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(A) Consider a hollow cone of mass $M$,height $H$,and base radius $R$. The slant height is $L = \sqrt{R^2 + H^2}$.Let $	heta$ be the semi-vertical an

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$\frac{MR^2}{2}$

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(A) Consider a hollow cone of mass $M$,height $H$,and base radius $R$. The slant height is $L = \sqrt{R^2 + H^2}$.Let $	heta$ be the semi-vertical angle,so $	an 	heta = R/H$.Consider a thin circular ring element at a distance $y$ from the apex along the axis,with thickness $dy$ along the slant height.The radius of this ring is $r = y 	an 	heta$.The slant length of the element is $dl = dy / \cos 	heta$.The surface area of the cone is $A = \pi R L = \pi R \sqrt{R^2 + H^2}$.The mass per unit area is $\sigma = M / A = M / (\pi R \sqrt{R^2 + H^2})$.The area of the ring element is $dA = 2 \pi r dl = 2 \pi (y 	an 	heta) (dy / \cos 	heta)$.The mass of the element is $dm = \sigma dA = \frac{M}{\pi R \sqrt{R^2 + H^2}} \cdot 2 \pi (y 	an 	heta) \frac{dy}{\cos 	heta} = \frac{2M}{R \sqrt{R^2 + H^2}} \cdot \frac{	an 	heta}{\cos 	heta} y dy$.Since $	an 	heta = R/H$ and $\cos 	heta = H/L = H/\sqrt{R^2 + H^2}$,we have $\frac{	an 	heta}{\cos 	heta} = \frac{R/H}{H/L} = \frac{RL}{H^2} = \frac{R\sqrt{R^2+H^2}}{H^2}$.Thus,$dm = \frac{2M}{R\sqrt{R^2+H^2}} \cdot \frac{R\sqrt{R^2+H^2}}{H^2} y dy = \frac{2M}{H^2} y dy$.The moment of inertia of this ring about the axis is $dI = dm \cdot r^2 = (\frac{2M}{H^2} y dy) (y 	an 	heta)^2 = \frac{2M}{H^2} 	an^2 	heta \cdot y^3 dy$.Integrating from $y=0$ to $y=H$:$I = \int_0^H \frac{2M}{H^2} (R/H)^2 y^3 dy = \frac{2MR^2}{H^4} \int_0^H y^3 dy = \frac{2MR^2}{H^4} [\frac{y^4}{4}]_0^H = \frac{2MR^2}{H^4} \cdot \frac{H^4}{4} = \frac{MR^2}{2}$.

Shown in the figure is a hollow ice cream cone (it is open at the top). If its mass is $M$,radius of its top is $R$,and height is $H$,then its moment of inertia about its axis is

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