WriMini lab report that includes: 1)Title of the Experiment 2)Introduction/Purpose. It should be about half of page long and describe what was the purpose of the experiment, which specific physics laws and concepts the experiment was related to, and what you were expected to observe in the experiment. 5) Conclusion should be about half of a page long and should describe your experimental results and explain whether these results were in an agreement or in disagreement with theoretical prediction.

Formatting ~ Styles ~ Painter Styles Cells Forit Alignment Number Clipboard C18 X fx =SLOPE(B7:B14,C7:C14) N O D E F G H K M Experiment M25a Rotational Dynamics Disk y = 0.009099x + 0.000654 R2 = 0.999294 Torque Angular Acceleration Disk Nm) (rad/s') 0.025 0.008496199 0.87119 0.012844408 1.3509 0.019133275 2.0245 0.009780882 0.02 0.98376 Torque (Nm) 0.015906809 1.6815 0.015 0.01 slope 0.009098853|kg*m^2 stav of slope 0.000139589 kg*m^2 0.005 0.5 Angular Acceleration (rad/s?) 2.5 26 27 28 29 calculations Disk graph Disk & Hoop graph + Ready 1% Accessibility: Investigate KI Type here to search DELLX Cut ap Wrap Text General Calibri 11 - A A [ Copy Conditional Format as Cell Insert Delete Form BIU - - E Merge & Center ~ $ ~ % 9 00 Formatting ~ Table ~ Styles ~ Format Painter Styles Cells Clipboard Font Alignment Number X V =C18-'Disk graph'!C18 D E F G H K L M N O Disk & Hoop y = 0 015978x + 0.000897 R = 0.999072 Torque Angular Acceleration Disk & Hoop (Nm) (rad/s') 0.025 0.008498458 0.55698 0.012852578 0.85465 0.019159472 1.3105 0.009785606 0.62322 0.02 0.015919265 1.0709 0.015 Torque (Nm) 0.01 slope 10.01397 7682 kg mm2 stav of slope 0.000246017 kg"m 2 20 0.005 0.004878829 22 23 24 0.2 25 0.4 0.6 0.8 1.2 26 Angular Acceleration (rad/s?) 27 28 29 30 31 32 calculations Disk graph Disk & Hoop graph Ready x Accessibility: Investigate Type here to searchData Sheets: M25a: Rotational Equilibrium and Rotational Dynamics NAME: DATE: Table 3: Rotational Dynamics - Disk a= Ra applied ITT : W-ma Q. T = TI masks radius angular stand m acc ac devise linear acc. tension torque 2.104868 0.00828 0.87 /19 7.6x10 2-007213 45 110261 1069/ 0-008416179 varying radius 10 .104868 0 .01253 13509 0. 00 12 1 09 364 6 7 32 0 .91 1 75 94 84 820268220 . 20.24083189 10 . 109868 10. 018765 2.0245 0 . 0017 0.03 796 79 27 10 .0191332 7 applied angular mass radius stand. acc. dev. linear acc. tension torque varying mass 0.07982 0.01253 1. 0002 9.6X 10 78059709 1.8059709 0. 09780 381 10. 129926 0.01 253 1068 15 10. 00 19 1. 26922394 1.26 9 77 0 410 . 0142 8 88 Table 4: Rotational Dynamics - Disk and Hoop applied stand. radius angular mass acc. dev. linear acc. tension torque 0.104868 200 828 16.58698 13 6 x 10 6.00 46 11 7 6 / 1. 0263838 2 0.00 84 9BY 5B varying radius 6.109868 0.0125 3 6.85 965 8.2x10% 10 .010 7 087 65 1 . 025 2 4 49 0 01 28 325- 7 8 0.10 4868 0.0187205 1.3 05 0.0012 10 02457203 1 . 0242968370. 019,59492 applied radius angular stand. mass dev. linear acc. acc. tension torque varying mass 0.07982 0.01 253 6-62322 16 2 x10 9 10. 00780894 70. 78097913 6. 0097 05606 10.129926 0.01) 53|1. 0709 16-2 x 10 0 . 013 41837 7 1. 27 0491 9920. 01591 924 Date Modified 03/16/22 12Table 1: Rotational Equilibrium Position - Location = L Location of point of suspension = 0 . 5 0O (m) Trial One mass (kg) position (m) lever arm (m) force (N) torque (Nm) left side 0.114794 6 . 758 0. 258 1:12 4062 0 29 000 799 6 right side 10.064923 10 045 0. 455 106359260 1 28925 533 Table 2: Rotational Equilibrium - Unknown Mass Unknown mass (recorded from lab balance) =0.051257 (kg) F=m g Trial Two mass (kg) position (m) lever arm (m) force (N) torque (Nm) left side 0.164887 6. 400 0 . 100 1 . 614735 16. 16 45735 position (m) lever arm (m) torque (Nm) force (N) calc. mass (kg) right side 10. 