MIE100H1 20191 661555434949MIE100 Formula Booklet

Velocity and acceleration ds v= dt dv 𝑑𝑣 a= =𝑣 dt 𝑑𝑠 Special cases (a = 0) 𝑠 = 𝑠0 + 𝑣0 𝑡 (const a) v = v0 + a0 t 1 s = s0 + 𝑣0 𝑡 + 𝑎0 𝑡 2 2 𝑣 2 = v02 + 2𝑎0 (𝑠 − 𝑠0 ) ρ= 𝑣2 𝜌 𝑣02 𝑣 = + 2 ∫ 𝑎(𝑠) 𝑑𝑠 𝑡 𝑠0 𝑣 = 𝑣0 + ∫0 𝑎(𝑡) 𝑑𝑡 s = s0 + 𝑣0 𝑡 + 𝑡 𝑡 ∫0 ∫0 𝑎(𝑡) 𝑑𝑡 𝑑𝑡 𝑟⃑ = 𝑥𝑖⃑ + 𝑦𝑗⃑ + 𝑧𝑘⃑⃑ 𝑣⃑ = 𝑥̇ 𝑖⃑ + 𝑦̇ 𝑗⃑ + 𝑧̇ 𝑘⃑⃑ ⃑⃑ 𝑎⃑ = 𝑥̈ 𝑖⃑ + 𝑦̈ 𝑗⃑ + 𝑧̈ 𝑘 |𝑣⃑| = √𝑥̇ 2 + 𝑦̇ 2 + 𝑧̇ 2 SUVAT equation (a const) v = v0 + a 0 t 1 s = s0 + 𝑣0 𝑡 + 𝑎0 𝑡 2 2 𝑣 2 = v02 + 2𝑎0 (𝑠 − 𝑠0 ) (𝑣 + 𝑣0 )𝑡 𝑠= 2 Projectile motion 𝑣0𝑥 = 𝑣0 cos 𝜃 𝑣0𝑦 = v0 sin 𝜃 𝑎𝑥 = 0 𝑥 = 𝑥0 + 𝑣0 𝑡 𝑎𝑦 = 𝑎0𝑦 1 y = y0 + 𝑣0𝑦 𝑡 + 𝑎0𝑦 𝑡 2 Circular motion Dependent motion 2 Cylindrical 𝐫 − 𝛉 system Normal-tangential system 𝑣𝑛−𝑡 = 𝑣𝑢 ⃑⃑⃑⃑⃑⃑⃑⃑ ⃑⃑⃑𝑡 𝑢𝑏 = ⃑⃑⃑ ⃑⃑⃑⃑ 𝑢𝑡 × ⃑⃑⃑⃑ 𝑢𝑛 ds v= dt 𝑎 = 𝑣̇ ⃑⃑⃑ 𝑢𝑡 + 𝑣𝜃̇ ⃑⃑⃑⃑ 𝑢𝑛 = 𝑣̇ ⃑⃑⃑ 𝑢𝑡 + Rectangular coordinates 𝑠 2 - Rope has constant length - Define good datum lines (fixed position) - Find fixed length if possible - Divide the rope into sections if needed 𝑟⃑ = 𝑟𝑢 ⃑⃑⃑⃑⃑𝑟 𝑣⃑ = 𝑟̇ ⃑⃑⃑⃑⃑ 𝑢𝑟 + 𝑟𝜃̇⃑⃑⃑⃑⃑ 𝑢𝜃 𝑎⃑ = (𝑟̈ − 𝑟𝜃̇ 2 )𝑢 ⃑⃑⃑⃑⃑𝑟 + (2𝑟̇ 𝜃̇ + 𝑟𝜃̈)𝑢 ⃑⃑⃑⃑⃑ 𝜃 𝑢𝑛 ⃑⃑⃑⃑ dy 2 (1+(dx) ) 𝑑2 𝑦 | 2| 𝑑𝑥 Relative motion 𝑟𝐵 = ⃑⃑⃑ ⃑⃑⃑ 𝑟𝐴 + ⃑⃑⃑⃑⃑⃑⃑ 𝑟𝐵/𝐴 𝑣𝐵 = ⃑⃑⃑⃑ ⃑⃑⃑⃑ 𝑣𝐴 + ⃑⃑⃑⃑⃑⃑⃑⃑ 𝑣𝐵/𝐴 𝑎𝐵 = ⃑⃑⃑⃑ ⃑⃑⃑⃑ 𝑎𝐴 + ⃑⃑⃑⃑⃑⃑⃑⃑ 𝑎𝐵/𝐴 LT = 𝑠𝐴 + 𝑠𝐵 Then 