Carleton University Laboratory Report Course #: PHYS 1007-L9 Experiment #: 4 Atwood’s Machine Experiment Nicholas Adeyeye 101010922 Date Performed: November 9th, 2016 Date Submitted: November 23th, 2016 Lab Period: L9 Partner: Abdullahi Station #: 28 TA: Matt. E Purpose The purpose of the conducted experiment was to find the frictional torque and acceleration due to gravity using the Atwood’s Machine. Theory In this experiment the acceleration due to gravity and the frictional torque were calculated using Atwood’s machine. The equation used to calculate the acceleration due to gravity was: g = 2mhA (1) where g is the acceleration due to gravity, m is the slope of the graph of 1/t2 vs βm in 1/s2/g, h is the distance travelled by the first mass in m, and A is the total mass plus the related inertia. The equation to determine the error on the acceleration due to gravity calculated is as follows: πg=g×√(π2m/m2)+(π2h/h2) +(π2A/A2) (2) where πg is the statistical error on the acceleration due to gravity, g is the calculated acceleration due to gravity in m/s2, πm is the error on the slope, m is the slope in 1/s2/g, πA is the error on the value determined for A, A is the total mass plus the related inertia, πh is the error on the measurement of the distance travelled by the first mass and h is the distance travelled in metres. The equation used to calculate the frictional torque was: Γ = -2bhrA (3) where Γ is the frictional torque in N/m, b is the y-intercept of the graph of 1/t 2 vs βm in 1/s2, and h is the distance travelled by first mass in metres, r is the radius of the pulley in metres. The equation for the statistical error on the frictional torque is: πΓ= Γ×√(π2b/b2) + (π2h/h2) +(π2A/A2) +(π2r/r2) (4) where πΓ is the statistical error on the frictional torque, Γ is the frictional torque in N/m, πb is the error on the y-intercept, b is the y-intercept of the graph of 1/t 2 vs βm in 1/s2, A is the total mass plus the related inertia, πA is the error for the A value, h is the distance travelled by the first mass in metres, πh is the error for h. The equation used to determine the density using the Archimedes principle is: M= m1 + m2 + 10mwasher (5) Where M is the total mass of the machine in g, m1 is the mass of the iron core and the screw in g, m2 is the mass of the weight without the iron core and the screw in g, and 10mwasher is the mass of ten washers in g. The error on the total mass is calculated using the following equation: πM = √(π2m1+π2m2+100π2mw) (6) where πM is the statistical error on the total mass, πm1 is the error on the first mass, πm2 is the error on the second mass, πmw is the error on the mass of ten washers. The equation for the A value is as follows: A = M + I/r2 (7) where A is the total mass plus the related inertia, M is the total mass of the machine, I is the inertia, r is the radius in m, and the value for I/r2 of the apparatus is 80 ± 1 g. The equation for the error on A is the following: πA = √ (π2M+π2I/r2) (8) where πA is the error for the A value, πM is the error on the mass, πI/r2 is the error on the related inertia. The reciprocal of the average time squared was calculated using the following: 1/(tav2) (9) where tav is the average time it takes for m1 to hit the switch at the top of the apparatus. The error on the reciprocal of the average time squared was calculate by the following equation: 2 σ1/(tav2) = π‘ 3 σtav ππ£ (10) The equation used to calculate Δm is as follows: Δmn-x = (m2 + nmw) – (m1 + xmw) (11) where Δm is the change in mass between the two masses connected to the apparatus in g, n is the number of washers on m2, x is the number of washers on m1, and mw is the mass of one washer in g. The equation for the error on Δm is as follows: σΔm = √(π2m1+π2m2+100π2mw) (12) where σΔm is the error on Δm, πm1 is the error on the first mass, πm2 is the error on the second mass, πmw is the error on the mass of ten washers. The equation of the graph of 1/t2 vs βm for the data collected is: 1/tav2 = 0.007695x – 0.004906 (13) The standard deviation of a collection of measurements is found using the inefficient statistic equation: xmax−xmin √N ππΌπ= (14) where ππΌπ is the approximation of the standard deviation, π₯πππ₯ is the maximum value, π₯πππ is the minimum value, and N is the number of measurements. The averages were determined using the following equation: π₯ +π₯ +π₯ +π₯ Average = 1 2 π 3 π (15) The standard deviation of the mean is calculated using the following equation: πmean = σπΌπ √π (16) The consistency test equation is as follows: |x1−x2| π‘= 2 2) (√σ1 +σ2 (17) The