If the linear ensity of a rod of length L varies as lambda = A+Bx, find the position of its centre of mass.

If the linear ensity of a rod of length L varies as lambda = A+Bx, fin

1.	If the linear ensity of a rod of length L varies as lambda = A+Bx, find the position of its centre of mass.
Answer» Solution :Let the x-axis be along the length of the ROD and ORIGIN at ONE of its ends. As rod is along x-axis, for all points on it y and Z co - ordinates are zero. Centre of mass will be on the rod. Now CONSIDER an element of rod of length dx at a distance x from the origin, then `dm = lambda dx = (A+Bx)dx` `X_(CM)=(int_(0)^(L) xdm)/(int_(0)^(L)dm)=(int_(0)^(L)x(A+Bx)dx)/(int_(0)^(L)(A+Bx)dx)` or `X_(CM)=((AL^(2))/(2)+(BL^(3))/(3))/(AL+(BL^(2))/(2))=(3Al+2BL^(2))/(3(2A+BL))=(L(3A+2BL))/(3(2A+BL))`