Question

Find $I=?$

Answer 1

1. $I = \frac{1}{2}MR^2$, 2) $I = \frac{3}{10}MR^2$

Answer 2

The problem asks to find the moment of inertia ($I$) for a hollow cone and a solid cone, both of mass $M$, height $H$, and base radius $R$, about their central axis.

1. Hollow Cone

A hollow cone can be considered as a collection of infinitesimally thin rings. Let's consider an elemental ring at a distance $x$ from the apex along the height. The radius of this elemental ring is $r$, and its width along the slant height is $dl$.

From similar triangles (referring to the cone's cross-section): The ratio of radius to height is constant: $\frac{r}{x} = \frac{R}{H} \implies r = \frac{R}{H}x$.

Let $L$ be the slant height of the cone, $L = \sqrt{R^2 + H^2}$. The slant height $l$ to the elemental ring is related by $\frac{l}{x} = \frac{L}{H} \implies l = \frac{L}{H}x$. The elemental width along the slant height is $dl = \frac{L}{H}dx$.

The total surface area of the hollow cone is $A = \pi R L$. The surface mass density $\sigma = \frac{M}{A} = \frac{M}{\pi R L}$.

The area of the elemental ring is $dA = (2\pi r)dl$. Substituting $r$ and $dl$: $dA = 2\pi \left(\frac{R}{H}x\right) \left(\frac{L}{H}dx\right) = \frac{2\pi R L}{H^2} x \,dx$.

The mass of the elemental ring is $dm = \sigma \, dA$. $dm = \frac{M}{\pi R L} \cdot \frac{2\pi R L}{H^2} x \,dx = \frac{2M}{H^2} x \,dx$.

The moment of inertia of a thin ring of mass $dm$ and radius $r$ about its central axis is $dI = dm \cdot r^2$. $dI = \left(\frac{2M}{H^2} x \,dx\right) \left(\frac{R}{H}x\right)^2 = \frac{2M}{H^2} x \,dx \cdot \frac{R^2}{H^2}x^2 = \frac{2MR^2}{H^4} x^3 \,dx$.

To find the total moment of inertia, integrate $dI$ from $x=0$ to $x=H$: $I = \int_0^H \frac{2MR^2}{H^4} x^3 \,dx = \frac{2MR^2}{H^4} \int_0^H x^3 \,dx$ $I = \frac{2MR^2}{H^4} \left[\frac{x^4}{4}\right]_0^H = \frac{2MR^2}{H^4} \left(\frac{H^4}{4} - 0\right)$ $I = \frac{2MR^2}{4} = \frac{1}{2}MR^2$.

2. Solid Cone

A solid cone can be considered as a stack of infinitesimally thin disks. Let's consider an elemental disk at a distance $x$ from the apex along the height. The radius of this elemental disk is $r$, and its thickness is $dx$.

From similar triangles: $\frac{r}{x} = \frac{R}{H} \implies r = \frac{R}{H}x$.

The total volume of the solid cone is $V = \frac{1}{3}\pi R^2 H$. The volume mass density $\rho = \frac{M}{V} = \frac{M}{\frac{1}{3}\pi R^2 H} = \frac{3M}{\pi R^2 H}$.

The volume of the elemental disk is $dV = \pi r^2 \,dx$. Substituting $r$: $dV = \pi \left(\frac{R}{H}x\right)^2 \,dx = \frac{\pi R^2}{H^2} x^2 \,dx$.

The mass of the elemental disk is $dm = \rho \, dV$. $dm = \frac{3M}{\pi R^2 H} \cdot \frac{\pi R^2}{H^2} x^2 \,dx = \frac{3M}{H^3} x^2 \,dx$.

The moment of inertia of a thin disk of mass $dm$ and radius $r$ about its central axis is $dI = \frac{1}{2} dm \cdot r^2$. $dI = \frac{1}{2} \left(\frac{3M}{H^3} x^2 \,dx\right) \left(\frac{R}{H}x\right)^2 = \frac{1}{2} \frac{3M}{H^3} x^2 \,dx \cdot \frac{R^2}{H^2}x^2 = \frac{3MR^2}{2H^5} x^4 \,dx$.

To find the total moment of inertia, integrate $dI$ from $x=0$ to $x=H$: $I = \int_0^H \frac{3MR^2}{2H^5} x^4 \,dx = \frac{3MR^2}{2H^5} \int_0^H x^4 \,dx$ $I = \frac{3MR^2}{2H^5} \left[\frac{x^5}{5}\right]_0^H = \frac{3MR^2}{2H^5} \left(\frac{H^5}{5} - 0\right)$ $I = \frac{3MR^2}{10}$.

The moments of inertia are:

1. Hollow cone: $I = \frac{1}{2}MR^2$
2. Solid cone: $I = \frac{3}{10}MR^2$