Question

Calculate the moment of inertia of:

(a) A ring of mass $M$ and radius $R$ about an axis coinciding with a diameter of the ring. (b) A thin disc of same mass and radius about an axis coinciding with a diameter.

Answer 1

For (a), the moment of inertia of a ring about a diameter is $\frac{1}{2}MR^2$. For (b), the moment of inertia of a thin disc about a diameter is $\frac{1}{4}MR^2$.

Answer 2

The Perpendicular Axis Theorem states that for a planar object, $I_z = I_x + I_y$, where $I_x$ and $I_y$ are moments of inertia about two perpendicular axes in the plane, and $I_z$ is the moment of inertia about the axis perpendicular to the plane. Due to symmetry, $I_x = I_y = I_d$ for diameters. Thus, $I_d = \frac{1}{2}I_z$.

(a) For a ring, $I_z = MR^2$. Therefore, $I_d = \frac{1}{2}MR^2$. (b) For a disc, $I_z = \frac{1}{2}MR^2$. Therefore, $I_d = \frac{1}{2}\left(\frac{1}{2}MR^2\right) = \frac{1}{4}MR^2$.