Equation Sheet for Final Exam-124-s21.final

Final Exam Equation Sheet 750:124 Spring 2021 Kinematic Equations linear, constant acceleration 1 𝑥 = 𝑥𝑖 + 𝑣𝑖 𝑡 + 𝑎𝑡 2 2 𝑣 = 𝑣𝑖 + 𝑎𝑡 𝑣 2 = 𝑣𝑖2 + 2𝑎(𝑥𝑓 − 𝑥𝑖 ) Rotational, constant acceleration 1 revolution = 2π radian 𝜔 = 𝜔𝑖 + 𝛼𝑡 1 𝜃𝑓 − 𝜃𝑖 = 𝜔𝑖 𝑡 + 𝛼𝑡 2 2 2 2 𝜔 = 𝜔𝑖 + 2𝛼(𝜃𝑓 − 𝜃𝑖 ) Circular motion 𝑠 = 𝑟𝜃 𝑣 = 𝑟𝜔 𝑎 = 𝑟𝛼 1 2 𝐾𝐸𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 = 𝐼𝑐𝑚 𝜔𝑐𝑚 2 Rolling without Slipping 𝑣𝐶𝑀 = 𝑅𝜔 𝑎𝐶𝑀 = 𝑅𝛼 1 1 2 2 𝐾𝐸𝑟𝑜𝑙𝑙𝑖𝑛𝑔 = 𝑚𝑣𝑐𝑚 + 𝐼𝑐𝑚 𝜔𝑐𝑚 2 2 Vector Products ⃗⃗ | = |𝐴⃗||𝐵 ⃗⃗ | 𝑠𝑖𝑛 𝜃 |𝐶⃗| = |𝐴⃗ × 𝐵 ⃗⃗ = (𝑎2 𝑏3 − 𝑎3 𝑏2 )𝑖̂ − (𝑎1 𝑏3 − 𝑎3 𝑏1 )𝑗̂ + 𝐴⃗ × 𝐵 (𝑎1 𝑏2 − 𝑎2 𝑏1 )𝑘̂ Torque and angular momentum 𝜏⃗ = 𝑟⃗ × 𝐹⃗ ⃗⃗ = 𝑟⃗ × 𝑝⃗ 𝐿 ⃗⃗ 𝑑𝐿 ∑ 𝜏⃗𝑒𝑥𝑡 = 𝑑𝑡 ⃗⃗ = 𝐼𝜔 𝐿 ⃗⃗ 𝜏 = 𝐼𝛼 Conservation of Angular Momentum: 𝐼𝑖 𝜔𝑖 = 𝐼𝑓 𝜔𝑓 Static equilibrium: ∑𝐹⃗𝑒𝑥𝑡 = 0 ∑𝜏⃗𝑒𝑥𝑡 = 0 Simple Harmonic Oscillator 𝑥(𝑡) = 𝐴 𝑐𝑜𝑠(𝜔𝑡 + 𝜑) 2𝜋 𝑇= 𝜔 1 𝑓= 𝑇 𝑘 𝜔 = √𝑚 𝑔 𝜔 = √𝐿 (spring) (simple pendulum) 𝑣(𝑡) = −𝜔𝐴 𝑠𝑖𝑛(𝜔𝑡 + 𝜑) 𝑎(𝑡) = −𝜔2 𝐴 𝑐𝑜𝑠(𝜔𝑡 + 𝜑) 𝐹𝑠𝑝𝑟𝑖𝑛𝑔 = −𝑘𝑥 Waves & Sound 𝑠(𝑥, 𝑡) = 𝑠0 𝑐𝑜𝑠(𝑘𝑥 ± 𝜔𝑡)𝑚𝑎𝑥 𝑊 Newton’s Laws for Rotational Motion Moment of Inertia 𝐼 = Σ 𝑚𝑖 𝑟𝑖2 for point masses 𝐼ℎ𝑜𝑜𝑝 = 𝑚𝑅 2 for hoop 𝐼𝑑𝑖𝑠𝑘 = 12𝑚𝑅 2 for disk 𝐼𝑠𝑜𝑙𝑖𝑑 𝑠𝑝ℎ𝑒𝑟𝑒 = 25 𝑚𝑅 2 for solid sphere 𝐼ℎ𝑜𝑙𝑙𝑜𝑤 𝑠𝑝ℎ𝑒𝑟𝑒 = 23 𝑚𝑅 2 for hollow sphere Threshold of hearing: 𝐼𝑜 = 10−12 𝑚2 𝐼 𝛽 = 10 𝑙𝑜𝑔 ( ) 𝐼0 𝑃 𝑃 𝑎𝑣 𝑎𝑣 𝐼 = 𝐴𝑟𝑒𝑎 ; 𝐼 = 4𝜋𝑟 2 for spherical waves 𝐹 𝑣 = √ 𝜇𝑇 , μ is linear mass density 𝑦(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛 [ 𝑣 = ±𝜆𝑓 = ± 𝜔 𝑘 (taut string) 2𝜋 (𝑥 ± 𝑣𝑡)] = 𝐴 𝑠𝑖𝑛(𝑘𝑥 ± 𝜔𝑡) 𝜆 2𝜋 𝜆 2𝜋 𝜔= 𝑇 3𝑘𝐵 𝑇 𝑘= 𝑣𝑟𝑚𝑠 = √ 𝜆 Destructive: |𝑟1 − 𝑟2 | = 𝑛 2 , 𝑛 = 1,3,5 … Constructive: |𝑟1 − 𝑟2 | = 𝑛𝜆, 𝑛 = 1,2,3 … Standing Waves in Pipes 𝑠𝑛 (𝑥, 𝑡) = 𝑠0𝑛 𝑠𝑖𝑛( 𝑘𝑛 𝑥) 𝑐𝑜𝑠( 𝜔𝑛 𝑡) 2 , 𝑛 = 1,2,3, . .. 