Determination of Moment of Inertia - Lab Report Example

Determination of moment of Inertia Student’s Name Instructor’s Name Institutional Affiliation Course City Date Abstract Discovery and Innovations are part of development. Therefore, engineering deals with experimental design development and manufacturing process. This is because it helps in determining many factors that positively contributes to the growth of economy and improvement of technology in general. This work aims at the determination of the moment of inertia of a flywheel. This experiment will be carried out in the laboratory to help in determining the moment of inertia of the flywheel. The method used here was composed of several attempts with the result being recorded. From the results, it is clear that the practical or calculated inertia is almost close to the theoretical value and the difference in the value is brought as a result which is being bought by a defect in the measuring instruments. The at construction testing of a simple laboratory made a device for establishing the moment of inertia in the flywheel has been carried out successfully. A specific method for rotating the wheel is developed specifically for this experiment. As stated earlier, the calculated value and experimental value are in agreement, and variation might be as a result of several factors which were assumed in the experiment process. Introduction One important section of engineering has been the part of experimental design development and manufacturing process. This is because it helps in determining many factors that positively contributes to the growth of economy and improvement of technology in general (Braune and Fischer 2013). This work aims at the determination of the moment of inertia of a flywheel. This experiment will be carried out in the laboratory to help in determining the moment of inertia of the flywheel. The experiment helps in testing the fundamental laws and analytic methods being applied in mechanical problems to help coming up with a practical solution in real life. Description of the machine The general presentation of the machine is shown in figure 1 below which is the conceived device for determining the moment of inertia in flywheels. This instrument is distinct from the Norwood instrument version, though the principles and operational methods are heavily borrowed from the Norwood instrument (Hanay, et al., 2015). Here instead of setting the flywheel in oscillation, the wheel rotation is what is used in calculating the moment of inertia. Figure 1: Schematic presentation of Freewheel diagram From the diagram, it can be seen that the device is composed of two curved side named (A) which bears housing to Flange (B). This assist in allowing free disc rotation. Axle is labeled (C) which is well fitted on the wheel. (D) Is an inextensible string of considerable length, Section (E) is weight while (F) is attached string at one end. It is clear that the string is usually passed through the hole in the axle and then wound on it. Lastly, a flat base (G) helps in maintaining the equipment in a stable state during the operation. The construction of this simple machines is heavily borrowed from the Galileo assumption in which he states how stresses in the beam normally vary from the top to the bottom. Theoretical perspective The moment of inertia is defined as the limit of the sum of the product of all masses of every particle into which the body is conceived to be divided and the distance square from the given axis (Pluta and Hryniewicz 2013). This can be expressed mathematically as:- This indicates the mass of an elementary particle and y its distance from a given axis. Therefore, the moment of inertia I of a given body about a given axis can be expressed as:- This can also be expressed as:- In circumstances where the body of an instrument is perceived to be having different parts, to get the moment of the inertia about the axis is normally equals to the total sum of the moment of inertia of each part about the same axis. That means that we have to calculate the moment of inertia separately and then sum them up to get a total moment of inertia. This can be expressed mathematically as IT = I1 + I2 + I3 ………. +In Where I’s is the moment of inertia of each part of the body. There are circumstances where the first two equations can be expressed regarding the area. However, we will limit ourselves to weight in line with this experiment. Brief analysis of the Flywheel device From figure 1 above, the disc is set to go on the motion as different weights are drop. The equation of the energy for the motion for the above flywheel can be given as:- And it should be noted that From the equation I- is the moment of inertia, f is the work done against the friction per very revolution, w is the angular velocity of the wheel, and n1 &n2 are some revolution of the wheel. In the setting we