Select the correct answer.
2 answers:
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Answer:
ω' = 0.815 rad/s
Explanation:
Given,
R = 1.20 m
Inertia of merry-go- round= 240 kg.m²
Rotating speed = 9 rpm =
=0.9424 rad/s
mass of the child, m = 26 kg
angular speed of the merry-go-round=?
we know
Angular momentum, L = I ω
Moment of inertia of the child
I' = m r² = 26 x 1.2² = 37.44 kgm²
Conservation of angular momentum
initial angular momentum = Final angular momentum
I ω = (I+I')ω'
240 x 0.9424 = (240+37.44) ω'
226.176= 277.44 ω'
ω' = 0.815 rad/s
new angular speed of the merry-go- round is equal to 0.815 rad/s