# Answer to Question #1508 in Mechanics | Relativity for mike

Question #1508

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

Answer in units of degrees C.

^{−5}(degrees C)^{−1}and the temperature coefficient of expansion for aluminum is 2.4 × 10^{−5}(degrees C)^{−1}, to what temperature must the ring be heated so that it will just slip overthe rod?Answer in units of degrees C.

Expert's answer

Denote the diameter of the ring as D

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:

D

D

Δ T = (D

Δ T = T - T

T = Δ T +T

_{1}and the the diameter of the rod as D_{2}, the temperature coefficient of expansion for aluminium as α_{1}, for the brass as α_{2}.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:

D

_{1}+ΔD_{1}= D_{2}+ ΔD_{2}D

_{1}+ D_{1}* α_{1}* ΔT = D_{2}+ D_{2}* α_{2}* ΔTΔ T = (D

_{1}-D_{2})/(α_{2}D_{2}-α_{1}D_{1})= (8-8.06)/(1.9*8.06-2.4*8) x10^{5}= 154.4 C.Δ T = T - T

_{initial}T = Δ T +T

_{initial}= 154.4 + 21 = 175.4 C.Need a fast expert's response?

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