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ques no 41

Asked by chandanbr6004 26th November 2017, 10:24 AM
Answered by Expert
Answer:
 
Let us consider small area of half disc as shown in figure whose radius is r and width dr.
 
mass of this shaded portion,  m = ρ × Π × r × dr
 
where ρ is mass per unit area.
 
moment of inertia of this shaded area of disc about the axis passing through the centre of disc is mr2 = ρ × Π × r × dr × r2
 
Hence moment of inertia of whole disc is given by

begin mathsize 12px style I space equals space integral subscript R 1 end subscript superscript R 2 end superscript rho pi r cubed d r space equals space fraction numerator rho pi open parentheses R subscript 2 superscript 4 minus R subscript 1 superscript 4 close parentheses over denominator 4 end fraction space equals space fraction numerator rho pi open parentheses R subscript 2 superscript 2 minus R subscript 1 superscript 2 close parentheses open parentheses R subscript 2 superscript 2 plus R subscript 1 superscript 2 close parentheses over denominator 4 end fraction end style
mass of the given disc M is begin mathsize 12px style rho pi open parentheses R subscript 2 superscript 2 minus R subscript 1 superscript 2 close parentheses end style
Hence begin mathsize 12px style I space equals space M over 4 open parentheses R subscript 1 superscript 2 plus R subscript 2 superscript 2 close parentheses end style
Answered by Expert 27th November 2017, 2:54 PM
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