A straight rod of length L has one of its end at the origin and the other at `X=L`. If the mass per unit length of the rod is given by `Ax` where A is constant, where is its centre of mass?

A straight of length L extends from x=a to x=L+a. the gravitational force it exerts on a point mass 'm' at x=0, if the mass per unit length of the rod is A+Bx^(2), is given by :

A light rod of length l has two masses m_1 and m_2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is.

A rod of mass m and length L is non- uniform.The mass of rod per unit length (mu)], varies as "(mu)=bx" ,where b is a constant.The , centre of mass of the ,rod is given by

A straight rod of length 3l units slides with its ends A,B always on the x and y axes respectively,then the locus of the centriod of Delta OAB is

A non–uniform thin rod of length L is placed along x-axis as such its one of ends at the origin. The linear mass density of rod is lambda=lambda_(0)x . The distance of centre of mass of rod from the origin is :

The centre of mass of a non-uniform rod of length L whose mass per unit length lambda is proportional to x^2 , where x is distance from one end

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

A thin rod AB of length a has variable mass per unit length P_(0) (1 + (x)/(a)) where x is the distance measured from A and rho_(0) is a constant. Find the position of centre of mass of the rod.