A Rotational motion experiments

A Rotational motion experiment is the simplest method of finding the Moment of Inertia. Minimum equipment is required to perform this experiment. For the purposes of increasing the accuracy of the results, the procedure should be repeated three times, making our conclusion more reliable. While load is moving downwards it’s potential energy converts to kinetic. Load is accelerating because weight(Fg=mg) of the load is bigger than tension on a string so load is not in equilibrium and by Newton’s Second Law (F=ma) resultant force creates an acceleration. Resultant force can be calculated by the equation S=0.

5at2+ut to find acceleration and F=ma. String rotates the spindle which rotates the disc by creating a ?? torque(T=Fr). Torque accelerates the disc and it can be found by ? = ??. To find moment of inertia now T=I? equation is used. 1 Method and observation: Apparatus: 3 different size discs, spindle, ruler, set of weights, stopwatch, stand. Disc is attached to one end of the spindle and string with load is attached to the other end. Disc’s weight, diameter and radius are required to be measured before experiment. Length of the string (L), number of loops on the spindle (n) and horizontal distance of loops (H) were measured before experiment.

Using equation below r is found. ?? = ??????? 2???? Spindle Disc String Stopwatch Weights Stand After setting all the equipment up the experiment starts. The string is then wrapped around the spindle. Time was measured for load pass the distance of length of the string. To plot graph 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 reduce 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 observed the string with vibrating and load was moving a little which can cause some systematic error. While spindle is spinning there 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. 14×10-3 DISC 1 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 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) Average Time (s) 1/t? (s-2) K (m s? ) I (kg m2) experimental I (kg m2) theoretical ?I (kg m2) 0. 1 6. 49 | 6. 16 | 6. 33 6. 33 0. 0250

0. 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.

14×10? 3 )? I1e= 2×1. 439×0. 26 =0. 000346ms? 9. 8x(5. 14×10? 3 )? I2e= 2×0. 686×0. 26 =0. 000726ms? 9. 8x(5. 14×10? 3 )? I3e= 2×0. 290×0. 26 =0. 00172 ms? I1t=0. 5×0. 052×0. 314=0. 000393 ms? I2t=0. 5×0. 06352×0. 490=0. 000988 ms? I3t=0. 5×0. 07602×0. 696=0. 00201 ms? ?I1=|0. 000393-0. 000346|=0. 000047 ms? ?I2=|0. 000988-0. 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. 14×10-3=1. 79×10-5 ?s=0. 00192 ?k=0. 176 ?I=1. 79×10-5×0. 00192×0. 176=6. 05×10-9 6 Graphical representation

Disc 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 larger 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 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 human reaction error. It could be decreased by using light gate 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 have larger moment of inertia so they require more torque to be accelerated.