Solving the Stuck Ring Problem Using Expansion Coefficients

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SUMMARY

The discussion centers on solving the Stuck Ring Problem involving a brass ring with a diameter of 10.00 cm and an aluminum rod with a diameter of 10.01 cm at an initial temperature of 19.5°C. The average coefficients of linear expansion are given as B(brass) = 19x10^-6 and B(alum) = 24x10^-6. To separate the two metals, calculations must determine the necessary cooling temperature, factoring in their respective thermal contractions. The formula L = L0(1 + α ΔT is essential for establishing the relationship between temperature change and diameter reduction.

PREREQUISITES
  • Understanding of linear thermal expansion coefficients
  • Familiarity with the formula L = L0(1 + α ΔT
  • Basic knowledge of material properties of brass and aluminum
  • Concept of thermal contraction in metals
NEXT STEPS
  • Calculate the cooling temperature required to separate a brass ring from an aluminum rod using linear expansion principles
  • Explore the effects of varying diameters on thermal contraction for different metals
  • Study the implications of thermal expansion in engineering applications
  • Investigate the maximum temperature change (ΔT) applicable for safe material handling
USEFUL FOR

Students, engineers, and physicists interested in thermal expansion concepts, particularly those dealing with metal fittings and material properties in mechanical design.

skwerl
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A brass ring of diameter 10.00 cm at 19.5C is heated and slipped over an aluminum rod of diamter 10.01 cm at 19.5C. Assume the average coefficients of linear expansion are constant.

a) To what temperature must this combination be cooled to separate them? is this possible?
b) If the aluminum rod were 10.03 cm in diameter, what would be the required temperature? Is this possible?

B(brass) = 19x10^-6
B(alum) = 24x10^-6

Well...let's see... first I thought of heating the brass until it was wide enough to slip over the tube, and then I tried to set up some kind of relationship and failed miserably.

oh and by the way I'm new! hi guys!
 
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skwerl said:
Well...let's see... first I thought of heating the brass until it was wide enough to slip over the tube, and then I tried to set up some kind of relationship and failed miserably.
Welcome to Physics Forums!

When the temperature is reduced, both metals will contract. The question is will they contract enough to separate?

The length is given by: L = L0(1 + α ΔT). Set it up and solve for the temperature that will make both metals the same diameter.

Hints: You know the "unstressed" diameter of both metals at 19.5°C, so start there. Also, what is the maximum possible ΔT?
 
neat. so much more elegant than what I thought of

thanks!
 

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