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CBSE Class 12-science Answered

pls explain
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Asked by mufeedatvp2000 | 18 Apr, 2020, 02:22: PM
answered-by-expert Expert Answer
Let us find the centre of mass of semi circular ring of radius R .
From the given figure above, if we consider origin of X-Y coordinate system is at center of semi-circular ring,
then centre of mass is in y-axis by symmetry, i.e. x-coordinate Cx of centre of mass is zero.
 
y-coordinate Cy of centre of mass is given by
 
begin mathsize 14px style C subscript y space equals space 1 over M integral d m space y space equals space 1 over M integral subscript 0 superscript theta rho space R d theta space R sin theta space equals fraction numerator rho space R squared over denominator M end fraction space integral subscript 0 superscript theta sin theta space d theta space equals space fraction numerator 2 space rho space R squared over denominator M end fraction end style  ........................(1)
where M is mass of semi-circular ring, dm is mass of a small element of length dl .
 
Let radial line connecting the element dl makes angle θ with x-axis and dθ be angle subtended  by element dl at centre.
 
ρ is mass per unit length of wire.  Mass M of wire is obtained as M = ρ ( π R )
 
Using the substitution  ( ρ/M ) = 1/( π R ) ,  equation (1) for y-coordinate of centre of mass becomes
 
Cy = 2R/π
 
Moment of inertia I about centre of mass is given by
 
begin mathsize 14px style I space equals space integral d m space r squared space equals space integral subscript 0 superscript pi R space d theta space rho space open curly brackets R squared cos squared theta space plus space open parentheses R space sin theta space minus space fraction numerator 2 R over denominator pi end fraction close parentheses squared close curly brackets end style
we get from above integration, begin mathsize 14px style I space equals space M space open curly brackets 1 space minus space 4 over pi squared close curly brackets R squared end style
Answered by Thiyagarajan K | 19 Apr, 2020, 07:54: AM

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