# Problem in a circular motion (find the max Frequency)

Giannakoulis
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In the figure below we put a small coin on a disc, which rotates at constant frequency f, at a distance D=0.2m by the disc rotation axis. If the coefficient of friction is μ=0.5, then what is is the maximum frequency that can have the disc so that the currency does not slip? g=10 m/sec^2

Anyone could help me how to start? I know that i can take Fk=Ts=m*v^2/D and replace the velocity with v=2πDf but i dont know the mass.. am i on the right way? Ts=static friction

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## Answers and Replies

Homework Helper
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You are on the right track. Call the mass "m" and proceed with the algebraic solution. Maybe you will discover that you don't need the mass after all.

Also, I don't know what Fk=Ts=v^2/D is all about, but it does not look correct.

Giannakoulis
its the centripetal force my mistake, Fc=m*v^2/D which equals to static friction if i am correct.

Homework Helper
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You are correct. As the speed increases, the force of static friction has to increase to keep the coin going around in the circle. Can the force of static friction increase indefinitely?

Homework Helper
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@kuruman, do any of the offered answers look right to you? I get the reciprocal of one of them.

Giannakoulis
@haruspex me too.. My math show me f=5π/2..am I doing something wrong?

Homework Helper
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My math show me f=5π/2
That's not what I get. Please post your working.

Homework Helper
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@kuruman, do any of the offered answers look right to you? I get the reciprocal of one of them.
None of the answers look right to me and I too get the reciprocal of one of them.

Giannakoulis
I'm saying that static friction equals to maximum friction, also the static friction equals to centripetal force. So i have Ts=Fc=m*v^2/D and Tmax=μ*Ν (N=weight)..
If you replace the speed in the first equation you get Ts=4π²Df²m.. So my equation is: μN=4π²Df²m =>μ*m*g=4π²Df²m.. if i solve this for f and replace the numbers, i know i get fmax=5/2π.. what am i doing wrong then?

Homework Helper
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I also get 5/(2π). Are we both wrong in exactly the same way or is the correct answer missing from the list that you are given?

Giannakoulis
no its not.. anyway I will wait until the tuesday.. thanks both of you guys