기타 풀이/회전관성

속이 빈 얇은 구 껍질의 회전 관성

짱세디럭스 2024. 5. 8. 01:24

[Rotational Inertia of Hollow Sphere]\title{Rotational Inertia of Hollow Sphere} put {x=Latitude Liney=Longitude Line\put \begin{cases} x&=\text{Latitude Line}\\ y&=\text{Longitude Line}\\ \end{cases} {rx=Rsinθry=R \begin{cases} r_x&=R\sin\theta\\ r_y&=R \end{cases} {x=rxϕ=(Rsinθ)ϕy=ryθ=Rθ \begin{cases} x&=r_x \phi=(R\sin\theta)\phi\\ y&=r_y \theta=R\theta \end{cases} { ⁣dx=Rsinθ ⁣dϕ ⁣dy=R ⁣dθ \begin{cases} \dd x&=R\sin\theta \dd \phi\\ \dd y&=R \dd \theta \end{cases} r=rx,r=r_x, σ= ⁣dm ⁣da=MA=M4πR2(1) \sigma=\frac{\dd m}{\dd a}=\frac{M}{A}=\frac{M}{4\pi R^2} \taag1  ⁣da= ⁣dx ⁣dy=(Rsinθ ⁣dϕ)(R ⁣dθ)=R2sinθ ⁣dθ ⁣dϕ(2) \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}  ⁣dm=σ ⁣da=σ(R2sinθ ⁣dθ ⁣dϕ)=σR2sinθ ⁣dθ ⁣dϕ(3) \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}  ⁣dI=r2 ⁣dm=(Rsinθ)2(σR2sinθ ⁣dθ ⁣dϕ)=σR4sin3θ ⁣dθ ⁣dϕ(4) \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} IHollow Sphere=A ⁣dI=02π0πσR4sin3θ ⁣dθ ⁣dϕ=σR4(0πsin3θ ⁣dθ)(02π ⁣dϕ)=(M4πR2)R4[13cos3θcosθ]0π(2π)=23MR2 \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}