Minimum Coefficient of Friction calculation using Centripetal force

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Homework Help Overview

The problem involves calculating the minimum coefficient of static friction required for a child standing on a rotating merry-go-round. The context includes centripetal acceleration, forces acting on the child, and the relationship between friction and normal force.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of centripetal acceleration and the force required to maintain circular motion. There is confusion regarding the determination of the minimum coefficient of static friction, particularly in relation to the equations for static versus kinetic friction.

Discussion Status

Some participants have provided insights into the relationship between friction force and the coefficient of friction. There is an ongoing exploration of the definitions and equations relevant to static friction, with no explicit consensus reached on the approach to finding the minimum coefficient.

Contextual Notes

Participants note the distinction between static and kinetic friction, and there is a reference to the potential confusion surrounding the term "minimum" in the context of friction calculations.

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A() 61 kg child stands at the rim of a merrygo-round of radius 1.05 m, rotating with an
angular speed of 1.87 rad/s
The acceleration of gravity is 9.8 m/s^2

What is the child’s centripetal acceleration?
Answer in units of m/s

What minimum force between her feet and
the floor of the carousel is required to keep
her in the circular path?
Answer in units of N


What minimum coefficient of static friction is
required?

The Attempt at a Solution



1)
I found the centripetal acceleration since a = v^2/r and v = 1.9635 and r = 1.05.
The acceleration is 3.671745

2) I found the force which is F=ma
F = 61(3.671745) = 223.976445 N

3) I'm confused as to how I should solve for the minimum of coefficient of friction.

Help is appreciated!
 

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swag said:
A() 61 kg child stands at the rim of a merrygo-round of radius 1.05 m, rotating with an
angular speed of 1.87 rad/s
The acceleration of gravity is 9.8 m/s^2

What is the child’s centripetal acceleration?
Answer in units of m/s

What minimum force between her feet and
the floor of the carousel is required to keep
her in the circular path?
Answer in units of N


What minimum coefficient of static friction is
required?

The Attempt at a Solution



1)
I found the centripetal acceleration since a = v^2/r and v = 1.9635 and r = 1.05.
The acceleration is 3.671745

2) I found the force which is F=ma
F = 61(3.671745) = 223.976445 N

3) I'm confused as to how I should solve for the minimum of coefficient of friction.

Help is appreciated!

Assumin it is the word "minimum"that is confusing you, remember that the usual formula F = μN refers to the maximum friction possible, so if the coefficient of friction was 2.3, only a fraction of the potential friction would be necessary.
If the friction force required matches the maximum friction available, then we have the minimum coefficient.

If it is the idea of friction that is confusing you, you better re-read the section of your text on friction
 
But, I have to solve for the coefficient of static friction. The equation for static friction is f ≤ μN , the one you showed is for kinetic friction.
 
swag said:
But, I have to solve for the coefficient of static friction. The equation for static friction is f ≤ μN , the one you showed is for kinetic friction.

OK, replace the = in my response with ≤ and re-read.

EDIT: I don't distinguish between static and kinetic - since I am always aware which one I am using.
 

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