Error estimation of moment of inertia of a flywheel

[angelina]

now i get the formula I = mr^2 [n2/(n1+n2)] [(gt^2)/(2h) - 1]
where m = mass of the hanging wieght; r = radius; n1 = no. of revolutions before the weight reaches the floor; n2 = no. of revolutions before the flywheel stops; t = time taken for the weight to reach the floor; h = height of which the weight had dropped.

and I'm to calculate the % error of I.

- what is the error of n1 and n2? 1 revolution??
- in my practical book, before asking me to calculate the respective % error of all the variables, it states "since gt^2/2h >> 1, by approximation, % error of ... = ... ", what the actual meaning of this statement??
- the formula of % error of I stated is the book is 2x<r> + <n1> + <n2> + 2x<t> + <h>, where <*> = % error of *. but i think % error of n2 should be multiplied by a 2, is this a mistake of me or the book??

 

[haruspex]

it states "since gt^2/2h >> 1,

That means you can ignore the -1 in (gt^2)/(2h) - 1

i think % error of n2 should be multiplied by a 2

Although it occurs twice, it does not occur as n₂². E.g. consider n₂/n₂; the errors would cancel.
However, the sum in the denominator makes it rather tricky. There is no simple formula. If n₂>>n₁ then it would be <n₁>+<n₂>; if n₂<<n₁ then it would be 0. So <n₁>+<n₂> is worst case.