Problem in a circular motion (find the max Frequency)

Giannakoulis
Messages
8
Reaction score
0
Member advised to use the homework template for posts in the homework sections of PF.
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 don't know the mass.. am i on the right way? Ts=static friction
 

Attachments

  • 14686180_967343080042526_1138366334_n.jpg
    14686180_967343080042526_1138366334_n.jpg
    39.3 KB · Views: 474
  • 14741594_967344550042379_2108361043_n.jpg
    14741594_967344550042379_2108361043_n.jpg
    15.4 KB · Views: 409
Last edited:
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.
 
its the centripetal force my mistake, Fc=m*v^2/D which equals to static friction if i am correct.
 
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?
 
@kuruman, do any of the offered answers look right to you? I get the reciprocal of one of them.
 
@haruspex me too.. My math show me f=5π/2..am I doing something wrong?
 
Giannakoulis said:
My math show me f=5π/2
That's not what I get. Please post your working.
 
haruspex said:
@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.
 
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?
 
  • #10
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?
 
  • #11
no its not.. anyway I will wait until the tuesday.. thanks both of you guys
 
  • #12
Giannakoulis said:
i get fmax=5/2π
Me too. In post #6 you wrote 5π/2.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 17 ·
Replies
17
Views
9K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 14 ·
Replies
14
Views
5K