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

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

1.	If the linear density 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))`