The moment of inertia of a semicircular disc | vedclass

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(B) For a semicircular disc of mass $M$ and radius $R$,the moment of inertia about the diameter (axis $XX'$) is $I_x = \frac{1}{4} M R^2$.Since the di

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$\frac{M R^2}{4}$

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(B) For a semicircular disc of mass $M$ and radius $R$,the moment of inertia about the diameter (axis $XX'$) is $I_x = \frac{1}{4} M R^2$.Since the disc is symmetric about the axis perpendicular to the diameter in its plane (axis $AA'$),the moment of inertia about this axis is also $I_y = \frac{1}{4} M R^2$.The product of inertia $I_{xy}$ for a semicircular disc about these axes is zero due to symmetry.The moment of inertia $I$ about an axis passing through the center and making an angle $	heta$ with the diameter is given by the formula:$I = I_x \sin^2 	heta + I_y \cos^2 	heta - 2 I_{xy} \sin 	heta \cos 	heta$Substituting $I_x = \frac{1}{4} M R^2$,$I_y = \frac{1}{4} M R^2$,and $I_{xy} = 0$:$I = \frac{1}{4} M R^2 \sin^2 	heta + \frac{1}{4} M R^2 \cos^2 	heta$$I = \frac{1}{4} M R^2 (\sin^2 	heta + \cos^2 	heta)$$I = \frac{1}{4} M R^2$

The moment of inertia of a semicircular disc of mass $M$ and radius $R$ about the shown axis is:

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