11판/11. 굴림운동, 토크, 각운동량

11-20 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 5. 16. 07:01

(풀이자 주: v = ω × r의 증명은 별도 링크로 분리했습니다. 이에 관한 이해는 현재과정에서의 필수는 아니니 결론만 보고 건너뛰어도 무방합니다.)

https://solutionpia.tistory.com/864

 

v=ω × r 의 증명

θ=ωt, \begin{aligned}\theta&=\omega t,\end{aligned} r=rcosθi^+rsinθj^=rcos(ωt)i^+rsin(ωt)j^(1) \begin{aligned}\vec r&=r\cos\theta\i+r\sin\theta\j\\&=r\cos(\omega t)\i+r\sin(\omega t)\j\taag1\\\end{aligned} $$ \begin{aligned}\vec v&=\dyt{\vec r}\\&=\dt(r\cos(\omega t)\i+r\sin(\omega t)\j)\\&=-r\

solutionpia.tistory.com

put r^=rr,  r=rr^ \begin{aligned} \put \r &=\frac{\vec r}{r},\\ \therefore ~~\vec r&=r\r\\ \end{aligned} {θ=30°ω=38r^[rev/s]=76πr^[rad/s]m=0.41[kg]I=5.0×104[kgm2]r=4.0[cm]=4.0×102[m] \begin{cases} \theta&=30\degree\\ \vec \omega&=-38\r\ut{rev/s}=-76\pi\r\ut{rad/s}\\ m&=0.41\ut{kg}\\ I&=5.0\times10^{-4}\ut{kg\cdot m^2}\\ r&=4.0\ut{cm}=4.0\times10^{-2}\ut{m}\\ \end{cases}  ⁣dL ⁣dt=τnet =r×mg=rr^×mg(k^)=mgr(k^×r^)(1) \begin{aligned} \dyt{\vec L}&=\vec \tau_\net\\ &=\vec r\times m\vec g\\ &=r\r\times mg(-\k)\\ &=mgr(\k\times\r)\taag1 \end{aligned} L=Iωr^,\vec L=I\omega {\r} , put Ω=ωr,\put \vec \Omega=\vec \omega_{\vec r}, v= ⁣dr ⁣dt=ω×r, \vec v=\dyt {\vec r}=\vec \omega \times \vec r,  ⁣dL ⁣dt=Ω×L=Ωk^×Iωr^=ΩIω(k^×r^)(2) \begin{aligned} \dyt{ \vec L}&=\vec \Omega\times \vec L\\ &=\Omega\k\times I\omega\r\\ &=\Omega I \omega(\k\times\r)\taag2 \end{aligned} ΩIω(k^×r^)=mgr(k^×r^)ΩIω=mgrΩ=mgrIω \begin{aligned} \Omega I \omega(\k\times\r)&=mgr(\k\times\r)\\ \Omega I \omega&=mgr\\ \Omega&=\frac{mgr}{I \omega}\\ \end{aligned} Ω=mgrIωk^\therefore \vec \Omega=\frac{mgr}{I \omega}\k Ω=mgrIωk^=41g95πk^ \begin{aligned} \vec \Omega &=\frac{mgr}{I \omega}\k\\ &=-\frac{41 g}{95 \pi }\k\\ \end{aligned} (a)\ab{a} Ω=41g95π1.3471968364117382[rad/s]1.3[rad/s] \begin{aligned} \Omega &=\frac{41 g}{95 \pi }\\ &\approx 1.3471968364117382\ut{rad/s}\\ &\approx 1.3\ut{rad/s}\\ \end{aligned} (b)\ab{b} Ω<0Clockwise \begin{aligned} \vec \Omega\lt0\Rarr\text{Clockwise} \end{aligned}