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anonymous

  • one year ago

1. Modulus of rigidity G=(R^4) T/L where R is the radius, T- theta the angle of twist and L the length a) Determine the approximate percentage error in G when R is increased by 2%, T is reduced by 5% and L is increased by 4% b) Solve a) by another method.

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  1. anonymous
    • one year ago
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    For a) i can solve by using small increment formula (partial differential) but for question b i cannot think of other method.

  2. IrishBoy123
    • one year ago
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    the simple and accurate way, & poss what they wanted for part a) \[\frac{1.02^4 \times 1.05}{1.04} - 1\] these deltas are far too big for a total differential ... which i assume is the sought b) solution

  3. anonymous
    • one year ago
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    can i ask why you substitute 1.02,1.05 and 1.04 inside the variable and minus with one.

  4. IrishBoy123
    • one year ago
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    you are increasing them (R,T,L) by those percentages, right?

  5. anonymous
    • one year ago
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    \[\delta G=\frac{ dG }{ dR } \delta R+\frac{ dG }{ dL } \delta L +\frac{ dG }{ dT } \delta T\] when i substitute all the information given, i get the aswe as G decrease by 1%

  6. anonymous
    • one year ago
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    i dont think we can simply substitute the given percentage.. thes the question said R increased by 2% meaning 0.02R not just 0.02.

  7. IrishBoy123
    • one year ago
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    ok, i think we are in agreement but maybe this will show how. what i wrote was shorthand for \[\Delta G = \frac{(1.02R)^4 (0.95T)}{(1.04L)} - \frac{R^4 T}{L} = -0.0112 G\] NB: i had a typo in previous version as T decreases and not increases, but this is the way to do it using differentials gives ∆G = -0.01 G, so we do agree the first method is most accurate and simplest. PS i know i haven't helped by misreading the question; but the principle still holds

  8. anonymous
    • one year ago
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    yaaaa....now i understand...thanks for your response..i really appreciated..

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