First slide

Moment of intertia

Question

A circular disc X of radius R is made from an iron plate of thickness t and another disc f of radius 4R is made from an iron plate of thickness $\frac{t}{4}$ . Then the relation between the moment of inertia I_X and I_Y is

Moderate

A

I_Y = 32I_X
B

I_Y = 16I_X
C

I_Y = I_X
D

I_Y = 64I_X

Solution

$\begin{matrix} M_{X} = ρ ({πR}^{2}) t and M_{Y} = ρ (16 {πR}^{2}) (t / 4) = 4 M_{X} \\ \frac{I_{Y}}{I_{X}} = \frac{M_{Y} R_{Y}^{2}}{M_{X} R_{X}^{2}} = \frac{64}{1} \end{matrix}$

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