Answer:
77.76729 rad/s or 12.37704 rev/s
[tex]\frac{K_n}{K_o}=6.18852[/tex]
Explanation:
[tex]I_1[/tex] = Rotational inertia of the system consisting of the man, bricks, and platform about the central axis = 7.55 kgm²
[tex]I_2[/tex] = Decreased the rotational inertia of the system = 1.22 kgm²
[tex]\omega_1[/tex] = Angular velocity of the old system = [tex]4\pi[/tex]
[tex]\omega_1[/tex] = Angular velocity of the new system
[tex]K_n[/tex] = Kinetic energy of the new system
[tex]K_o[/tex] = Kinetic energy of the old system
Here, angular momentum is conserved
[tex]L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \omega_2=\frac{I_1\omega_1}{I_2}\\\Rightarrow \omega_2=\frac{7.55\times 2\times 2\pi}{1.22}\\\Rightarrow \omega_2=77.76729\ rad/s\\ =\frac{\frac{7.55\times 2\times 2\pi}{1.22}}{2\pi}=12.37704\ rev/s[/tex]
The resulting angular speed of the platform is 77.76729 rad/s or 12.37704 rev/s
[tex]\frac{K_n}{K_o}=\frac{\frac{1}{2}I_2\omega_2^2}{\frac{1}{2}I_1\omega_1^2}\\\Rightarrow \frac{K_n}{K_o}=\frac{I_2\omega_2^2}{I_1\omega_1^2}\\\Rightarrow \frac{K_n}{K_o}=\frac{1.22\times 77.76729^2}{7.55\times (2\times 2\pi)^2}\\\Rightarrow \frac{K_n}{K_o}=6.18852[/tex]
The ratio of the new kinetic energy of the system to the original kinetic energy is [tex]\frac{K_n}{K_o}=6.18852[/tex]
