Four identical point masses (each of mass m) | vedclass

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(B) The moment of inertia of the disc about an axis passing through its center and perpendicular to its plane is $I_{disc} = \frac{1}{2}MR^2$.Since th

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$\frac{1}{2}MR^2 + 4mR^2$

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(B) The moment of inertia of the disc about an axis passing through its center and perpendicular to its plane is $I_{disc} = \frac{1}{2}MR^2$.Since the four point masses $m$ are placed on the circumference,each is at a distance $R$ from the axis of rotation.The moment of inertia of each point mass is $I_{point} = mR^2$.Since there are four such point masses,their total moment of inertia is $I_{total\,points} = 4 	imes mR^2 = 4mR^2$.The total moment of inertia of the system is the sum of the moment of inertia of the disc and the four point masses:$I = I_{disc} + I_{total\,points} = \frac{1}{2}MR^2 + 4mR^2$.

Four identical point masses (each of mass $m$) are placed on the circumference of a disc. The disc has mass $M$ and radius $R$. What will be the moment of inertia about an axis passing through the center $O$ and perpendicular to the plane of the disc?

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