Two discs have densities d_1 = 7.2 g/cm^3 an | vedclass

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(A) The moment of inertia of a disc about its central axis is $I = \frac{1}{2}MR^2$.Since the mass $M$ and thickness $t$ are the same for both discs,w

Vedclass · Accepted Answer

$8.9 / 7.2$

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(A) The moment of inertia of a disc about its central axis is $I = \frac{1}{2}MR^2$.Since the mass $M$ and thickness $t$ are the same for both discs,we have $M = ho V = ho (\pi R^2 t)$.Since $M$ and $t$ are constant,$ho R^2 = 	ext{constant}$,which implies $R^2 \propto \frac{1}{ho}$.Therefore,the ratio of the radii squared is $\frac{R_1^2}{R_2^2} = \frac{d_2}{d_1}$.The ratio of the moments of inertia is $\frac{I_1}{I_2} = \frac{\frac{1}{2} M R_1^2}{\frac{1}{2} M R_2^2} = \frac{R_1^2}{R_2^2}$.Substituting the density relationship,we get $\frac{I_1}{I_2} = \frac{d_2}{d_1} = \frac{8.9}{7.2}$.

Two discs have densities $d_1 = 7.2 \ g/cm^3$ and $d_2 = 8.9 \ g/cm^3$. Both discs have the same mass and the same thickness. Find the ratio of their moments of inertia.

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