Calculating Coefficient of Friction for a Coin on a Turntable

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The discussion revolves around calculating the coefficient of static friction for a coin on a turntable, where the user suspects a typo in the textbook answer due to an extra zero. The user calculated the coefficient as 0.0085, while the book states ".085 0," leading to confusion about the correct value. A participant pointed out that the user may have incorrectly converted units from centimeters to meters, which contributed to the discrepancy. The user acknowledged this mistake, attributing it to a habit formed from previous coursework. Overall, the conversation highlights the importance of accurate unit conversion in physics calculations.
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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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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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