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??