Moment of Inertia of the Cylinder about the Axis Perpendicular to the Length of the Cylinder and Passing Through the Center
According to the figure a cylinder of mass M, length l and radius R is shown; which is made up of coaxial discs of radius R. We have to calculate the moment of inertia about the YY’ axis perpendicular to XX’ and passing through the centre O. If there is a disc of width dx and at a distance x from the YY’ axis then;
Volume of the disc = Area × thickness = πR2dx
If ρ is the density of the cylinder, then ρ = \(\frac{M}{\pi R^{2} l}\)
dm = (Volume) × mass per unit volume of the cylinder
Therefore, mass of the disc (dm) = πR2dxρ
Fig: Moment of inertia with respect of perpendicular axis along the length of the cylinder
We know the moment of inertia of this disc about the diameter PQ;
\((dI)_{PQ} = \frac{1}{4}dmR^2
\)
\(= \frac{1}{4}\frac{M}{l}dxR^2\)
By the theorem of parallel axis, the moment of inertia of this disc about the YY’ axis.
The moment of inertia of the whole cylinder about the YY’ axis will be equal to the sum of moment of inertia of all these discs which are between