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Joey, whose mass is 40kg , stands at rest at the outer edge of the frictionless 250kg merry-go-round, 2.0m from its center. The merry-go-round is also at rest. Joey then begins to run around the perimeter of the merry-go-round, finally reaching a constant speed, measured relative to the ground, of 5.0 m/s.Joey, whose mass is 40kg , stands at rest at the outer edge of the frictionless 250kg merry-go-round, 2.0m from its center. The merry-go-round is also at rest. Joey then begins to run around the perimeter of the merry-go-round, finally reaching a constant speed, measured relative to the ground, of 5.0 m/s.Joey, whose mass is 40kg , stands at rest at the outer edge of the frictionless 250kg merry-go-round, 2.0m from its center. The merry-go-round is also at rest. Joey then begins to run around the perimeter of the merry-go-round, finally reaching a constant speed, measured relative to the ground, of 5.0 m/s.

What is the final angular speed of the merry-go-round? Assume that Joey runs in positive direction.

Respuesta :

Answer:

[tex]\omega_{f}=5.787\ rpm[/tex]

Explanation:

given,  

radius of merry - go - round = 2 m  

mass of the disk = 250 kg  

initial angular speed = 0 rpm

speed = 5 m/s  

mass of Joey = 40 kg  

[tex]I_{disk} = \dfrac{1}{2}MR^2[/tex]  

[tex]I_{disk} = \dfrac{1}{2}\times 250 \times 2^2[/tex]  

[tex]I_{disk} = 500 kg.m^2[/tex]  

initial angular momentum of the system  

[tex]L_i = I\omega_i + mvR[/tex]  

[tex]L_i =500 \times 0 + 40 \times 5 \times 2[/tex]  

[tex]L_i =400\ kg.m^2/s[/tex]  

final angular momentum of the system  

[tex]L_f = (I_{disk}+mR^2)\omega_{f}[/tex]  

[tex]L_f = (500 + 40\times 2^2)\omega_{f}[/tex]  

[tex]L_f= (660)\omega_{f}[/tex]  

from conservation of angular momentum  

[tex]L_i = L_f[/tex]  

[tex]400= (660)\omega_{f}[/tex]  

[tex]\omega_{f}=0.606 \times \dfrac{60}{2\pi}[/tex]  

[tex]\omega_{f}=5.787\ rpm[/tex]  

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