physics Help………………..?

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A & B)http://oi43.tinypic.com/23vjgas.jpgC)http://oi41.tinypic.com/2znu88o.jpgTHANKS

Everardo Feil · Accepted Answer

They asked you to do the problem by integration.Here is how that is done.Consider the rod divided into thin slices of thickness dlthe mass of each slice is M/L dland the moment of inertia of the slice is M/L * x^2 dl where x is the distance from the pivot.There are two portions of the rod. One side of the point d has a length of d and the other has a length of (L-d)So we need to integrate each portion.integral M/L x^2 dl is 1/3 M/L x^3now for one part x increases from 0 to d and the other part from 0 to (L-d)hence the moment of intertia is1/3 M/L d^3 + 1/3 M/L (L-d)^3 (equation 1)The right hand term could be expanded to produce a power series if desired.Now where d = 0d^3 = 0 so the first term is zero(L-d) = Lso the whole equation reduces to 1/3 M/L * L^3= 1/3 ML^2 which is exactly what is expected. (derived equation 2)And where d = L/2d^3 = (L/2)^3 = L^3 /8(L-d)^3 = (L-L/2)^3 = (L/2)^3 = L^3/8hence the moment of inertia = 1/3 M/L L^3 /8 +1/3 M/L L^3 /8=1/12 ML^2 (derived equation 3)Now I don't have your book or table 12.2 but from one of these formulae you would be expected to substitute values and confirm that the result of the formula agrees with the table in your book.

Everardo Feil · Answer

M(d^2-dL+L^2/3)1/12*ML^21/3*ML^2

Mr. Durward Cummerata

physics Help………………..?

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