Respuesta :
Answer:
[tex]\omega_{2} = 1.107\,\frac{rad}{s}[/tex]
Explanation:
Let assume that circular platform is a solid cylinder. Given the absence of external forces, the situation can be analyzed by applying the Principle of Angular Momentum, which states that:
[tex]I_{1}\cdot \omega_{1} = I_{2}\cdot \omega_{2}[/tex]
The initial moment of inertia is:
[tex]I_{1} = \frac{1}{2}\cdot (89.3\,kg)\cdot (1.69\,m)^{2}[/tex]
[tex]I_{1} = 127.525\,kg\cdot m^{2}[/tex]
Likewise, the final moment of inertia is:
[tex]I_{2} = 127.525\,kg\cdot m^{2} + (8.83\,kg)\cdot (1.352\,m)^{2} + (21.1\,kg)\cdot (1.69\,m)^{2}[/tex]
[tex]I_{2} = 203.929\,kg\cdot m^{2}[/tex]
The final angular speed is:
[tex]\omega_{2} = \frac{I_{1}}{I_{2}} \cdot \omega_{1}[/tex]
[tex]\omega_{2} = \frac{127.525\,kg\cdot m^{2}}{203.929\,kg\cdot m^{2}} \cdot (1.77\,\frac{rad}{s} )[/tex]
[tex]\omega_{2} = 1.107\,\frac{rad}{s}[/tex]
Answer:
The angular velocity of the platform is 1.11 rad/s
Explanation:
The initial angular momentum is equal to:
[tex]L_{o} =\frac{1}{2} m_{o} R_{o} ^{2} w_{o}[/tex]
Where
m₀ = 89.3 kg
R₀ = 1.69 m
w₀ = 1.77 rad/s
When the moment is conserved, then:
L₀ = L₁
[tex]L_{1} =\frac{1}{2} m_{o} R_{o} ^{2} w_{1} +(m_{1} R_{1}^{2} )w_{1}[/tex]
Where
m₁ = 8.83 kg
R₁ = (4/5)R₀
The monkey drops vertically, the moment is conserved, then:
L₀ = L₂
[tex]L_{2} =\frac{1}{2} m_{o} R_{o} ^{2} w_{2} +(m_{1} R_{1}^{2} )w_{2}+(m_{2} R_{o}^{2} )w_{2}\\\frac{1}{2} m_{o} R_{o} ^{2} w_{o}=\frac{1}{2} m_{o} R_{o} ^{2} w_{2}+(m_{1}(4/5)^{2} R_{o} ^{2})w_{2}+(m_{2} R_{o}^{2} )w_{2}\\w_{2}=\frac{m_{o}w_{o}}{m_{o}+(32/25)m_{1}+2m_{2}}[/tex]
Replacing:
[tex]w_{2}=\frac{89.3*1.77}{89.3+(32/25)*8.83+(2*21.1)} =1.11rad/s[/tex]
