Answer:
The total angular momentum is 292.59 kg.m/s
Explanation:
Given that :
Rotation of the horizontal circular platform [tex]\omega[/tex] = 0.919 rad/s
mass of the platform (m) = 90.7 kg
radius (R) = 1.91 m
mass of the poodle [tex]m_p[/tex] = 20.5 kg
Your mass [tex]m'[/tex] = 73.5 kg
speed v = 1.05 m/s with respect to the platform
[tex]V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525 \ m /s \\ \\r = \frac{R}{2}[/tex]
r = 0.955
Mass of the mutt [tex]m_m[/tex] = 18.5 kg
[tex]r' = \frac{3}{4} \ R[/tex]
Your angular momentum is calculated as:
Your angular velocity relative to the platform is [tex]\omega' = \frac{v}{R} = \frac{1.05}{1.91 } = 0.5497 \ rad/s[/tex]
[tex]Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s[/tex]
[tex]I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2[/tex]
[tex]L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s[/tex]
For poodle :
[tex]Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \ rad/s[/tex]
Actual [tex]\omega_p = \omega - \omega' = 0.919 -0.550 = 0.369 \ rad/s[/tex]
[tex]I_p = m_p(\frac{R}{2} )^2 = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2[/tex]
[tex]L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \ kgm/s[/tex]
[tex]I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \ kgm^2[/tex]
[tex]L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s[/tex]
Disk [tex]I = \frac{mr^2}{2} = \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2[/tex]
[tex]L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s[/tex]
Total angular momentum of system is:
L = [tex]L_D +L_Y+L_P+L_M[/tex]
= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s
= 292.59 kg.m/s
