Question

A group of students, performing the same "Uniform Circular Motion experiment that you performed in lab, obtained the f following results r (m) linh (kg) t (s) 0.1825 0.08215 69.67 0.2430 0.11966 63.35 0.2980 015376 83.92 0.3425 0.18135 60.38 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), m is the total hanging mass (including the hanger), and t is the time required for 50 complete rotations (a) usa Excal tn canstruct a spreadsheet that show the following. (You will not submit this spreadsheat. Howevar, tha rnsults will be nAAdad later in this prohlem.) (the above data (ii) all calcul eded to compute the acceleration, , for each trialHNT ii) a graph of vs HINT (iv) Use the trandline option to draw tha hast fit lina and detarmine Its slope. T (b) Report this value below. mass of the bob-003316 ka-%

Answer 1

NOTE - VERY VERY IMPORTANT

THE VALUES IN THE GIVEN TABLE ARE NOT VERY CLEAR. SOME VALUES ARE HALF VISIBLE.  I HAVE TAKEN UTMOST CARE TO READ CORRECT VALUES. STILL IF THERE IS ANY MISTAKE, PLEASE LET ME , I WILL CORRECT IT.

PLEASE DO NOT GIVE THUMBS DOWN WITHOUT ANY REASON. MY METHOD IS 100 % FINE.

The a_c (centripetal acceleration ) is found as

w = 2$\pi$ / T

where T = t/50

w = 100$\pi$ / t

a_c = w²r

a_c = 10000$\pi$² r / t²

r	t	mhg	ac
0.1825	89.67	0.80507	2.2401
0.2430	83.35	1.172668	3.4521
0.2610	82.61	1.282036	3.7746
0.2980	83.92	1.506848	4.1762
0.3425	80.38	1.77723	5.2319

Now, plot a graph between m_hg (ON Y AXIS) and a_c ( ON X AXIS)

Here is the graph

2.2401 0.80507 3.4521 1.172668 3.7746 1.282036 4.1762 1.506848 52319 1 77723г 1.6 Linear Fit for: Data Set IY m (Slope)0.3329 b (Y-Intercept): 0.05218 Correlation: 0.9943 RMSE: 0.04478 1.4- 1.2

as we can see that

slope = 0.3329

therefore,

mass of the bob = 0.3329 Kg

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PLEASE LET ME KNOW IF YOU HAVE ANY DOUBT. THERE IS NO MISTAKE IN THE METHOD.