1) Assuming the Eq.(3) in the text is valid, what period of rotation of a mass o

Question

1) Assuming the Eq.(3) in the text is valid, what period of rotation of a mass of 0.365 kg is to be expected for rotation radius of 0.140 m, if the outward pull which counters the spring inward pull for vertical mass-hanging string position at 0.140 m is delivered by a mass of 0.680 kg (pan included)? Use 9.810 m/s^2 for the acceleration of gravity.

Eq.(3) F=4pi^2mr/T^2

2. If the quantities measured in order to verify Eq.(3) were obtained with values as in Problem 1, estimate the fractional error (%) on the left side and on the right side of the equation. Use the uncertainties originating from the precision of the measuring equipment: the pulling mass is measured by slotted weights with 10 gram increment (uncertainty +/- 5 gram), the rotating mass is measured by a beam balance with uncertainty 0.1 gram, the rotating radius is measured by a meter stick with 1mm divisions (uncertainty +/- 1 mm), and the period of rotation is determined by measuring the time for 10 revolutions, with 0.5 s uncertainty of that time due to the human reaction at operating of stop watch.

Explanation / Answer

1) Here,   F = 4pi2 * m * r/T2

=>   0.680 * 9.81 = 4 * 3.142 * 0.365 * 0.140/T2

=>   T   =   0.5496 sec

=>    period of rotation of a mass =   0.5496 sec

2)    the fractional error (%) on the left side   = (0.005/10) * 100 %

                                                                      =   0.05 %

       the fractional error (%) on the right side   = (0.001/10) * 100 %

                                                                      =   0.01 %

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