A uniform rod of mass `m` and length `l_(0)` is rotating with a constant angular speed `omega` about a vertical axis passing through its point of suspension. Find the moment of inertia of the rod about the axis of rotation if it make an angle `theta` to the vertical (axis of rotation).

Options

A. `(ml_0^2 sin^2 theta)/(12)`
B. `(ml_0^2 sin^2 theta)/(6)`
C. `(2 ml_0^2 sin^2 theta)/(3)`
D. `(ml_0^2 sin^2 theta)/(3)`

Answer

D

Explanation

(d) We can observe each and every element of rod in rotating with different radius about the axis of rotation.
Take an elementary mass `dm` of the rod.
`dm = (m)/(l_0) dl`
The moment of inertia of the elementary mass gives as `d I = (dm) r^2`
The moment of inertia of the rod
=`I = int d I rArr I = int r^2 dm`
Substituting `r = 1 sin theta and dm = (m)/(l_0). dl`, we obtain
`I = int(l^2 sin^2 theta) (m)/(l_0) d I`
=`(m sin^2 theta)/(l_0) int_0^(l_0) l^2 d l = (ml_0^3)/(3 l_0) sin^2 theta`
`rArr I = (m l_0^2 sin^2 theta)/(3)`.
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