Question

The figure shows two cones $A$ and $B$ with the conditions $h_A<h_B$; ${\rho }_A>{\rho }_B;$ $R_A=R_B;m_A=m_B$ Identify the correct statement about their axis of symmetry :

• A. both have same moment of inertia
• B. $A$ has greater moment of inertia
• C. $B$ has greater moment of inertia
• D. nothing can be said

Answer 1

The correct option is A both have same moment of inertia

$\frac{x}{r}=\frac{h}{R}\Rightarrow r=\frac{x}{h}R$
Volume of disc $=\left(\pi r^2\right)dX$
Mass of disc $=\left(\pi r^2\right)\left(dX\right)\rho =dm$
$dI=\frac{1}{2}\left(dm\right)r^2$

$=\frac{1}{2}\left(\pi r^{2}dx\right)\rho r^2$

$=\frac{\pi \rho }{2}{r}^{4}dx$

$=\left(\frac{\pi \rho }{2}\right)\left(\frac{R^4}{h^4}\right)x^{4}dx$

$\therefore I={\int }_0^{h}dI=\left(\frac{\pi \rho }{2}\right)\left(\frac{R^4}{h^4}\right)\left(\frac{h^5}{5}\right)$

$=\left(\frac{\pi \rho }{10}\right)\left(h{R}^4\right)$

$\rho =\frac{m}{\left(\frac{1}{3}\pi R^{2}h\right)}$

$\therefore I=\left(\frac{\pi }{10}\right)\left(\frac{m}{\frac{1}{3}\pi R^{2}h}\right)\left(h{R}^4\right)$
So, $I\propto m{R}^2$
Since $m$ and $R$ are same. Therefore $I$ is same.