Nonuniform circular motion merry-go-round

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SUMMARY

The discussion focuses on the physics of a merry-go-round with a diameter of 5.0 meters and an initial period of 4.0 seconds, which slows to a stop over 20 seconds. The speed of a child on the rim is calculated to be 3.93 m/s using the formula V = 2(π)r/T. Additionally, the user attempts to determine the total number of revolutions made during the stopping period using the angular displacement equation, but struggles to find the initial angular velocity (ωo) and angular acceleration (α).

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Homework Statement


A 5.0-m-diameter merry-go-round is initially turning with a 4.0 s period. It slows down and stops in 20 s.
a. Before slowing, what is the speed of a child on the rim?
b. How many revoluotions does the merry-go-round make as it stops?
( Can i know what is a merry-go-round)


Homework Equations





The Attempt at a Solution



I solved a) like that:

V= 2(Pi)r/T
V= 2(pi)(2.5)/4
V= 3.93 m/s

b) I think this is the correct equation:
(\theta)f = \thetao + \omegao t + a/2r ( t^2)

but i can't find \omegao or a
 
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http://www.worldsfinestshows.com/equipment/images/MerryGoRound_1.jpg
 
Last edited by a moderator:
initially turning with a 4.0 s period ---> T
ω=2pi/T
α = △ω/△T

θf = θo + ωo t + α/2( t^2)
 

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