Question

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 is equal to $R$ is $I=\frac{3M{R}^2}{y}$. Find the value of $y$.

Answer 1

Step 1: Given data

Moment of inertia of a uniform solid cone relative to its symmetry axis, $I=\frac{3M{R}^2}{y}$

Where $M$ is mass of the come and $R$ is the radius of its base

We have to find the value of $y$.

Step 2: Writing equations from the figure

Finding moment of inertia of a solid cone

Small cross sections of the cone are considered to be discs.

Mass of elemental disc,

From the figure, consider triangle $OAB$ and triangle $OCD$.

Since the two triangles are similar$\left(CD\parallel AB,\angle COD=\angle AOB\right)$, their corresponding sides are proportional

$$\begin{gathered} \therefore \frac{r}{R}=\frac{x}{H} \\ \Rightarrow r=\frac{Rx}{H}\to \left(1\right) \end{gathered}$$

If $dm$ is the mass of the small cross-section with thickness $dx$, then

$$\begin{gathered} dm=\frac{M}{{\displaystyle \frac{1}{3}}{\mathrm{\pi R}}^2\mathrm{H}}\times {\mathrm{\pi r}}^2\mathrm{dx} \\ From\left(1\right) \\ \Rightarrow \mathrm{dm}=\frac{3\mathrm{M}}{{\mathrm{\pi R}}^2\mathrm{H}}\times \mathrm{\pi }\left(\frac{{\mathrm{R}}^2{\mathrm{x}}^2}{{\mathrm{H}}^2}\right)\mathrm{dx} \\ \Rightarrow \mathrm{dm}=\frac{3{\mathrm{Mx}}^2}{{\mathrm{H}}^3}\mathrm{dx}\to \left(2\right) \end{gathered}$$

Step 3: Finding the moment of inertia of the elemental disc

Moment of inertia of a disc is $\frac{1}{2}M{R}^2$

where $M$ is mass and $R$ is radius.

We consider the cone to be made up of thin discs of mass $dm$ and radius $r$.

Moment of inertia of the small cross-section of the cone,

$$\begin{gathered} dI=\frac{1}{2}dm{r}^2 \\ From\left(1\right)and\left(2\right), \\ \Rightarrow dI=\frac{1}{2}\times \frac{3M{x}^2}{H^3}dx\times \frac{R^2{x}^2}{H^2} \end{gathered}$$

Step 4: Finding moment of inertia of the solid cone

Moment of inertia of the cone is found by integrating the moment of inertias of the small cross sections of the cone.

Moment of inertia of the cross sections are integrated

$$\begin{gathered} I=\int dI=\frac{3M{R}^2}{2H^5}{\int }_0^H{x}^{4}dx \\ \Rightarrow I=\frac{3M{R}^2}{2H^5}{\left[\frac{x^5}{5}\right]}_o^H \\ \Rightarrow I=\frac{3M{R}^2}{2H^5}\times \frac{H^5}{5} \\ \Rightarrow I=\frac{3M{R}^2}{10} \end{gathered}$$

The given equation is $I=\frac{3M{R}^2}{y}$ and the equation we got is $I=\frac{3M{R}^2}{10}$

Comparing the equations, we get $y=10$.