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Energy Conservation (Ch 18) All types of energy are measured in Joules Potential Energy PE = mgh m = mass (kg) g = gravity accel (m/s2) h = height (m) Spring Energy SE = 0.5kx2 k = Spring constant (N/m) x = displacement or deformation (m) Kinetic Energy (Linear) KE = 0.5mv2 m = mass (kg) v = velocity (m/s) Kinetic Energy (Rotational) KE = 0.5I2 I = Mass moment of Inertia (kgm2) = velocity (rad/s) Work (Linear) W=FS F = Force (N) S = displacement (m) Work (Rotation) W = T T = Torque (Nm) = Angular displacement (rad) Power = Work / time (Watts) = (Joules)/(second) P = F v (Linear) F = Force (N) v = velocity (m/s) P = T (Rotary) T = Torque (Nm) = angular velocity (rad/s) Conservation of Energy. (Energy at point 1) = (Energy at point 2) PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2 +W = Positive work = Force moves the object (motor). Force in same direction as motion. -W = Negative work = Brakes (the motor is working as a generator). Friction is always negative. Force opposes motion. Energy-Conservation Page 1 Rail Car Spring Thursday, 10 May 2012 11:12 AM PE1 + KE1 + SE1 = PE2 + KE2 + SE2 KE1 = SE2 0.5mv^2 = 0.5kx^2 KE = 0.5mv^2 = 0.5*9400*(12/3.6)^2 = 52222 J In engineering we usually use kJ and kW. Find k in SE = 0.5kx^2 But KE1 = SE2 k = SE/0.5x^2 = 52222 / (0.5*0.37^2) = 762922 N/m This is the spring modulus (k) which is normally given in N/mm, but you must calculate in N/m in formulas. Energy-Conservation Page 2 Charpy Energy Tuesday, 8 May 2012 6:44 PM Job for CAD... 1341.4078 mm PE = mgh = 33*9.81*1.3414 = 434.2514 J Velocity of hammer before impact? v = (2gh)^0.5 = (2*9.81*1.3414)^0.5 = 5.1301 m/s Velocity of hammer after impact? PE after: PE = mgh = 33*9.81*0.32253 = 104.41264 J KE after = 104.41264 v = (2KE/m)^0.5 = (2*104.41264/33)^0.5 = 2.51556 m/s Check: v = (2gh)^0.5 = (2*9.81*0.32253)^0.5 = 2.51556 m/s Finding height by CAD. Energy-Conservation Page 3 Charpy apparatus Energy-Conservation Page 4 Weight falls on a spring. Tuesday, 17 May 2011 6:26 PM Example 18.5. Mass 15kg, falls 350mm to spring. Modulus 10N/mm. Find max compression. 1. At rest and height = 350mm. 2. At speed and touching spring. 3. At rest and spring compressed. PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2 PE1 = KE2 PE1 = KE2 mg(h) = 0.5mv2 15*9.81*0.35 = 0.5 *15*v2 v2 = 51.5025 /(0.5*15) = 6.867 v = 6.867 ^0.5 = 2.6205 m/s (Check (2*9.81*0.35)^0.5 = 2.6205m/s) PE2 + KE2 + SE2 +/- W = PE3 + KE3 + SE3 PE2 + KE2 = SE3 mgx + 0.5mv2 = 0.5kx2 Quadratic equation... 0.5kx2 - mgx - 0.5mv2 = 0 0.5*10*1000x2 - 15*9.81x - 0.5*15*2.62052 = 0 5000 x2 - 147.15 x - 51.5027 = 0 a = 5000, b = -147.15, c = - 51.5027 Quadratic equation solution… Energy-Conservation Page 5 Quadratic equation solution… Read the book. Page 258 So answers are x = 117mm or x = -87mm (????) This quadratic formula assumes the mass and spring are stuck together. Imagine dropping the mass, then it latches onto the spring for good. -87mm is how high the stuck mass will pull the spring ABOVE the initial start position of the spring. If it doesn't latch, the mass will bounce back to the start (350mm). Energy-Conservation Page 6 Spring and Mass Tuesday, 31 March 2015 6:48 PM