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## Q. 20.3

Calculate the moment of inertia of a uniform circular disc of radius r, thickness t and mass M about an axis through its centre perpendicular to its plane.

## Verified Solution

$\boxed{\text{Decide on appropriate elements.}}$

The disc is divided into elementary rings. A typical ring has radius x, width δx and thickness t. Its volume is approximately (2 πx δx) × t = 2 πtx δx.

It is useful to use the density, ρ, of the disc so that the mass of each part can be obtained.
It can be written in terms of M at the end of the calculation.

The mass of the ring is then approximately

$\begin{matrix} \boxed{\text{Find the mass of a typical element in terms of the density.}} & \delta m = (2\pi tx\delta x)\times \rho & \boxed{\text{Area of ring } = \pi (x + \delta x)^{2} – \pi x^{2} \approx 2\pi x\delta x\text{ for negligible } \delta x^{2}.} \\ & = 2\pi t\rho x\delta x. \end{matrix}$

For small δx every particle of such a ring is approximately the same distance, x, from the axis. The moment of inertia of the ring about the axis is therefore approximately

$\begin{matrix}\delta mx^{2} = (2\pi t\rho x\delta x)x^{2} & \boxed{\text{Determine the moment of inertia of the element about the axis.}} \\ = 2\pi t\rho x^{3}\delta x & \end{matrix}$

and the moment of inertia of the whole disc about the axis is approximately

$\begin{matrix} \boxed{\text{Sum for all elements.}} & I = \sum{(2\pi t\rho x^{3}\delta x)} \\ & = 2\pi t\rho \sum{(x^{3}\delta x)} \text{(since ρ and t are the same for all rings).} \end{matrix}$

In the limit as δx → 0, this gives

$\begin{matrix} \boxed{\text{Write as an integral.}} & I = 2\pi t\rho \int_{0}^{r}{x^{3} dx} \end{matrix}$ $\begin{matrix} \boxed{\text{Evaluate the integral.}} ⇒ & I = 2\pi t\rho \left[\frac{x^{4}}{4}\right]^{r}_{0} & \boxed{\text{Note the limits : x takes values between 0 and r .}}\end{matrix}$

$= \frac{1}{2}πtρr^{4}.$

It is now necessary to replace ρ in terms of the mass of the disc, which has volume πr²t, so

$\begin{matrix} M = πr^{2}tρ. & \boxed{\text{Substitute for the density.}} \end{matrix}$

Hence the moment of inertia of a disc about a perpendicular axis through its centre is

$\begin{matrix} I = \frac{1}{2}(\pi t\rho r^{2})r^{2} & \boxed{\text{or you could write} \frac{1}{M} = \frac{\pi t\rho r^{4}}{2\pi r^{2}t\rho } = \frac{1}{2}r^{2}.} \\ \Rightarrow I = \frac{1}{2}Mr^{2}. & \end{matrix}$ 