Answer to Question #135151 in Mechanics | Relativity for alfraid

Question

find moment of inertia of a circular ring about x-axis.

Answer 1

Explanations & Calculations

• Moment of inertia (I) is calculated by the relationship $I = \int r^2\delta m$ about the axis in consideration.

When the x axis lies on the middle, (perpendicular to the plane of the ring) as the first figure,

• Consider a small segment of mass $\small \delta m$ of the ring then,

$\qquad\qquad \begin{aligned} \small I_1 &= \small \int r^2 \delta m\\ &= \small r^2\int_{0}^M \delta m\\ &= \small r^2 (M-0)\\&= \small Mr^2 \end{aligned}$ : distance to the axis from each $\small \delta m$ is the same ; r

When the x axis lies on a diameter of the ring consider a $\small \delta m_1$ and linear mass density ($\small \rho = \frac{M}{2 \pi r}$ ). Then,

$\qquad\qquad \begin{aligned} \small \delta m_1 &= \small \rho \delta l=\rho r\delta \theta\cdots (\because S=r\theta)\\ \small I_2 &= \small \int y^2 \delta m\\ &= \small \int (r\sin\theta)^2 \times \rho r\delta \theta \\ &= \small r^3\rho \int_{0}^{2\pi} \sin^2\theta \delta \theta\\ &= \small r^3\rho \int_{0}^{2\pi} \frac{1-\cos2\theta}{2} \delta \theta \\ &= \small \frac{r^3\rho}{2} \Bigg[\theta-\frac{\sin2 \theta}{2} \Bigg]_{0}^{2\pi}\\ &= \small \frac{r^3}{2}\times \frac{M}{2\pi r}\times 2\pi\\ &= \small \frac{Mr^2}{2} \end{aligned}$