Q10: (cont) Spring modulus 1.1N/mm and a mass 2.9kg hooked on and then released suddenly. (b) What is the maximum stretch of the spring? PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2 PE1 = SE2 mgh = 0.5kx2 mgx = 0.5kx2 mg = 0.5kx x= 2mg/k = 2*2.9*9.81/(1.1*1000) = 0.0517 m = 0.0517 * 1000 = 51.7 mm Energy-Conservation Page 7 Rollercoaster Tuesday, 17 May 2011 5:33 PM Example 18.2: Roller Coaster. Mass = 500kg, starting height A=18m, lowest point B=0m, up to C=13m. From A to B PEA + KEA = PEB + KEB From B to C PEB + KEB = PEC + KEC Straight from A to C PEA + KEA = PEC + KEC PEA = PEC + KEC mgha = mghc + 0.5*m*vc2 500*9.81*18 = 500*9.81*13 + 0.5*500*v2 88290 = 63765 + 250*v2 250*v2 = 88290 - 63765 = 24525 J v2 = 24525/250 = 98.1 v = 98.1 ^ 0.5 = 9.9045 m/s (9.9045*3.6 = 35.6562 km/h) Check with h=5... v = (2*9.81*5)^0.5 = 9.9045 m/s PE = KE is a common problem: mgh = 0.5mv2 mgh = 0.5mv2 v = 2gh Energy-Conservation Page 8 Rail-Car Buffer Tuesday, 8 May 2012 8:16 PM PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2 PE1= SE2 mgh = 0.5*k*x^2 x = (2mgh/k)^0.5 = (2*9400*9.81*2.6/640000)^0.5 = 0.8656 m Note: We could not do this using linear motion & F=ma because the spring does not give a constant force (therefore not constant accel) e.g. velocity before impact..(2*9.81*2.6)^0.5 = 7.1423 m/s Assume this stops in 0.8656m, by linear motion... v2-vo2 = 2as, so a = (0-7.1423^2)/(2*0.8656) = -29.4665 m/s2 From F=ma, F = 9400*29.4665 = 276.98 kN from spring How many mm? F=kx x=F/k = 276980/640 = 432.7813 mm This shows the average is half of the maximum. Energy-Conservation Page 9 Rolling Inertia Tuesday, 17 May 2011 6:02 PM Example 18.3. Cylinder 100mm, Mass = 30kg, height difference 1.2m Find linear/rotary speed at bottom. PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2 PE1 = KE2 mgh = 0.5mv2 + 0.5I2 Calculate mass moment of inertia I = 0.5mr2 (solid cylinder) = 0.5*30*0.05^2 = 0.0375 kgm2 30*9.81*1.2 = 0.5*30*v2 + 0.5*0.0375*2 Oh no! We have 2 variables in 1 equation!!!! No probs. Use v=r 30*9.81*1.2 = 0.5*30*v2 + 0.5*0.0375* (v/r)2 353.16 = 15v2 + 0.5*0.0375* (v/0.05)2 353.16 = 15v2 + 0.5*0.0375* (v2/0.052) 353.16 = v2(15 + 0.5*0.0375/0.052) v2 = 353.16/ 22.5 = 15.696 v= 15.696 ^0.5 = 3.9618 m/s =v/r 3.9618/0.05 = 79.236 rad/s Energy-Conservation Page 10 Note: Compare to not spinning. v=(2gh)^0.5 = (2*9.81*1.2)^0.5 = 4.8522 m/s (or with a zero mass moment of inertia I) Energy-Conservation Page 11 Multiple Bodies - energy Tuesday, 17 May 2011 7:03 PM First solve this question WITHOUT friction. PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2 PE1 = PE2 + KE2 mAgh1 + mBgh1 = mAgh2 + mBgh2 + 0.5mAv^2 + 0.5mBv^2 PE1 = (1.3+3.3)*9.81*1.6 = 72.2016 J PE2 = 1.3*3.2*9.81 = 40.8096 J KE2 = PE1-PE2 = 72.2016 - 40.8096 = 31.392 J 0.5*v2*(1.3+3.3) = 31.392 V2 = 31.392/(0.5*(1.3+3.3)) = 13.6487 V = 13.6487 ^0.5 = 3.6944 m/s Compare to Freefall: v= (2*9.81*1.6)^0.5 = 5.6029 m/s Include friction this time… PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2 PE1 - W = PE2 + KE2 PE1 = (1.3*9.81*1.6+3.3*9.81*1.6) = 72.2016 J PE2 = (1.3*9.81*3.2+3.3*9.81*0) = 40.8096 J W = T Find angle: s=r so = s/r = 1.6/0.04 = 40.0 rad In linear W=FS or in rotary motion; W = T Energy-Conservation