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Error estimation of moment of inertia of a flywheel

  1. Dec 10, 2004 #1
    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??
     
  2. jcsd
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