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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

Respuesta :

Answer:

Final angular velocity, [tex]\omega_f=0.61\ rad/s[/tex]

Explanation:

Given that,

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

Inertia of the merry- go- round, [tex]I=600\ kg-m^2[/tex]

Angular speed, [tex]\omega=0.8\ rad/s[/tex]

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,

[tex]L_f=L_o[/tex]

[tex]I_f\omega_f=I_o\omega_o[/tex]

[tex]\omega_f=(\dfrac{I}{I+mr^2})\omega_o[/tex]

[tex]\omega_f=(\dfrac{600}{600+20\times 3^2})\times 0.8[/tex]

[tex]\omega_f=0.61\ rad/s[/tex]

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.

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