Example 10.3.2. Circular Ring.
Use the parallel axis theorem to find the moment of inertia of the circular ring about the \(y\) axis.
The dimensions of the ring are \(R_i = \mm{30}\text{,}\) \(R_o = \mm{45}\text{,}\) and \(a = \mm{80}.\)
Answer.
\begin{equation*}
I_y = \mm{57.8 \times 10^6}^4
\end{equation*}
Solution.
To apply the parallel axis theorem, we need three pieces of information
- The centroidal moment of inertia of the ring, \(I_y\text{,}\)
- the area of the ring, \(A\text{,}\)
- the distance between the two parallel axes, \(d\text{.}\)
The area of the ring is found by subtracting the area of the inner circle from the area of the outer circle. The centroidal moment of inertia is calculated similarly using (10.2.10). The distance between the \(y\) and \(y'\) axis is available from the diagram. Inserting these values into the parallel axis theorem gives,
\begin{align*}
I_y \amp = I_y + A d^2\\
\amp = \underbrace{\frac{\pi}{4}\Big(r_o^4-r_i^4 \Big )}_{\bar{I}_y} \quad + \quad \underbrace{\pi \Big(r_o^2 - r_i^2 \Big)}_A\ \underbrace{\Big ( a + r_o \Big)^2}_{d^2}\\
\amp = \frac{\pi}{4}\left(45^4-30^4 \right ) + \pi \left(45^2 -30^2 \right )\left(80 + 45 \right)^2\\
\amp = \mm{2.58 \times 10^6}^4 + \mm{55.2 \times 10^6}^4\\
I_y \amp = \mm{57.8 \times 10^6}^4
\end{align*}
It is interesting that the ‘correction factor’ is more than 20 times greater than the centroidal moment of inertia of the ring. This indicates the importance of the distance squared term on the moment of inertia of a shape.
