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CBSE Class 11-science Answered

find the centre for mass for given system of a mass
Asked by js05 | 01 Feb, 2021, 08:51: PM
Expert Answer
Let us assume the masses are placed at the vertex of equilateral triangle of side a.
 
Let base of triangle is parallel to x-axis .
 
X-coordinate of centre of mass  XCM  and Y-coordinate of centre of mass YCM are given as
 
begin mathsize 14px style X subscript C M end subscript space equals space fraction numerator open parentheses 4 cross times 0 space plus space 4 cross times a cos 60 space plus space 3 cross times a close parentheses over denominator open parentheses 4 plus 4 plus 3 close parentheses end fraction space equals space fraction numerator 5 a over denominator 11 end fraction end style  ;   begin mathsize 14px style Y subscript C M end subscript space equals space fraction numerator open parentheses 4 cross times 0 space plus space 4 cross times a sin 60 space plus space 3 cross times 0 close parentheses over denominator open parentheses 4 plus 4 plus 3 close parentheses end fraction space equals space fraction numerator 2 square root of 3 a over denominator 11 end fraction end style
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Part-(b)
 
Let mass m1 is moving with velocity v1 = v1x begin mathsize 14px style i with hat on top end style + v1y begin mathsize 14px style j with hat on top end style ;
Let mass m2 is moving with velocity v2 = v2x begin mathsize 14px style i with hat on top end style + v2y begin mathsize 14px style j with hat on top end style ;
similarly for other masses
 
X-coordinate and Y-coordinate of centre of mass are given as
 
begin mathsize 14px style X subscript C M end subscript space equals space fraction numerator begin display style sum from i equals 1 to n of end style m subscript i space x subscript i over denominator begin display style sum from i equals 1 to n of end style m subscript i end fraction space space space semicolon space space Y subscript C M end subscript space equals space fraction numerator begin display style sum from i equals 1 to n of m subscript i space y subscript i end style over denominator begin display style sum from i equals 1 to n of m subscript i end style end fraction space space space end style......................... (1)
velocity components of centre of mass are obtained by differentiating eqn.(1)
 
begin mathsize 14px style fraction numerator d over denominator d t end fraction X subscript C M end subscript space equals V subscript x C M end subscript space equals space space fraction numerator begin display style sum from i equals 1 to n of end style m subscript i space begin display style fraction numerator d over denominator d t end fraction end style x subscript i over denominator begin display style sum from i equals 1 to n of end style m subscript i end fraction space space equals space fraction numerator begin display style sum from i equals 1 to n of m subscript i space v subscript x i end subscript end style over denominator begin display style sum from i equals 1 to n of m subscript i end style end fraction space space space semicolon
space fraction numerator d over denominator d t end fraction space Y subscript C M end subscript space equals V subscript y C M end subscript space equals space fraction numerator begin display style sum from i equals 1 to n of m subscript i space fraction numerator d over denominator d t end fraction y subscript i end style over denominator begin display style sum from i equals 1 to n of m subscript i end style end fraction space space equals space space fraction numerator begin display style sum from i equals 1 to n of m subscript i space v subscript y i end subscript end style over denominator begin display style sum from i equals 1 to n of m subscript i end style end fraction
end style
 
 
Answered by Thiyagarajan K | 01 Feb, 2021, 10:51: PM
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