How Does Temperature Affect the Fit Between a Brass Ring and an Aluminum Rod?

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Heating a brass ring and an aluminum rod causes both to expand, but the brass ring expands less than the aluminum rod due to differing coefficients of linear expansion. To separate a brass ring of 10.00 cm diameter from an aluminum rod of 10.01 cm diameter at 20.0°C, cooling is necessary, as both materials will contract, potentially allowing separation. The discussion raises questions about whether heating could also achieve separation, but it is clarified that both materials would expand, complicating the fit. The equation Lfrod = Lfring indicates that the final lengths of both materials must be equal for separation to occur. Understanding the thermal expansion properties of both metals is crucial in determining the feasibility of separation.
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1.A brass ring of diameter 10.00 cm at 20.0°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 20.0°C. Assuming the average coefficients of linear expansion are constant, (a) to what temperature must this combination be cooled to separate them? Is this attainable? (b) What if the aluminum rod were 10.02 cm in diameter?
Lfrod = Lfring


2. Linear expansion of solids Δl = l0αΔT



3. Shouldn't the temperature be heated instead to separate the ring and the rod because then it would expand? Also why is Lfrod = Lfring?
 
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If you heat the combined metals, both expand - what is the thermal expansion of aluminium, compared to the expansion of brass? Does heating (both together) work then?

What does Lfrod = Lfring mean?
To find the point where they get separable, both should have the same diameter.
 
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