It is desired to slip an aluminum ring over a steel bar (see Figure). At 10.00 °

Question

It is desired to slip an aluminum ring over a
steel bar (see Figure). At 10.00 °C the inside diameter of
the ring is 4.000 cm and the diameter of the rod is 4.040
cm. (a) Find the temperature of the ring at which it fits
over the bar. The bar remains at 10.0 °C. Once the
aluminum ring is slipped over the bar, the ring and bar
are allowed to equilibrate at a temperature of 22 °C. The
ring is now stuck on the bar. (b) If the temperature of
both the ring and the bar are changed together, should
the system be heated or cooled to remove the ring? (c)
Find the temperature at which the ring can be removed.

Explanation / Answer

In this case, we must consult the coefficients of linear expansion, or, a. If the aluminum expands at a faster rate than the steel as we heat them, then we should heat the aluminum. Otherwise, cool it. a. a(Al) = 24 x 10- 6 per degree K; and a(Steel) = 12 x 10- 6 per degree K. This means that Aluminum expands faster, so we heat the aluminum. b. The Aluminum ring will NOT fit over the Steel bar at this temperature (10° C), so we need it to expand to exceed the diameter of the steel, i.e. Da > ds. So, Da > 4.040 cm, such that Da = (4.000 cm )+ Dd. According to the linear expansion formula, DL = aL0DT. So, for this problem, Dd = a(da DT). And, of course, Dd = ds – da = 0.040 cm; DT = Tf – Ti = Tf - 10.00° C. Where We are looking for Tf. So, let's re-write this as: (Dd)/ (ada ) = DT = (Tf – Ti). And now, we can re-arrange that to get Tf. Tf = Ti + (Dd)/ (ada ) = 10.00° C + (0.040 cm) / [(24 x 10- 6)(4.000 cm)] = 10.00° C + (0.040 cm) / [(96 x 10- 6)] = 10.00° C + (4.17 x 102 °C) = 427°C.

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