Calculating Thickness of Copper Cladding on Zinc Pennies | US Mint 1983-Present

  • Thread starter Thread starter knowNothing23
  • Start date Start date
  • Tags Tags
    Density Fluids
Click For Summary
The discussion centers on calculating the thickness of the copper cladding on zinc pennies minted by the US Mint since 1983. Participants are modeling the penny as a uniform cylinder and using the densities of zinc and copper to derive the mass equation. The equation incorporates the volumes of both the copper and zinc components, with specific attention to the geometry of the penny. There is some confusion regarding the volume calculations, particularly the distinction between the total volume and the volume of the zinc core. The conversation emphasizes the need for clarity in the volume formulas to accurately determine the copper cladding thickness.
knowNothing23
Messages
35
Reaction score
0
26 ••• Since 1983, the US Mint has coined pennies that are made out of zinc
with a copper cladding. The mass of this type of penny is 2.50 g. Model the penny
as a uniform cylinder of height 1.23 mm and radius 9.50 mm. Assume the copper
cladding is uniformly thick on all surfaces. If the density of zinc is 7140 kg/m3
and that of copper is 8930 kg/m3, what is the thickness of the copper cladding?

The problem is similar to a coin inside a coin. We need to find the outer coin's height- the inner coin's height or thickness of the outer coin.

The inner coin is made of Zinc.

m = (ρ Cu)Vcu + (ρ Zn) Vzn, where p is rho of Cu and Zn respectively and Vcu and Vzn are volumes of copper and zinc respectively.

m = ρ Cu[ 2π(r)square d + 2πr (h − 2d )d ] + ρ Zn π (r − d )(square) (h − 2d ), where d is the thickness sought.

Could someone help? I don't undestand why the V covering the zinc coin is, what's in bold.
The volume I get is total volume- volume of zinc coin: πr(square)h-π(r-d)(square)(h-2d)
 
Physics news on Phys.org
one of the Cu π(r)square d is the top disk , the other π(r)square d is the bottom disk.
2πr (h − 2d) d is the edge band , in between the top and bottom disks "approximately".
Your way is exact , so it's better ... if you can solve it.
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
View full post »