combined weighted average equation is the following: π₯π€πππβπ‘ππ = (π22π₯1 + π12π₯2) / (π12 + π22 ) (18) The combined weighted error equation is the following: ππ€πππβπ‘ππ = π1 π2 / √ π12 + π22 (19) Apparatus Figure 1: Atwood’s Machine Apparatus Electronic Balance Error: ±0.005 g Range: 0 - __ g Vernier Caliper Range length: 0-15 cm Range width: 0-10mm Error: ±0.0025 mm Observations Table 1. Masses Used, Diameter and Radius of Pulley in Apparatus Masses Lengths Readings Iron core Other weight one washer Distance + screw + screw (mass of travelled mβ(g) mβ(g) all/10) by mβ mw (g) h(cm) 254.47 255.44 0.995 99.4 254.47 255.45 0.994 99.3 254.44 255.44 0.994 99.3 Average 254.46 255.44 0.994 99.33 RE 0.02 0.02 0.002 0.05 0.02 0.005 0.0005 0.05 ππΌπ 0.01 0.002 0.0003 0.02 πππππ Error RE RE RE RE chosen Final 254.46 ± 255.44 ± 0.994 ± 99.33 ± Value ± 0.02 g 0.02 g 0.002 g 0.05 cm error Radius of the pulley r(cm) 6.18 6.045 6.065 6.097 0.025 0.07 0.04 RE 6.097 ± 0.025 cm βm (g) σΔm (g) Time 1 (s) 10.92 8.94 6.95 4.96 2.97 0.011 0.011 0.011 0.011 0.011 3.580 3.911 4.477 5.558 7.552 Table 2. Change in Mass, Time and Error Measurements Time Time Time Time Reading σ (s) Average 2 (s) 3 (s) 4 (s) 5 (s) Error Time (s) (s) 3.567 3.547 3.554 3.526 0.002 0.024 3.555 3.919 3.889 3.893 3.897 0.002 0.013 3.902 4.450 4.467 4.487 4.477 0.002 0.017 4.472 5.428 5.472 5.594 5.488 0.002 0.074 5.508 7.550 7.515 7.691 7.852 0.002 0.151 7.632 Calculations The calculation for Δm was done using Equation 11: Δmn-x = (m2 + nmw) – (m1 + xmw) Sample Calculation: Δm9-1 = (255.44 + 9(0.994)) – (254.46 + 0.994) Δm9-1 = (264.386g) – (255.454g) Δm9-1 = 9.532g The calculation for the average times were done using Equation 15: π₯ +π₯ +π₯ +π₯ Average = 1 2 π 3 π Sample Calculation: 3.580+3.567+3.547+3.554+3.526 Average time for Δm10-0 = 5 Average time for Δm10-0 = 3.555 s The calculation for the error on the average time was done using Equation 14 & 16: xmax−xmin ππΌπ= √N 3.580−3.526 ππΌπ= ππΌπ= √5 0.024 πmean = σπΌπ √π = 0.024 √5 πmean = 0.011 πmean Therefore, the average time for the Δm10-0 data was 3.555 ± 0.011 s σmean (s) 1/Tav2 (1/s2) 0.011 0.006 0.007 0.033 0.067 0.079 0.066 0.050 0.033 0.017 Error on 1/Tav2 (1/s2) 0.000481 0.000202 0.000166 0.000397 0.000303 The calculation for the 1/t av2 was done using Equation 9: Sample Calculation for Δm10-0 times: 1/tav2 = 1 / (3.555s)2 1/tav2 = 0.079 1/s2 The error on 1/tav2 was calculated using Equation 10: σ1/(tav2) = σ1/(tav2) = 2 3 π‘ππ£ σtav 2 (3.555)3 (0.011 s) σ1/(tav2) = 0.000 Therefore, the value for 1/tav2 for the Δm10-0 data was 0.079 ± 0.00 1/s2 The calculation for the total mass was done using Equation 5: M= m1 + m2 + 10mwasher M= 254.46 g + 255.44 g + 10(0.994 g) M = 519.84 g The calculation for π for the total mass was done using Equation 6: πM = √(π2m1+π2m2+100π2mw) πM = √[(0.01)2+(0.003)2+100(0.003)2] πM = 0.011 Therefore, the total mass is 519.84 ± 0.011 g The calculation for the A value was done using Equation 7: A = M + I/r2 A = 519.84 g + 80 g A = 599.84 1/s2g The calculation for the error on the A value was done using Equation 8: πA = √ (π2M+π2I/r2) πA = √ (0.011)2+(1)2) πA = 1.00006 Therefore, the value of A is 599.84 ± 1.000 1/ s2g The calculation for the acceleration due to gravity was done using Equation 1: g = 2mhA The slope of the graph was: 0.007880 ± 1.797e-41/s2/g g = 2(0.007880 1/s2/g) (0.9933 m) (599.84 1/s2g) g = 9.39 m/s2 The error for the calculated acceleration due to gravity was determined using Equation 2: πg= (9.39 m/s2)×√(1.797e-4)2/(0.007880)2+(0.05)2/(0.9933)2 +(1.00006)2/(599.842) πg= 0.519 Therefore, the calculated value for the acceleration due to gravity was 9.39 ± 0.519 m/s2 The calculation for the frictional torque was done using Equation 3: Γ = -2bhrA The y-intercept of the graph was -0.005761 ± 0.001347 1/s2 Γ = -2(-0.005761 1/s2)(0.9933 m)(0.06091 m)(0.59984 1/s2kg) Γ = 0.000418 Nm The error for the frictional torque was determined using Equation 4: πΓ= Γ×√(π2b/b2) + (π2h/h2) +(π2A/A2) +(π2r/r2) πΓ= (0.000418 Nm) ×√(0.001347)2/(-0.005761)2 + (0.0005)2/(0.9933)2 +(1.00006)2/(0.59984)2) +(0.025)2/(0.0697)2 πΓ= 0.0001 Therefore, the value of Γ was 0.000418 ± 0.0001 Nm The calculation for the consistency test was done using Equation 17: π‘= π‘= |x1−x2| √(σ21 +σ22 ) |9.81−9.39| √(0.01)2 +(0.519)2 ) π‘= 0.81 Since π‘ < 2, the results are consistent. The calculation for combining the accepted value of g with the calculated value was done using Equation 18: π₯π€πππβπ‘ππ = [(0.519)2(9.81) + (0.01)2(9.39)] / (0.519)2 + (0.01)2 π₯π€πππβπ‘ππ = 9.81 m/s2 