Coin Weight Variation Tolerance Calculation

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Find variation - really need help!

guys i really need your help on this. I don't even know which section to look for here. can some one please get me started for this problem.

Problem: A manufacturer contracts to mint coins for the federal government. how much variation dr in the radius of the coins can be tolerated if the coins are to weigh within 1/50 of their ideal weight? Assume the thickness does not vary.
 
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I do not know if this is correct, but what I'm thinking is:

1) assume the material is homogeneous and isotropic
2) assume the thickness is constant
3) assume the volume may be written as the product of the area of a circle and its constant height

Then, we can write the mass as the usual density, i.e. m=\rho V then the weight is just g times the mass. So, note that
dm = \rho dV=\rho (2\pi r dr) t
where t is the constant thickness. So, if we compare to its ideal weight, note the constant g drops out, along with a lot of other stuff. So
\frac{dm}{m} = \frac{\rho dV}{\rho V}=\frac{2dr}{r}
So, plugging in what we know we have
\frac{1}{50} =\frac{2dr}{r} \Rightarrow \frac{dr}{r}=\frac{1}{100}
So the variation in the radius must be within 1/100. Does this make sense? I think it works.
 
Thanks xman... yes it looks correct...

can anyone else look at it and confirm it? I want to make sure i do the homework right.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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