Solving the Temperature Difference for Steel & Aluminum Rings

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To determine the temperature at which a steel sphere falls through an aluminum ring, the coefficients of linear expansion for both materials must be applied to their respective diameters. The steel sphere has a diameter of 4 cm and the aluminum ring has an inside diameter of 3.9940 cm at 0°C. The correct approach involves calculating the change in diameter for both materials as a function of temperature increase (ΔT) using the formula D = D_0 + D_0 * α * T. By setting the diameters equal and solving for temperature, the accurate threshold can be found. This method clarifies the relationship between temperature and diameter expansion for both materials.
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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
 
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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 \Delta T. Then solve for the temperature at which the diameters are equal.
 
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
 
The diameter at temperature T equals the original diameter (at 0 degrees) plus the change in diameter from the temperature increase (\Delta T = T):
D = D_0 + D_0 \alpha T

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