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A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 240 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?

Respuesta :

Answer:

ω' = 0.815 rad/s

Explanation:

Given,

R = 1.20 m

Inertia of merry-go- round= 240 kg.m²

Rotating speed  = 9 rpm = [tex] 9\times \dfrac{2\pi}{60}[/tex]

                           =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

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