기타 풀이/회전관성

속이 빈 얇은 구 껍질의 회전 관성

짱세디럭스 2024. 5. 8. 01:24

$$\title{Rotational Inertia of Hollow Sphere}$$ $$\put \begin{cases} x&=\text{Latitude Line}\\ y&=\text{Longitude Line}\\ \end{cases} $$ $$ \begin{cases} r_x&=R\sin\theta\\ r_y&=R \end{cases} $$ $$ \begin{cases} x&=r_x \phi=(R\sin\theta)\phi\\ y&=r_y \theta=R\theta \end{cases} $$ $$ \begin{cases} \dd x&=R\sin\theta \dd \phi\\ \dd y&=R \dd \theta \end{cases} $$ $$r=r_x,$$ $$ \sigma=\frac{\dd m}{\dd a}=\frac{M}{A}=\frac{M}{4\pi R^2} \taag1$$ $$ \begin{aligned} \dd a&=\dd x \cdot \dd y\\ &=(R \sin\theta\dd \phi) \cdot (R \dd \theta)\\ &=R^2\sin\theta\cdot \dd \theta\dd \phi\taag2\\ \end{aligned} $$ $$ \begin{aligned} \dd m&=\sigma\cdot\dd a\\ &=\sigma\cdot (R^2\sin\theta\cdot \dd \theta\dd \phi)\\ &=\sigma R^2\sin\theta\cdot\dd\theta\dd\phi\taag3\\ \end{aligned} $$ $$ \begin{aligned} \dd I&={r}^2\cdot \dd m\\ &=\(R\sin\theta\)^2\cdot \(\sigma R^2\sin\theta\cdot\dd\theta\dd\phi\)\\ &=\sigma R^4\sin^3\theta\cdot\dd\theta\dd\phi\taag4\\ \end{aligned} $$ $$ \begin{aligned} I_{\text{Hollow Sphere}}&=\oiint_A \dd I\\ &=\int_0^{2\pi}\int_0^{\pi} \sigma R^4\sin^3\theta\cdot\dd\theta\dd\phi\\ &=\sigma \cdot R^4\cdot\(\int_0^{\pi}\sin^3\theta \dd\theta\)\cdot \(\int_0^{2\pi}\dd\phi\)\\ &=\(\frac{M}{4\pi R^2}\)\cdot R^4\cdot\[\frac{1}{3}\cos^3\theta-\cos\theta\]_0^{\pi}\cdot (2\pi)\\ &=\frac{2}{3}MR^2 \end{aligned} $$