Mass is non-uniformly distributed over the rod of length l, its linear mass density varies linearly with length as `lamda=kx^(2)`. The position of centre of mass (from lighter end) is given by-

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

Mass M is distributed over the rod of length L. If linear mass density (lambda) of the rod is linearly increasing with length as l = Kx, where x is measured from one end as shown in figure & K is contant. The moment of inertia. of the rod about the end perpendicular to rod where linear mass density is zero.

There is a rod of length l , its density varies with length as distance of lambda=ax , where x is distance from left end. Find center of mass from left end.

The density of a thin rod of length l varies with the distance x from one end as rho=rho_0(x^2)/(l^2) . Find the position of centre of mass of rod.

The density of a thin rod of length l varies with the distance x from one end as rho=rho_0(x^2)/(l^2) . Find the position of centre of mass of rod.

If linear density of a rod of length 3m varies as lamda=2+x , then the position of the centre of mass of the rod is P/7m . Find the value of P.

A thin bar of length L has a mass per unit length lambda , that increases linerarly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is lambda_(0) , then the distance of the centre of mass from the lighter end is