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## Q. 20.9

Find the moment of inertia of a uniform solid cylinder of mass M, radius r and height h about an axis which is perpendicular to the axis of the cylinder and which passes through the centre of one end.

## Verified Solution

Choose axes as shown in figure 20.24 so that the required moment of inertia is about the x axis.
Let ρ be the density of the cylinder, so that M = πr²hρ.
Subdivide the cylinder into elementary discs as shown.

A typical disc has thickness δy and centre at C(0, y).

Its mass δm is approximately ρπr²δy.

The moment of inertia of the element about an axis through C parallel to the x axis (i.e. a diameter) is $\frac{1}{4}δmr².$

C is the centre of mass of the disc, so by the parallel axes theorem, its moment of inertia about the x axis is

$\frac{1}{4}δmr² + δmy² = \frac{1}{4}(ρπr²δy)r² + (ρπr²δy)y²$

$= \frac{1}{4}ρπr²(r² + 4y²)δy.$

For the whole cylinder, the moment of inertia about the x axis, $I_{x}$, is approximately

$\sum{\frac{1}{4}ρπr²(r² + 4y²)δy = \frac{1}{4}ρπr²} \sum{(r² + 4y²)δy.}$

In the limit as δy → 0, this gives

$I_{x} = \frac{1}{4}ρπr² \int_{0}^{h}{(r² + 4y²)dy}$

$= \frac{1}{4}ρπr²\left[r²y + 4\frac{y^{3}}{3}\right]^{h}_{0}$

$= \frac{1}{4}ρπr^{4}h + \frac{1}{3}ρπr²h^{3}$

$= \frac{1}{4}Mr^{2} + \frac{1}{3}Mh^{2}$         (M = ρπr²h).