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The moment of inertia of a circular ring of mass M and radius R about diameter is\\[\\begin{align} A)\\dfrac{1}{2}M{{R}^{2}} \\\\ B)\\dfrac{3}{2}M{{R}^{2}} \\\\ C)\\dfrac{M{{R}^{2}}}{2} \\\\ D)2M{{R}^{2}} \\\\ \\end{align}\\] – #74

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Two rings have their moment of inertia in the ratio 2:1 and their diameters are in the ratio 2:1. The ratio of their masses will be:a)1:2b)2:1c)1:4d)1:1Correct answer is option ‘A’. Can you – #99

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• moment of inertia of half ring
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• moment of inertia of a thin ring

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The moment of inertia of a circular ring (radius R, mass M) about an axis which passes through tangentially and perpendicular to its plane will be?(A) $ \\dfrac{{M{R^2}}}{2} $ (B) $ M{R^2} $ ( – #118

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The moment of inertia of a circular ring with mass M and radius R about an axis passing through its centre and perpendicular to its plane is:A. $\\dfrac{MR^2}{4}$B. $MR^2$C. $\\dfrac{MR^2}{2}$D. $\\dfrac{3}{4}MR^2$ – #122

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• moment of inertia of semicircular ring about diameter
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• moment of inertia of thin circular ring
• area moment of inertia of a ring
• moment of inertia table

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