Answer to Question #1508 in Mechanics | Relativity for mike
At 21C, an aluminum ring has an inner diameter of 8 cm and a brass rod has a diameter of 8.06 cm. If the temperature coefficient of expansion for brass is 1.9 × 10[sup]−5[/sup] (degrees C)[sup]−1[/sup] and the temperature coefficient of expansion for aluminum is 2.4 × 10[sup]−5[/sup] (degrees C)[sup]−1[/sup], to what temperature must the ring be heated so that it will just slip overthe rod?
Answer in units of degrees C.
Denote the diameter of the ring as D1 and the the diameter of the rod as D2, the temperature coefficient of expansion for aluminium as α1, for the brass as α2. The change in the linear dimension can be estimated by the formula: Δ L = α * L * Δ T
In our case to slip over the rod the ring must have minimum diameter equal to the diameter of the rod: D1+ΔD1 = D2 + ΔD2 D1 + D1* α1 * ΔT = D2 + D2* α2* ΔT Δ T = (D1-D2)/(α2D2-α1D1)= (8-8.06)/(1.9*8.06-2.4*8) x105 = 154.4 C. Δ T = T - Tinitial T = Δ T +Tinitial = 154.4 + 21 = 175.4 C.