Nonuniform circular motion merry-go-round

AI Thread Summary
A 5.0-meter diameter merry-go-round initially has a period of 4.0 seconds, resulting in a tangential speed of approximately 3.93 m/s for a child on the rim. To calculate the total revolutions made while stopping over 20 seconds, the user seeks to determine initial angular velocity and angular acceleration. The equations for angular motion are applied, including the relationship between angular velocity and period. The discussion highlights the need for further calculations to find the angular acceleration and total revolutions. Understanding these concepts is essential for solving the problem effectively.
AvrGang
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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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