The Theoretical Value of the Moment of Inertia of a Flywheel - Lab Report Example

MOMENT OF INERTIA OF A FLYWHEEL Abstract In this experiment the aim was establishing the moment of inertia of the disc ( flywheel) by use of a combination of linear motion and circular motion equations. The procedure of experiment involved a mass being hanged on the flywheel through the spindle where the diameter of spindle was 38mm (radius=0.019m). The force causing acceleration was increased from 5N to 25N and for each mass 3 values of time were recorded with the average being used in calculations. Using the relevant measurement that that had been obtained there was calculation of the torque Pr and the angular acceleration  . Pr was plotted against  and the gradient of the graph gave the moment of inertia of the flywheel as 0.469 kgm2 while the value obtained theoretically through calculation was  kgm2. The conclusion made was that the experiment was successful in obtaining the moment of moment of inertia. Introduction Bodies under linear motion usually will move in a straight line even though either at constant speed or may be in acceleration with the acceleration being either constant or changing. A good example depicting linear uniform acceleration motion is that involving a trolley motion on a frictionless plane where the acceleration will be approximately constant. Another scenario featuring uniform acceleration is the case where compact metal piece or a piece of stone is undergoing free fall in the air (Tipler P.A. 2003). A part from linear motion, in real life objects undergo circular motion. Some of the scenarios that feature circular motion are in the leisure park equipments where we have the pirate ship and wave swinger. Other circular motion example include negotiation of a corner by a vehicle and household equipment such as washing machines (Knudsen, M.& Hjorth, G. ,2000). Theory Linear motion Under linear motion there is analysis of bodies with uniform velocity (constant velocity) or where the velocity is increasing constantly per unit time ie there is constant acceleration. As seen in introduction a good example of linear motion where there is constant acceleration is where there is free fall of objects with the assumption being that gravity is the sole force acting on the object with the gravity force being approximately 9.81m/s2 near the earth surface (Nainan K. ,2009). The impact of air resistance is neglected when dealing with high density objects such as stones or metals and when the distance of drop is relatively small (Resnick, R. & Halliday, D.,1966). The equation that are dealt with under linear motion are i.  ii.  iii.  iv.  Where v= final velocity, u = initial velocity, t= time taken, s= the distance covered a = acceleration Circular motion objects under circular motion, will have their velocity constantly changing despite the faction that the speed could be constant. This can be seen in case where a mass attached at the end of a string is whirled to describe a circle whose radius is equal to the length of string and the speed of whirling is kept constant. Since the direction of the mass will be constantly changing it means that there will also be changing velocity because velocity is a vector quantity with both magnitude and direction and thus it will change due to the change in direction even if magnitude remains the same. With the direction changing it means then that there is acceleration referred to as centripetal acceleration  given by and the acceleration is given by Figure 1 The equation that are dealt with under circular motion are i.  ii.  iii.  iv.  Where = final velocity,  = initial velocity, t= time taken, s= the distance covered  = circular acceleration Moment of inertia Moment of inertia for a flat disc is given by Experiment set up and quantities M=mass of disc R= flywheel (disc radius) (m) m= mass attached on spindle (kg) h=distance of fall (m) t=time of fall of mass (s) d= spindle diameter (m) r= spindle radius , (m) (r=d/2) a=mass acceleration m/s2  =angular velocity (rad/s)  =angular acceleration (rad/s2) P= cord tension ( N) g =is gravitational acceleration (9,.81m/s2) Method Relationship in formulars The centripetal force F=ma Now if we consider the inertia of a flywheel we will have  (1) Where  frictional torque Moment of inertia Applying Newton second law of motion to falling mass we have  (2) From  Acceleration  (3) Putting (3) into (2)  (4) From (v=u+at) and by use of the expression for acceleration given by equation 3 , the speed of mass m when it hits the ground is given by (4) Since  and substitution of v using (5) the angular velocity  of the flywheel when mass hits ground will be  (6)  (7) Substituting for  in (7) using (6)  ( 8) By rewriting equation (1) we have Results Theoretical moment of inertia The theoretical moment of inertia is given by Where M = 59.31lb= = 59.31x0.453592=26.9kg R=0.19m Table 1 gives a summary of results obtained in the experiment. The table shows that for a mass of 0.51kg (5N) the time required to cover the 1m height if drop is between 23.28 to 23.69s with the average of the 3 values being 23.43s. The heaviest mass used was 2.55kg (25N) with the average time to drop 1m being 10.21m. Calculating angular acceleration The calculation of  involved the use of the equation So for the 5kg mass t=24.3s 0.2151 Calculating P The tension in the string P was calculated from the eq  (equation 4) For mg= 5 4.998N Pr for the same mass = 4.998x0.019=0.094965 With the same procedure being used the result were as shown in table 1 column 7 From the table it can be seen that as the force applied increased from 5N to 25N there was an increase in angular acceleration from 0.191639 m/s2 to 1.008458 m/s2 . With regard to the torque Pr there was an increase from 0.094654Nm to 0.474072Nm. Plotting Pr against the angular acceleration resulted to a linear relationship graph given by figure 2. From the line the linear equation associated with the line graph it can be seen that the gradient is 0.469 and that translates to the moment of inertia of the flywheel being 0.469kgm3 . The frictional torque for the system is given by the y-intercept which according to the graph is 0.006N. Weight(N) Time 1 Time 2 Time 3 Average Pr 5 23.28 23.34 23.69 23.4367 0.191639 0.094965 10 16.59 16.22 16.13 16.3133 0.395541 0.189854 15 13.44 13.25 13.28 13.3233 0.592994 0.284673 20 11.75 11.57 11.53 11.6167 0.780034 0.379426 25 10.19 10.34 10.12 10.2167 1.008458 0.474072 Figure 2: Relationship between angular acceleration vs Pr Discussion This experiment involved the force causing acceleration in a flywheel (disc) being increased from 5N to 25N. The results revealed that increasing the force causing acceleration (increasing the mass) resulted in faster falling rate in mass where the time taken to cover the 1m reduced. This is in obedience to the Newton second law that stipulates that acceleration of object is proportional to the force causing the acceleration. In order to reduce error in timing, for each mass used there were three runs so that this resulted to having 3 time values with the average being used in the calculation. Apart from the error in timing having a number of runs for the same mass ensured that any variation resistance in motion of the system was evened out. Plotting Pr against the angular acceleration  resulted to a straight line graph being obtained and this was in agreement with what was expected. The line y-intercept of the graph was 0.006 which is the frictional torque of the system. This shows that apart from the resistance of motion of the flywheel that is majorly due to its inertia, the is additional resistant that is caused by friction between the rotating parts of the system. Theoretically the flywheel moment of inertia was found to be  which is slightly above the value obtained graphically of 0.469 is slightly lower but close enough to the theoretical value. This could be attributed to error in timing and also in taking dimensions used in calculation. Conclusion The conclusion from the results is that it is possible to obtain the moment of inertia of a flywheel experimentally. The theoretical value was found to be slightly larger than the value obtained graphically. The difference in the values could be attributed error in measurements made to the but it also means that the formula used in the theoretical calculation is not exact but gives a value that is close to the actually theoretical value. References Knudsen, M.& Hjorth, G. (2000). Elements of Newtonian mechanics: including nonlinear dynamics (3 ed.). Springer. p. 96. ISBN 3-540-67652-X Nainan K. (2009).,Gravitation Resnick, R. & Halliday, D. 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