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(C) The circular frame is massless,so we only consider the moment of inertia of the four point masses located at a distance $a$ from the center $O$.Th

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$a$

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(C) The circular frame is massless,so we only consider the moment of inertia of the four point masses located at a distance $a$ from the center $O$.The moment of inertia $I$ of the system about an axis passing through the center $O$ and perpendicular to the plane is given by the sum of $mr^2$ for each mass:$I = (3m)a^2 + (2m)a^2 + (m)a^2 + (2m)a^2 = 8ma^2$The total mass of the system is $M = 3m + 2m + m + 2m = 8m$.By the definition of the radius of gyration $k$,we have $I = Mk^2$.Substituting the values:$8ma^2 = (8m)k^2$$k^2 = a^2$$k = a$

Four masses are attached to a light circular frame of radius $a$ as shown in the figure. The radius of gyration of this system about an axis passing through the center $O$ of the circular frame and perpendicular to its plane is:

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