Assuming the Eq.(3) in the text is valid, what period of rotation of a mass of 0

Question

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

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.