Mass Momentum of Inertia of Capsule

To calculate the mmoi of a capsule, we consider it as the sum of a rectangle and two semi-circles. Then we can add the mmoi of each of the components and add them together.

We already know the mmoi of a rectangle, around the z-axis through its centre of mass.

\[I=\frac{1}{12}M(w^2+h^2\]

But what of a semi-circle? It is in fact, half the mmoi of a full circle - if this is not obvious, go back to how we derived the mmoi of a full circle in the previous circle, consider a semi-circle as consisting of very thin semi-circular rings. The "length" of a semi-circular ring is obviously half the length of a full "hoop". The rest of the derivation is the same - hence the mmoi is:

\[I_z=\frac{1}{4}MR^2\]

Then we can use the parallel axis theorem to add the mmoi of the elements together to arrive at the I of the total.

Parallel Axis Theorem

Suppose a body of mass M is rotated about an axis z passing through the body's center of mass. The body has a moment of inertia Icm with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis z′, which is parallel to the first axis and displaced from it by a distance d, then the moment of inertia I with respect to the new axis is related to Icm by:

\[I=I_{cm}+Md^2\]

where, d is the perpendicular distance between the axes z and z′.

Hence:

\[\begin{equation}\begin{split} I &=\frac{1}{12}M(w^2+h^2)+\left(\frac{1}{4}MR^2 \right)(\frac{w}{2})^2+\left(\frac{1}{4}MR^2 \right)(\frac{w}{2})^2\\&=\frac{1}{12}M(w^2+h^2)+ \frac{1}{2}MR^2 \left(\frac{w}{2}\right)^2\end{split}\end{equation}\]