Centripetal merry-go-round problem

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Homework Help Overview

The problem involves a carnival merry-go-round rotating about a vertical axis, with a focus on calculating the distance of a man standing at the edge based on his speed and centripetal acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal acceleration and radius, with one participant attempting to calculate the radius using the given values. Questions arise regarding the interpretation of terms like "centripetal acceleration of magnitude" and whether the speed can be directly used in calculations.

Discussion Status

Some participants provide affirmations regarding the calculations and clarify the meanings of terms related to speed and acceleration. Multiple interpretations of the problem's wording are being explored, but no consensus has been reached on the approach to the solution.

Contextual Notes

The original poster expresses uncertainty about their calculations and the implications of the terms used in the problem statement, indicating potential gaps in understanding the concepts involved.

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Homework Statement


A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.43 m/s and a centripetal acceleration of magnitude 1.87 m/s2. How far is the man from the center of the merry-go-round?


Homework Equations


Ac= v^2/r


The Attempt at a Solution


I understand I am looking for the radius in the equation, I am just not sure about the other aspects of it. I get a radius of 6.29 the way i do it by: (3.43^2/1.87). Not confident though.
 
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Looks fine to me.
 
alright, so the constant speed of 3.43 m/s can be plugged in as the velocity? and I am confused on the wording "centripetal acceleration of magnitude... "
 
"...centripetal acceleration of magnitude 1.87 m/s2" means that the magnitude of the centripetal acceleration vector is 1.87.
 
Similarly the speed is the magnitude of the velocity vector, so yes you use the speed for v in the equation.
 

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