PHYS 2310 Engineering Physics I Formula Sheets Chapters 1-18 Chapter 1/Important Numbers Chapter 2 Velocity Quantity Units for SI Base Quantities Unit Name Unit Symbol Length Meter M Time Second s Mass (not weight) Kilogram kg 1 kg or 1 m 1m 1m 1 second 1m Common Conversions 1000 g or m 1m 100 cm 1 inch 1000 mm 1 day 1000 milliseconds 1 hour 3.281 ft 360° Average Velocity Average Speed 1 × 106 ππ 2.54 cm 86400 seconds 3600 seconds 2π rad Important Constants/Measurements Mass of Earth 5.98 × 1024 kg Radius of Earth 6.38 × 106 m 1 u (Atomic Mass Unit) 1.661 × 10−27 kg Density of water 1 π/ππ3 or 1000 ππ/π3 g (on earth) 9.8 m/s 2 Circumference Surface area (sphere) Density Common geometric Formulas Area circle πΆ = 2ππ ππ΄ = 4ππ 2 Volume (rectangular solid) Instantaneous Velocity πππ£π = πππ πππππππππ‘ βπ₯ = π‘πππ βπ‘ π‘ππ‘ππ πππ π‘ππππ π‘πππ Μ ππ₯ βπ₯ π£ = lim = βπ‘→0 βπ‘ ππ‘ π ππ£π = Average Acceleration βπ£ βπ‘ ππ£ π2 π₯ π= = ππ‘ ππ‘ 2 πππ£π = Motion of a particle with constant acceleration π΄ = ππ 4 Volume (sphere) π = ππ 3 3 π =πβπ€ββ π = ππππ β π‘βππππππ π 2.3 2.4 Acceleration Instantaneous Acceleration 2 2.2 π£ = π£0 + ππ‘ 1 βπ₯ = (π£0 + π£)π‘ 2 1 βπ₯ = π£0 π‘ + ππ‘ 2 2 π£ 2 = π£02 + 2πβπ₯ 2.11 2.17 2.15 2.16 2.7 2.8 2.9 Chapter 3 Adding Vectors Geometrically Adding Vectors Geometrically (Associative Law) Chapter 4 πβ + πββ = πββ + πβ 3.2 (πβ + πββ) + πβ = πβ + (πββ + πβ) 3.3 3.5 Magnitude of vector |π| = π = √ππ₯2 + ππ¦2 3.6 Angle between x axis and vector ππ¦ π‘πππ = ππ₯ 3.6 Unit vector notation πβ = ππ₯ πΜ + ππ¦ πΜ + ππ§ πΜ 3.7 ππ₯ = ππ₯ + ππ₯ ππ¦ = ππ¦ + ππ¦ ππ§ = ππ§ + ππ§ 3.10 3.11 3.12 πβ β πββ = πππππ π 3.20 Adding vectors in Component Form Scalar (dot product) Scalar (dot product) Projection of πβ ππ πββ or component of πβ ππ πββ Vector (cross) product magnitude πβ β πββ = (ππ₯ πΜ + ππ¦ πΜ + ππ§ πΜ) β (ππ₯ πΜ + ππ¦ πΜ + ππ§ πΜ ) πβ β πββ = ππ₯ ππ₯ + ππ¦ ππ¦ + ππ§ ππ§ displacement Average Acceleration Instantaneous Acceleration 4.4 βπ₯ βπ‘ 4.8 ππβ = π£π₯ πΜ + π£π¦ πΜ + π£π§ πΜ ππ‘ βπ£β πβππ£π = βπ‘ ππ£β πβ = ππ‘ πβ = ππ₯ πΜ + ππ¦ πΜ + ππ§ πΜ π£β = Projectile Motion π£π¦ = π£0 π πππ0 − ππ‘ 3.24 1 βπ¦ = π£0 π ππππ‘ − ππ‘ 2 2 2 2 π£π¦ = (π£0 π πππ0 ) − 2πβy πΜ ππ§ | ππ§ 4.16 4.17 4.22 4.24 4.23 2 Trajectory Range 3.26 π ππ¦ ππ¦ 4.15 4.21 π£π¦ = π£0 π πππ0 − ππ‘ π = πππ πππ 4.10 4.11 4.23 1 βπ₯ = π£0 πππ ππ‘ + ππ₯ π‘ 2 2 or βπ₯ = π£0 πππ ππ‘ if ππ₯ =0 πβ β πββ |π| or πΜ ββ πβπ₯π = πππ‘ |ππ₯ ππ₯ βπβ = βπ₯πΜ + βπ¦πΜ + βπ§πΜ ββππ£π = π Instantaneous Velocity 3.22 πβπ₯πββ = (ππ₯ πΜ + ππ¦ πΜ + ππ§ πΜ)π₯(ππ₯ πΜ + ππ¦ πΜ + ππ§ πΜ ) = (ππ¦ ππ§ − ππ¦ ππ§ )πΜ + (ππ§ ππ₯ − ππ§ ππ₯ )πΜ + (ππ₯ ππ¦ − ππ₯ ππ¦ )πΜ Vector (cross product) 4.4 Average Velocity ππ₯ = ππππ π ππ¦ = ππ πππ Components of Vectors πβ = π₯πΜ + π¦πΜ + π§πΜ Position vector Relative Motion Uniform Circular Motion π¦ = (π‘πππ0 )π₯ − π = ππ₯ 2(π£0 πππ π0 )2 π£02 sin(2π0 ) π βββββββ π£π΄πΆ = βββββββ π£π΄π΅ + βββββββ π£π΅πΆ ππ΄π΅ = ββββββββ βββββββ ππ΅π΄ π= π£2 π π= 2ππ π£ 4.25 4.26 4.44 4.45 4.34 4.35 General Component form Chapter 5 Chapter 6 Newton’s Second Law Friction πΉβπππ‘ = ππβ πΉπππ‘,π₯ = πππ₯ πΉπππ‘,π¦ = πππ¦ πΉπππ‘,π§ = πππ¦ 5.1 Kinetic Frictional Weight πΉπ = ππ π = ππ πβπ ,πππ₯ = ππ πΉπ 6.1 πβπ = ππ πΉπ 6.2 5.2 Drag Force Gravitational Force Gravitational Force Static Friction (maximum) Terminal velocity 1 π· = πΆππ΄π£ 2 2 2πΉπ π£π‘ = √ πΆππ΄ 6.14 6.16 5.8 5.12 Centripetal acceleration Centripetal Force π£2 π= π πΉ= ππ£ 2 π 6.17 6.18 Chapter 7 Work- Kinetic Energy Theorem 7.1 π = πΉππππ π = πΉβ β πβ 7.7 7.8 Spring Force (Hooke’s law) Work done by spring Work done by Variable Force Average Power (rate at which that force does work on an object) Instantaneous Power π₯π π¦π π₯π π¦π πππ£π = Mechanical Energy πΈπππ = πΎ + π 8.12 7.15 Principle of conservation of mechanical energy πΎ1 + π1 = πΎ2 + π2 πΈπππ = βπΎ + βπ = 0 8.18 8.17 7.20 7.21 Force acting on particle π§π π= 7.36 π§π π βπ‘ ππ = πΉππππ π = πΉβ β π£β ππ‘ 8.7 8.11 7.25 π = ∫ πΉπ₯ ππ₯ + ∫ πΉπ¦ ππ¦ + ∫ πΉπ§ ππ§ βπ = ππβπ¦ 1 2 ππ₯ 2 7.12 βπΎ = ππ + ππ ππ = πππππππ πΉππππ πΉβπ = −ππβ πΉπ₯ = −ππ₯ (along x-axis) Gravitational Potential Energy 8.1 8.6 π(π₯) = ππ = ππππππ π 1 2 1 2 ππ₯ − ππ₯ 2 π 2 π βπ = −π = − ∫ πΉ(π₯)ππ₯ Elastic Potential Energy 7.10 ππ = Potential Energy π₯π βπΎ = πΎπ − πΎ0 = π Work done by gravity Work done by lifting/lowering object π₯π 1 πΎ = ππ£ 2 2 Kinetic Energy Work done by constant Force Chapter 8 7.42 7.43 7.47 Work on System by external force With no friction Work on System by external force With friction Change in thermal energy Conservation of Energy *if isolated W=0 πΉ(π₯) = − ππ(π₯) ππ₯ 8.22 π = βπΈπππ = βπΎ + βπ 8.25 8.26 π = βπΈπππ + βπΈπ‘β 8.33 βπΈπ‘β = ππ ππππ π 8.31 π = βπΈ = βπΈπππ + βπΈπ‘β + βπΈπππ‘ 8.35 Average Power Instantaneous Power **In General