Calculating Coefficient of Friction for a Coin on a Turntable

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SUMMARY

The discussion centers on calculating the coefficient of static friction for a coin on a turntable, specifically when the coin slips at a speed of 50.0 cm/s from a distance of 30.0 cm from the center. The correct formula used is μs = v² / (g * r), leading to a computed value of μs = 0.0085. The user suspects a typographical error in the textbook answer, which states ".085 0", due to improper unit conversion from centimeters to meters. The final conclusion confirms that the user's calculation is accurate, but highlights the importance of proper unit conversion.

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  • Basic algebra for manipulating equations
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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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