Answer to Question #1508 in Mechanics | Relativity for mike

Question

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.

Answer 1

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 - T_initial
T = Δ T +T_initial = 154.4 + 21 = 175.4 C.