Question

A merry-go-round is a common piece of playground equip- ment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 30 kg. What is the merry-go- round’s angular velocity, in rpm, after John jumps on?

Answer 1

`\omega_(f)=22.31\ rpm`

Step-by-step explanation:

given,

diameter of merry - go - round = 3 m

mass of the disk = 250 kg

speed of the merry- go-round = 20 rpm

speed = 5 m/s

mass of John = 30 kg

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

`I_(disk) = (1)/(2)* 250 * 1.5^2`

`I_(disk) = 281.25 kg.m^2`

initial angular momentum of the system

`L_i = I\omega_i + mvR`

`L_i =281.25 * 20 * (2\pi)/(60) + 30 * 5 * 1.5`

`L_i =814.57\ kg.m^2/s`

final angular momentum of the system

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

`L_f = (281.25 + 30* 1.5^2)\omega_(f)`

`L_f= (348.75)\omega_(f)`

from conservation of angular momentum

`L_i = L_f`

`814.57 = (348.75)\omega_(f)`

`\omega_(f)=2.336 * (60)/(2\pi)`

`\omega_(f)=22.31\ rpm`