Four identical solid spheres each of mass m a | vedclass

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(C) Let the four spheres be located at the corners of a square of side $b$. Let the axis of rotation be one of the sides of the square.For the two sph

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$\frac{8}{5} ma^{2} + 2 mb^{2}$

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(C) Let the four spheres be located at the corners of a square of side $b$. Let the axis of rotation be one of the sides of the square.For the two spheres whose centers lie on the axis of rotation,the distance from the axis is $0$. The moment of inertia of each of these spheres about its own diameter is $I_{cm} = \frac{2}{5} ma^{2}$. Since the axis passes through their centers,their moment of inertia about the axis is $I_1 = I_2 = \frac{2}{5} ma^{2}$.For the other two spheres,the distance of their centers from the axis is $b$. Using the parallel axis theorem,the moment of inertia of each of these spheres about the axis is $I_3 = I_4 = I_{cm} + mb^{2} = \frac{2}{5} ma^{2} + mb^{2}$.The total moment of inertia of the system is $I = I_1 + I_2 + I_3 + I_4 = 2 	imes (\frac{2}{5} ma^{2}) + 2 	imes (\frac{2}{5} ma^{2} + mb^{2})$.$I = \frac{4}{5} ma^{2} + \frac{4}{5} ma^{2} + 2 mb^{2} = \frac{8}{5} ma^{2} + 2 mb^{2}$.

Four identical solid spheres each of mass $m$ and radius $a$ are placed with their centres on the four corners of a square of side $b$. The moment of inertia of the system about one side of the square is:

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