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(A) The moment of inertia of a disc about its central axis is given by $I = \frac{1}{2} M R^2$. Since the discs are of the same material and thickness

Vedclass · Accepted Answer

$1 : 81$

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(A) The moment of inertia of a disc about its central axis is given by $I = \frac{1}{2} M R^2$. Since the discs are of the same material and thickness $(t)$,their mass $M$ can be expressed as $M = 	ext{density} (ho) 	imes 	ext{volume} = ho 	imes (\pi R^2 t)$. Substituting this into the formula for $I$: $I = \frac{1}{2} (ho \pi R^2 t) R^2 = \frac{1}{2} ho \pi t R^4$. Since $ho$,$\pi$,and $t$ are constant for both discs,we have $I \propto R^4$. Therefore,the ratio of the moments of inertia is $\frac{I_1}{I_2} = \left( \frac{R_1}{R_2} ight)^4$. Given $R_1 = 0.2\, m$ and $R_2 = 0.6\, m$,the ratio is $\frac{I_1}{I_2} = \left( \frac{0.2}{0.6} ight)^4 = \left( \frac{1}{3} ight)^4 = \frac{1}{81}$. Thus,the ratio is $1 : 81$.

Two discs of the same material and thickness have radii $0.2\, m$ and $0.6\, m.$ Their moments of inertia about their axes will be in the ratio

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