Moment Of Inertia Of Hollow Cone - Derivation Using Integration

Moment of inertia of hollow cone can be determined using the following expression;

I = MR² / 2

Here students will learn and understand how the formula is derived as well as its application.

Moment Of Inertia Of Hollow Cone Formula Derivation

For deriving the moment of inertia formula of a hollow cone we can basically follow some general guidelines which are;

Determining the mass density in terms of per unit area.
Defining parameters and finding the mass of small parts.
Finding the moment of inertia of the small parts corresponding to a ring.
Integration

Moment Of Inertia Of Hollow Cone

1. There is a hollow cone with radius R, height H and mass M.

Now, we will take the element disc at a slant height l and | radius r having thickness dl and mass dm.

Using the similarity of the triangle we get;

\(\begin{array}{l}\frac{r}{x}= \frac{R}{\sqrt{R^{2}+H^{2}}}\end{array} \)

\(\begin{array}{l}r= \frac{R}{\sqrt{R^{2}+H^{2}}}x\end{array} \)

The mass of the elemental disc mass is given by;

\(\begin{array}{l}dm = \frac{M}{\pi R\sqrt{R^{2}+H^{2}}}2\pi rdx\end{array} \)

\(\begin{array}{l}dm = \frac{M}{\pi R\sqrt{R^{2}+H^{2}}}2\pi \frac{R}{\sqrt{R^{2}+H^{2}}}xdx\end{array} \)

\(\begin{array}{l}dm = \frac{2Mxdx}{R^{2}+H^{2}}\end{array} \)

2. Calculating the moment of inertia of the elemental disk. It will be given as;

Moment of inertia of element is, dI = dmr²

\(\begin{array}{l}dI= \frac{2Mxdx}{R^{2}+H^{2}}\frac{R^{2}}{{R^{2}+H^{2}}}x^{2}\end{array} \)

3. Finding the moment of inertia through integration.

\(\begin{array}{l}I= \int \frac{2Mxdx}{R^{2}+H^{2}}\frac{R^{2}}{{R^{2}+H^{2}}}x^{2}dx\end{array} \)

I = [2MR² / (R² + H²)²] ∫ x³dx

\(\begin{array}{l}\text{Using the limits of x where x usually varies from 0 to}\ \sqrt{R^{2}+H^{2}}\end{array} \)

We now get;

I = [2MR² / (R² + H²)² ] x [(R² + H²)² / 4]

I = ½ MR²

Therefore, I = MR² / 2

⇒ Check Other Object’s Moment of Inertia:

Parallel Axis Theorem

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