기타 풀이/회전관성

두꺼운 원형 고리의 회전 관성

짱세디럭스 2024. 5. 8. 02:06
$$\title{Rotational Inertia of Ring}$$ $$ \put \begin{cases} R_i:\text{Inner Radius}\\ R_o:\text{Outer Radius} \end{cases} $$ $$ \sigma=\frac{\dd m}{\dd a}=\frac{M}{A}=\frac{M}{\pi {R_o}^2-\pi {R_i}^2} \taag1$$ $$ \begin{aligned} l&=r\theta,\\ \dd l&=r \dd \theta\\ \end{aligned} $$ $$ \begin{aligned} \dd a&=\dd r \cdot \dd l\\ &=\dd r \cdot (r \dd\theta)\\ &=r\cdot\dd r \dd\theta\taag2\\ \end{aligned} $$ $$ \begin{aligned} \dd m&=\sigma\cdot\dd a\\ &=\sigma\cdot (r\dd r \dd\theta)\\ &=\sigma r\cdot\dd r \dd\theta\taag3\\ \end{aligned} $$ $$ \begin{aligned} \dd I&=r^2\cdot \dd m\\ &=r^2\cdot \(\sigma r\dd r \dd\theta\)\\ &=\sigma r^3\cdot\dd r \dd\theta\taag4\\ \end{aligned} $$ $$ \begin{aligned} I_{\text{Ring}}&=\oiint_A \dd I\\ &=\int_0^{2\pi}\int_{R_i}^{R_o} \sigma r^3\dd r \dd\theta\\ &=\sigma\cdot\(\int_{R_i}^{R_o} r^3\dd r\)\cdot\(\int_0^{2\pi} \dd\theta\)\\ &=\(\frac{M}{\pi {R_o}^2-\pi {R_i}^2}\)\cdot\(\frac{{R_o}^4-{R_i}^4}{4}\)\cdot(2\pi)\\ &=\frac{1}{2}M({R_i}^2+{R_o}^2)\\ \end{aligned} $$