𝑣A + 𝑣𝐵 = 0 𝑎A + 𝑎𝐵 = 0 ⃑⃑⃑⃑⃑𝑟̇ = 𝜃̇ ⃑⃑⃑⃑⃑ 𝑢 𝑢𝜃 𝑢𝜃̇ = −𝜃̇⃑⃑⃑⃑⃑ ⃑⃑⃑⃑⃑ 𝑢𝑟 1.5 Work and motion dU = 𝐹 ∙ d𝑟 = 𝐹 cos 𝜃 𝑑𝑟 𝑃′ UP>P′ = ∫𝑃 𝑑𝑈 = Frictional force (oppose motion) Static |𝐹𝑓𝑠𝑚𝑎𝑥 | = 𝜇𝑠 𝐹𝑁 Kinetic |𝐹𝑓𝑘 | = 𝜇𝑘 𝐹𝑁 𝐹𝑓𝑘 > 𝜇𝑘 𝐹𝑁 velocity decrease 𝐹𝑓𝑘 = 𝜇𝑘 𝐹𝑁 velocity same 𝐹𝑓𝑘 < 𝜇𝑘 𝐹𝑁 velocity increase Work by gravitational F Ug = −𝑊∆𝑦 = −𝑚𝑔(𝑦2 − 𝑦1 ) ∫𝑃 𝐹 d𝑟 = ∫𝑃 𝐹 cos 𝜃 ds *Always negative Internal force is zero If particles connected by inextensible cable 𝑠𝑖2 𝑓𝑖𝑡 ds = 0 ∫𝑠 ⃑⃑⃑⃑ Work done by force Ug = −𝑚𝑔∆𝑦 1 𝑈𝑠 = − 𝑘(𝑠22 − 𝑠12 ) 2 Uf = −𝐹𝑓𝑘 ∆𝑠 Potential energy Vg = 𝑚𝑔ℎ 1 𝑉𝑠 = 𝑘𝑥 2 2 Conservation of energy T1 + 𝑉1 + 𝑈1>2 = 𝑇2 + 𝑉2 If (𝐔𝟏>𝟐 = 𝟎) T1 + 𝑉1 = 𝑇2 + 𝑉2 Linear momentum ⃑ = mv L ⃑ Elastic collision Inelastic collision Conservation of momentum: Constant force Conservation of momentum: Avg force Conservation of momentum: ∑ 𝑭 = 𝟎 || ∆𝐭 = 𝟎 Multiple particles Moment Equilibrium (𝐫 − 𝛉) ∑ 𝐹𝑟 = 𝑚𝑎𝑟 = 𝑚 (𝑟̈ − 𝑟𝜃̇ 2 ) ∑ 𝐹𝜃 = 𝑚𝑎𝜃 = 𝑚 (2𝑟̇ 𝜃̇ + 𝑟𝜃̈) Gravitational force g⃑ = −9.81𝑗⃑ms−2 F = mg⃑ 𝑃′ Spring force Fs = −𝑘𝑥 k is spring constant x is deviation from rest Equilibrium (x-y-z) Equilibrium (n-t) ∑ 𝐹𝑥 = 𝑚𝑎𝑥 = 𝑚 𝑥̈ ∑ 𝐹𝑛 = 𝑚𝑎𝑛 = 𝑚𝑣𝜃̇ ∑ 𝐹𝑦 = 𝑚𝑎𝑦 = 𝑚 𝑦̈ = ∑ 𝐹𝑧 = 𝑚𝑎𝑧 = 𝑚 𝑧̈ Work by kinetic friction *Against motion> negative Uf = −𝐹𝑓 ∆𝑥 Work