Physics Lab 10

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Massachusetts Institute of Technology *

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8.02

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Physics

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Oct 30, 2023

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Moment of Inertia and Energy in Rotational Motion Lab Number and Title: 9a1 Moment of Inertia and Energy in Rotational Motion Name: Aaron Hsu Group Id: 6 Date of Experiment: 4/6/22 Date of Submission: 4/13/22 Course and Section Number: PHYS 111A006 Instructor’s Name: Bhairavi Apte Partners’ Names: John Pryce IV and Lucas Velerio Mala 1. Introduction: The goals and objectives of this lab are to find and figure out the moment of inertia of an object experimentally and compare it to the one that we have calculated. Examine and explore the rotational motion of still and rigid bodies with respect to angular velocity and angular position both at a constant angular acceleration. Show the conversation of energy in a system that involves rotational motion. Evaluate the relation of those angular quantities to the linear velocity and linear position in a system with a bound motion that includes both rotational and translational motion. Get a better understanding and learn how to deal and use equations related to rotational motion and moments of inertia and energy. 2. Experimental Procedure: The equipment that we used for this lab are a rotating platform with an adaptor, 50g mass hanger, digital scale on lab counter, photogate sensor, a base, 300g square mass with a thumbscrew and a nut, lab computer with capstone software installed, a step pulley with a string, 850 universal interface, rotational system, disk, 50g or 100g weights, vernier caliper, ring, bubble level, a spoke pulley on a rod. To set this experiment up we first set up the rotational system apparatus and then we would adjust the leg screws on the rotational base so we could level it using a bubble level. Then we turn on and log on to the computer and connect the 850 universal interface. We then would get tne disk, the ring, a square mass, a mass hanger, and the given weight and measure the masses of each one of them on the digital scale. Then we would use the vernier caliper to measure the radius of the disk and the inner and outer radii of the ring and we would also measure the radius of the rim of the second pulley from the top of the step pulley where the string is. We then would have to calculate the rotational inertia when it rotates around the disk, the ring, and the two point masses under certain circumstances. We

then go back to our setup and hang the mass hanger with either a weight of 50g or 100g at the end of the string. Then we would start to bind the string around the rim of the second pulley from top of the step pulley by slowly turning the rotating platform until the mass hanger is relatively close to the spoke pulley. We then would start to record our data from the computer software from different tests of just the hanging mass, the two point masses and the hanging mass, the disk centered and the hanging mass, and the disk towards the end and the hanging mass. 3. Results: Mass of weighted mass (kg) 0.1505 A Table of Masses and moment of inertia by object Object Mass (kg) Radius (m) Theoretical Inertia (kg/ms 2 ) Measured Inertia (kg/ms 2 )t % Difference Disk Center Mass 1.4424 0.1135 0.00929 0.00903 2.80% Disk Off-centered 1.4424 0.1135 0.0237 0.0239 0.837% Square Weight (Avg) 0.2766 - 0.02766 0.02496 9.76% Ring 1.4297 Inner: 0.05365 Outer: 0.06380 0.0199 0.0195 1.61% These are our results and calculations that we did: Moments of Inertia: Cylinder/Disk about the center: 1 2 ?? 2 Mass at distance r from rotational axis: ?? 2 Disk with a set of radii: 1 2 ?(? 𝑖??𝑖?? 2 + ? ????𝑖?? 2 ) Finding a displaced center of mass: 𝐼 ?? + ?? 2 = 𝐼 ???−???????? Solving for 𝐼 ??ℎ = 1 2 ?? 2 + 1 2 𝐼ω 2 𝐼 = 2 ??ℎ− 1 2 ?? 2 ω 2 General

Avg: (𝑥 1 +𝑥 2 +...+𝑥 ? ) ? = 𝐴???𝑎?? %Diff: 𝑉𝑎??? ? −𝑉𝑎??? 𝑖 | | 𝑉𝑎??? ? * 100 = % 𝐷𝑖???????? Disk CM Point A: 2 (0.1505)(9.8)(0.1294)− 1 2 (0.1505)(0.0573) 2 (4.6021) 2 = 0. 009 Disk CM Point B: 2 (0.1505)(9.8)(0.2680)− 1 2 (0.1505)(0.0830) 2 (6.6412) 2 = 0. 00895 Disk CM Point C: 2 (0.1505)(9.8)(0.4571)− 1 2 (0.1505)(0.1073) 2 (8.5804) 2 = 0. 00915 Disk CM Inertia Avg: (0.009+0.00895+0.00915) 3 = 0. 00903 0.00929−0.00903 | | 0.00929 * 100 = 2. 80% Disk OC Point A: 2 (0.1505)(9.8)(0.0826)− 1 2 (0.1505)(0.0206) 2 (1.647) 2 = 0. 0299 Disk OC Point B: 2 (0.1505)(9.8)(0.1676)− 1 2 (0.1505)(0.0344) 2 (2.752) 2 = 0. 0218 Disk OC Point C: 2 (0.1505)(9.8)(0.2278)− 1 2 (0.1505)(0.0419) 2 (3.352) 2 = 0. 0199 Disk OC Inertia Avg: (0.0299+0.0218+0.199) 3 = 0. 0239 0.0237−0.0239 | | 0.00239 * 100 = 0. 837% Square (2 mass) A: Point C: 2 (0.1505)(9.8)(0.1009)− 1 2 (0.1505)(0.0341) 2 (2.7286) 2 = 0. 03995 Square (2 mass) B: 2 (0.1505)(9.8)(0.1733)− 1 2 (0.1505)(0.0673) 2 (5.3841) 2 = 0. 0176 Square (2 mass) C: 2 (0.1505)(9.8)(0.2506)− 1 2 (0.1505)(0.0816) 2 (6.53) 2 = 0. 0173 Square (2 mass) Inertia Avg: (0.03995+0.01760+0.01730) 3 = 0. 02496 0.02496−0.02766 | | 0.02766 * 100 = 9. 76% Ring Point A: 2 (0.1505)(9.8)(0.0982)− 1 2 (0.1505)(0.048) 2 (3.8369) 2 = 0. 01965 Ring Point B: 2 (0.1505)(9.8)(0.2160)− 1 2 (0.1505)(0.0718) 2 (5.7432) 2 = 0. 01929 Ring Point C: 2 (0.1505)(9.8)(0.3809)− 1 2 (0.1505)(0.0944) 2 (7.5539) 2 = 0. 01967 Ring Inertia Avg: (0.01965+0.01929+0.01967) 3 = 0. 01954

Physics Lab 10

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