Question

A cylindrical rod of mass $M$ and length $L$ is hinged about an end to swing freely in a vertical plane. However, its density is non uniform and varies linearly from hinged end to the free end doubling its value. What is moment of inertia of the rod about the rotation axis passing through the hinge point?

Answer 1

Given: a cylindrical rod of mass M and length L is hinged about an end to swing freely in a vertical plane. However, its density is non uniform and varies linearly from hinged end to the free end doubling its value.

To find the moment of inertia of the rod about the rotation axis passing through the hinge point

Solution:

Here as rod doesn`t have uniform density. We have to integrate the moment of inertia

$I=\int r^{2}dm$

where $dm=(1+\frac{x}{L})dx$ is mass of a section at a distance x from the hinge

Hence

On integrating we get,

$tI={\int }_0^L\left(x^{2}dm\right)⟹I={\int }_0^L(x^2\times (1+\frac{x}{L})dx)⟹I={[\frac{x^3}{3}+\frac{x^4}{4L}]}_0^L⟹I=[\frac{L^3}{3}+\frac{L^4}{4L}]-[0+0]⟹I=\frac{4L^3+3L^3}{12}⟹I=\frac{7}{12}{L}^3$

is the moment of inertia of the rod about the rotation axis passing through the hinge point.