Small block on a rotating table

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The discussion focuses on calculating the maximum radius at which a small block can be placed on a rotating table before it moves outward. The key equation derived is r = μsg/4π²f², where r is the radius, μs is the static friction coefficient, g is the acceleration due to gravity, and f is the frequency of rotation. The centripetal force must equal the frictional force to maintain the block's position, leading to this relationship. The analysis highlights that the maximum radius is primarily influenced by the rotation speed and the frictional coefficient of the materials involved. Overall, the derivation appears correct, with a note that different materials will yield varying frictional coefficients.
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Imagine a surface that rotates with frequency f about its center, if we set a small block (or a coin, or any flat object for that matter) on the table, I wanted to calculate the maximum radius that you can place this block from the center before it starts to move outward from the center. This is how I did it:

I figured the maximum centripetal force on block had to equal the maximum frictional force keeping it in place.

Fc = Ff

and since:

Fc = mv2/r = 4π2rf2m
Ff = μsmg


where v2 = 4π2r2f2

where f is the frequency of rotation in Hz

Now I set

2rf2m = μsmg

2rf2 = μsg

The mass variables m cancel, this is why we can add any object, as long as it's center of mass rests at a distance r from the center.

Now it's easy to see that

r = μsg/4π2f2
I think it's interesting that the distance is only dependent on how fast the table is rotating, and that there are no other factors(other than what planet you are on :p) that determine it.

Would this be considered a correct derivation? Basically I just want to know if I missed anything or if I have the correct basic intuition behind this type of situation.

Thanks PF! :)
 
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Well technically the object does matter since you will get different frictional coefficients depending on the materials you end up putting on the table. But yea, if you can get the same frictional coefficients then they would be the same.
 
Yea I forgot to include the friction coefficient in that sentence, oh well D:
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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