Phys 250 lab 1

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College of San Mateo *

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250

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Mechanical Engineering

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Jan 9, 2024

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Phys 250 2022 September 7th Chase Marangu (C.M.) Jaden Lee (J.L.) Jinesh Tandel (J.T.) Safa Kurdi (S.K.) Yosef Samara (Y.S.) Lab 1 - Measuring a Cylinder, Lab 1 Table of Contents Table of Contents 1 Purpose (J.L.) 2 Procedure Overview (J.L.) 2 Data (C.M.) (J.T.) (J.L.) (S.K.)(Y.S.) 3 Box 3 Cylinder 4 Paper 5 Marble 6 Calculations (C.M.) (J.L) (J.T.) (S.K) 7 Possible Error / “Uncertainty” (Differentials) 7 Percent Error (comparing to expected) 7 Questions 8 Sources of Error (J.L.) 10 Conclusion (Y.S.) 10 Raw Data 11 Chase Marangu’s Raw data 12 Jaden Lee data 14 Jinesh’s Raw Data 14 Safa Kurdi Raw data 16 Yosef Raw Data 17

Phys 250 2022 September 7th Purpose (J.L.) The purpose of this experiment is to learn to use measurement instruments for linear dimensions and mass, as well as how to make calculations with these measurements. Procedure Overview (J.L.) We used a meter stick to measure the width, length, and depth of the inside of a box. Vernier calipers were used to measure the width and height of a metal cylinder A micrometer measured the thickness of 10 A4 sheets of paper and the diameter of a marble The weight of the cylinder and Marble using the triple beam balance. (JL)

Phys 250 2022 September 7th Data (C.M.) (J.T.) (J.L.) (S.K.)(Y.S.) Tables full of data Box Box Tool Ruler Zero Correction 2.5 mm Differential Error (dL) 1 mm Measurement Attempt #1 Attempt #2 Attempt #3 Average Depth (Zero Correction) 7.25 cm 7.25 cm 7.25 cm 7.25 cm Length (Zero Correction) 17.85 cm 17.85 cm 17.9 cm 17.877 cm Width (Zero Correction) 16.35 cm 16.25 cm 16.35 cm 16.31 cm

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Phys 250 2022 September 7th Cylinder Cylinder (height) Tool Vernier Caliper Zero Correction Already accounted for Differential Error (dH) 0.1 mm Notes The actual density for the cylinder is 16.5 grams / cm³ Measurement Attempt #1 Attempt #2 Attempt #3 Average Height 2.55 cm 2.54cm 2.54 cm 2.543 cm Cylinder (diameter) Tool Micrometer Zero Correction 0.015 mm (Already accounted for) Differential Error (dD) 0.005 mm Doubtful Digits Hundreds and thousands place in millimeters. e.g. “41” in 25.441 Measurement Attempt #1 Attempt #2 Attempt #3 Average Diameter 25.441 mm 25.350 mm 25.435 mm 25.409 mm

Phys 250 2022 September 7th Cylinder (mass) Tool Triple Beam Balance Zero Correction 0 Differential Error (dM) 0.05 g ( https://www.physics.smu.edu/~scalise/apparatus/triplebeam/ ) Measurement Attempt #1 Attempt #2 Attempt #3 Average Mass 224.4 g 224.4 g 224.4 g 224.4 g Paper Papers Tool Micrometer Zero Correction 0.015 mm Differential Error (dH) 0.005 mm Measurement Attempt #1 Attempt #2 Attempt #3 Average Thickness 0.930 mm 0.930 mm 0.930 mm 0.930 mm

Phys 250 2022 September 7th Marble Glass Marble (diameter) Tool Micrometer Zero Correction 0.015 mm Differential Error (dD) 0.005 mm Measurement Attempt #1 Attempt #2 Attempt #3 Average Diameter 25.185 mm 25.255 mm 25.245 mm 25.228 mm Glass Marble (mass) Tool Triple Beam Balance Zero Correction 0 grams Differential Error (dM) 0.05 grams Measurement Attempt #1 Attempt #2 Attempt #3 Average Mass 20.09 g 20.09 g 20.09 g 20.09 g

