Inertia lab report Essay

A Rotational proceeding experiment is the simplest method of finding the Moment of Inertia. Minimum equipment is required to work out this experiment. For the purposes of increasing the accuracy of the results, the procedure should be tell three times, devising our conclusion more reliable. While load is moving downwards its potential energy converts to kinetic. Load is accelerating be arrest weight(Fg=mg) of the load is bigger than tension on a string so load is not in proportionality and by Newtons Second Law (F=ma) resultant king creates an acceleration. Resultant force digest be calculated by the comparability S=0.5at2+ut to find acceleration and F=ma. String rotates the mandril which rotates the disc by creating a torsion(T=Fr). Torque accelerates the disc and it can be found by ? = . To find moment of inactivity now T=I? equation is used. 1 Method and observation Apparatus 3 unalike size discs, pergola, ruler, set of weights, stopwatch, stand. magnetic disc is atta ched to one supplant of the spindle and string with load is attached to the former(a) end. Discs weight, diameter and radius be required to be measured onward experiment. Length of the string (L), number of loops on the spindle (n) and horizontal standoffishness of loops (H) were measured before experiment.Using equation below r is found. = 2 Spindle Disc String Stopwatch Weights Stand After circumstance all the equipment up the experiment starts. The string is then wrapped rough the spindle. Time was measured for load pass the distance of length of the string. To fleck graphical record one over time2 is required to be calculated. 4 different masses of the load are used in experiment is repeated 3 times every time mass is changed to make sink random error. After finishing all the experiments on one of the discs other disc is placed and experiment repeats.When all the experiments are done and measurements are recorded mass against one over time2 is plotted using results. 3 graphs are going to be plotted for each disc. Gradient of the graph is constant k which we could use to find I using formulae below. When notice the string with vibrating and load was moving a little which can example just about systematic error. While spindle is spinning at that place is some friction which is neglected and the disc is vibrating while it is spinning which also cause some systematic error. = ? 2 2 Results MEASUREMENTS ON THE SPINDLE MEASUREMENTS ON THE DISCS n= 8 L= 0. 26m DISC 1 (small)DISC 2 (medium) DISC 3 (large) Weight (kg) 0. 314 0. 490 Diameter (m) Radius (m) 0. 1 0. 127 0. 05 0. 0635 0. 696 0. 152 0. 076 H= 0. 026m r= 5. 1410-3 DISC 1 Weight (kg) Time (s) mean(a) Time (s) 1/t? (s-2) K (m s? ) I (kg m2) observational I (kg m2) theory-based ?I (kg m2) 0. 1 2. 93 2. 73 2. 62 2. 76 0. 131 0. 15 2 2. 1 2. 23 2. 11 0. 225 1. 439 0. 000346 0. 2 1. 87 1. 85 1. 86 1. 86 0. 287 0. 22 1. 81 1. 74 1. 78 1. 78 0. 317 0. 000393 0. 000047 DISC 2 Weight (kg ) Time (s) fairish Time (s) 1/t? (s-2) K (m s? ) I (kg m2) data-based I (kg m2) theoretical ?I (kg m2) 0. 1 6. 49 6. 16 6. 33 6. 33 0. 02500. 15 4. 97 4. 77 509 4. 92 0. 0413 0. 686 0. 000726 0. 000988 0. 000262 3 0. 17 4. 38 4. 97 Z 4. 43 4. 43 0. 0510 0. 20 4. 00 4. 13 4. 08 4. 07 0. 0604 DISC 3 Weight (kg) Time (s) Average Time (s) 1/t? (s-2) K (m s? ) I (kg m2) experimental I (kg m2) theoretical ?I (kg m2) 0. 1 4. 21 4. 13 4. 17 4. 17 0. 0575 0. 15 3. 13 3. 27 3. 00 3. 13 0. 102 0. 290 0. 00172 0. 00201 0. 00029 4 0. 2 2. 73 2. 75 2. 73 2. 74 0. 113 0. 17 3. 03 2. 77 2. 83 2. 9 0. 119 Calculations T=I? -3 T=Fr I = = = = ? = 2 1 = ? 1 =km k = 2 2 9. 8x(5.1410? 3 )? I1e= 21. 4390. 26 =0. 000346ms? 9. 8x(5. 1410? 3 )? I2e= 20. 6860. 26 =0. 000726ms? 9. 8x(5. 1410? 3 )? I3e= 20. 2900. 26 =0. 00172 ms? I1t=0. 50. 0520. 314=0. 000393 ms? I2t=0. 50. 063520. 490=0. 000988 ms? I3t=0. 50. 076020. 696=0. 00201 ms? ?I1=0. 000393-0. 000346=0. 000047 ms? ?I2=0. 0 00988-0. 000726=0. 000262 ms? ?I3=0. 00201-0. 00172=0. 000290 ms? 5 Error Analysis = 2 1(2 +2 ) + 2( + ) + = =? r= 0. 0005+0. 0005 ?r= 0. 26+0. 026 r x5. 1410-3=1. 7910-5 ?s=0. 00192 ?k=0. 176 ?I=1. 7910-50. 001920. 176=6. 0510-9 6 Graphical representationDisc 1. 1/s? 0. 35 0. 3 y = 1. 439x 0. 25 0. 2 0. 15 0. 1 0. 05 1/t? 0 0 0. 05 0. 1 0. 15 0. 2 0. 25 Weight kg 1/s? 0. 16 Disc 2. 0. 14 y = 0. 686x 0. 12 0. 1 0. 08 0. 06 1/t? 0. 04 0. 02 0 0 0. 05 0. 1 Weight 0. 15 0. 2 0. 25 kg 7 Disc 3. 1/s? 0. 07 0. 06 y = 0. 290x 0. 05 0. 04 0. 03 1/t? 0. 02 0. 01 0 0 0. 05 0. 1 Weight 0. 15 0. 2 0. 25 kg 8 Discussion From the results gained it can be concluded that bigger and heavier the disc is greater the moment of inertia of a body. As we can see the gradient on the graphs are larger at larger discs. From theoretical values which it can be concluded that experiment was right. more than time is taken to pass that distance for larger discs because the moment of inertia is bigger so it torque is required to accelerate the disc. However there were some random and systematic errors. One of the most effecting random errors is the clement reaction error. It could be decreased by using light gateway instead of stop watch. Using more accurate equipment for taking measurements of discs and spindle would decrease the error. Masses of the loads are not exact so more accurate loads would decrease the error. Conclusion Larger and heavier discs go larger moment of inertia so they require more torque to be accelerated.