First slide

Moment of intertia

Question

Two circular discs A and B of equal masses and thickness but made of metals with densities d_A and d_B $(d_{A} > d_{B}) .$ If their moments of inertia about an axis through the centre and normal to the circular forces be I_A and I_B, then.

Moderate

A

I_A =I_B
B

I_A > I_B
C

I_A < I_B
D

$I_{A} \geq I_{B}$

Solution

$I_{A} = \frac{m_{A} r_{A}^{2}}{2} and I_{B} = \frac{m_{B} r_{B}^{2}}{2}$
$∴ \frac{I_{A}}{I_{B}} = \frac{r_{A}^{2}}{r_{B}^{2}} (∵ m_{A} = m_{B})$ ………………(1)
Now, $m_{A} = {πr}_{A}^{2} {td}_{A}$
$m_{B} = {πr}_{B}^{2} {td}_{B}$
So, ${πr}_{A}^{2} {td}_{A} = {πr}_{B}^{2} {td}_{B}$
Or $\frac{{r_{A}}^{2}}{{r_{B}}^{2}} = \frac{d_{B}}{d_{A}}$ ……………………..(2)
$\frac{I_{A}}{I_{B}} = \frac{d_{B}}{d_{A}}$
As $d_{A} > d_{B} hence I_{A} < I_{B}$

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