Question

To find moment of inertia of a solid cylinder about transverse (Perpendicular) axis passing through its centre.

Answer 1

The moment of inertia of a cylinder of mass$M$and radius$R$about its central axis is given as

$I_Z=\frac{1}{2}M{R}^2$

The moment of inertia about any axis parallel to the axis of centre of mass is given as

$I_{parallel}=I_{C.O.M.}+Mass\times {\left(dis\tan ce\right)}^2$

On integrating the above equation with the limit as $\frac{L}{2}to\left(-\frac{L}{2}\right)$ we get the moment of inertia of a cylinder having centre of axis about$z$direction as $d{I}_x=\frac{1}{4}dm{R}^2+dm{z}^2$

By substituting $dm=\rho \times Volumeofdisk=\frac{M}{L}dz$

We get the Moment of inertia as

$I_x=\frac{1}{4}M{R}^2+\frac{1}{12}M{L}^2$