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

Area of element \(d A =2 \pi rd \ell==2 \pi r \frac{ dh }{\cos \theta}\)

mass of element \(dm =\frac{ M }{\pi R \sqrt{ H ^{2}+ R ^{2}}} \times \frac{2 \pi rdh }{\cos \theta}\)

\(dm =\frac{2 Mh \tan \theta dh }{ R \sqrt{ H ^{2}+ R ^{2}} \cos \theta} \quad\) (here \(\left. r = h \tan \theta\right)\)

\(I =\int( dm ) r ^{2}=\int \frac{ h ^{2} \tan ^{2} \theta}{\cos \theta}\left(\frac{2 m }{ R } \frac{ h \tan \theta}{\sqrt{ R ^{2}+ H ^{2}}}\right) dh\)

\(=\frac{2 M }{\cos \theta R } \frac{\tan ^{3} \theta}{\sqrt{ R ^{2}+ H ^{2}}}\)

\(\int_{0}^{ H } h ^{3} dh =\frac{ MR ^{2} H ^{4}}{2 RH ^{3} \sqrt{ R ^{2}+ H ^{2}} \cos \theta}\)

\(=\frac{ MR ^{2} H \sqrt{ R ^{2}+ H ^{2}}}{2 \sqrt{ R ^{2}+ H ^{2}} \times H }\)

\(=\frac{ MR ^{2}}{2}\)