by spring 𝑥2 Us = ∫ −𝑘𝑥 𝑑𝑥 𝑥1 1 = − 𝑘(𝑥22 − 𝑥12 ) 2 𝑃′ mv 2 𝜌 ∑ 𝐹𝑡 = 𝑚𝑎𝑡 = 𝑚𝑣̇ Kinetic energy 1 T = 𝑚𝑣 2 2 𝑇1 + 𝑈1>2 = 𝑇2 𝑠 1 𝑚 𝑣 2 + ∫𝑠 𝑖2 ⃑⃑⃑⃑ 𝐹𝑖𝑡 ds + 2 𝑖 𝑖1 𝑖1 𝑠𝑖2 2 ⃑⃑⃑⃑𝑖𝑡 ds = 1 𝑚𝑖 𝑣𝑖2 ∫ 𝑓 𝑠𝑖1 𝑖1 m1 vi1 + 𝑚2 𝑣𝑖2 = (𝑚1 + 𝑚2 )𝑣𝑓 Angular momentum ⃑⃑⃑⃑ 𝐻0 = ⃑⃑⃑ 𝑟0 × 𝑚𝑣 ⃑⃑⃑⃑0 | = 𝑟0 𝑚𝑣 sin 𝜃 |𝐻 = 𝑟0 𝑚𝑣𝜃 𝑡2 𝑡2 ∫ 𝐹 dt = 𝐹 ∆𝑡 𝑡1 𝑡1 Principle of angular momentum and impulse 𝑡 ⃑⃑⃑⃑⃑⃑ H01 + ∫𝑡 2 ∑ ⃑⃑⃑⃑ 𝜇0 𝑑𝑡 = ⃑⃑⃑⃑⃑⃑ H02 1 𝜇0 = ⃑⃑⃑⃑ ∫ 𝐹 dt = ⃑⃑⃑⃑⃑⃑⃑⃑ 𝐹𝑎𝑣𝑔 ∆𝑡 Conservation of linear momentum ∑ 𝐻01𝑖 = ∑ 𝐻02𝑖 ⃑⃑⃑⃑⃑0 𝑑H 𝑑𝑡 Fixed rotation ⃑ Angular displacement θ Angular velocity 𝜔 ⃑ Angular acceleration 𝛼 v⃑⃑⃑⃑𝑃 = 𝜔 ⃑ × 𝑟⃑ 𝑎𝑃 = 𝜔 ⃑⃑⃑⃑ ⃑ × ⃑⃑𝑣 + 𝛼 × 𝑟 If 𝛼 constant, ω = ω0 + 𝛼𝑐 𝑡 1 𝜃 = 𝜃0 + ω0 𝑡 + 𝛼𝑐 𝑡 2 2 ω2 = ω20 + 2𝛼𝑐 𝜃 General motion Decompose the motion Translation > rotation Force, Moment, Angular Momentum ∑ 𝐻 = 𝑤𝐼 𝐹 = m𝑎 ⃑⃑⃑⃑𝐺 = ∑𝑖 𝑚𝑖 ⃑⃑⃑ 𝑎𝑖 H0 = 𝑤𝐼𝑜 ⃑⃑⃑⃑⃑ HA = ⃑⃑⃑⃑⃑⃑⃑⃑ rp⁄A × mv ⃑ H 𝐺 = 𝑤𝐼𝐺 ⃑ = ⃑⃑⃑⃑⃑⃑⃑ 𝑟𝑝⁄𝐴 × 𝐿 ∑ 𝑀 = 𝐼𝛼 ̇ ⃑⃑⃑⃑⃑𝐺 = ∑ 𝑀 ⃑⃑⃑⃑⃑𝐺 𝐻 𝑀𝐺 = 𝐼𝐺 α 𝑀0 = 𝐼0 α (pinned at O) ⃑⃑⃑⃑⃑𝐺̇ = ∑ ⃑⃑⃑⃑⃑ (𝐻 𝑀𝐺 X) 𝑀𝐴 = 𝐼𝐴 α (rolling, no slip) Kinetic Energy (𝐓𝟏 + 𝑼𝟏→𝟐 = 𝑻𝟐 ) 1 T = Trotate + 𝑇𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑒 TBody = 𝐼𝐼𝐶 𝑤 2 2 1 1 1 2 2 = 𝐼𝐺 𝑤 2 + 𝑚𝑣𝐺2 = 𝑤 (𝐼𝐺 + 𝑚𝑑 ) 2 2 2 Parallel Axis Theorem Ip = I𝐺 + 𝑚𝑑 2 