## Chapter 20

## 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.

## Step-by-Step

## 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).