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. $\frac{M{R}^2}{2}$
• B. $\frac{M{H}^2}{3}$
• C. $\frac{M{R}^2}{3}$
• 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 $dA=2\pi rd\ell =2\pi r\frac{dh}{\cos \theta }$
mass of element $dm=\frac{M}{\pi R\sqrt{H^2+R^2}}\times \frac{2\pi rdh}{\cos \theta }$
$dm=\frac{2Mh\tan \theta dh}{R\sqrt{H^2+R^2}\cos \theta }$ (here $r=h\tan \theta )$
$I=\int (dm)r^2=\int \frac{h^2{\tan }^2\theta }{\cos \theta }\left(\frac{2m}{R}\frac{h\tan \theta }{\sqrt{R^2+H^2}}\right)dh$
$=\frac{2M}{\cos \theta R}\frac{{\tan }^3\theta }{\sqrt{R^2+H^2}}$
$\underset{0}{\overset{H}{\int }}{h}^{3}dh=\frac{M{R}^2{H}^4}{2R{H}^3\sqrt{R^2+H^2}\cos \theta }$
$=\frac{M{R}^{2}H\sqrt{R^2+H^2}}{2\sqrt{R^2+H^2\times H}}$
$=\frac{M{R}^2}{2}$