A disc X of radius R and thickness t is made | vedclass

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(D) The moment of inertia of a disc about its central axis is given by $I = \frac{1}{2}MR^2$. Since both discs are made of the same material,their mas

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$I_y = 64I_x$

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(D) The moment of inertia of a disc about its central axis is given by $I = \frac{1}{2}MR^2$. Since both discs are made of the same material,their mass $M$ is given by $M = 	ext{density} 	imes 	ext{volume} = ho 	imes (\pi R^2 t)$. For disc $X$: $M_x = ho \pi R^2 t$,so $I_x = \frac{1}{2} (ho \pi R^2 t) R^2 = \frac{1}{2} ho \pi R^4 t$. For disc $Y$: $M_y = ho \pi (4R)^2 (t/4) = ho \pi (16R^2) (t/4) = 4 ho \pi R^2 t$. The moment of inertia $I_y = \frac{1}{2} M_y (4R)^2 = \frac{1}{2} (4 ho \pi R^2 t) (16R^2) = 32 ho \pi R^4 t$. Comparing $I_x$ and $I_y$: $I_y = 64 	imes (\frac{1}{2} ho \pi R^4 t) = 64 I_x$.

$A$ disc $X$ of radius $R$ and thickness $t$ is made of an iron plate. Another disc $Y$ of radius $4R$ and thickness $t/4$ is also made of the same iron plate. What is the relationship between their moments of inertia $I_x$ and $I_y$?

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