Page 12 In linear W=FS or in rotary motion; W = T W = 0.340*40 = 13.6 J PE1 - W - PE2 = KE2 72.2016 - 13.6 - 40.8096 = 17.792 KE = 0.5*mv^2 v = (2KE/m)^0.5 = (2*17.792/(1.3+3.3) )^0.5 = 2.7813 m/s Check: Freefall drop 1.6m v = (2*9.81*1.6)^0.5 = 5.6029 Our answer is slower than freefall, so looks good. Energy-Conservation Page 13 Multiple Bodies - force method Tuesday, 17 May 2011 7:03 PM Compare to solving by the FORCE METHOD… \First solve this question WITHOUT friction. Set up FBD for each body: Bodies are NOT in equilibrium!!! Body A: Fy = ma (Newton's 2nd law) Ta - mg = ma Ta = ma+mg = 1.3a + 1.3*9.81 = 1.3a + 12.753 Body B: Fy = ma (Newton's 2nd law) Tb - mg = -ma (Accel DOWN) Tb = mg - ma = 3.3*9.81 - 3.3a = 32.373 -3.3a Because the pulley has no friction, Ta=Tb So 1.3a + 12.753 = 32.373 -3.3a 4.6a = 32.373 -12.753 = 19.62 a = 19.62/4.6 = 4.2652 m/s2 Linear motion question: S = 1.6m V0 = 0 m/s V1 = ? a = 4.2652 m/s2 t= Energy-Conservation Page 14 t= Linear motion (no time): V2 - V02 = 2as V2 = 2*4.2652*1.6 = 13.6486 V = 13.6486^0.5 = 3.6944 m/s This time the friction will alter the tension across the pulley Find force due to friction on pulley; Torque = force x radius F = T/r = 0.34/0.04 = 8.5 N So all the same until Ta=Tb - 8.5 Etc.. Energy-Conservation Page 15 Coal Loader Tuesday, 15 May 2012 4:47 PM This is NOT an inclined plane problem. The conveyor is curved, and even if it was a straight line, we don't know the angle! ...but no worries, we can solve using conservation of energy! Coal loaded at 560 t/hr = 560/60 = 9.3333 tonnes/min Tonnes/sec = 560/3600 = 0.1556 or 155.6 kg/s Belt carries 44kg per m 1 second = 155.6/44 = 3.5364 m Velocity = 3.5364 m/s Do as 1 hour's worth... PE1 + KE1 +/- W = PE2 + KE2 The coal gains KE at the start and has the same KE at the end, so... KE1 = KE2 so... PE1 + KE1 +/- W = PE2 + KE2 Energy-Conservation Page 16 KE1 = KE2 so... PE1 + KE1 +/- W = PE2 + KE2 or Work: W = PE2 = mg(h2-h1) = 560 *1000* 9.81*(25-2.5) = 123606000 J (123.606 MJ) 1 second's worth... Work: W = 123606000/3600 = 34335 J (34.335kJ) Power: P = 34.335 kJ/s = 34.335 kW 3 phase motor: I = 34.335*1000/(3^0.5*415) = 47.767 Amps For 1 second: (m=155.6m, v=3.5364 m/s) KE = 0.5*m*v^2 = 0.5*155.6*3.5364^2 = 972.9765 J Power = 972.9765 W Total Power = gravity + KE + friction = 34.335 + 0.973 + 8.1 = 43.408 kW PE1 + KE1 +/- W = PE2 + KE2 We know: Power, Diam, Belt Velocity P, D, v... Find T From W = T, so P = Tor T = P/ Find : From v=r, =v/r = 3.5364/0.43 = 8.2242 rad/s So from T = P/43408 / 8.2242 = 5278.08 Nm Energy-Conservation Page 17 Car on a hill Tuesday, 15 May 2012 6:31 PM Total Force = Fdrag + Frolling + Fgravity Velocity = 95/3.6 = 26.3889 Fdrag = 0.5*1.2*26.3889^2*0.27*2.19 = 247.0596 N Convert to power: P = Fv = 247.0596 * 26.3889 = 6.519 kW Frolling = FN = 0.013*1700*9.81 = 216.801 N Convert to power: P = Fv = 216.801 * 26.3889 = 5.72114 kW Actual value: FN cos = 0.013*1700*9.81*cos(3.5)= 216.397 N Fgravity = mgsin = 1700*9.81*sin(3.5) = 1018.106 N Convert to power: P = Fv = 1018.106 * 26.3889 = 26.8667 kW Total Force = 247.0596 + 216.801 + 1018.106 = 1481.9666 N Convert to power: P=Fv = 1481.9666 * 26.3889 = 39107.47 W (39.107 kW) Energy-Conservation Page 18