Question

A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg∙m2 about a vertical axis through its center. The system (including the child) is initially rotating at 1.80 rad/s.

What is its angular velocity if the child moves to a new position 1.20 m from the center of the merry-go-round?

Answer 1

2.3 rad/s

Step-by-step explanation:

moment of inertia of disc, I = 356 kgm^2

mass of child, m = 25 kg

radius of disc, r = 2.4 m

initial angular velocity, ω = 1.8 rad/s

r' = 1.2 m

Let the new angular velocity is ω'

Angular momentum remains constant

I x ω = I' x ω'

(356 + 25 x 2.4 x 2.44 ) x 1.8 = ( 356 + 25 x 1.2 x 1.2) x ω'

900 = 392ω'

ω' = 2.3 rad/s

Answer 2

`\omega_(f)=1.634\ rad/s`

Step-by-step explanation:

given,

diameter of merry - go - round = 2.40 m

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg

initial angular momentum of the system

`L_i = I\omega_i`

`L_i =356* 1.80`

`L_i =640.8\ kg.m^2/s`

final angular momentum of the system

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

`L_f = (356 + 25* 1.2^2)\omega_(f)`

`L_f= (392)\omega_(f)`

from conservation of angular momentum

`L_i = L_f`

`640.8= (392)\omega_(f)`

`\omega_(f)=1.634\ rad/s`