The ratio of the masses of two circular rings | vedclass

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(B) The moment of inertia $I$ of a circular ring about its central axis is given by $I = MR^2$,where $M$ is the mass and $R$ is the radius.Since the r

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$2:1$

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(B) The moment of inertia $I$ of a circular ring about its central axis is given by $I = MR^2$,where $M$ is the mass and $R$ is the radius.Since the ratio of diameters is $2:1$,the ratio of radii $R_1/R_2$ is also $2:1$.Given $M_1/M_2 = 1/2$ and $R_1/R_2 = 2/1$.The ratio of moments of inertia is $\frac{I_1}{I_2} = \frac{M_1 R_1^2}{M_2 R_2^2} = \left( \frac{M_1}{M_2} ight) \left( \frac{R_1}{R_2} ight)^2$.Substituting the values: $\frac{I_1}{I_2} = \left( \frac{1}{2} ight) 	imes \left( \frac{2}{1} ight)^2 = \frac{1}{2} 	imes 4 = 2$.Thus,the ratio is $2:1$.

The ratio of the masses of two circular rings is $1:2$ and the ratio of their diameters is $2:1$. The ratio of their moments of inertia is:

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