Question

A rod of length 1.00m has a mass per unit length given by λ = 2.00 + 1.00x^2, where λ is in kg/m. The rod is placed on the x axis going from x = 0.00 m to x = 1.00 m. What is the moment of inertia of the rod in kg•m^2 about the y axis?

Answer 1

`I=(13)/(15) =0.867\ kg.m^2`

Step-by-step explanation:

Given:

• length of rod,
  `l=1\ m`

• mass density of rod,
  `\lambda=2+x^2\ kg.m^(-1)`

• initial point of rod length,
  `x_0=0\ m`

• final point of rod length,
  `x=1\ m`

We know, moment of inertia from its basic definition:

`I=\int\limits^x_(x_0) {x^2}\lambda \, dx`

`I=\int\limits^1_0 {x^2}(2+x^2) \, dx`

`I=\int\limits^1_0 (2x^2+x^4) \, dx`

`I=[(2)/(3) x^3+(x^5)/(5) ]\limi_0^1`

`I=(13)/(15) =0.867\ kg.m^2`