Question

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round, which can be modeled as a disk with a mass of 230 kg , is spinning at 17 rpm. John runs tangent to the merry-go-round at 4.2 m/s. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

Answer 1

`\omega_(f)=19.01\ rpm`

Step-by-step 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

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

`I_(disk) = (1)/(2)* 230 * 1.5^2`

`I_(disk) = 258.75 kg.m^2`

initial angular momentum of the system

`L_i = I\omega_i + mvR`

`L_i =258.75* 17 * (2\pi)/(60) + 30 * 4.2 * 1.5`

`L_i =649.64\ kg.m^2/s`

final angular momentum of the system

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

`L_f = (258.75 + 30* 1.5^2)\omega_(f)`

`L_f= (326.25)\omega_(f)`

from conservation of angular momentum

`L_i = L_f`

`649.64 = (326.25)\omega_(f)`

`\omega_(f)=1.99 * (60)/(2\pi)`

`\omega_(f)=19.01\ rpm`