Centripetal acceleration and static friction

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Homework Help Overview

The problem involves a coin placed on a rotating record, specifically examining the conditions under which the coin will not slip off due to static friction. The subject area includes concepts of centripetal acceleration and friction, with a focus on the relationship between angular velocity and linear velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conversion of rotational speed from rpm to radians per second and the correct application of equations relating centripetal acceleration and friction. There is confusion regarding the distinction between angular velocity and linear velocity, with attempts to clarify the use of variables in the equations.

Discussion Status

The discussion is active, with participants offering guidance on the correct formulas and questioning the assumptions made in the original poster's approach. There is a mix of interpretations regarding the application of the equations, and some participants suggest working with the variables before substituting numerical values.

Contextual Notes

There is an emphasis on understanding the relationships between angular and linear quantities, as well as the need to derive a formula for the unknown radius before substituting specific values. The original poster expresses uncertainty about their calculations and the correct application of the equations.

shaanyr
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Homework Statement




A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.2, how far from the center of the record can the coin be placed without having it slip off?

Homework Equations



mv^2/r = mg(mu)

The Attempt at a Solution



I can't figure out what I'm doing wrong. I converted the 33.3 rpm to radians /sec and pluged that in for v then solved for R.

I keep getting 6.204254889


 
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Welcome to PF!

Hi shaanyr! Welcome to PF! :smile:

(have a mu: µ and an omega: ω :wink:)
shaanyr said:
mv^2/r = mg(mu)

I can't figure out what I'm doing wrong. I converted the 33.3 rpm to radians /sec and pluged that in for v

Nooo … that's ω, not v isn't it? :cry:

Use ω2r. :smile:
 
im sorry i don't understand

use

[tex]\omega^2/r[/tex] = g x [tex]\mu[/tex]
 
High school physics to the rescue!
Make a proportion to find the period:
33 turns/60 s = T turns/1 second
Once you have the period you can use v = circumference/period.

This is an interesting problem because r is unknown, yet needed to find v. In this kind of problem, you have to work with the letters and put in numbers only at the last stage when you have an r = formula.

Show your equations if you would like more help or a check!
 
Just got up :zzz: …
shaanyr said:
[tex]\omega^2/r[/tex] = g x [tex]\mu[/tex]

No …

look at the dimensions (or the units) …

v = ωr,

so centripetal acceleration = v2/r = ω2r. :smile:
 

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