𝑓𝑛 = 𝑛𝑓1 , 𝑛 = 1,2,3, . .. If one end is open and one is closed, 𝐿=𝑛 𝜆𝑛 4 𝑄 = 𝑚𝑐(𝑇𝑓 − 𝑇𝑖 ) = 𝑛𝑐(𝑇𝑓 − 𝑇𝑖 ) 3 𝑐𝑉 = 2 𝑅 monatomic ideal gas 5 𝑐𝑃 = 2 𝑅 monatomic ideal gas Δ𝐸𝑖𝑛𝑡 = 𝑄 + 𝑊𝑜𝑛 𝑔𝑎𝑠 Δ𝐸𝑖𝑛𝑡 = 𝑄 − 𝑊𝑏𝑦 𝑔𝑎𝑠 The work done by the gas: 𝑊𝑖𝑠𝑜𝑏𝑎𝑟𝑖𝑐 = 𝑃(𝑉𝑓 − 𝑉𝑖 )( ) 𝑉𝑓 𝑊𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 = 𝑛𝑅𝑇 ln ( ) 𝑉𝑖 If both ends are open or closed, 𝜆𝑛 𝑀 𝑃𝑉 = 𝑛𝑅𝑇 = 𝑁𝑘𝐵 𝑇 Interference (Path Length Difference of Two In-Phase Sources) 𝐿=𝑛 3𝑅𝑇 =√ 𝑚 , 𝑛 = 1,3,5, . .. Heat Engines 𝑊 = 𝑄ℎ − 𝑄𝑐 𝑊 𝑄 𝑒 = 𝑄 = 1 − 𝑄𝑐 𝑓𝑛 = 𝑛𝑓1 , 𝑛 = 1,3,5, . .. ℎ 𝑄𝑐 𝑄ℎ Fluids ℎ 𝑇 = 𝑇𝑐 for an ideal cycle ℎ 𝑇 𝑒𝑖𝑑𝑒𝑎𝑙 = 1 − 𝑇𝑐 𝑑𝑆 = 𝑓 𝑖 𝑑𝑄 𝑇 Entropy of an Ideal Gas 1 𝑇 𝑓 𝑉 ∆𝑆 = 2 𝑛𝑅 ln ( 𝑇𝑓 ) + 𝑛𝑅 ln ( 𝑉𝑓 ) ; f = number of 𝐹𝑏𝑜𝑢𝑦𝑎𝑛𝑡 = 𝜌𝑓𝑙𝑢𝑖𝑑 𝑔𝑉𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 𝑘𝑔 𝑑𝑄 𝑇 𝛥𝑆 = ∫ 𝑃 + 𝜌𝑔ℎ + 2 𝜌𝑣 2 =constant 𝜌𝑤𝑎𝑡𝑒𝑟 = 103 𝑚3 Carnot efficiency ℎ 𝑀 𝜌= 𝑉 𝐹 𝑃= 𝐴 1 atm = 1.01x105 Pa 𝐴1 𝑣1 = 𝐴2 𝑣2 real efficiency 𝑖 𝑖 degrees of freedom 𝑘𝑔 𝜌𝑎𝑖𝑟 = 1.2 𝑚3 𝑇 𝛥𝑆 = 𝑛𝑐𝑉 ln ( 𝑇𝑓 ) Isochoric Process 𝑖 𝑄 Thermodynamics 𝛥𝑆 = 𝑇 𝑅 ≈ 8.3 𝐽/𝑚𝑜𝑙 ⋅ 𝐾 𝑁𝐴 ≈ 6.02 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠/𝑚𝑜𝑙𝑒 𝑘𝐵 ≈ 1.38 𝑥 10−23 𝐽/𝐾 5 𝑇𝐶 = (𝑇𝐹 − 32∘ ) 9 𝑇 = 𝑇𝐶 + 273.15𝐾 𝛥𝑆 = 𝑛𝑐𝑃 ln ( 𝑇 ) 3 𝐾𝐸𝑡𝑟𝑎𝑛𝑠𝑙 = 2 𝑘𝐵 𝑇 ; average 𝐾𝐸𝑡𝑟𝑎𝑛𝑠𝑙 for molecules in gas Isothermal Process 𝑇𝑓 𝑖 Isobaric Process 𝑇 𝛥𝑆 = 𝑚𝑐 ln ( 𝑇𝑓 ) Heating solid/liquid 𝑖 Phase Transformations 𝑚𝐿 𝛥𝑆 = 𝑇 𝑄𝑓 = 𝑚𝐿𝑓 = 𝑇𝑓 (𝑆𝐿𝑖𝑞𝑢𝑖𝑑 − 𝑆𝑆𝑜𝑙𝑖𝑑 ) 𝑄𝑣 = 𝑚𝐿𝑣 = 𝑇𝑣 (𝑆𝐺𝑎𝑠 − 𝑆𝐿𝑖𝑞𝑢𝑖𝑑 )

Equation Sheet for Final Exam-124-s21.final

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