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Phys 250 2022 September 7th Calculations (C.M.) (J.L) (J.T.) (S.K) For this first lab we were only tasked with a few simple calculations, as its main purpose was to introduce us to the strategies for error propagation, measurement, and the like, rather than to actually experiment (which is what we will do in future labs). Possible Error / “Uncertainty” (Differentials) The volume of the cylinder is V = πr²h When you propagate the differentials to solve for the Possible Relative Error (PRE) in V, you get dV/V = π ⋅ ( 2 ⋅ dr/r + dH/H ) Since the radius of the cylinder r = D/2 = 12.705mm then dr = dD/2 = 0.005/2 mm = 0.0025 mm Since the height of the cylinder is 2.543cm = 25.43 mm, and the Possible Error for the cylinder’s height is dH = 0.1mm we compute the Possible Relative Error in V to be dV/V = π ⋅ ( 2 ⋅ dr/r + dH/H ) = π ⋅ ( 2 ⋅ 0.0025mm/12.705mm + 0.1mm/25.43mm ) = π ⋅ (0.039354584% + 0.393236335%) = π ⋅ 0.432590919% = 1.35902445314% ≈ 1.359% To clarify, the Possible Relative Error (PRE) in volume of the cylinder is dV/V = 1.359% Given our measurements earlier, the Possible Relative Error (PRE) in mass is dM/M = 0.05g/224.4g = 0.022282% dM/M = 0.022% Density is mass divided by volume. P = M/V When you propagate the differentials to solve for the Possible Relative Error (PRE) in Density (P) you get, dP/P = dM/M + dV/V dP/P = 0.022% + 1.359% = 1.381% dP/P = 1.381% Rounding to two significant figures, dP/P = 1.4% Percent Error (comparing to expected) When giving us the lab instructions, Prof. Mohsen noted that the density of the cylinder was expected to be 16.5g/cm³ = 0.0165g/mm³

Phys 250 2022 September 7th When we took measurements and did calculations we did for the cylinder, we found r = 12.075mm H = 25.43mm M = 224.4g V = π ⋅ r² ⋅ H = 11648.51mm³ P (density) = M/V = 224.4g / 11648.51mm³ = 0.01926426641 g/mm³ = 19.26426641 g/cm³ Computing the Percent Error |Experimental - Standard|/Standard = |19.26426641 - 16.5|/16.5 = 0.16753129757 = 16.753129757% Rounding to three significant figures, The Percent Error for the density of the cylinder was 16.8% Questions (from https://collegeofsanmateo.edu/physics/docs/physics250/lab01.pdf ) 1. Calculate the volume of the inside of the laboratory equipment box in cm³. Convert the volume to mm³ 7.25x17.877x16.31 = 2113.91 cm^3 = 2113910 mm³ 2. Calculate the volume of the metal cylinder in cm³. V= pi.r^2.h = 11648.51mm^3 = 11.648 cm^3 3. Calculate the density of the metal cylinder in g/cm³. P=224.4/pi*12.75²*2.55 0.00017231g/cm³ 4. Calculate the percent error in the density of the metal cylinder, using the following table of standard values: Aluminum:0.062% Brass: 0.019% copper:0.018% iron :0.018813% 5. Estimate possible absolute errors in your measurements of the diameter, the height and the mass of the metal cylinder. Give some justification for these estimates. Use the rules

Phys 250 2022 September 7th of error propagation to calculate the possible relative error in the density of the cylinder. Comment on this result. How does the possible relative error compare to the percent error? What does this imply about your measured density? dH/H = 0.1 mm / 2.543 cm = 0.1 mm / 25.43 mm = 0.00393236335 = 0.4% dD/D = 0.005mm / 25.409 mm = 0.00019678066 = 0.019% dM/M = 0.05g/224.4g = 0.022282% = 0.022% As shown in Possible Error section above dP/P = 1.381% 6. Calculate the volume of the glass marble in cm³ (4/3) ⋅ π ⋅ r³ (4/3) ⋅ π ⋅ (25.228/2)³ mm³ = 8407.116 mm³ = 8.407 cm³ 7. Calculate the density of the glass marble in g/cm³. One textbook lists the density of common glass as ranging from 2.4 to 2.8 g/cm³. Does your value fall within this range? 19.26426641 g/cm³ /1000 *100^3 = 19264.26641kg/m³ 8. Estimate possible absolute errors in your measurements of the diameter and the mass of the marble. Give some justification for these estimates. Use the rules of error propagation to find the possible absolute error in the density of your marble. What is the range of values for the density of your marble? Comment on this result. Possible Absolute Error (PAE) For Diameter (Micrometer): dD = 0.005 mm For Mass (Triple beam balance): dM = 0.05 grams 9. Convert the density of your marble into kg/m³ 19.26426641 g/cm³ *1000 = 19264.26641kg/m^3 10. How many marbles could fit into the equipment box? Consider 11% of the volume of the box to consist of the empty space between marbles, so that only 89% of the box volume actually contains glass. Calculate the mass of this many marbles in kilograms.