Physics, Kinetic Energy is abbreviated to KE and Potential Energy is PE πππ£π = π= βπΈ βπ‘ ππΈ ππ‘ 8.40 8.41 Chapter 9 Impulse and Momentum π‘π Impulse π½β = ∫ πΉβ (π‘)ππ‘ π‘π 9.35 πβ = ππ£β 9.22 π½β = Δπβ = πβπ − πβπ ππβ ππ‘ πΉβπππ‘ = πβπβπππ πββ = ππ£βπππ πβββ π πΉβπππ‘ = Newton’s 2nd law System of Particles 9.30 π½ = πΉπππ‘ βπ‘ Linear Momentum Impulse-Momentum Theorem Collision continued… πΉβπππ‘ = 9.31 9.32 Inelastic Collision Conservation of Linear Momentum (in 2D) Average force Elastic Collision π π βπ = − πβπ£ βπ‘ βπ‘ βπ πΉππ£π = − βπ£ βπ‘ 2π1 =( )π£ π1 + π2 1π π 9.14 9.25 9.27 Center of mass location πβπππ 1 = ∑ ππ πβπ π 9.67 π=1 π π£βπππ = 1 ∑ ππ π£βπ π Rocket Equations Thrust (Rvrel) π π£πππ = ππ 9.68 Change in velocity 9.42 9.43 πβ1π + πβ2π = πβ1π + πβ2π 9.50 9.51 9.78 πΎ1π + πΎ2π = πΎ1π + πΎ2π 9.8 π=1 πββ = ππππ π‘πππ‘ πββπ = πββπ π1 π£π1 + π2 π£12 = π1 π£π1 + π2 π£π2 9.37 9.40 Center of Mass ππ‘ π1 − π2 π£1π = ( )π£ π1 + π2 1π 9.77 9.22 Collision π£2π πββ1π + πββ2π = πββ1π + πββ2π πΉππ£π = − Center of mass velocity Final Velocity of 2 objects in a head-on collision where one object is initially at rest 1: moving object 2: object at rest Conservation of Linear Momentum (in 1D) π1 π£01 + π2 π£02 = (π1 + π2 )π£π Δπ£ = π£πππ ππ ππ ππ 9.88 9.88 Chapter 10 Angular displacement (in radians Average angular velocity Instantaneous Velocity Average angular acceleration Instantaneous angular acceleration π π Δπ = π2 − π1 βπ πππ£π = βπ‘ ππ π= ππ‘ βπ πΌππ£π = βπ‘ ππ πΌ= ππ‘ 10.1 10.4 π= Rotational Kinematics π = π0 + πΌπ‘ 1 Δπ = π0 π‘ + πΌπ‘ 2 2 2 2 π = π0 + 2πΌΔπ 1 Δπ = (π + π0 )π‘ 2 1 Δπ = ππ‘ − πΌπ‘ 2 2 10.5 10.6 10.7 10.8 10.13 10.14 10.15 πΌ = πΌπππ + πβ2 10.36 π = ππΉπ‘ = π⊥ πΉ = ππΉπ πππ 10.3910.41 Newton’s Second Law ππππ‘ = πΌπΌ 10.45 Rotational work done by a toque π = ∫ πππ Torque Rotational Kinetic Energy Work-kinetic energy theorem π£ = ππ 10.18 Tangential Acceleration ππ‘ = πΌπ 10.19 2 π£ = π2 π π 10.23 2ππ 2π = π£ π 10.19 10.20 π= 10.35 Power in rotational motion Velocity Period πΌ = ∫ π 2 ππ Rotation inertia (discrete particle system) Parallel Axis Theorem h=perpendicular distance between two axes ππ Relationship Between Angular and Linear Variables ππ = 10.34 10.12 10.16 Radical component of πβ πΌ = ∑ ππ ππ2 Rotation inertia ππ π = πβπ (π constant) ππ π= = ππ ππ‘ 1 πΎ = πΌπ2 2 1 1 βπΎ = πΎπ − πΎπ = πΌππ2 − πΌππ2 = π 2 2 10.53 10.54 10.55 10.34 