Calculating Coefficient of Friction for a Coin on a Turntable

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Homework Help Overview

The discussion revolves around calculating the coefficient of static friction for a coin placed on a rotating turntable. The original poster presents their calculations and expresses concern about a potential discrepancy with the answer provided in the textbook.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to verify their calculation of the coefficient of static friction, questioning the accuracy of the textbook answer. They express uncertainty about unit conversion and formatting issues. Other participants raise concerns about unit conversion from centimeters to meters and the proper representation of units.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's calculations and highlighting potential errors in unit conversion. There is an acknowledgment of common mistakes in handling units, suggesting a supportive environment for clarification.

Contextual Notes

Participants are discussing the implications of unit conversion errors and the formatting of answers in the context of homework guidelines. The original poster's calculations involve specific values for distance and speed, which are critical for determining the coefficient of friction.

atmega-ist
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I've got the answer to the problem but I'm one decimal place off from the back of the book and am suspecting a typo (no, lol... not that presumptuous...) the reason for the suspicion is that there's a zero after the answer where the units would go but this question is asking for a coefficient of friction so, of course, there should be no units... I'm just wondering if that zero was somehow misplaced in printing/formatting... I could be completely missing something though.

Homework Statement


A coin placed 30.0cm from the center of a rotating, horizontal turntable slips when its speed is 50.0cm/s. What is the coefficient of static friction between the coin and turntable?


Homework Equations


\mu_{s}=\frac{v^{2}}{gr}

\mu_{s}=\frac{.05m/s^{2}}{(9.81m/s^{2})*.03m/s^{2}}

\mu_{s}=.0085


The Attempt at a Solution



The book states ".085 0"

Is my solution correct or am I missing something?

Thanks!
 
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atmega-ist said:
I've got the answer to the problem but I'm one decimal place off from the back of the book and am suspecting a typo (no, lol... not that presumptuous...) the reason for the suspicion is that there's a zero after the answer where the units would go but this question is asking for a coefficient of friction so, of course, there should be no units... I'm just wondering if that zero was somehow misplaced in printing/formatting... I could be completely missing something though.

Homework Statement


A coin placed 30.0cm from the center of a rotating, horizontal turntable slips when its speed is 50.0cm/s. What is the coefficient of static friction between the coin and turntable?


Homework Equations


\mu_{s}=\frac{v^{2}}{gr}

\mu_{s}=\frac{.05m/s^{2}}{(9.81m/s^{2})*.03m/s^{2}}

\mu_{s}=.0085


The Attempt at a Solution



The book states ".085 0"

Is my solution correct or am I missing something?

Thanks!

You didn't convert from cm to meters correctly. The units aren't written properly, either. Is that just a typing error on your part?
 
You didn't convert from cm to meters correctly.

I have no words...

You're exactly right. Two semesters of chemistry with EVERYTHING in mm and mL started a pretty bad habit of glossing over conversions.

The units aren't written properly, either. Is that just a typing error on your part?

Sure is. I think I got a bit carried away with the ol' CTRL+V.

Thanks so much for the second look (a.k.a. "smack-in-the-back-of-the-head").

Lol... Apologies for taking up server space for this one :rolleyes:
 
Don't worry, I think we've all been there! :wink:
 

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