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Phys 250 2022 September 7th N ⋅ (volume of marble) = (volume of box) ⋅ (1 - 0.11) N ⋅ (8.407 cm³) = (2113.91 cm³) ⋅ 0.89 N = 2113.91 ⋅ 0.89 /8.407 = 223.787308196 = 224 marbles . 224 marbles would have a mass of (25.228 g) ⋅ 224 = 5651.072g = 5.651 kilograms . The mass is 74575 ⋅ (20.09g) = 1,498,211.75g = 1,498.21 kg 11. Calculate the expected thickness of one ream (500 sheets) of paper in mm. Convert this thickness to inches. Does your answer seam reasonable? (You have probably all seen a ream of paper for a computer printer. If not, a textbook which has pages numbered 1-1000 is made up of 500 sheets of paper.) 10 sheets of paper is 0.93 mm thick, so one sheet of paper is .093 mm. 500 x .093 mm = 46.5 mm = 1.83 inches A book with 600 labeled pages measured to be 1.375 inches thick, but these pages are a lot thinner. 600/1.375=1000(pages)/x → 1375=600x, x=2.29 inches in a 1000 labeled page book. Textbook with 500 sheets of paper is 2.29 inches. Calculation based on 10 pieces of paper measurement for a 500 sheets of paper book is 1.83 inches. This seems somewhat reasonable at a 20% percent error. Sources of Error (J.L.) The major sources of error in this lab come from misreading the measurement devices; the micrometer, vernier caliper, ruler, and triple beam balance. The ruler could have been slightly bent, leading to an inconsistent measurement. The surfaces of each item are not equal on each side, so the calculations of the object’s size may be slightly off. The zero corrections may still have been a bit off, as we can only really round it to the nearest tenth of a millimeter for the vernier caliper. (JL) Conclusion (Y.S.) Our 15 % error seems a bit large to be solely accounted for by the sources of error mentioned above. Other factors, such as inaccurate measurement and reaction time may have led to our high error. Our graph of Velocity vs. Time is very linear indicating that the acceleration was nearly uniform in our experiment. (ys)

Phys 250 2022 September 7th Raw Data Each of us will submit our own raw data

Phys 250 2022 September 7th Chase Marangu’s Raw data 10:49 Now we start taking the data. 14:00 imagine we're going to measure the interior of a box the Zero Correction is, in Mohsen's case, the 2.5 millimeters at the end of his ruler many instruments have a zero correction depth: measured 7 + correction 2.5mm = 7.25cm in science you never trust one measurement, so he's going to do 5, one for each corner, one in the center of the edge length: 17.6 + 0.25 = 17.85 middle length 17.6 other side 17.65 + 0.25 = 17.9 width: 16.1 + 0.25 middle width 16.0 + 0.25 far side 16.1 + 0.25 The Differential Error for the Ruler is 1millimeter because of the error of the length, because 1mm is the smallest Graduation of the ruler dL = 1 millimeter 22:43 the actual density of the cylinder is 16.5 grams / cm³ to measure, need the length and the width . height: Don't trust one measurement, so he does 3 about 120° apart 2.5 < x < 2.6. best line at 5. 2.55 because 2.5 and 2.6 are the best lining up another 2.54

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Phys 250 2022 September 7th for the Height measurement the error is 0.10 mm dH = 0.1 millimeters for the Diameter going to use the micrometer 25 has passed but 25.5 is not reached measurement 25.4 41 mm (41 is doubtful digits) 25.350 25.435 dD = 0.005 mm) (53:40 half of the divisions of the micrometer is 0.005 mm) the zero correction for the micrometer is 0.015 millimeter thickness of papers 0.5 passed, 1 not reached 43 microns 0.930 mm glass marble 25 is passed + 0.095 = 25.095 mm second measurement (thanks Guadalupe for help) 25.245 mm third measurement 25.245 mm first measurement redo 25.185 mm remember the 0.015 millimeter zero correction for the micrometer for the marble too Now we're measuring mass using triple beam balance. mass of the cylinder 224.4 grams mass of marble 20.09 grams

Phys 250 2022 September 7th Jaden Lee data https://drive.google.com/file/d/1tFUcKF-wRt02aKTLlskNiiJ_FFTB 81-c/view?usp=sharing Jinesh’s Raw Data Measurement of Box Ruler Zero Correction 2.5 mm Depth 1) 7cm + 2.5mm = 7.25cm 2) 7cm 3) 7cm Length 1) 17.6 + 0.25 = 17.85 cm 2) 17.6 middle length 3) 17.65 + 0.25 = 17.9 cm Width 1) 16.1 + 0.25 = 16.35 cm 2) 16 cm 3) 16.1cm Measurement of Cylinder

Phys 250 2022 September 7th Density 16.5 g/cm^3 Height 1) 2.55 mm 2) 2.54 mm 3) 2.54 mm dH = 0.1 mm Diameter (micrometer) 1) 25.43 mm 2) 25.35 mm dD = 0.005 mm Mass of Cylinder 1) 224.4 grams Thickness of Paper 0.43 + 0.5 = 0.930 mm Glass Marble Diameter 1) 25 + 0.095 = 25.095 mm 2) 25.245 mm 3) 25.18 mm Mass (triple beam balance) 20.09 grams

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