Question

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:

• A. $\frac{ MR ^{2}}{2}$
• B. $\frac{ MH ^{2}}{3}$
• C. $\frac{ MR ^{2}}{3}$
• D. $\frac{ M \left( R ^{2}+ H ^{2}\right)}{4}$

Answer 1

Answer: (A)

Area $=\pi R \ell=\pi R \left(\sqrt{ H ^{2}+ R ^{2}}\right)$
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 $\left. r = h \tan \theta\right)$
$I =\int( dm ) r ^{2}=\int \frac{ h ^{2} \tan ^{2} \theta}{\cos \theta}\left(\frac{2 m }{ R } \frac{ h \tan \theta}{\sqrt{ R ^{2}+ H ^{2}}}\right) dh$
$=\frac{2 M }{\cos \theta R } \frac{\tan ^{3} \theta}{\sqrt{ R ^{2}+ H ^{2}}}$
$\int\limits_{0}^{ H } h ^{3} dh =\frac{ MR ^{2} H ^{4}}{2 RH ^{3} \sqrt{ R ^{2}+ H ^{2}} \cos \theta}$
$=\frac{ MR ^{2} H \sqrt{ R ^{2}+ H ^{2}}}{2 \sqrt{ R ^{2}+ H ^{2} \times H }}$
$=\frac{ MR ^{2}}{2}$