iN parallel axis theo MI=sigma m .sq(d+x) =sigma m.sq(d) +sigma m.sq(x) + 2d sigma(x) =m.sq(d) + Icg +0 why is 2d sigma(x)=0

Asked by timesnow | 21st Nov, 2010, 11:22: PM

Expert Answer:

Dear student
A/C parallel axis theorem,
I = Ig + Md2
 I = Σm ( x2 + d2 + 2dh)
      = Σmx2  + Σmd   + Σ2mdh............(i)
Now if the body rotates about a given axis such that the chosen particle rotates in a circle of radius x;
Ig  = Σmx2
Σmd2 = d2 Σm = Md2
The term,  Σ2mxd = 2d Σmx
'Σmx' is a quantity which is proportional to the sum of moments of weights of various particles about the center of gravity G. Since a body can be balanced about its center of gravity ,
2d Σmx = 0
and substituiting this value in eqn (i) , we have
I = Ig + Md2
We hope this clarifies your doubt.

Answered by  | 23rd Nov, 2010, 01:33: PM

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