Moment of Inertia Formulas for Annular Discs and Spheres

Moment of Inertia of a Flat Annular Disc

A flat annular disc is defined by its mass $M$, an inner radius $R_1$, and an outer radius $R_2$. The moment of inertia ($M. I.$) for this object varies based on the axis of rotation chosen. For an axis passing through its centre and perpendicular to the plane of the disc, the moment of inertia is given by the formula:

$\frac{M}{2}(R_1^2 + R_2^2)$

When calculating the moment of inertia for an axis that is a tangent and perpendicular to the plane of the disc, the parallel axes theorem is employed. This involves taking the moment of inertia about the center and adding the term $MR^2$, specifically using the outer radius $R_2$. The resulting formula is:

$\frac{M}{2}(R_1^2 + 3R_2^2)$

For rotation about an axis passing through its diameter, which lies within the plane of the disc, the perpendicular axes theorem is used. This theorem indicates that the moment of inertia about the diameter is half ($\times \frac{1}{2}$) of the value for the axis through the center perpendicular to the plane. The formula is:

$\frac{M}{4}(R_1^2 + R_2^2)$

Finally, for an axis that is a tangent and in the plane of the disc, the parallel axes theorem is applied again by adding $MR^2$ to the diameter formula. Using the outer radius $R_2$, the calculation yields:

$\frac{M}{4}(R_1^2 + 5R_2^2)$

Moment of Inertia of a Solid Sphere

A solid sphere of mass $M$ and radius $R$ has a uniform mass distribution. The moment of inertia about an axis passing through its diameter is defined by the standard constant for solid spherical bodies:

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

To find the moment of inertia about its tangent, the parallel axes theorem is utilized. This requires adding $MR^2$ to the diametric moment of inertia ($\frac{2}{5}MR^2 + MR^2$), resulting in the formula:

$\frac{7}{5}MR^2$

Moment of Inertia of a Hollow Sphere

A hollow sphere, characterized by mass $M$ and radius $R$, has its mass concentrated on its outer shell. For an axis passing through its diameter, the moment of inertia is expressed as:

$\frac{2}{3}MR^2$

Applying the parallel axes theorem to determine the moment of inertia about its tangent, the value is calculated by adding $MR^2$ to the value for the diameter ($\frac{2}{3}MR^2 + MR^2$). This results in the following expression:

$\frac{5}{3}MR^2$