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(B) The moment of inertia $I$ of a ring about its central axis is given by $I = MR^2$.Since the rings are made of the same material with the same thic

Vedclass · Accepted Answer

$1:27$

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(B) The moment of inertia $I$ of a ring about its central axis is given by $I = MR^2$.Since the rings are made of the same material with the same thickness $t$,the mass $M$ can be expressed as $M = 	ext{Volume} 	imes 	ext{Density} = (2\pi R \cdot A) \cdot ho$,where $A$ is the cross-sectional area and $ho$ is the density. Given the thickness is the same,$M \propto R$.Thus,$I = MR^2 \propto (R)R^2 = R^3$ is incorrect for a ring; for a ring,$M = 	ext{Area} 	imes 	ext{Circumference} 	imes ho = (t^2) 	imes (2\pi R) 	imes ho$. So $M \propto R$.Therefore,$I = MR^2 \propto R \cdot R^2 = R^3$.Wait,if the thickness $t$ is the cross-sectional radius,$M = \pi t^2 (2\pi R) ho \propto R$.Then $\frac{I_1}{I_2} = \frac{R_1^3}{R_2^3} = (\frac{0.2}{0.6})^3 = (\frac{1}{3})^3 = \frac{1}{27}$.However,if the rings are thin hoops where $M$ is proportional to the circumference $(M \propto R)$,then $I = MR^2 \propto R^3$. If the question implies the rings have the same cross-sectional area,$M \propto R$,then $I \propto R^3$. If the question implies the rings have the same mass,$I \propto R^2$. Given the standard interpretation of such problems where $M \propto R$,the ratio is $1:27$.

Two rings made of the same material and having the same thickness have radii of $0.2 \ m$ and $0.6 \ m$. The ratio of their moments of inertia is:

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