Question

If moment of inertia of a point particle at a distance r from an axis would have been defined as I=mr,then what will be the M.I of an uniform rod of mass M and length L about an axis passing through its centre and perpendicular to the rod?

Answer 1

`I = 1/12ML^2`

Explanation:

Little bit of calculus needed.

Let's consider a very tiny segment `dl` of the rod as `dx`.

You claim the rod is uniform. That means the mass density is not changing along the rod. If we denote the linear mass density as `\lambda`, then `\lambda = (dm)/(dl) = M/L`, where `M` is the mass of the rod and `L` is the length of the rod.

Furthermore, considering our small segment `dl = dx`, the differential relation `dm = M/L dx` will allow us to solve for the moment of inertia.

For a point, the moment inertia given is `I = mr^2`, not `I = mr`!!!

Moment of inertia is defined as:
`I = int_(solid) r^2 dm`

So for a point mass, this would become `I = mr^2`.

We can use the differential relation we found to convert to an integral over the length of the rod:

`I = int_(-L/2)^(L/2)x^2 M/L dx`

We consider half of the length of the rod on either side, because we are determining moment of inertia for the rod where the axis is through the center of the rod.

`I = M/L int_(-L/2)^(L/2)x^2 dx`

`= M/L[(x^3)/(3)]_(-L/2)^(L/2)`

`= M/(3L)[(L^3)/(8) + (L^3)/(8)]`

`= M/(3L)(2L^3)/(8) = 1/12ML^2`

Hence, `I = 1/12ML^2`