솔루션피아

$$ \begin{cases} \Delta t&=220\ut{ms}=0.220\ut{s}\\\omega_i&=0\\\omega_f&=7.00\ut{rad/s}\\I&=16.0\ut{kg\cdot m^2}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\bar\alpha&=\frac{\Delta \omega}{\Delta t}\\&=\frac{350}{11}\ut{rad/s^2}\\&\approx 31.818181818181817\ut{rad/s^2}\\&\approx 31.8\ut{rad/s^2}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\bar\tau &= I \alpha\\&=\frac{5600}{11}\ut{N\cdot m}\\&..

$$ \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\\end{cases} $$$$ \begin{cases} \omega_0&=-4\ut{rad/s}\\\omega_4&=8\ut{rad/s}\\\RE_4&=2.20\ut{J}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\alpha&=\frac{\Delta \omega}{\Delta t}\\&=3.0\ut{rad/s^2}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\frac{\RE_0}{\RE_4}&=\frac{\frac{1}{2}I{\omega_0}^2}{\frac{1}{2}I{\omega_4}^2}=\(\frac{\omega_..

$$ \begin{cases} m&=1.6\ut{kg}\\L&=0.60\ut{m}\\\theta&=30\degree\\\omega&=1.2\ut{rad/s}\\\end{cases} $$$$ \begin{cases} m_1&=m_2=2m\\m_3&=m\\\end{cases} $$$$ \put \begin{cases} \RE : \text{Rotational Kinetic Energy}\\\end{cases} $$$$\ab{a}$$$$I=\Sigma mr^2,$$$$ \begin{cases} r_1&=r_2=L\\r_3&=2L\\\end{cases} $$$$ \begin{cases} I_{a1}&=I_{a2}=(2m)L^2=2mL^2\\I_3&=m(2L)^2=4mL^2\\\end{cases} $$$$ \be..

(풀이자주:바퀴의 정보가 부족하여 풀이자 임의로 균일분포 질량의 원기둥 형태로 가정했습니다. 만일 아닐경우 답은 달라집니다.)(풀이자주:풀이에 도형의 회전관성이 필요합니다. 관련한 내용은 별도의 링크로 분리했습니다. 해당 도형의 회전관성을 구하는 법에 관한 이해는 현재과정에서의 필수는 아니니 결론만 보고 건너뛰어도 무방합니다.)https://solutionpia.tistory.com/795 속이 채워진 원기둥의 회전 관성$$\title{Rotational Inertia of Solid Cylinder}$$ $$ \rho=\frac{\dd m}{\dd v}=\frac{M}{V}=\frac{M}{AH}=\frac{M}{\pi R^2 H} \taag1$$ $$ \begin{aligned} l&=r\theta,..

$$ \begin{cases} \omega_i&=33\frac{1}{3}\ut{rev/min}\\t&=15.0\ut{s}=\frac{1}{4}\ut{min}\\\omega_f&=0\\\end{cases} $$$$\ab{a}$$$$\omega=\omega_0+\alpha t,$$$$ \begin{aligned}\alpha&=\frac{\omega_f-\omega_i}{t}\\&=\frac{0-\(33+\frac{1}{3}\ut{rev/min}\)}{\frac{1}{4}\ut{min}}\\&=-\frac{400}{3}\ut{rev/min^2}\\&\approx -133.33333333333334\ut{rev/min^2}\\&\approx -133\ut{rev/min^2}\\\end{aligned} $$$$\..

$$ \begin{cases} R_A&=15\ut{cm}\\\omega_A&=10\ut{rad/s}\\R_B&=10\ut{cm}\\R_{B'}&=5\ut{cm}\\R_C&=25\ut{cm}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}v_1&=R_A\cdot\omega_A\\&=150\ut{cm/s}\\&=1.5\ut{m/s}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\omega_B&=\frac{v_1}{R_B}\\&=15\ut{rad/s}\\\end{aligned} $$$$\ab{c}$$$$ \begin{aligned}\omega_{B'}&=\omega_B\\&=15\ut{rad/s}\\\end{aligned} $$$$\ab{d}$..

$$ \begin{cases} \omega &= 33\frac{1}{3}\ut{rev/min}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\omega &= \(33+\frac{1}{3}\)\ut{rev/min}\cdot\frac{1\ut{min}}{60\ut{s}}\cdot\frac{2\pi\ut{rad}}{1\ut{rev}}\\&=\frac{10\pi}{9}\ut{rad/s}\\&\approx 3.490658503988659\ut{rad/s}\\&\approx 3.5\ut{rad/s}\\\end{aligned} $$$$\ab{b,c}$$$$ \begin{aligned}v(r)&=r\omega\\&=\frac{10\pi}{9}r\\\end{aligned} $$$$\ab{..

$$ \begin{cases} m&=0.855\ut{kg}\\L&=r=0.780\ut{m}\\\omega&=3160\ut{rev/min}\\F&=2.30\times10^{-2}\ut{N}\end{cases} $$$$\ab{a}$$$$I=\Sigma mr^2,$$$$ \begin{aligned}I&=0.520182\ut{kg\cdot m^2}\\&\approx 0.520\ut{kg\cdot m^2}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}\tau&=R\times F\\&=1.794\times10^{-2}\ut{N\cdot m}\\&\approx 1.79\times10^{-2}\ut{N\cdot m}\\&\approx 17.9\ut{mN\cdot m}\\\end{al..

$$ \begin{cases} d&=0.50\ut{m}\\m&=0.30\ut{kg}\\M&=\Sigma m=4m\end{cases} $$$$\ab{a}$$$$I=\Sigma mr^2,$$$$ \begin{cases} r_1&=0\\r_2&=d\\r_3&=d\\r_4&=\sqrt2 d\\\end{cases} $$$$ \begin{cases} I_1&=0\\I_2&=md^2\\I_3&=md^2\\I_4&=m(\sqrt2 d)^2=2md^2\\\end{cases} $$$$ \begin{aligned}\Sigma I&=I_1+I_2+I_3+I_4\\&=4md^2\\&=0.3\ut{kg\cdot m^2}\end{aligned} $$$$\ab{b}$$$$ \put \begin{cases} \RE : \text{Ro..

$$ \begin{cases} m&=85\ut{kg}\\R&=0.40\ut{m}\\g&=9.80665\ut{m/s^2}\\\end{cases} $$$$\tau=FR\sin\theta,$$$$ \begin{aligned}\max(\tau)&=F\cdot R\\&=34\ut{N\cdot m}\end{aligned} $$