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Steel Sphere

  • Thread starter pkossak
  • Start date
  • #1
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A steel sphere sits on top of an aluminum ring. The steel sphere (a (average
coefficient of linear expansion) = 1.1*10^-5/C) has a diameter of 4 cm at 0
C. The aluminum ring (a = 2.4*10^-5/C) has an inside diameter of 3.9940 cm at 0 C. Closest to which temperature given will the sphere just fall through the ring?

I thought that if I took the a*L*deltaT(aluminum ring) - a*L*deltaT(steel) =
1*10^-4 (negative difference of diameters, sphere - ring), I would get the
right answer. However, I wasn't all that close (ended up being 208 C, and I
got an answer much less than that). I was wondering if there was a better way to approach this problem? Thanks a lot
 

Answers and Replies

  • #2
Doc Al
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You know the diameters of the aluminum and steel at T = 0. Figure out the diameter of each as a function of [itex]\Delta T[/itex]. Then solve for the temperature at which the diameters are equal.
 
  • #3
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This has definitely helped me, but I'm stumped on what to do about the deltaL in thermal expansion equation. And if I do figure that out, what do I do to figure out T?

Would it be deltaL/L*a(steel) + deltaL/L*a(aluminum) = T?

Thank you a ton
 
  • #4
Doc Al
Mentor
44,912
1,170
The diameter at temperature T equals the original diameter (at 0 degrees) plus the change in diameter from the temperature increase ([itex]\Delta T = T[/itex]):
[tex]D = D_0 + D_0 \alpha T[/tex]

Use this to solve for the temperature at which both diameters are equal.
 

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