A thin brass ring has an inner diameter of at

PHYSICS · Multi-part · AP · Physics 2: Algebra-based · Thermodynamics · Thermodynamics

Problem

A thin brass ring has an inner diameter of 60.00 mm60.00\,\text{mm} at 15∘C15^{\circ}\text{C}. A steel disc of diameter 60.05 mm60.05\,\text{mm} at the same temperature is to be mounted inside the ring by a heat-and-shrink process. (i) Obtain an expression for the common temperature TT at which the diameters of the ring and the disc become equal, taking 15∘C15^{\circ}\text{C} as the initial reference temperature. (ii) Using αbrass=1.9×10−5  K−1\alpha_{\text{brass}}=1.9\times10^{-5}\;\text{K}^{-1} and αsteel=1.2×10−5  K−1\alpha_{\text{steel}}=1.2\times10^{-5}\;\text{K}^{-1}, calculate the required temperature TT to the nearest degree Celsius. (iii) After the assembly is allowed to cool back to 0∘C0^{\circ}\text{C}, determine the interference (difference in diameters) between the disc and the ring at this low temperature. (iv) Discuss briefly, with calculation, whether the assembly will remain secure if the ambient temperature subsequently fluctuates between 0∘C0^{\circ}\text{C} and 40∘C40^{\circ}\text{C}. Support your answer with quantitative reasoning.

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