Assuming the Eq.(3) in the text is valid, what period of rotation of a mass of 0
ID: 2136696 • Letter: A
Question
Assuming the Eq.(3) in the text is valid, what period of rotation of a mass of 0.435 kg is to be expected for rotation radius of 0.200 m, if the outward pull which counters the spring inward pull for vertical mass-hanging string position at 0.200 m is delivered by a mass of 1.400 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 based on the precision of the available equipment in class: the pulling mass is measured by slotted weights with precision of 10 gram, the rotating mass is measured by a beam balance with precision of 0.1 gram, the rotating radius is measured by a meter stick with 1mm divisions, and the period of rotation is determined by measuring the time for 20 revolutions, with 0.5 s uncertainty of that time due to the human reaction at operating a stop watch.
Explanation / Answer
1)
Mg = 4pi^2(mr)/T^2
==> T = sqrt(4pi^2(mr)/Mg)
==> T = sqrt(4*3.1416*3.1416*0.435*0.200/(1.4*9.81))
==> T = 0.50008288
==> T = 0.5001 s
2)
left side:
Mg
==> fractional error = (10e-3/1.4) * 100 = 0.7143 %
Right side:
R = 4pi^2(mr)/T^2
==> dR = 0.01+0.001+0.05 = 0.061%