822 0 . 322 10. 16 / 45735 0.5014 2034 0. 051 207/42 % Difference between calculated and measured mass = 0.0973129476 Date Modified 03/16/22Data Sheets: M25a: Rotational Equilibrium and Rothfinal dynamics Ex NAME: DATE: APR 12 2023 Experimental Determination Moments of Inertia: Valencia College WC Moment of Inertia for the Disk: 0. 00 909 853 Kgym - Moment of Inertia for Disk and Hoop: 0 . 01 3 9 2 7 6 82 Kg X m Z Moment of Inertia for the Hoop: 6 048 78929 KSXm 2 Theoretical Determination: 0. 22901 M 1= 0. 1 14505 Diameter of the Disk: 1 . 4057 159 Mass of the Disk: Theoretical Moment of Inertia of the Disk: 0 009 215 3 4 3 790 1.264098 TZEL MRZ Percent Error: 0. 10735 m x = 0.053675 Inside Diameter of the Hoop: Outside Diameter of the Hoop: 0 . 127 30 Kq x = 0. 56365 Mass of the Hoop: 1.4250 Theoretical Moment of Inertia of the Hoop: 0.004789283 Percent Error: 1 2239 % Date Modified 03/16/22 14\fFor additional information on these concepts please read review the following sections in your textbook Torque & Equilibrium: Walker. Physics. Chapter 11 section 1 & 3 Cutnell & Johnson. Physics. Chapter 9 section 1 & 2 Rotational Dynamics & Moment of Inertia: Walker. Physics. Chapter 11 section 5 Chapter 10 section 5 Cutnell & Johnson. Physics. Chapter 9 section 4Procedures: This experiment consists of two separate parts. The first part is conducted in order to calculate torque and verify rotational equilibrium using the meter stick and the set of masses as shown in Figure 2. The second part is conducted in order to examine rotational dynamics by varying the torque applied to the disk and hoop, determining the resulting acceleration and ultimately calculating the moment of inertia. This part uses the apparatus shown in Figure 3. Part I: Rotational Equilibrium 1. Record the position of the point of suspension where the meter stick is balanced with no masses attached to it. This represents the center of mass of the meter stick, and can be found by looking at the center, point of suspension, of the clamp. The position should be recorded using units of meters with precision to the millimeter in Table 1. 1. For each trial measure the total mass placed on each side (combined mass and mass hanger) using the most precise balance available. Remember to convert to kilograms. 3. Place a 100-gram mass and hanger on one side of the meter stick by hanging it from the string at some convenient location. Hang a 50-gram mass and hanger on the other side of the meter stick, again from the string but now moving it along the stick until the meter stick reaches equilibrium and is level to the table. Record the position for each mass, using meters, in Table 1. 4. Calculate the lever arm for each mass. This is the distance from the mass to the point of suspension. Next calculate the force and then the torque for each side. Note: the system has reached equilibrium; therefore both values for torque should be identical, or at least very close. 5. For trial 2 place one of the known masses, from the mass kit, on the left side and the unknown mass provided for the right side. Note: Make sure the masses differ from those used in trial I. Again position them so equilibrium is reached. 6. Complete all the calculations for your known mass, entering the information on Table 2. 7. For the unknown mass, record its position and lever arm. Since we are dealing with a mass of unknown quantity, the torque cannot be directly calculated. But since the system is at equilibrium, we can assume their torques to be identical. Record the same torque value obtained for the known mass as also being the unknown's torque. 8. Use this torque to calculate the force and then the mass of the unknown. Record your results on Table 2.