mu) x mass x gravity x cos(theta)
    - which gives surface friction coefficient (mu) x mass x gravity x cos(theta) = 1/2 x linear acceleration
    - hence mu = 1/2 x linear acceleration / (mass x gravity x cos(theta) )
  - force = mass x linear acceleration
    - force = downward gravitation force - frictional force = (mass x gravity x sin(theta) ) - (surface friction coefficient (mu) x mass x gravity x cos(theta) ) = mass x linear acceleration
    - hence (gravity x sin(theta) ) - (surface friction coefficient (mu) x gravity x cos(theta) ) = linear acceleration
    - eliminate mu by substituting in mu = 1/2 x linear acceleration / (mass x gravity x cos(theta) )
    - (mass x gravity x sin(theta) ) - (1/2 x linear acceleration / (mass x gravity x cos(theta) ) x mass x gravity x cos(theta) ) = mass x linear acceleration
    - which simplifies to: linear acceleration = 2/3 x gravity x sin(theta)
    - ie. the linear acceleration down the slope is INDEPENDENT on the surface friction coefficient (mu) AS LONG as the cylnder does not start slipping due to inadequate friction

electricity

see also electricity in Australia
electrical power
- watts = current in amps x voltage
energy capacity of a battery
- capacity is measured in Watt-hours (Wh) ie. how many watts of power is available over 1 hour
- watt-hours = amp-hours x nominal voltage
- ie. a 100Ah 12.8V lithium battery will have capacity of 100 x 12.8 = 1280Wh and thus will last 12.8 hours if using 100W of power
Ohm's law
- voltage difference between two points in a circuit (volts) = current (amps) x resistance (ohms)

heat

heat energy required to heat a substance
- heat energy = mass x specific heat x change in temperature in deg C
- eg. water has specific heat of 4,180J/kg°C
- eg. cast iron and stainless steel has specific heat of ~500J/kg°C
Stefan–Boltzmann law for rate of radiant heat loss from a surface
- radiant heat loss in watts = ϵσA(T₁⁴−T₂⁴)
- where,
  - ϵ is the emissivity of the surface (dimensionless, between 0 and 1)
  - σ is the Stefan-Boltzmann constant (5.67×10⁻⁸ W m⁻²K⁻⁴)
  - A is the surface area (in square meters)
  - T₁ is the absolute temperature of the radiating surface (in kelvin)
  - T₂ is the absolute temperature of the surroundings (in kelvin)
Wien's displacement law for frequency of light emitted from a black body
- peak frequency of light =2.82k_BT/h
- where,
  - k_B is Boltzmann's constant (1.38×10⁻²³J/K
  - T is the absolute temperature in kelvin
  - h is Planck's constant (6.63×10⁻³⁴ J·s)

flow of fluids

laminar flow
- there is minimal mixing between layers of the fluid and there is very little fluctuation of the speed of the flow
- blood flow in arteries is generally laminar
turbulent flow
- there is increasing mixing of the layers of flow with eddy formation resulting in varying flow rates - the greater the variation in flow, the greater the turbulence
- energy cascade: turbulence converts kinetic energy into heat via the creation of ever smaller eddies and eventually dissipates due to viscosity
- turbulence is complex and varies over both spatial and time dimensions
  - direct numerical simulation (DNS) can be used to model this but it is computationally complex to calculate, instead large eddy simulation (LES) is often used instead which uses a sub-grid scale model to address the small eddies
  - the Reynolds-averaged Navier-Stokes technique is the least computationally expensive model technique but this only models the effects of eddies using turbulent viscosity, not the actual eddies
  - given the computational complexity, wind tunnels are often used to assess the effect of wind on objects such as aerodynamic vehicles, tents, etc
- turbulent flow requires more energy and results in a greater pressure drop
- most flows in nature are turbulent
Reynold's number
- a value for a given system which means turbulent flow is likely to occur if Reynold's number is high for a given system
- Reynold's number (Re) = fluid density x velocity x characteristic length / dynamic viscosity = velocity x length / kinematic viscosity = inertial forces / viscous forces
  - for flow past a cylinder, characteristic length = cylinder diameter
  - for flow past a air foil (wing of an airplane), characteristic length = length
  - for flow through a pipe:
    - characteristic length = pipe diameter
    - turbulent flow is likely if Re > 2000-4000
    - in laminar flow, flow rate at the pipe wall is zero in “no-slip condition” and velocity increases to a maximum at the centre of the pipe and the velocity profile is parabolic
    - in turbulent flow, flow rate at the pipe wall is zero in “no-slip condition” (there is a laminar sublayer here but its thickness decreases as Re increases) however, away from the wall, flow rate is more even throughout the pipe due to the mixing of layers creating a more homeogenous time averaged flow velocity but with more random variations
    - if the pipe wall roughness thickness is less than the laminar sublayer thickness then the pipe is said to be hydraulically smooth as it will not effect the turbulent flow
Darcy-Weisbach equation to calculate pressure drop along a pipe
- pressure drop = length of pipe x Darcy friction factor x (density/2) x (avg velocity)²/hydraulic diameter
- for laminar flow, Darcy friction factor = 64/Re
- for turbulent flow, there is a complicated Colebrook equation to determine the friction factor which requires an iterative calculation hence it is generally looked up on a Moody Diagram, but simplified pressure drop is proportional to velocity squared and pipe surface roughness
Bernoulli principle and equation
- for incompressible, non-viscous fluids:
  - fluid flow speed²/2 +gz + static pressure/density = constant
    - g = acceleration due to gravity
    - z = the elevation of the point above a reference plane
  - when applied to determine the static pressure drop when fluid flows faster through a narrowing in a pipe (the Venturi effect), the equation becomes:
    - pressure drop = density x (velocity₂² - velocity₁²)/2
    - as flow rate is equal in each pipe section, velocity₂ x pipe area₂ = velocity₁ x pipe area₁ (see also https://en.wikipedia.org/wiki/Venturi_effect))
- see https://en.wikipedia.org/wiki/Bernoulli%27s_principle for more detail

Gary Ayton's camping and photography wiki

Table of Contents

physics

Introduction

a few basic equations

electricity

heat

flow of fluids