The calculation for the error of the combined value was done using Equation 19: ππ€πππβπ‘ππ = (0.519) (0.01) / √ (0.519)2 + (0.01)2 ππ€πππβπ‘ππ = 0.001 Therefore, the combined value for g is 9.81 ± 0.001 m/s2 Results The goal of the conducted experiment was to calculate the value of acceleration due to gravity as well as the frictional torque of the Atwood’s machine. The value of acceleration due to gravity was calculated to be 9.39 ± 0.519 m/s2 which was combined with the value of 9.81 ± 0.01 m/s2 to yield a value of 9.81 ± 0.001 the value for the frictional torque was calculated to be 4.18 x 10-4 ± 0.0001 Nm. Discussion The results of the conducted experiment produced a value of acceleration due to gravity equal to 9.39 ± 1.003 m/s2 and a value of frictional torque equal to 4.18 x 10-4 ± 0.0001 Nm. The accepted value for acceleration due to gravity is 9.81 ± 0.01 m/s2, compared to this value the calculated value has a much more significant error value, however the two values are consistent with each other seeing as t = 0.81. Since the values are consistent, they are combined and the smaller error, that of the accepted in this case, becomes the more important one. The experiment could be improved if the apparatus was no as long, because if the string was shorter it would be easier to control and thus easier to avoid a centripetal force that would cause error in collected data. The friction of the pulley that exists contributes to the error of the calculation of the acceleration of gravity because it opposes the motion that is being observed, slowing down the time and thus causing error in data. The air resistance could also contribute to the error of the measurement of g because the heavier mass that is descending, propelling the other mass upwards, would be experiencing air resistance slowing it down, so the time measurements would be recorded as being lower than they should be in the ideal system, increasing the error. The elasticity and the weight of the string on the pulley is important because if the string has considerable weight and low elasticity, the friction will be increased which means that the tension force would not be the same throughout the string, which is an assumption made when the calculations were done, therefore increasing risk of error. Also, if the string had very low elasticity, then two masses of different weight would be able to balance on the pulley due to increased friction and the timing for one mass to reach the top and the other to reach the bottom would be measured as being slower than it really is, increasing the error of the results. Thus, another improvement for this experiment would be to use the lightest and most elastic string possible to be able to sustain the weight of the masses used. Questions 1. How does this experiment compare to one where you would calculate π by measuring the time it takes an object to fall on the ground from a certain height? Measuring the time it takes an object to fall to the ground would have to take into account visual and physical reaction times when starting the clock when the object is let go and stopping it when the object hits the ground; yielding less accurate results. This experiment was done using an apparatus that hit stop immediately without anyone have to react in time to do anything, so the results yielded by the experiment conducted would be more accurate. 2. Geologists are able to discover oil deposits by measuring variations on the gravitational acceleration in the order of ~0.00005 m/s2. Is your measurement precise enough to find a variation of such magnitude? The measurement calculated is not precise enough to calculate gravitational acceleration in the order of 0.00005 m/s2 as the calculated value only goes to two decimal places, which means a variation of magnitude with anything more than two decimal places would not be found. 3. Under what circumstances the acceleration of the Atwood’s machine is equal to zero? Explain. The Atwood’s machine’s acceleration would equal to zero if there were equal masses placed on both sides either at rest, or at constant velocity. In both cases the force of gravity is the same on both sides, which means the tension in the string is equal to the force opposing gravity; thus, the net force is zero. If the net force is zero, then there is no acceleration. In the scenario with constant velocity, the upward velocity is the same in magnitude as the downwards velocity but the directions are different, this means the net velocity is zero and the acceleration is therefore also equal to zero.