# Calculating Mss of a Rod using Integration

1. Feb 26, 2012

### mm391

1. The problem statement, all variables and given/known data

Let a rod with length L and constant cross-sectional area A have the density
ρ(x) = ρ((3.x^2 + 2.x.L)/L^2)

0 ≤ x ≤ L

where x is the distance from one end of the rod and ρ0 is a real constant.
(a) Find the total mass M of the rod.
(b) Find the x-coordinate xc of the centre of mass of the rod.
(c) Find, in terms of M, the moment of inertia of the rod about the vertical axis through x = 0.

2. Relevant equations

M = ∫Aρ(x).dx

3. The attempt at a solution

M = ∫Aρ(x).dx

M= A.ρ∫((3.x^2 + 2.x.L)/L^2).dx

I am slightly confused where to go from here! Do I use integration by parts?

2. Feb 26, 2012

### HallsofIvy

Staff Emeritus
Why? That is just the integral of a polynomial. What is the integral of $3x^2 dx$? What is the integral of $2x dx$?