기타 풀이/회전관성

속이 채워진 구의 회전 관성

짱세디럭스 2024. 5. 8. 01:25
[Rotational Inertia of Solid Sphere]\title{Rotational Inertia of Solid 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, ρ= ⁣dm ⁣dv=MV=M43πR3=3M4πR3(1) \rho=\frac{\dd m}{\dd v}=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}=\frac{3M}{4\pi R^3} \taag1  ⁣dv= ⁣dr ⁣dx ⁣dy= ⁣dr(rsinθ ⁣dϕ)(r ⁣dθ)=r2sinθ ⁣dr ⁣dθ ⁣dϕ(2) \begin{aligned} \dd v&=\dd r \cdot \dd x \cdot \dd y\\ &=\dd r \cdot (r \sin\theta\dd \phi) \cdot (r \dd \theta)\\ &=r^2\sin\theta\cdot \dd r \dd \theta\dd \phi\taag2\\ \end{aligned}  ⁣dm=ρ ⁣dv=ρ(r2sinθ ⁣dr ⁣dθ ⁣dϕ)=ρr2sinθ ⁣dr ⁣dθ ⁣dϕ(3) \begin{aligned} \dd m&=\rho\cdot\dd v\\ &=\rho\cdot (r^2\sin\theta\cdot \dd r \dd \theta\dd \phi)\\ &=\rho r^2\sin\theta\cdot\dd r \dd\theta\dd\phi\taag3\\ \end{aligned}  ⁣dI=r2 ⁣dm=(rsinθ)2(ρr2sinθ ⁣dr ⁣dθ ⁣dϕ)=ρr4sin3θ ⁣dr ⁣dθ ⁣dϕ(6) \begin{aligned} \dd I&={r}^2\cdot \dd m\\ &=\(r\sin\theta\)^2\cdot \(\rho r^2\sin\theta\cdot\dd r \dd\theta\dd\phi\)\\ &=\rho r^4\sin^3\theta\cdot\dd r \dd\theta\dd\phi\taag6\\ \end{aligned} ISolid Sphere=V ⁣dI=02π0π0Rρr4sin3θ ⁣dr ⁣dθ ⁣dϕ=ρ(0Rr4 ⁣dr)(0πsin3θ ⁣dθ)(02π ⁣dϕ)=(3M4πR3)(15R5)[13cos3θcosθ]0π(2π)=25MR2 \begin{aligned} I_{\text{Solid Sphere}}&=\oiiint_V \dd I\\ &=\int_0^{2\pi}\int_0^{\pi}\int_0^{R} \rho r^4\sin^3\theta\cdot\dd r \dd\theta\dd\phi\\ &=\rho\cdot\(\int_0^{R} r^4\dd r\)\cdot\(\int_0^{\pi}\sin^3\theta\dd\theta\)\cdot\(\int_0^{2\pi} \dd\phi\)\\ &=\(\frac{3M}{4\pi R^3}\)\cdot\(\frac{1}{5}R^5\)\cdot\[\frac{1}{3}\cos^3\theta-\cos\theta\]_0^{\pi}\cdot\(2\pi\)\\ &=\frac{2}{5}MR^2 \end{aligned}