Question

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.

Answer 1

`\omega_(f)=5.787\ rpm`

Step-by-step 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

`I_(disk) = (1)/(2)MR^2`

`I_(disk) = (1)/(2)* 250 * 2^2`

`I_(disk) = 500 kg.m^2`

initial angular momentum of the system

`L_i = I\omega_i + mvR`

`L_i =500 * 0 + 40 * 5 * 2`

`L_i =400\ kg.m^2/s`

final angular momentum of the system

`L_f = (I_(disk)+mR^2)\omega_(f)`

`L_f = (500 + 40* 2^2)\omega_(f)`

`L_f= (660)\omega_(f)`

from conservation of angular momentum

`L_i = L_f`

`400= (660)\omega_(f)`

`\omega_(f)=0.606 * (60)/(2\pi)`

`\omega_(f)=5.787\ rpm`