Answer to Question #109473 in Mechanics | Relativity for wantmeknow

Question

[img]https://upload.cc/i1/2020/04/13/32I8fV.jpg[/img]



question in photo,d8 is 4

Answer 1

(a)

$L_1=I_1\omega_1=(I_0+m_1d^2)\omega_1=(\frac{1}{2}m_1r_1^2+m_1(1.2-0.8)^2)\omega_1=$

$=(\frac{1}{2}\cdot 150\cdot 1.2^2+150\cdot(1.2-0.8)^2)\cdot2.5=330$ $\frac{kg\cdot m^2\cdot rad}{s}$

$L_2=-m_2(1.4\cdot\sin58°)^2\cdot\frac{u}{1.4\cdot\sin58°}=$

$=-30\cdot(1.4\cdot\sin58°)^2\cdot\frac{2}{1.4\cdot\sin58°}\approx-71$ $\frac{kg\cdot m^2\cdot rad}{s}$

$L_\Sigma=330-71=259$ $\frac{kg\cdot m^2\cdot rad}{s}$

(b)

$I_1\omega_1+I_2\omega_0=I\omega_2$

$I=I_1+30\cdot1.4^2=132+58.8\approx191$ $kg\cdot m^2$

$\omega_2=\frac{L_1+L_2}{191}=\frac{259}{191}=1.36$ $rad/s$

(c)

$(I_1+m\cdot1.4^2)\omega_2=I_1\omega_3 \to \omega_3=\frac{(132+30\cdot1.4^2)\cdot1.36}{132}=1.96$ $rad/s$