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 X_CM and Y-coordinate of centre of mass Y_CM 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 = v_1x

begin mathsize 14px style i with hat on top end style

+ v_1y

begin mathsize 14px style j with hat on top end style

;

Let mass m₂ is moving with velocity v₂ = v_2x

begin mathsize 14px style i with hat on top end style

+ v_2y

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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