10.52 Moments of Inertia I for various rigid objects of Mass M Thin walled hollow cylinder or hoop about central axis Annular cylinder (or ring) about central axis πΌ = ππ 2 1 πΌ = π(π 12 + π 22 ) 2 Solid Sphere, axis through center 2 πΌ = ππ 2 5 Thin rod, axis perpendicular to rod and passing though end Solid Sphere, axis tangent to surface 7 πΌ = ππ 2 5 Thin Rectangular sheet (slab), axis parallel to sheet and passing though center of the other edge Solid cylinder or disk about central axis 1 πΌ = ππ 2 2 Thin Walled spherical shell, axis through center Solid cylinder or disk about central diameter 1 1 πΌ = ππ 2 + ππΏ2 4 12 Thin rod, axis perpendicular to rod and passing though center 2 πΌ = ππ 2 3 Thin Rectangular sheet (slab_, axis along one edge πΌ= 1 ππΏ2 12 Thin rectangular sheet (slab) about perpendicular axis through center 1 πΌ = ππΏ2 3 πΌ= 1 ππΏ2 12 1 πΌ = ππΏ2 3 πΌ= 1 π(π2 + π 2 ) 12 Chapter 11 Rolling Bodies (wheel) Speed of rolling wheel Kinetic Energy of Rolling Wheel Acceleration of rolling wheel Acceleration along x-axis extending up the ramp π£πππ = ππ 11.2 Angular Momentum 1 1 2 πΎ = πΌπππ π2 + ππ£πππ 2 2 11.5 Magnitude of Angular Momentum ππππ = πΌπ 11.6 ππππ,π₯ ππ πππ =− πΌ 1 + πππ2 ππ 11.10 Magnitude of torque Newton’s 2nd Law πβ = πβ × πΉβ 11.14 π = ππΉ⊥ = π⊥ πΉ = ππΉπ πππ 11.1511.17 πβπππ‘ = ββ πβ ππ‘ β = πππ£π πππ β = ππ⊥ = πππ£⊥ 11.18 11.1911.21 π Angular momentum of a system of particles ββ = ∑ ββ πΏ βπ π=1 πβπππ‘ = Torque as a vector Torque Angular Momentum ββ × βπ£β) π£ βββ = βπβ × βπβ = π(π ββ ππΏ ππ‘ Angular Momentum continued Angular Momentum of a πΏ = πΌπ rotating rigid body Conservation of angular πΏββ = ππππ π‘πππ‘ momentum πΏββπ = πΏββπ 11.23 11.26 11.29 11.31 11.32 11.33 Precession of a Gyroscope Precession rate Ω= πππ πΌπ 11.31 Chapter 12 Chapter 13 Static Equilibrium 12.3 Gravitational Force (Newton’s law of gravitation) πβπππ‘ = 0 12.5 Principle of Superposition πΉβπππ‘,π₯ = 0, πΉβπππ‘,π¦ = 0 12.7 12.8 πΉβπππ‘ = 0 If forces lie on the xy-plane Stress (force per unit area) Strain (fractional change in length) Stress (pressure) Tension/Compression E: Young’s modulus Shearing Stress G: Shear modulus Hydraulic Stress B: Bulk modulus πβπππ‘,π§ = 0 π π‘πππ π = ππππ’ππ’π × π π‘ππππ πΉ π= π΄ πΉ βπΏ =πΈ π΄ πΏ πΉ βπ₯ =πΊ π΄ πΏ βπ π=π΅ π 12.9 12.22 πΉ=πΊ πΉβ1,πππ‘ = ∑ πΉβ1π πΉβ1 = ∫ ππΉβ Escape Speed Kepler’s 3rd Law (law of