Moment of Inertia of a Drilled Sphere

Question

A solid sphere of mass $$M$$ and radius $$R$$ has a cylindrical hole of radius $$R/2$$ drilled through its center. The axis of the hole coincides with a diameter of the sphere. What is the moment of inertia of the resulting object with respect to this axis?

Accepted Answer

$$\frac{31}{80}MR^2$$

Answer

$$\frac{MR^2}{2}$$

Answer

$$\frac{2MR^2}{5}$$

Answer

$$\frac{2MR^2}{5} - \frac{MR^2}{4}$$

Moment of Inertia of a Drilled Sphere

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