# The Moment Of Inertia Of A Uniform Solid Cone Relative To Its Symmetry Axis, If The Mass Of The Cone Is Equal To M And The Radius Of Its Base To R Is I=3mr2y. Find The Value Of Y.

Consider an element disc of radius r and thickness dx at a distance x from the point O.

$$r = c tan \alpha$$

The volume of the disc =$$\Pi x^{2}tan^{2}\alpha d x$$

Hence, its mass dm = $$\pi x^{2} tan \alpha dx * \rho$$

Moment of inertia of this element, about the axis OA,

$$dI = dm \frac{r^{2}}{2}$$ $$\Rightarrow (\Pi x^{2} tan^{2} \alpha dx) \frac{x^{2}tan^{2} x}{2}$$ $$\Rightarrow \frac{\Pi \rho}{2} x^{4} tan^{4}\alpha dx$$

Thus the sought moment of inertia.

$$\Rightarrow \frac{\Pi \rho}{2} tan^{4} \alpha \int_{h}^{0} dx$$ $$\Rightarrow\frac{\Pi \rho R^{4} * h^{5}}{10h^{4}}$$

Therefore, I = \frac{3mR^{2}}{10} \)

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