d – distance between P and G ω= 𝑣𝐵 −𝑣𝐴 𝑟𝐵/𝐴 mv ⃑⃑⃑1 = mv⃑⃑⃑2 ⃑⃑⃑ ⃑⃑⃑⃑2 L1 = L Rigid body motions 2 m1 vi1 + 𝑚2 𝑣𝑖2 = 𝑚1 𝑣𝑓1 + 𝑚2 𝑣𝑓2 𝑡 ⃑⃑⃑⃑⃑ 𝑀0 = ⃑⃑⃑ 𝑟0 × 𝐹 ∑ 𝑚𝑖 ( v⃑⃑⃑⃑⃑𝑖1 ) + ∑ ∫𝑡 2 ⃑⃑𝐹𝑖 dt = 1 ∑ 𝑚𝑖 ( v⃑⃑⃑⃑⃑𝑖2 ) Instantaneous centre of zero velocity (Point where perpendicular vectors of velocities meet) - Translation - Fixed rotation - General motion ⃑⃑⃑⃑B 𝑎 = ⃑⃑⃑⃑ 𝑎𝐴 + 𝜔 ⃑ × ⃑⃑⃑⃑⃑⃑⃑⃑⃑ 𝑣𝐵/𝐴 + 𝛼 × ⃑⃑⃑⃑⃑⃑⃑⃑ 𝑟𝐵/𝐴 = ⃑⃑⃑⃑ 𝑎𝐴 − 𝜔2 ⃑⃑⃑⃑⃑⃑⃑ 𝑟𝐵/𝐴 + 𝛼𝑟⃑⃑⃑⃑⃑⃑⃑⃑ 𝐵/𝐴 Translation 𝑟𝐵 = ⃑⃑⃑ ⃑⃑⃑ 𝑟𝐴 + ⃑⃑⃑⃑⃑⃑⃑ 𝑟𝐵/𝐴 𝑣𝐵 = ⃑⃑⃑⃑ ⃑⃑⃑⃑ 𝑣𝐴 𝑎𝐵 = ⃑⃑⃑⃑ ⃑⃑⃑⃑ 𝑎𝐴 Magnitude don’t change Direction don’t change |𝑎𝐵/𝐴𝑡 | = 𝑟𝛼 |𝑎𝐵/𝐴𝑛 | = 𝜔2 𝑟 vB = ⃑⃑⃑⃑ ⃑⃑⃑⃑ 𝑣𝐴 + 𝜔 ⃑ × ⃑⃑⃑⃑⃑⃑⃑⃑ 𝑟𝐵/𝐴 Momentum, impulse and angular momentum ⃑ = 𝑚𝑣 ⃑⃑ = r × mv 𝐿 H ⃑ = Iw 𝑡 ⃑⃑ = 𝑟 × 𝐹 = 𝐼α ⃑⃑⃑ ⃑⃑⃑⃑2 𝑀 L1 + ∑ ∫𝑡 2 ⃑⃑𝐹𝑖 dt = L 1 𝑡 ⃑⃑⃑⃑⃑⃑⃑ 𝐻𝐺 1 + ∑ ∫𝑡 2 ⃑⃑⃑⃑⃑ 𝑀𝐺 dt = ⃑⃑⃑⃑⃑⃑⃑ 𝐻𝐺 2 1 𝑡2 ⃑⃑⃑⃑⃑⃑⃑ 𝐻𝑂 1 + ∑ ∫𝑡 ⃑⃑⃑⃑⃑ 𝑀𝑂 dt = ⃑⃑⃑⃑⃑⃑⃑ 𝐻𝑂 2 1 Moment of inertia can be added/subtracted Work by forces Ug = −𝑚𝑔∆ℎ 1 𝑈𝑒 = − 𝑘(𝑠22 − 𝑠12 ) 2 ⃑⃑⃑⃑2 L1 = L ∫ 𝐹 𝑑𝑡 = 0 → ⃑⃑⃑ ⃑⃑⃑⃑⃑𝐺 𝑑𝑡 = 0 → ⃑⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑⃑⃑⃑ 𝐻𝐺 1 = 𝐻 ∫ Σ𝑀 𝐺2 ⃑⃑⃑⃑⃑𝑂 𝑑𝑡 = 0 → 𝐻 ⃑⃑⃑⃑⃑⃑⃑ ⃑⃑⃑⃑⃑⃑⃑ ∫ Σ𝑀 𝑂 1 = 𝐻𝑂 2 Square moment of inertia 1 I𝐺 = 𝑚𝑙2 12 1 I𝐴 = 𝑚𝑙2 3 G – centre of gravity A – end part of square ⃑⃑ = 𝑟 × 