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A circular ring of mass 6 kg and radius a is placed such that its centre lies at the origin.two particles of masses 2 kg each are placed at the

intersecting points of the circle with positive x and y axis.find angle made by position vector of centre of mass of entire system


X CM = ma/2m = a/2 ; Y CM = ma/2m = a/2 ; tan theta = X CM/Y CM = (a/2)/(a/2)=1 So, theta = 45 degree
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After\ placing\ the\ masses;\\ \\ x-coordinate\ of\ CM= \frac{2*a+2*0+6*0}{2+2+6} = \frac{2a}{10} =\frac{a}{5} \\  \\ y-coordinate\ of\ CM= \frac{2*0+2*a+6*0}{2+2+6} = \frac{2a}{10} =\frac{a}{5} \\  \\ position\ of\ CM\ is\ \boxed{( \frac{a}{5} , \frac{a}{5} ) }
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