Minimum Coefficient of Friction calculation using Centripetal force

AI Thread Summary
The discussion focuses on calculating the minimum coefficient of static friction required for a 61 kg child on a rotating merry-go-round with a radius of 1.05 m and an angular speed of 1.87 rad/s. The centripetal acceleration was calculated to be approximately 3.67 m/s², leading to a required force of about 224 N to maintain the child's circular path. Participants clarify that the minimum coefficient of friction can be determined using the relationship between friction force and normal force, emphasizing that the maximum friction must equal the required friction for stability. Confusion arises regarding the distinction between static and kinetic friction, but it is noted that the relevant equation for static friction should be applied. Understanding these concepts is crucial for accurately solving the problem.
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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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