Respuesta :
Answer:
[tex]\omega_{f}=19.01\ rpm[/tex]
Explanation:
given,
diameter of merry - go - round = 3 m
mass of the disk = 230 kg
speed of the merry- go-round = 17 rpm
speed = 4.2 m/s
assuming mass of John = 30 kg
[tex]I_{disk} = \dfrac{1}{2}MR^2[/tex]
[tex]I_{disk} = \dfrac{1}{2}\times 230 \times 1.5^2[/tex]
[tex]I_{disk} = 258.75 kg.m^2[/tex]
initial angular momentum of the system
[tex]L_i = I\omega_i + mvR[/tex]
[tex]L_i =258.75\times 17 \times \dfrac{2\pi}{60} + 30 \times 4.2 \times 1.5[/tex]
[tex]L_i =649.64\ kg.m^2/s[/tex]
final angular momentum of the system
[tex]L_f = (I_{disk}+mR^2)\omega_{f}[/tex]
[tex]L_f = (258.75 + 30\times 1.5^2)\omega_{f}[/tex]
[tex]L_f= (326.25)\omega_{f}[/tex]
from conservation of angular momentum
[tex]L_i = L_f[/tex]
[tex]649.64 = (326.25)\omega_{f}[/tex]
[tex]\omega_{f}=1.99 \times \dfrac{60}{2\pi}[/tex]
[tex]\omega_{f}=19.01\ rpm[/tex]
