A merry-go-round is a common piece of playground equip- ment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 30 kg. What is the merry-go- round’s angular velocity, in rpm, after John jumps on?

Respuesta :

Answer:

[tex]\omega_{f}=22.31\ rpm[/tex]  

Explanation:

given,  

diameter of merry - go - round = 3 m  

mass of the disk = 250 kg  

speed of the merry- go-round = 20 rpm  

speed = 5 m/s  

mass of John = 30 kg  

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

[tex]I_{disk} = \dfrac{1}{2}\times 250 \times 1.5^2[/tex]  

[tex]I_{disk} = 281.25 kg.m^2[/tex]  

initial angular momentum of the system  

[tex]L_i = I\omega_i + mvR[/tex]  

[tex]L_i =281.25 \times 20 \times \dfrac{2\pi}{60} + 30 \times 5 \times 1.5[/tex]  

[tex]L_i =814.57\ kg.m^2/s[/tex]  

final angular momentum of the system  

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

[tex]L_f = (281.25 + 30\times 1.5^2)\omega_{f}[/tex]  

[tex]L_f= (348.75)\omega_{f}[/tex]  

from conservation of angular momentum  

[tex]L_i = L_f[/tex]  

[tex]814.57 = (348.75)\omega_{f}[/tex]  

[tex]\omega_{f}=2.336 \times \dfrac{60}{2\pi}[/tex]  

[tex]\omega_{f}=22.31\ rpm[/tex]