In: Physics
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.
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.
Eq.(3) --> F= 4pi^2(mr)/T^2
F = 4(pi^2)*mr/(T^2)
F = mg = 0.680 * 9.810
force equal so
0.680 * 9.810 = 4(pi^2) * .365 x .140 /T^2
T = sqrt(4pi^2 * .365 *.140 / (.680*9.810))
T = 0.5499 s
part b )
fractional error = (Actual - experimental)/actual
set m = 1.001kg difference of 1 gram = .001 kg actual 1.000 kg
set r = 1.001 m difeerence of 1 mm = .001 m actual 1.000 m
set T to be (0.5s/10) to determine 1 revolution difference = 0.5/10 rev
Right side
F = 4pi^2 mr/T^2
Fractional error = [(4pi^2*1.001*1.001 / ((0.5/10)^2)) - 4pi^2*1*1/(1/10)^2) ] / (4pi^2*1*1/(1/10)^2))
Fractional error = (400.8004 - 100 )/100
Fraction error = 3.008004
for left side
I set m = 1.001 kg, difference of 1 gram = .001 kg, actual put 1.000kg.
I set meters = 9.811, difference of 1 mm = .001 meter, actual i put 9.810
Time I put 0.5 s, difference of 0.5 s = 0.5 s, actual I put 1 s
F=mg(1.001kg * 9.811 m/ (0.5s)^2 - 1.000kg & 9.810m/ (1s)^2) / 1.000kg & 9.810m/ (1s)^2
Fractional Error = 3.004408155 = ~ 3