Moment of Inertia Report

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Apr 3, 2024

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Moment of Inertia October 30th, 2023 Introduction The purpose of this lab is to dynamically measure the moment of inertia of a rotating system. Then, students compared the observed value to a predicted value using physic equations. In this experiment, a torque is applied to the rotational apparatus by a string that is wrapped around an axle. The tension is supplied by a hanging weight that is attached to the string. Then, the rotational apparatus has a moment of inertia. Theoretical Analysis There were several different equations used for this experiment. To start with the force of friction, students will multiply the frictional mass times gravity. The frictional torque is calculated by multiplying the force of friction by the radius of the apparatus. To find the acceleration we multiplied used the y= ½ (a)(t) equation. Plugging in our knowns for our height and time, we solved for acceleration. The tension force was calculated by subtracting the mass times acceleration from the gravitational force. The torque was calculated by multiplying the tension force and the radius of the apparatus. The net torque was calculated by subtracting the torque and frictional torque. Data Plots/Charts Figure 1

Figure 2 Results Table 1 Apparatus Configuration Hanging Mass Calc. Friction Calc. Friction Torque Calc Tensio n Calc. Torque Net Torque Calc. acceleration Calc Angular Acceleration

1st Run (wheel only) 50g .0392N .001Nm .483N .0132N m .012Nm .133m/s^2 .003 rad/s^2 2nd Run (wheel only) 100g .0392N .001 Nm .961 N .025 Nm .024Nm .186m/s^2 .004 rad/s^2 1st Run (wheel and added mass) 50g .069N .0018N m .487 N .0132 Nm .0112N m .059m/s^2 .002 rad/s^2 2nd Run (wheel and added masses) 100g .069N .0018N m .972 N .025 Nm .023 Nm .084 m/s^2 .002 rads/s^2 Discussion of Results For part I of this experiment, the first value we found was the frictional force. From the frictional force, we were able to calculate the frictional torque. Next, we timed how long it took the weight to fall to the ground. With the time and distance traveled, we were able to calculate the acceleration of the hanging mass. Knowing the acceleration, we could find the tension force by multiplying the hanging mass by the difference between gravity and the acceleration of the mass. This tension force, when multiplied by the radius of the rotational apparatus, is equal to the applied torque. We found the net torque by subtracting the frictional torque from the applied torque. Angular acceleration was calculated by multiplying the acceleration by the radius. After finding all these values, we were able to find the moment of inertia due to its inverse relationship with angular acceleration and its direct relationship with net torque. We plotted angular acceleration versus net torque and the slope of the best fit line is equal to the moment of inertia. When the equation torque = moment of inertia x angular acceleration is solved for the moment of

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inertia graph proves their relationship. Knowing the moment of inertia before the mass was added, we were able to predict the moment of inertia for the added mass. After we added mass to rotational apparatus, we were able to examine the effects on the moment of inertia. The moment of inertia was much higher with the added mass. This experiment confirmed the inverse relationship between angular acceleration and moment of inertia, as well as the direct relationship with torque and moment of acceleration. Post Lab Questions 1. What is the maximum kinetic energy that the inertia device (wheel) used is given by the hanging mass (just before the mass reaches the floor). Compare this value to the gravitational energy that the hanging mass has just before you release it. Choose only one of your 4 runs and say which you used. Show all work. For the first run, height was .84 m, and time was 12.60 seconds. V= V(initial) +at V= 0 + .133(12.6) —> v = 1.68m/s ½(.05kg)(1.68m/s)= .042 J PE= mgh (.05)(9.8)(.84)= .412J The gravitational potential energy is much greater than the kinetic energy. 2. If the torque applied to a rigid body is doubled, what happens to the moment of inertia of the body? What happens to the angular acceleration? Explain your answers.

When the applied torque is doubled, the net torque is doubled, and the angular acceleration is also doubled. The moment of inertia remains the same because it solely depends on the axis of rotation and the mass of the object. Furthermore, in the equation torque = moment of inertia x angular acceleration, when torque is doubled and angular acceleration are doubled, then they cancel each other out and moment of inertia remains the same. 3. In the theoretical determination of the moment of inertia Inew with the additional masses, it was assumed that the masses are points. Using the parallel-axis theorem, calculate the moment of inertia such that the diameter of the masses is taken into account. Determine the percentage difference between this and the previous value. Is it a good approximation to assume that the masses are points in this particular case? Parallel axis theorem: I= I(mass) +MR^2 I(mass)= ½ MR^2= ½(1.362)(.026)^2= 4.6 *10^-4 I= 3(4.6*10^-4)+ .140= .141 Percent difference: 20% 4. What is the moment of inertia of a drum which is rotated about the vertical axis which runs through the center of the drum. Assume the drum is a hoop. You should ignore the top and bottom pieces of the drum and assume the drum is a hoop. Section 1- Section 1 – diameter is 0.36 m, height is 0.15 m, mass is 4 kg I = MR^2 = 4(0.18)^2 = 0.1296 kg m^2 = 0.13 kg m^2 Raw Data/Calculations See attached.

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