𝐹 𝑀 𝑈𝑀 = 𝑀(𝜃2 − 𝜃1 ) Circle moment of inertia 1 IG = 𝑚𝑅 2 2 IG = 𝑚𝑘𝐺2 1 I = 𝜌𝑡𝜋𝑅 4 2 𝐼 𝑅 𝑚 2 𝑘𝐺 = √ 𝐺 = √ 𝑘𝐺 − radius of gyration Conservation of energy T1 + 𝑉1 + U1→2 T1 + 𝑉1 = 𝑇2 + 𝑉2 = 𝑇2 + 𝑉2 U1→2 – work of nonconservative F Second order differential equations m𝑥̈ + c𝑥̇ + 𝑘𝑥 = 0 𝑐 2 − 4𝑘𝑚 > 0 𝑥 = 𝑒 𝜆𝑡 |c| > √4𝑘𝑚 𝑒 𝜆𝑡 (𝑚𝜆2 + 𝑐𝜆 + 𝑘) 𝑥 =0 = 𝐴𝑒 𝜆1𝑡 + 𝐵𝑒 𝜆2 𝑡 −𝑐±√𝑐 2 −4𝑘𝑚 2 real λs 𝜆= 2𝑚 Donut moment of inertia 1 IG = 𝑚(𝑅0 + 𝑅𝑖 )2 2 𝑅0 – outer radius 𝑅𝑖 – inner radius 1 IG = 𝜌𝑡𝜋(𝑅04 − 𝑅𝑖4 ) 2 m = 𝜌𝑡𝜋(𝑅02 − 𝑅𝑖2 ) If no nonconservative F 𝑐 2 − 4𝑘𝑚 = 0 |c| = √4𝑘𝑚 𝑥 = (𝐴 + 𝐵𝑡)𝑒 𝜆𝑡 one real λ V = V𝑒 + 𝑉𝑔 1 V𝑒 = 𝑘𝑠 2 2 𝑉𝑔 = 𝑚𝑔ℎ𝐺 𝑐 2 − 4𝑘𝑚 < 0 2 complex λs λ1 = 𝑎 + 𝑏𝑖 λ2 = 𝑎 − 𝑏𝑖 𝑥 = 𝐴𝑒 𝑎𝑡 𝑒 𝑏𝑡𝑖 + 𝐵𝑒 𝑎𝑡 𝑒 −𝑏𝑡𝑖 = 𝑒 𝑎𝑡 (𝛼𝑐𝑜𝑠𝑏𝑡 + 𝛽𝑠𝑖𝑛𝑏𝑡) Formula Booklet MIE100 Final Exam | Revision 2 Apr 14, 2019 | 1 Undamped free vibration – spring motion (horizontal) 𝑘 𝑥̈ + x = 0 𝑚 𝑥̈ + 𝑤𝑛2 x = 0 wn = √ x = A sin 𝑤𝑛 𝑡 + 𝐵 cos 𝑤𝑛 𝑡 = 𝐶 sin(𝑤𝑛 𝑡 + 𝜃) 2𝜋 1 C = √𝐴2 + 𝐵 2 𝜏 = = 𝑘 𝑚 𝜃= Bar pendulum 𝑤𝑛 wn = √ 3𝑔 2√2𝑎 2𝑙 𝑓 𝑚𝑔 − 𝑘(𝑙 − 𝑙0 ) = 0 = 𝛿𝑒𝑞 𝑚𝑔 = 𝑙 − 𝑙0 𝑘 Parallel / Series spring Parallel Series 1 = keq k eq = ∑𝑖 𝑘𝑖 1 ∑𝑖 𝑘𝑒𝑞 𝑥̈ + x=0 𝑚 wn = 2𝜋𝑓 Square pendulum 3𝑔 wn = √ 𝐵 tan−1 𝐴 Vertical spring ∑ 𝐹𝑦 = 0 𝑒 𝜆𝑡 (𝑚𝜆2 + 𝑐𝜆 + 𝑘) = 0 𝜆= −𝑐±√𝑐 2 −4𝑘𝑚 𝑥 = 𝐴𝑒 𝜆1 𝑡 2𝑚 + 