Question

Calculate the moment of inertia of thin rod about the following axis of rotation :
1. About an axis passing through its centre and perpendicular to its length.
2. About an axis passing through its one end and perpendicular to its length.

Answer 1

Dear student.

Let us solve this by integration

let an elemental mass dm of length dx be located at a distance x from centre of rod.

Length of rod = L and mass = m

Mass per unit length of rod = m/l

Mass of dm = (m/l) x dx

MI of dm about centre of rod = $dm{x}^2$
1)...MI of rod about the central axis =

$$\begin{gathered} {\int }_{-\frac{l}{2}}^{\frac{l}{2}}dm{x}^2 \\ {\int }_{-\frac{l}{2}}^{\frac{l}{2}}\frac{m}{l}dx{x}^2 \\ \frac{m}{l}\left[\frac{x^3}{3}\right];-l/2tol/2 \\ =\frac{m{l}^2}{12} \end{gathered}$$


2)...MI of rod about pivotal axis, perpendicular to one end.


$$\begin{gathered} {\int }_0^{l}dm{x}^2 \\ {\int }_0^l\frac{m}{l}dx{x}^2 \\ \frac{m}{l}\left[\frac{x^3}{3}\right];0tol \\ =\frac{m{l}^2}{3} \end{gathered}$$


Regards