Coin on Turntable: Find Max Distance w/o Slipping

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SUMMARY

The problem involves calculating the maximum distance a coin can be placed from the center of a turntable rotating at 45.0 rpm without slipping, given a coefficient of static friction of 0.186. The correct approach requires converting angular velocity from radians per second to linear velocity, using the relationship v = ωr. The final calculation shows that the maximum radius (r) is 0.082 meters, confirming the correct application of the formula mv²/r ≤ μs mg.

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Homework Statement


A coin is placed on a turntable that is rotating at 45.0 rpm. If the coefficient of static friction between the coin and the turntable is 0.186, how far from the center of the turntable can the coin be placed without having it slip off?


Homework Equations



mv^2/r <= usmg
us= coefficient of static friction

The Attempt at a Solution


I tried using the equation and I converted 45 rpms to rad/sec and I got r=12.17 but this is wrong so I don't know what to do. Maybe I am using the wrong equation?
 
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Can you show more details of your calculations? We can't see what you've done wrong if you don't show your work.
 
mv^2/r =usmg
v^2/r = usg
(4.71^2)/r <= .186g
22.184 = 1.8228r
r= 12.17

This seems logical to me but it says it is wrong.
 
Note: the 4.71 rad/s, is angular velocity. You need to convert that to m/s, which is linear velocity. Do you know the relationship between these two? It should be in your textbook.
 
I think v = omega*r

Where v is the velocity, omega is angular speed (rad/s), and r is the radius.
so v= 4.71r
then (4.71r)^2/r= 22.18r = 1.8228
so r= .082

which is right! thank you!
 
You're welcome. :smile:
 

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