Sanchayan Ghosh
Last Activity: 10 Years ago
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.
