Question

A playground merry-go-round has a radius of 3.0 m and arotational inertia of 600 kg ⋅ m2. It

is initially spinning at 0.80 rad/s when a 20-kg child crawlsfrom the center to the rim. When
the child reaches the rim the angular velocity of themerry-go-round is:_________
A) 0.61 rad/s
B) 0.73 rad/s
C) 0.80 rad/s
D) 0.89 rad/s
E) 1.1rad/s

Answer 1

Final angular velocity,
`\omega_f=0.61\ rad/s`

Step-by-step explanation:

Given that,

Radius of merry- go- round, r = 3 m

Inertia of the merry- go- round,
`I=600\ kg-m^2`

Angular speed,
`\omega=0.8\ rad/s`

Mass, m = 20 kg

Let I' is the new rotational inertia of the merry- go- round. Here, the angular momentum of the system remains conserved. So,

`L_f=L_o`

`I_f\omega_f=I_o\omega_o`

`\omega_f=((I)/(I+mr^2))\omega_o`

`\omega_f=((600)/(600+20* 3^2))* 0.8`

`\omega_f=0.61\ rad/s`

So, the angular velocity of the merry-go-round is 0.61 rad/s. So, the correct option is (A). Hence, this is the required solution.