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I know the answer to the first one, however I keep getting different answers for

ID: 2113974 • Letter: I

Question

I know the answer to the first one, however I keep getting different answers for the second problem.



1. 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.


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

2.

Fractional error = ( error in m) + (error in r) + 2*(error in T)

= (0.1/435)*100 + ( 1/200)*100 + 2*(0.025/157.86)*100

= 0.023 + 0.5 + 0.032

= 0.555 percent

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