Question

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 $H$ then, its moment of inertia about its axis is:

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

Answer 1

The correct option is A $\frac{M{R}^2}{2}$

Hollow ice-cream cone can be assumed as stack of rings having different radius, so

$I=\int dI=\int dm\left(r^2\right)$ $....\left(i\right)$


From diagram,
$\frac{r}{h}=\tan \theta =\frac{R}{H}\text{or}r=\frac{R}{H}h.....\left(ii\right)$
and, $\cos \theta =\frac{dh}{dl}........\left(iii\right)$
Now,
$dm=\frac{M}{\pi Rl}\times \left(2\pi rdl\right)$, where $l$ is slant height of cone given by $l=\sqrt{H^2+R^2}$
so, from eq $\left(i\right)$ we have
$I=\int \frac{M}{\pi Rl}\times \left(2\pi rdl\right).r^2=\int \frac{2M{r}^{3}dl}{Rl}$
$\Rightarrow I={\int }_0^H\frac{2M(h\tan \theta {)}^{3}dh}{R\sqrt{H^2+R^2}\cos \theta }$
$\Rightarrow I=\left(\frac{2M{\tan }^3\theta }{R\sqrt{H^2+R^2}\cos \theta }\right){\int }_0^H{h}^{3}dh$
$\Rightarrow I=\left(\frac{2M{\tan }^3\theta }{R\sqrt{H^2+R^2}\cos \theta }\right)\times \frac{H^4}{4}$

$\Rightarrow I=\left[\frac{2M{\left(\frac{R}{H}\right)}^3}{\left(R\sqrt{H^2+R^2}\right)\times \left(\frac{H}{\sqrt{H^2+R^2}}\right)}\right]\times \frac{H^4}{4}$
$\Rightarrow I=\frac{M{R}^2}{2}$