Q. Find the ratio of moments of inertia of the two disks of same mass and thickness but one of density $7.2g/cm^{3}$ and the other of density $8.9g/cm^{3}$ .
NTA AbhyasNTA Abhyas 2020
Solution:
We know that moment of inertia $I=\frac{MR^{2}}{2}$
(of disk wrt principle axis)
Given $M_{1}=M_{2}$
$V_{1}d_{1}=V_{2}d_{2}$
$\pi R_{1}^{2}t_{1}d_{1}=\pi R_{2}^{2}t_{2}d_{2}...\left(\right.i\left.\right)$
Given $t_{1}=t_{2}$ $\left(\right.$ thickness $\left.$
So $\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{d_{2}}{d_{1}}\left(\right.fromeq.\left(\right.i\left.\right)\left.\right)$
So $\frac{M_{1} R_{1}^{2}}{M_{2} R_{2}^{2}}=\frac{d_{2}}{d_{1}}\left(\right.asM_{1}=M_{2}\left.\right)$
$\Rightarrow \frac{I_{1}}{I_{2}}=\frac{d_{2}}{d_{1}}$
$\Rightarrow \frac{I_{1}}{I_{2}}=\frac{8 . 9}{7 . 2}$
(of disk wrt principle axis)
Given $M_{1}=M_{2}$
$V_{1}d_{1}=V_{2}d_{2}$
$\pi R_{1}^{2}t_{1}d_{1}=\pi R_{2}^{2}t_{2}d_{2}...\left(\right.i\left.\right)$
Given $t_{1}=t_{2}$ $\left(\right.$ thickness $\left.$
So $\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{d_{2}}{d_{1}}\left(\right.fromeq.\left(\right.i\left.\right)\left.\right)$
So $\frac{M_{1} R_{1}^{2}}{M_{2} R_{2}^{2}}=\frac{d_{2}}{d_{1}}\left(\right.asM_{1}=M_{2}\left.\right)$
$\Rightarrow \frac{I_{1}}{I_{2}}=\frac{d_{2}}{d_{1}}$
$\Rightarrow \frac{I_{1}}{I_{2}}=\frac{8 . 9}{7 . 2}$