𝐵𝑒 𝜆2 𝑡 Overdamped (𝐜 > 𝐜𝐜 ) 2 real, negative λs No vibration c 2 𝑘 ( ) − >0 2m 𝑚 m𝑙𝜃 = −mg sin 𝜃 = −𝑚𝑔𝜃 𝑔 𝜃̈ + 𝜃 = 0 𝑙 𝜃̈ + 𝑤𝑛2 𝜃 = 0 Undamped forced vibration w0 forcing frequency 𝐹 𝛿0 = 0 static deflection 𝑘 𝑘 F0 𝑥̈ + x = sin 𝑤0 𝑡 𝑚 m x = x𝑐 + x𝑝 x𝑐 = 𝐴 sin 𝑤𝑛 𝑡 + 𝐵 cos 𝑤𝑛 𝑡 (Transient) x𝑝 = C sin 𝑤0 𝑡 (steady) 𝐹 x𝑝𝑚𝑎𝑥 = C = (𝑤 2 0 2 )𝑚 = 𝐹0 /𝑘 = 2 𝑤 (1−(𝑤 0 ) ) 𝑛 Damping equation m𝑥̈ + c𝑥̇ + 𝑘𝑥 = 0 𝑥 = 𝑒 𝜆𝑡 𝑘𝑖 Undamped free vibration – pendulum motion −mg sin 𝜃 = 𝑚𝑎𝑡 𝑔 wn = √ 𝑠 = 𝑙θ 𝑙 ̈ Critically damped (𝐜 = 𝐜𝐜 ) one real λ No vibration 𝑐c is smallest c which system won’t vibrate c 2 𝑘 ( ) − =0 2m 𝑚 𝑥 = (𝐴 + 𝐵𝑡)𝑒 𝜆𝑡 𝑛 −𝑤0 𝛿0 𝑤 2 (1−(𝑤 0 ) ) 2 c 2m ) − 𝑘 𝑚 <0 c x = D [e−2m𝑡 sin(𝑤𝑑 𝑡 + 𝜃)] 𝑐 2 𝑤𝑑 = 𝑤𝑛 √1 − ( ) 𝑐𝑐 x 2𝜋𝑐𝑐 𝑥2 √1−( ) ln ( 1 ) = 𝜏𝑑 = 2𝜋 𝑤𝑑 1 𝑤 2 1 − ( 0) 𝑤𝑘 Magnification factor Damping coefficient Fd = −𝑐𝑥̇ (c is damping coefficient) 𝑘 𝑐c = √ 2𝑚 = 2𝑚𝑤𝑛 𝑚 (critical damping coeff) 𝑛 Underdamped (𝐜 < 𝐜𝐜 ) 2 complex λs ( 𝑥𝑝𝑚𝑎𝑥 M= = 𝐹0 /𝑘 Electrical circuit analogy Mass <-> inductance Damp coeff. <-> resistance Spring con. <-> 1/capacitance Force <-> voltage Displacement <-> charge Velocity <-> current 𝑐 2 𝑐𝑐 >𝜏 Formula Booklet MIE100 Final Exam | Revision 2 Apr 14, 2019 | 2

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