Question

State an expression for moment of inertia of a solid cylinder of uniform cross section about
(i) an axis passing through its centre and perpendicular to its length.
(ii) its own axis of symmetry.

Answer 1

moment of inertia of solid cylinder of uniform cross section about an axis passing through its centre and perpendicular to its length.

cut an element dx, x distance from the centre of cylinder. let radius of cylinder is R.
then, mass of element , $dm=\frac{m}{\pi R^2l}\times\pi R^2dx$
=$\frac{m}{l}dx$

cutting element seems as disc.
so, moment of inertia of element, I = $\frac{dm. R^2}{2}$
so, moment of inertia about perpendicular to its axis. $I_1=\frac{dm.R^2}{4}$

now moment of inertia of solid cylinder , $I=I_1+\frac{m}{l}dx.x^2$

$I=\frac{m}{4l}\int\limits^{l/2}_{-l/2}{dx}+\frac{m}{l}\int\limits^{l/2}_{-l/2}{x^2}\,dx$

= $\frac{m}{12}(2R^2+l^2)$


now, moment of inertia of solid cylinder about its own axis of symmetry.
cut an element , dx at x distance from the centre of cylinder.
element seems as disc .
so, moment of inertia of disc , $I=\frac{dm.R^2}{2}$

$=\frac{R^2}{2}\int\limits^m_0{dm}$

= $\frac{1}{2}mR^2$