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Lynna [10]
3 years ago
15

A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic

al axle. As a child of mass 25.0 kg stands at a distance of 1.00 m from the axle, the system (merry-go-round and child) rotates at a rate of 14.0 rev/min. The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child teaches the edge?
Physics
1 answer:
ryzh [129]3 years ago
7 0

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

I_i = Initial angular momentum = 275\ kgm^2

\omega_i = Initial angular velocity = 14 rev/min

I_f = Final angular momentum

\omega_f = Final angular velocity

Here the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s

The final angular velocity is 0.2932 rad/s

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