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

`\omega_(f)=0.429\ rev/s`

Step-by-step 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) = (1)/(2)MR^2`

`I_(disk) = (1)/(2)* 125 * 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* 0.661 * 2\pi + 59.4 * 3.51 * 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* 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 * (1)/(2\pi)`

`\omega_(f)=0.429\ rev/s`