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(A) The moment of inertia of a circular disc about its central axis is given by $I = \frac{1}{2}MR^2$.Since mass $M = 	ext{Volume} 	imes 	ext{densi

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$I_y = 64I_x$

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(A) The moment of inertia of a circular disc about its central axis is given by $I = \frac{1}{2}MR^2$.Since mass $M = 	ext{Volume} 	imes 	ext{density} = (\pi R^2 t) ho$,where $t$ is the thickness and $ho$ is the density.Substituting $M$ into the formula,we get $I = \frac{1}{2}(\pi R^2 t ho) R^2 = \frac{1}{2} \pi ho t R^4$.Assuming the density $ho$ is the same for both discs,the ratio of the moments of inertia is $\frac{I_y}{I_x} = \frac{t_y}{t_x} \left( \frac{R_y}{R_x} ight)^4$.Given $R_y = 4R$ and $R_x = R$,so $\frac{R_y}{R_x} = 4$.Given $t_y = \frac{t}{4}$ and $t_x = t$,so $\frac{t_y}{t_x} = \frac{1}{4}$.Substituting these values: $\frac{I_y}{I_x} = \frac{1}{4} 	imes (4)^4 = \frac{256}{4} = 64$.Therefore,$I_y = 64I_x$.

$A$ circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$,and another disc $Y$ 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

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