Question

A group of students, performing the same "Uniform Circular Motion" experiment that you performed in lab, obtained the following results. r(m) mh (kg) t (s) 0.1825 0.10335 62.11 0.2430 0.15054 56.99 0.2610 0.16458 56.49 0.2980 0.19344 55.68 0.3425 0.22815 54.96 For this table, r is the distance from the center of rotation to the radius indicator (i.e., the post that marks the position of the center of the bob during rotation), mh is the total hanging mass (including the hanger), and t is the time required for 50 complete rotations (a) Use Excel to construct a spreadsheet that show the following. (You will not submit this spreadsheet. However, the results will be needed later in this problem.) (i) the above data HINT (i) all calculations needed to compute the acceleration, a, for each trial HINT (iii) a graph of mhg vs ac (iv) Use the trendline option to draw the best fit line and determine its slope HINT (b) Report this value below. mass of the bob =

Answer 1

trial 1 = 0.1825 m, m = 0.106 kg, t(s) = 62.37 sec. Time period - 218. uva 2 - 2 211 x 0.1825 62.37 - 0.018375 m/s. acceleration ar CO-01 8375) = 0.00185 rad/52 0.1825 toial 2 Na 2018 2ix 0.2430 - 59. 85 = 0.02549 N m/s. (0.0254902 0.2430 - 0.002673 rad /s2 trials 211 x 0.2610 DE 2+ = = 0.0276 ms 59.37 (0.027652 0.2610 -0.002918 rad /s2 Trial 4 : 2 x 0.2980 53.58 = 0.0349 mis. - COD 349) D. 2980 - - 0.00408 sad 1.2 Trials : 2 T X 0.3425 57.89 = 0.087155 nys a = Co.0371552 0.3425 = 0.00403 Bad 152

mug 0100185 1002673 0.106 x 9.8 = 1.0398 151312 1.65424 194432 2.2932 0.002918 0.00408 0.00403 Im Slope - 17 x0.00 045. 1940.018 = 0.02236

0.00408 set 0.00363 0.00319 0.002741 0.0023- 0.001865 HV 1.04 1.22 1.58 1.76 1.94 mg