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A thin bar of length "L" has a mass per unit length "a" that increases linearly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is "b", then the distance of the centre of the mass from lighter end is ?
So this is how you do it. Imagine that the left end of the rod has a mass per unit length of b, which increases to the right end as a. Consider a small element dx at a distance x from the lighter end. let mass per unit length be ? now, ? a x. therefore, ? = kx. at the heavier end, ? = a. so, a = kL. or, k = a/L Now, centre of mass = 1/M (? xdm). dm = ?dx. dm = kxdx. therefore, xdm = x^2dx Integrating and taking limits 0 to L. ? xdm =k ? x^2dx = x^3 /3 So, simplifying, we get, distance of centre of mass = 1/M (a/L) (L^3/3) = a(L^2/3M). Since the mass of the rod is given: b becomes unnecessary.
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