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

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.

If linear density of a rod of length 3m varies as lambda = 2 + x, them the position of the centre of gravity of the rod is

The linear density of a thin rod of length 1m lies as lambda = (1+2x) , where x is the distance from its one end. Find the distance of its center of mass from this end.

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 mass density of a rod of length 3 m (lying from y=0 to y=3 m) varies as lambda =(2+y)kg/m,then position of the centre of mass of the rod from y=0 is nearly at

A thin rod of length 6 m is lying along the x-axis with its ends at x=0 and x=6m. Its linear density *mass/length ) varies with x as kx^(4) . Find the position of centre of mass of rod in meters.

A thin rod of length 6 m is lying along the x-axis with its ends at x = 0 and x = 6m . It linear density (mass/length) varies with x as kx^(4) . Find the position of centre of mass of rod in meters.

The density of a linear rod of length L varies as rho=A+Bx where x is the distance from theleft end. Locate the centre of mass.

The linear mass density of a thin rod AB of length L varies from A to B as lambda (x) =lambda_0 (1 + x/L) . Where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicualr to the rod is :

A rod of length L is placed along the x-axis between x = 0 and x = L. The linear density (mass/length) lamda of the rod varies with the distance x from the origin as lamda = Rx. Here, R is a positive constant. Find the position of centre of mass of this rod.