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Three identical spheres , each of mass M , are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 2 m . Taking their point of intersection as the origin , find the position vector of centre of mass |
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Answer» SOLUTION :`x = (m_(1) x_(1) + m_(2) x_(2) + m_(3) x_(3))/(m_(1) + m_(2) + m_(3))` `therefore= ((M XX 0) + (M xx 0) + (M xx 2))/(M + M + M) = (2M)/(3 M) = (2)/(3) m` ![]() `therefore`x-coordination = `2//3 hati` Again `y = ((M xx 2) + ( M xx 0) + ( M xx 0))/(M + M + M)` or `y = (2M)/(3M) = (2)/(3)` m `therefore` y-coordinate = `2//3 hatj` `therefore` Position vector of CENTRE of mass= `(2)/(3) hati + (2)/(3) hatj = (2)/(3) (hati + hatj)` |
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