periods) Energy for bject in circular orbit 13.6 πΊπ π2 πΊππ πΉ= π π 3 πΊππ π=− π 13.11 ππ = Gravitational Potential Energy 12.24 π = −( 13.19 13.21 πΊπ1 π2 πΊπ1 π3 πΊπ2 π3 + + ) π12 π13 π23 2πΊπ π£=√ π π2 = ( π=− 4π 2 3 )π πΊπ πΊππ π πΎ= πΈ=− 13.22 13.28 πΊππ 2π πΊππ 2π πΊππ Mechanical Energy πΈ = − (elliptical orbit) 2π −11 *Note: πΊ = 6.6704 × 10 π β π2 /ππ2 Mechanical Energy (circular orbit) 13.5 π=2 Gravitation within a spherical Shell 12.23 13.1 π Gravitational Force acting on a particle from an extended body Gravitational acceleration Potential energy on a system (3 particles) π1 π2 π2 13.34 13.21 13.38 13.40 13.42 Chapter 14 Density βπ βπ π π= π π= Pressure and depth in a static Fluid P1 is higher than P2 Gauge Pressure Archimedes’ principle Mass Flow Rate Volume flow rate Bernoulli’s Equation Equation of continuity Equation of continuity when 14.1 14.2 βπΉ βπ΄ πΉ π= π΄ 14.3 14.4 π2 = π1 + ππ(π¦1 − π¦2 ) π = π0 + ππβ 14.7 14.8 π= Pressure Chapter 15 Angular frequency Acceleration Kinetic and Potential Energy πΉπ = ππ π 14.16 π π = ππ π = ππ΄π£ 14.25 π π = π΄π£ 14.24 π π = π΄π£ = ππππ π‘πππ‘ displacement Velocity ππβ 1 π + ππ£ 2 + πππ¦ = ππππ π‘πππ‘ 2 π π = ππ π = ππ΄π£ = ππππ π‘πππ‘ Frequency cycles per time 14.29 Angular frequency π= 1 π 15.2 π₯ = π₯π cos(ππ‘ + π) π= 2π = 2ππ π 15.3 15.5 π£ = −ππ₯π sin(ππ‘ + π) 15.6 π = −π2 π₯π cos(ππ‘ + π) 15.7 1 1 πΎ = ππ£ 2 π = ππ₯ 2 2 2 π π=√ π 15.12 Period π = 2π√ π π 15.13 Torsion pendulum πΌ π = 2π√ π 15.23 Simple Pendulum πΏ π = 2π√ π 15.28 Physical Pendulum πΌ π = 2π√ πππΏ 15.29 14.25 14.24 Damping force πΉβπ = −ππ£β displacement π₯(π‘) = π₯π π −2π cos(π′ π‘ + π) 15.42 Angular frequency π π2 π′ = √ − π 4π2 15.43 Mechanical Energy 1 2 −ππ‘ πΈ(π‘) ≈ ππ₯π π π 2 15.44 ππ‘ Chapter 16 Sinusoidal Waves Mathematical form (positive direction) Angular wave number Angular frequency Wave speed Average Power π¦(π₯, π‘) = π¦π sin(ππ₯ − ππ‘) 2π π= π 2π π= = 2ππ π π£= π π = = ππ π π 1 2 πππ£π = ππ£π2 π¦π 2 Traveling Wave Form 16.2 16.5 π¦(π₯, π‘) = β(ππ₯ ± ππ‘) 16.17 π π£=√ π 16.26 1 1 π¦ ′ (π₯, π‘) = [2π¦π cos ( π)] sin (ππ₯ − ππ‘ + π) 2 2 16.51 π¦ ′ (π₯, π‘) = [2π¦π sin(ππ₯)]cos(ππ‘) 16.60 Wave speed on stretched string 16.9 Resulting wave when 2 waves only differ by phase constant 16.13 Standing wave 16.33 Resonant frequency π£ π£ π = π = π 2πΏ for n=1,2,… 16.66 Chapter 17 Sound Waves Standing Waves Patterns in Pipes π΅ π£=√ π 17.3 π = π π cos(ππ₯ − ππ‘) 17.12 Change in pressure Δπ = Δππ sin(ππ₯ − ππ‘) 17.13 Standing wave frequency (open at both ends) Standing wave frequency (open at one end) Pressure amplitude Δππ = (π£ππ)π π 17.14 beats Speed of sound wave displacement π£ π=π= π£ π=π= ππ£ 2πΏ ππ£ 4πΏ for n=1,2,3 17.39 for n=1,3,5 17.41 πππππ‘ = π1 − π2 17.46 Interference Phase difference Fully Constructive Interference Full Destructive interference Mechanical Energy ΔπΏ 2π π π = π(2π) for m=0,1,2… ΔπΏ = 0,1,2 π π = (2π + 1)π for m=0,12 ΔπΏ = .5,1.5,2.5 … π π= 1 2 −ππ‘ πΈ(π‘) ≈ ππ₯π π π 2 17.21 17.22 17.23 17.24 17.25 15.44 Doppler Effect Source Moving toward stationary observer Source Moving away from stationary observer Observer moving toward stationary source Observer moving away from stationary source Sound Intensity πΌ= Intensity Intensity -uniform in all directions π π΄ 1 2 πΌ = ππ£π2 π π 2 πΌ= ππ 4ππ 2 π′ = π π£ π£ − π£π 17.53 π′ = π π£ π£ + π£π 17.54 π£ + π£π· π£ π£ − π£π· π′ = π π£ π′ = π Shockwave 17.26 17.27 17.29 Intensity level in decibels πΌ π½ = (10ππ΅) log ( ) πΌπ 17.29 Mechanical Energy 1 2 −ππ‘ πΈ(π‘) ≈ ππ₯π π π 2 15.44 Half-angle π of Mach cone π πππ = π£ π£π 17.49 17.51 17.57 Chapter 18 Temperature Scales Fahrenheit to Celsius Celsius to Fahrenheit 5 ππΆ = (ππΉ − 32) 9 9 ππΉ = ππΆ + 32 5 π = ππΆ + 273.15 Celsius to Kelvin First Law of Thermodynamics 18.8 18.8 18.7 βπΈπππ‘ = πΈπππ‘,π − πΈπππ‘,π = π − π First Law of 18.26 Thermodynamics 18.27 ππΈπππ‘ = ππ − ππ Note: βπΈπππ‘ Change in Internal Energy Q (heat) is positive when the system absorbs heat and negative when it loses heat. W (work) is work done by system. W is positive when expanding and negative contracts because of an external force Thermal Expansion Applications of First Law Linear Thermal Expansion βπΏ = πΏπΌβπ 18.9 Volume Thermal Expansion βπ = ππ½βπ 18.10 Q=0 βπΈπππ‘ = −π Adiabatic (no heat flow) W=0 βπΈπππ‘ = π βπΈπππ‘ = 0 Q=W π = π = βπΈπππ‘ = 0 (constant volume) Heat Heat and temperature change π = πΆ(ππ − ππ ) π = ππ(ππ − ππ ) 18.13 18.14 Heat and phase change π = πΏπ 18.16 Power (Conducted) πππππ π ππ» − ππΆ = = ππ΄ π‘ πΏ Free expansions Misc. ππ P=Q/t Power Cyclical process 18.32 Rate objects absorbs energy 4 ππππ = πππ΄ππππ£ 18.39 Power from radiation ππππ = πππ΄π 4 18.38 Work Associated with Volume Change π = ∫ ππ = ∫ πππ ππ 18.25 π = πβπ£ π = 5.6704 × 10−8 π/π2 β πΎ 4 Revised 7/20/17