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Q.
A solid cylinder has mass ‘$M$’, radius ‘$R$’ and length ‘$l$’. Its moment of inertia about an axis passing through its centre and perpendicular to its own axis is
Let, $X X'$ be the axis of symmetry and $Y Y'$ be the axis perpendicular to $X X'$. Let us consider a circular disc $S$ of width $d x$ at a distance $x$ from $Y Y'$ axis. Mass per unit length of the cylinder is $\frac{M}{I} .$ Thus, the mass of disc is $\frac{M}{I} d x$
Moment of inertia of this disc about the diameter of the rod
$=\left(\frac{M}{I} d x\right) \frac{R^{2}}{4}$
Moment of inertia of disc about $Y Y'$ axis given by parallel axes theorem is
$=\left(\frac{M}{I} d x\right) \frac{R^{2}}{4}+\left(\frac{M}{I} d x\right) x^{2}$
$\therefore $ Moment of inertia of cylinder,
$I=\int \limits_{-I / 2}^{I / 2}\left(\frac{M}{I} d x\right) \frac{R^{2}}{4}+\int \limits_{-I / 2}^{I / 2}\left(\frac{M}{I} d x\right) x^{2}$
$=\frac{M}{I}\left[\int \limits_{-I / 2}^{I / 2}\left(\frac{R^{2}}{4}+x^{2}\right) d x\right]=\frac{M}{I}\left[\frac{R^{2} x}{4}+\frac{x^{3}}{3}\right]_{-I / 2}^{I / 2}$
$\Rightarrow \,\,\, I=M\left[\frac{R^{2}}{4}+\frac{I^{2}}{12}\right]$
