Question

A circular disc $A$ of radius $r$ is made from an iron plate of thickness $t$ and another circular disc $B$ of radius $4r$is made from an iron plate of thickness t$/4$. The relation between the moments of inertia $I_A$ and $I_B$ is

• A. $I_A>I_B$
• B. $I_A=I_B$
• C. $I_A<I_B$
• D. depends on the actual values of $t$ and $r$.

Answer 1

The correct option is A $I_A>I_B$

Given: A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness $\frac{t}{4}$.

To find the relation between the moments of inertia $I_A$ and $I_B$

Solution:

Mass of the disc (X),

$m_A=\pi r^{2}t\rho .........\left(i\right)$

where $m=V\rho =At\rho =\pi r^2-\rho$

where V is the volume, m is the mass, A is the area, t is thickness, $\rho$ is the density and r is the radius of the circular disc.

$\therefore I_A=\frac{1}{2}{m}_A{r}^2⟹I_A=\frac{1}{2}\pi r^{4}t\rho ........\left(ii\right)$

Mass of the disc (B),

$m_B=\pi (4r{)}^2\times \frac{t}{4}\rho =4\pi r^{2}t\rho .......(iii)$

$\therefore I_B=\frac{1}{2}{m}_B(4r{)}^2⟹I_B=\frac{1}{2}\times (4\pi r^{2}t\rho \left)\right(16r^2)⟹I_B=32\pi r^{4}t\rho ........(iv)$

From eqn(ii) and eqn(iv), we get

$\frac{I_A}{I_B}=\frac{\frac{1}{2}\pi r^{4}t\rho }{32\pi r^{4}t\rho }⟹64I_A=I_B⟹I_A>I_B$

is the relation between the moments of inertia $I_A$ and $I_B$