The moments of inertia of a non-uniform circu | vedclass

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(A) Let $O$ be the geometrical centre of the disc and $C(x, y)$ be its centre of mass.Let $I_{CM}$ be the moment of inertia about an axis passing thro

Vedclass · Accepted Answer

$\frac{1}{4 M R} \sqrt{(I_1-I_3)^2+(I_2-I_4)^2}$

Vedclass · Answer

(A) Let $O$ be the geometrical centre of the disc and $C(x, y)$ be its centre of mass.Let $I_{CM}$ be the moment of inertia about an axis passing through the centre of mass of the disc parallel to the tangents.By the parallel axes theorem,the moment of inertia about a tangent is $I = I_{CM} + M d^2$,where $d$ is the distance between the axes.For tangents $AB$ and $CD$ (parallel to the $x$-axis),the distances from the centre of mass are $(R-y)$ and $(R+y)$ respectively:$I_1 = I_{CM} + M(R-y)^2$$I_3 = I_{CM} + M(R+y)^2$Subtracting these equations:$I_1 - I_3 = M[(R-y)^2 - (R+y)^2] = M[R^2 - 2Ry + y^2 - (R^2 + 2Ry + y^2)] = -4MRy$Similarly,for tangents $BC$ and $DA$ (parallel to the $y$-axis),the distances from the centre of mass are $(R-x)$ and $(R+x)$ respectively:$I_2 = I_{CM} + M(R-x)^2$$I_4 = I_{CM} + M(R+x)^2$$I_2 - I_4 = -4MRx$Squaring and adding the two results:$(I_1 - I_3)^2 + (I_2 - I_4)^2 = 16M^2R^2y^2 + 16M^2R^2x^2 = 16M^2R^2(x^2 + y^2)$Therefore,the distance $d = \sqrt{x^2 + y^2} = \frac{1}{4MR} \sqrt{(I_1 - I_3)^2 + (I_2 - I_4)^2}$.

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