A group of students, performing the same \"Uniform Circular Motion\" experiment

Question

A group of students, performing the same "Uniform Circular Motion" experiment that you performed in lab, obtained the following results.

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

  (ii) all calculations needed to compute the acceleration, a, for each trial

(iii) a graph of mhg vs ac

(iv) Use the trendline option to draw the best fit line and determine its slope.

(b) Report this value below.

mass of the bob =

I'm following all the directions but still cannot seem to get the right answer. So it would be helpful to have an explanation as well as the numerical answer. Thank you.

r (m) mh (kg) t (s) 0.1825 0.07685 95.13 0.2430 0.11194 86.17 0.2610 0.12238 85.41 0.2980 0.14384 86.68 0.3425 0.16965 83.10

Explanation / Answer

All we can help with in this forum is
(ii) acceleration a = v^2 / r
v = 50*2pi*r / t
so a = 10000*pi^2*r^2 / (t^2*r)
a = 10000pi^2*r / t^2
a = 98696*r / t^2

e.g. for the first trial
a = 98696*0.1825/ (95.13)^2 = 1.99 m/s^2

for the second trial
a = 98696*0.2430/ (86.17)^2 = 3.23 m/s^2

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