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} \)

Explore more such questions and answers at BYJU’S.

Was this answer helpful?


0 (0)


Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *




Free Class