Question

A disk-shaped merry-go-round of radius 3.03 m and mass 125 kg rotates freely with an angular speed of 0.661 rev/s . A 59.4 kg person running tangential to the rim of the merry-go-round at 3.51 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.

Answer 1

Answer:

$\omega_{f}=0.429\ rev/s$

Explanation:

given,  

radius of merry - go - round = 3.03 m  

mass of the disk = 125 kg  

speed of the merry- go-round = 0.661 rev/s

speed = 3.51 m/s  

mass of person =59.4 kg  

$I_{disk} = \dfrac{1}{2}MR^2$  

$I_{disk} = \dfrac{1}{2}\times 125 \times 3.03^2$  

$I_{disk} = 573.81 kg.m^2$  

initial angular momentum of the system  

$L_i = I\omega_i + mvR$  

$L_i =573.81\times 0.661 \times 2\pi + 59.4 \times 3.51 \times 3.03$  

$L_i =3014.86\ kg.m^2/s$  

final angular momentum of the system  

$L_f = (I_{disk}+mR^2)\omega_{f}$  

$L_f = (573.81 + 59.4\times 3.03^2)\omega_{f}$  

$L_f= (1119.16)\omega_{f}$  

from conservation of angular momentum  

$L_i = L_f$  

$3014.86 = (1119.16)\omega_{f}$  

$\omega_{f}=2.694 \times \dfrac{1}{2\pi}$  

$\omega_{f}=0.429\ rev/s$