can have:- Where h is the height through which the weight is dropped, and r is the radius, t is the mean time of the revolution recorded. These are the major concept and calculation formula required in this experiment and determination of the law of inertia. Experiment requirements The apparatus here include M- Falling mass (kg) h- The distance which is fallen by the mass t – This is the time of fall of the mass d- This will be the diameter of the instrument spindle r –in this case will be the radius of the spindle. w- Angular velocity α – this will be the angular acceleration of the flywheel P- The tension in the cord g- This will be the gravitational acceleration normally given as 9.82/ms2 Methodology used in the experiment To establish the moment of inertia, the following procedure will be followed:- 1. For a given mass, time the rate of fall and distance of fall is measured and recorded 2. We use the various formulae which are given below inserting them in Excel to help in determining the moment of inertia of the flywheel by plotting the line of best fit 3. This is done using different masses, and the result is recorded 4. Estimate the inertia using the formulae for the flywheel of the solid disc The result of the experiment Experiment from mass Weight Time 1 Time 2 Time 3 Average 5 23.28 23.34 23.69 23.43 10 16.59 16.22 16.13 16.31 15 13.44 13.25 13.28 13.32 20 11.75 11.57 11.53 11.65 25 10.19 10.34 10.12 10.21 The diameter is also given as 38 mm Height is give as 1, for all the experiment carried out in this case. The results is carried out at different instances Calculation of the values From the experiment, we are now able to calculate the moment of the inertia. We will apply the general formulae in the calculation. The formulae is given by:- From the above result in the table, m = h = 1 r = (38/2)/1000 g = 9.81 m/s2 n1=19 n2 = 8 For the first weight 0.5 kg I = 0.5* 0.0192 [9.81*23.422 /2*1)-1] [8/ (19 +8)] =0.0001805 (2689.40) (0.296) = 1.44*10-1 kg For the weight of 1 kg We have I = 1* 0.0192 [9.81*16.312 /2*1)-1] [8/ (19 +8)] I = 0.000361* 1303.80*(0.296) I = 1.393 X 10-1 kg For the third weight of 1.5 kg. I = 1.5* 0.0192 [9.81*13.322 /2*1)-1] [8/ (19 +8)] I = 0.0005415*869.30*(0.296) I = 1.393 X 10-1 kg For the fourth case the weight 2.0 kg I = 2* 0.0192 [9.81*11.652 /2*1)-1] [8/ (19 +8)] = 0.000722 * 664.70 *0.296 1.421*10-1 kg The last weight is 2.5 I = 2.5* 0.0192 [9.81*10.212 /2*1)-1] [8/ (19 +8)] = 0.0009025 *510.30 * 0.296 = 1.363 * 10-1 kg The total inertia is given as the product of all the inertia of different body parts. IT = I1 +I2 + I2 + I3 + I 4 + 15 = 1.44*10-1 kg + 1.393 X 10-1 kg + 1.393 X 10-1 kg +1.421*10-1 kg + 1.363 * 10-1 kg In this case we will not add all the inertia since they are calculated differently. We can plot the graph on time against inertia and this will be shown below:- Discussion and conclusion From the results, it is clear that the practical or calculated inertia is almost close to the theoretical value and the difference in the value is brought as a result which is being bought by defect in the measuring instruments. The development and testing of a simple laboratory made device for establishing the moment of inertia in the flywheel has been carried out successfully. A specific method for rotating the wheel is developed specifically for this experiment. As stated earlier, the calculated value and experimental value are in agreement, and variation might be as a result of several factors which were assumed in the experiment process. For more accurate results to be obtained from the experiment, there should be more effort in ensuring that a well-lubricated bearing is used to reduce the friction, and some errors which result due to parallax should be avoided from reading in the stopwatch. Reference Braune, W. and Fischer, O., 2013. Determination of the moments of inertia of the human body and its limbs. Springer Science & Business Media. Hanay, M.S., Keller, S.I., O'Connell, C.D., Mulvaney, P., Sader, J.E. and Roukes, M.L., 2015. Inertial imaging with nanomechanical systems. Nature Nanotechnology, 10(4), pp.339-344. Pluta, Z. and Hryniewicz, T., 2013. Mass Moment Determination Using Compound Pendulum. International Letters of Chemistry, Physics, and Astronomy, 3, pp.85-100. Read More

This can be expressed mathematically as:- This indicates the mass of an elementary particle and y its distance from a given axis. Therefore, the moment of inertia I of a given body about a given axis can be expressed as:- This can also be expressed as:- In circumstances where the body of an instrument is perceived to be having different parts, to get the moment of the inertia about the axis is normally equals to the total sum of the moment of inertia of each part about the same axis. That means that we have to calculate the moment of inertia separately and then sum them up to get a total moment of inertia.

