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CBSE Class 11-science Answered

kindly answer

Asked by arup.isro | 19 Dec, 2018, 10:54: PM
Expert Answer
Let us first get the Centre of mass of semicircular plate.
As shown in figure,  let us divide the semi circular plates into strips of width dr parallel to diameter.
Let us consider a strip of length 2l at a distance r from center as shown in figure.
Mass of this strip = ρ×2l×dr , where ρ is density which is expressed as mass per unit area.
 
Center of mass of semi-circular plate : begin mathsize 12px style y space equals space fraction numerator integral subscript 0 superscript R rho space cross times space 2 l space cross times space r space d r over denominator M end fraction space equals space fraction numerator integral subscript 0 superscript R rho cross times 2 square root of R squared minus r squared end root space r d r over denominator M end fraction space equals space fraction numerator rho cross times begin display style 2 over 3 end style R cubed over denominator rho cross times begin display style 1 half end style πR squared end fraction space equals space fraction numerator 4 over denominator 3 straight pi end fraction R end style
Let A be the centre of mass of rectangular plate,  B is the centre of mass of semicircular plate and
P is the center of mass of plate after cutting the semi-circle.
 
we have OA = r/2  and OB = [ 4/(3π) ] r
 
Center of mass of plate after cutting semi-circle is calculated as follows
 
mA × OA = mB × OB + mP × OP
 
where mA is mass of full rectangular plate = ρ×2r2
mB is mass of semi-circular plate = ρ×(π/2)×r2 
mP is mass of plate after cutting semi-circle = ρ× [ 2 - (π/2) ]×r2
 
Hence begin mathsize 12px style O P space equals space fraction numerator m subscript A cross times O A space minus space m subscript B cross times O B space over denominator m subscript P end fraction space equals space fraction numerator rho cross times 2 r squared cross times open parentheses begin display style r over 2 end style close parentheses space minus space rho cross times open parentheses begin display style straight pi over 2 end style close parentheses r squared cross times open parentheses begin display style fraction numerator 4 over denominator 3 straight pi end fraction end style close parentheses r over denominator rho cross times open parentheses 2 minus begin display style straight pi over 2 end style close parentheses cross times r squared end fraction space equals space fraction numerator begin display style 2 over 3 end style r over denominator open parentheses 4 minus straight pi close parentheses end fraction end style
Answered by Thiyagarajan K | 20 Dec, 2018, 12:06: PM
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