Small block on a rotating table

In summary, the conversation discusses how to calculate the maximum radius at which an object can be placed on a rotating surface before it starts to move outward. The formula for this is r = μsg/4π2f2, where μ is the friction coefficient, s is the maximum centripetal force on the block, g is the acceleration due to gravity, and f is the frequency of rotation. It is noted that the type of object placed on the surface can affect the friction coefficient, but otherwise the distance is only dependent on the frequency of rotation and gravity. The individual asks if this derivation is correct and if they have the correct understanding of the situation.
  • #1
Legaldose
74
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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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  • #2
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.
 
  • #3
Yea I forgot to include the friction coefficient in that sentence, oh well D:
 
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