This can be expressed mathematically as IT = I1 + I2 + I3 ………. +In Where I’s is the moment of inertia of each part of the body. There are circumstances where the first two equations can be expressed regarding the area. However, we will limit ourselves to weight in line with this experiment. Brief analysis of the Flywheel device From figure 1 above, the disc is set to go on the motion as different weights are drop. The equation of the energy for the motion for the above flywheel can be given as:- And it should be noted that From the equation I- is the moment of inertia, f is the work done against the friction per very revolution, w is the angular velocity of the wheel, and n1 &n2 are some revolution of the wheel.

In the setting we can have:- Where h is the height through which the weight is dropped, and r is the radius, t is the mean time of the revolution recorded. These are the major concept and calculation formula required in this experiment and determination of the law of inertia. Experiment requirements The apparatus here include M- Falling mass (kg) h- The distance which is fallen by the mass t – This is the time of fall of the mass d- This will be the diameter of the instrument spindle r –in this case will be the radius of the spindle.

w- Angular velocity α – this will be the angular acceleration of the flywheel P- The tension in the cord g- This will be the gravitational acceleration normally given as 9.82/ms2 Methodology used in the experiment To establish the moment of inertia, the following procedure will be followed:- 1. For a given mass, time the rate of fall and distance of fall is measured and recorded 2. We use the various formulae which are given below inserting them in Excel to help in determining the moment of inertia of the flywheel by plotting the line of best fit 3.

This is done using different masses, and the result is recorded 4. Estimate the inertia using the formulae for the flywheel of the solid disc The result of the experiment Experiment from mass Weight Time 1 Time 2 Time 3 Average 5 23.28 23.34 23.69 23.43 10 16.59 16.22 16.13 16.31 15 13.44 13.25 13.28 13.32 20 11.75 11.57 11.53 11.65 25 10.19 10.34 10.12 10.21 The diameter is also given as 38 mm Height is give as 1, for all the experiment carried out in this case. The results is carried out at different instances Calculation of the values From the experiment, we are now able to calculate the moment of the inertia.

We will apply the general formulae in the calculation. The formulae is given by:- From the above result in the table, m = h = 1 r = (38/2)/1000 g = 9.81 m/s2 n1=19 n2 = 8 For the first weight 0.5 kg I = 0.5* 0.0192 [9.81*23.422 /2*1)-1] [8/ (19 +8)] =0.0001805 (2689.40) (0.296) = 1.44*10-1 kg For the weight of 1 kg We have I = 1* 0.0192 [9.81*16.312 /2*1)-1] [8/ (19 +8)] I = 0.000361* 1303.80*(0.296) I = 1.393 X 10-1 kg For the third weight of 1.5 kg. I = 1.5* 0.0192 [9.81*13.322 /2*1)-1] [8/ (19 +8)] I = 0.

0005415*869.30*(0.296) I = 1.393 X 10-1 kg For the fourth case the weight 2.0 kg I = 2* 0.0192 [9.81*11.652 /2*1)-1] [8/ (19 +8)] = 0.000722 * 664.70 *0.296 1.421*10-1 kg The last weight is 2.5 I = 2.5* 0.0192 [9.81*10.212 /2*1)-1] [8/ (19 +8)] = 0.0009025 *510.30 * 0.296 = 1.363 * 10-1 kg The total inertia is given as the product of all the inertia of different body parts. IT = I1 +I2 + I2 + I3 + I 4 + 15 = 1.

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